Download Go FIGure

534
chapter 13 Properties of Solutions
Go Figure
How does the orientation of H2O molecules around Na+ differ from that around Cl-?
Solvent–solute interactions
between water molecules and
NaCl allow solid to dissolve
Ions hydrated in solution
Crystal of NaCl in water
Hydrated Cl− ion
Hydrated Na+ ion
▲ Figure 13.3 Dissolution of the ionic solid NaCl in water.
We can imagine the solution process as having three components, each with
an associated enthalpy change: A cluster of n solute particles must separate from
one another 1∆Hsolute2, a cluster of m solvent particles separate from one another
1∆Hsolvent2, and these solute and solvent particles mix 1∆Hmix2.
1.1solute2n ∆ n solute
2.1solvent2m ∆ m solvent
3.n solute + m solvent ∆ solution
∆Hsolute
∆Hsolvent
∆Hmix
4.1solute2n + 1solvent2m ∆ solution ∆Hsoln = ∆Hsolute + ∆Hsolvent + ∆Hmix
As seen above, the overall enthalpy change, ∆Hsoln, is the sum of the three steps:
∆Hsoln = ∆Hsolute + ∆Hsolvent + ∆Hmix[13.1]
Separation of the solute particles from one another always requires an input of
energy to overcome their attractive interactions. The process is therefore endothermic
1∆Hsolute 7 02. Likewise, separation of solvent molecules to accommodate the solute
always requires energy 1∆Hsolvent 7 02. The third component, which arises from the
attractive interactions between solute particles and solvent particles, is always exothermic 1∆Hmix 6 02.
The three enthalpy terms in Equation 13.1 can be added together to give either a
negative or a positive sum, depending on the actual numbers for the system being considered (▶ Figure 13.4). Thus, the formation of a solution can be either exothermic or
endothermic. For example, when magnesium sulfate 1MgSO42 is added to water, the
solution process is exothermic: ∆Hsoln = -91.2 kJ>mol. In contrast, the dissolution of
ammonium nitrate 1NH4NO32 is endothermic: ∆Hsoln = 26.4 kJ>mol. These particular salts are the main components in the instant heat packs and ice packs used to treat
535
section 13.1 The Solution Process
Go Figure
How does the magnitude of ∆Hmix compare with the magnitude of ∆Hsolvent + ∆Hsolute for exothermic solution processes?
∆Hsoln < 0
Solution
∆Hmix
Solution
Hfinal
∆Hsoln > 0
Solute
aggregated
∆Hsolute + ∆Hsolvent
Solvent
aggregated
∆Hsolute
∆Hsolvent
Hinitial
Enthalpy, H
Solute
separated
∆Hmix
∆Hsolute
∆Hsolvent
Enthalpy, H
Solute
aggregated
∆Hsolute + ∆Hsolvent
Solute
separated
Solvent
separated
Solvent
aggregated
Solvent
separated
Hinitial
Hfinal
Exothermic solution process
Endothermic solution process
▲ Figure 13.4 Enthalpy changes accompanying the solution process.
athletic injuries (▶ Figure 13.5). The packs consist of a pouch of water and the solid
salt sealed off from the water—MgSO41s2 for hot packs and NH4NO31s2 for cold packs.
When the pack is squeezed, the seal separating the solid from the water is broken and a
solution forms, either increasing or decreasing the temperature.
The enthalpy change for a process can provide insight into the extent to which
(Section 5.4) Exothermic processes tend to proceed spontanethe process occurs.
ously. On the other hand, if ∆Hsoln is too endothermic, the solute might not dissolve to
any significant extent in the chosen solvent. Thus, for solutions to form, the s­ olvent–
solute interaction must be strong enough to make ∆Hmix comparable in magnitude
to ∆Hsolute + ∆Hsolvent. This fact further explains why ionic solutes do not dissolve in
nonpolar solvents. The nonpolar solvent molecules experience only weak attractive
interactions with the ions, and these interactions do not compensate for the energies
required to separate the ions from one another.
By similar reasoning, a polar liquid solute, such as water, does not dissolve in a
nonpolar liquid solvent, such as octane 1C8H182. The water molecules experience strong
(Section 11.2)—attractive forces
hydrogen-bonding interactions with one another
that must be overcome if the water molecules are to be dispersed throughout the octane
solvent. The energy required to separate the H2O molecules from one another is not
recovered in the form of attractive interactions between the H2O and C8H18 molecules.
Give It Some Thought
Label the following processes as exothermic or endothermic:
(a) breaking solvent–solvent interactions to form separated particles
(b) forming solvent–solute interactions from separated particles
Solution Formation and Chemical Reactions
In discussing solutions, we must be careful to distinguish the physical process of solution formation from chemical reactions that lead to a solution. For example, nickel
metal dissolves on contact with an aqueous hydrochloric acid solution because the following reaction occurs:
Ni1s2 + 2 HCl1aq2 ¡ NiCl21aq2 + H21g2[13.2]
▲ Figure 13.5 Magnesium sulfate instant
hot pack.
536
chapter 13 Properties of Solutions
Go Figure
What is the molar mass of nickel chloride hexahydrate, NiCl2 # 6 H2O1s2?
Nickel metal and hydrochloric acid
Nickel reacts with hydrochloric acid,
forming NiCl2 (aq) and H2 (g). The
solution is of NiCl2 , not Ni metal
NiCl2·6 H2O(s) remains
when solvent evaporated
▲ Figure 13.6 The reaction between nickel metal and hydrochloric acid is not a simple
dissolution. The product is NiCl2 # 6 H2O1s2, nickel(II) chloride hexahydrate, which has exactly
6 waters of hydration in the crystal lattice for every nickel ion.
In this instance, one of the resulting solutes is not Ni metal but rather its salt NiCl2.
If the solution is evaporated to dryness, NiCl2 # 6 H2O1s2 is recovered (▲ Figure 13.6).
­Compounds such as NiCl2 # 6 H2O1s2 that have a defined number of water molecules
in the crystal lattice are known as hydrates. When NaCl(s) is dissolved in water, on the
other hand, no chemical reaction occurs. If the solution is evaporated to dryness, NaCl
is recovered. Our focus throughout this chapter is on solutions from which the solute
can be recovered unchanged from the solution.
13.2 | Saturated Solutions
and Solubility
As a solid solute begins to dissolve in a solvent, the concentration of solute particles
in solution increases, increasing the chances that some solute particles will collide
with the surface of the solid and reattach. This process, which is the opposite of the
solution process, is called crystallization. Thus, two opposing processes occur in
a solution in contact with undissolved solute. This situation is represented in this
chemical equation:
dissolve
Solute + solvent crystallize
∆ solution[13.3]
When the rates of these opposing processes become equal, a dynamic equilibrium
is established, and there is no further increase in the amount of solute in solution.
(Section 4.1)
A solution that is in equilibrium with undissolved solute is saturated. Additional
solute will not dissolve if added to a saturated solution. The amount of solute needed
to form a saturated solution in a given quantity of solvent is known as the solubility
section 13.2 Saturated Solutions and Solubility
of that solute. That is, the solubility of a particular solute in a particular solvent is the
maximum amount of the solute that can dissolve in a given amount of the solvent at a
specified temperature, assuming that excess solute is present. For example, the solubility
of NaCl in water at 0 °C is 35.7 g per 100 mL of water. This is the maximum amount
of NaCl that can be dissolved in water to give a stable equilibrium solution at that
temperature.
If we dissolve less solute than the amount needed to form a saturated solution, the
solution is unsaturated. Thus, a solution containing 10.0 g of NaCl per 100 mL of water
at 0 °C is unsaturated because it has the capacity to dissolve more solute.
Under suitable conditions, it is possible to form solutions that contain a greater
amount of solute than needed to form a saturated solution. Such solutions are supersaturated. For example, when a saturated solution of sodium acetate is made at a
high temperature and then slowly cooled, all of the solute may remain dissolved even
though its solubility decreases as the temperature decreases. Because the solute in
a supersaturated solution is present in a concentration higher than the equilibrium
concentration, supersaturated solutions are unstable. For crystallization to occur,
however, the solute particles must arrange themselves properly to form crystals. The
addition of a small crystal of the solute (a seed crystal) provides a template for crystallization of the excess solute, leading to a saturated solution in contact with excess
solid (▼ Figure 13.7).
Give It Some Thought
What happens if a solute is added to a saturated solution?
Go Figure
What is the evidence that the solution in the left photograph is supersaturated?
Amount of sodium acetate
dissolved is greater than its
solubility at this temperature
1
Seed crystal of sodium
acetate added to
supersaturated solution
2
Excess sodium acetate
crystallizes from solution
3
Solution arrives
at saturation
▲ Figure 13.7 Precipitation from a supersaturated sodium acetate solution. The solution on the
left was formed by dissolving about 170 g of the salt in 100 mL of water at 100 °C and then slowly
cooling it to 20 °C. Because the solubility of sodium acetate in water at 20 °C is 46 g per 100 mL
of water, the solution is supersaturated. Addition of a sodium acetate crystal causes the excess
solute to crystallize from solution.
537
538
chapter 13 Properties of Solutions
Table 13.1 Solubilities of Gases
in Water at 20 °C, with 1 atm Gas
Pressure
Molar
GasMass 1g>mol2
Solubility (M)
N228.0
0.69 * 10-3
O232.0
1.38 * 10-3
Ar39.9
1.50 * 10-3
Kr83.8
-3
2.79 * 10
13.3 | Factors Affecting Solubility
The extent to which one substance dissolves in another depends on the nature of
both substances.
(Section 13.1) It also depends on temperature and, at least for
gases, on pressure.
Solute–Solvent Interactions
The natural tendency of substances to mix and the various interactions among solute
and solvent particles are all involved in determining solubilities. Nevertheless, insight
into variations in solubility can often be gained by focusing on the interaction between
the solute and solvent. The data in ◀ Table 13.1 show that the solubilities of various
gases in water increase with increasing molecular mass. The attractive forces between
the gas m
­ olecules and solvent molecules are mainly dispersion forces, which increase
(Section 11.2) Thus, the data indicate that
with ­increasing size and molecular mass.
the ­solubilities of gases in water increase as the attraction between solute (gas) and solvent ­(water) increases. In general, when other factors are comparable, the stronger the
attractions between solute and solvent molecules, the greater the solubility of the solute in
that solvent.
Because of favorable dipole–dipole attractions between solvent molecules and
solute molecules, polar liquids tend to dissolve in polar solvents. Water is both polar
(Section 11.2) Thus, polar molecules, espeand able to form hydrogen bonds.
cially those that can form hydrogen bonds with water molecules, tend to be soluble
in water. For example, acetone, a polar molecule with the structural formula shown
below, mixes in all proportions with water. Acetone has a strongly polar C “ O bond
and pairs of nonbonding electrons on the O atom that can form hydrogen bonds with
water.
O
CH3CCH3
Acetone
Hexane
Water
▲ Figure 13.8 Hexane is immiscible with
water. Hexane is the top layer because it is
less dense than water.
Liquids that mix in all proportions, such as acetone and water, are miscible,
whereas those that do not dissolve in one another are immiscible. Gasoline, which is
a mixture of hydrocarbons, is immiscible with water. Hydrocarbons are nonpolar substances because of several factors: The C ¬ C bonds are nonpolar, the C ¬ H bonds are
nearly nonpolar, and the molecules are symmetrical enough to cancel much of the weak
C ¬ H bond dipoles. The attraction between the polar water molecules and the nonpolar hydrocarbon molecules is not sufficiently strong to allow the formation of a solution.
Nonpolar liquids tend to be insoluble in polar liquids, as ◀ Figure 13.8 shows for hexane
1C6H142 and water.
Many organic compounds have polar groups attached to a nonpolar framework
of carbon and hydrogen atoms. For example, the series of organic compounds in
▶ Table 13.2 all contain the polar OH group. Organic compounds with this molecular feature are called alcohols. The O ¬ H bond is able to form hydrogen bonds.
For example, ethanol 1CH3CH2OH2 molecules can form hydrogen bonds with water
­molecules as well as with each other (▶ Figure 13.9). As a result, the ­s olute–­s olute,
solvent–solvent, and solute–solvent forces are not greatly different in a mixture
of CH3CH2OH and H2O. No major change occurs in the environments of the
­m olecules as they are mixed. Therefore, the increased entropy when the components mix plays a significant role in solution formation, and ethanol is completely
miscible with water.
539
section 13.3 Factors Affecting Solubility
Table 13.2 Solubilities of Some Alcohols in Water and in Hexane*
Solubility
in h 2o
Solubility
in C6h 14
CH3OH 1methanol2
∞
0.12
CH3CH2OH 1ethanol2
∞
∞
CH3CH2CH2OH 1propanol2
∞
∞
CH3CH2CH2CH2OH 1butanol2
0.11
∞
0.030
∞
CH3CH2CH2CH2CH2CH2OH 1hexanol2
0.0058
∞
Alcohol
CH3CH2CH2CH2CH2OH 1pentanol2
*Expressed in mol alcohol/100 g solvent at 20 °C. The infinity symbol 1∞ 2 indicates that
the alcohol is completely miscible with the solvent.
H
H
H
C
H
H
C
C H
H
Hydrogen bond between
two ethanol molecules
Hydrogen bond between
ethanol molecule and
water molecule
▲ Figure 13.9 Hydrogen bonding involving OH groups. Notice in Table 13.2 that the number of carbon atoms in an alcohol affects its solubility in water. As this number increases, the polar OH group becomes an even smaller
part of the molecule, and the molecule behaves more like a hydrocarbon. The solubility
of the alcohol in water decreases correspondingly. On the other hand, the solubility of
alcohols in a nonpolar solvent like hexane 1C6H142 increases as the nonpolar hydrocarbon chain lengthens.
One way to enhance the solubility of a substance in water is to increase the n
­ umber of
polar groups the substance contains. For example, increasing the number of OH groups
in a solute increases the extent of hydrogen bonding between that ­solute and water,
thereby increasing solubility. Glucose (C6H12O6, ▶ Figure 13.10) has five OH groups
on a six-carbon framework, which makes the molecule very soluble in water: 830 g
dissolves in 1.00 L of water at 17.5 °C. In contrast, cyclohexane 1C6H122, which has a
similar structure to glucose but with all of the OH groups replaced by H, is essentially
insoluble in water (only 55 mg of cyclohexane can dissolve in 1.00 L of water at 25 °C).
H C
C
H
H
C
H
H
Cyclohexane, C6H12, which
has no polar OH groups, is
essentially insoluble in water
OH groups enhance the aqueous
solubility because of their ability
to hydrogen bond with H2O.
H CH2OH
H
O
HO C
C
C
HO C
OH
OH C
H H
H
Hydrogen-bonding
sites
▲ Figure 13.10 The correlation of
molecular structure with solubility.
Chemistry and Life
Fat-Soluble and
Water-Soluble Vitamins
Vitamins have unique chemical structures that affect their solubilities
in different parts of the human body. Vitamin C and the B vitamins
are soluble in water, for example, whereas vitamins A, D, E, and K are
soluble in nonpolar solvents and in fatty tissue (which is nonpolar).
­Because of their water solubility, vitamins B and C are not stored to any
appreciable extent in the body, and so foods containing these ­vitamins
should be included in the daily diet. In contrast, the fat-soluble
­vitamins are stored in sufficient quantities to keep vitamin-deficiency
diseases from appearing even after a person has subsisted for a long
period on a vitamin-deficient diet.
That some vitamins are soluble in water and others are not can be explained in terms of their structures. Notice in Figure 13.11 that ­vitamin
A (retinol) is an alcohol with a very long carbon chain. Because the OH
group is such a small part of the molecule, the molecule resembles the
long-chain alcohols listed in Table 13.2. This vitamin is nearly nonpolar.
In contrast, the vitamin C molecule is smaller and has several OH groups
that can form hydrogen bonds with water, somewhat like glucose.
Related Exercises: 13.7, 13.48
540
chapter 13 Properties of Solutions
Most of molecule nonpolar
Only one polar group
to interact with water
H
CH3 H
CH3 H
CH3
C
C
C
C
C
C
O
C
C
C
C
C
H2C
H
H
H
H
H
C
H2C
C
CH3
H2
Many polar groups to
interact with water
H3C
O
H
O
H
C
O
H
O
C C O
H
H
H
C
O H
C
C
H
H
Vitamin C
Vitamin A
▲ Figure 13.11 The molecular structures of vitamins A and C.
Over years of study, examination of different solvent–solute combinations has led
to an important generalization: Substances with similar intermolecular attractive forces
tend to be soluble in one another. This generalization is often simply stated as “like dissolves like.” Nonpolar substances are more likely to be soluble in nonpolar solvents; ionic
and polar solutes are more likely to be soluble in polar solvents. Network solids such as
diamond and quartz are not soluble in either polar or nonpolar solvents because of the
strong bonding within the solid.
Give It Some Thought
Suppose the hydrogens on the OH groups in glucose (Figure 13.10) were
replaced with methyl groups, CH3. Would you expect the water solubility of
the resulting molecule to be higher than, lower than, or about the same as
glucose?
S a mpl e
Exercise 13.1 Predicting Solubility Patterns
Predict whether each of the following substances is more likely to dissolve in the nonpolar
­solvent carbon tetrachloride 1CCl42 or in water: C7H16, Na2SO4, HCl, and I2.
Solution
Analyze We are given two solvents, one that is nonpolar 1CCl42 and the other that is
polar 1H2O2, and asked to determine which will be the better solvent for each
solute listed.
Plan By examining the formulas of the solutes, we can predict whether they are ionic or molecu-
lar. For those that are molecular, we can predict whether they are polar or nonpolar. We can then
apply the idea that the nonpolar solvent will be better for the nonpolar solutes, whereas the polar
solvent will be better for the ionic and polar solutes.
Solve C7H16 is a hydrocarbon, so it is molecular and nonpolar. Na2SO4, a compound containing
a metal and nonmetals, is ionic. HCl, a diatomic molecule containing two nonmetals that differ in electronegativity, is polar. I2, a diatomic molecule with atoms of equal electronegativity, is
nonpolar. We would therefore predict that C7H16 and I2 (the nonpolar solutes) would be more
soluble in the nonpolar CCl4 than in polar H2O, whereas water would be the better solvent for
Na2SO4 and HCl (the ionic and polar covalent solutes).
section 13.3 Factors Affecting Solubility
Practice Exercise 1
Which of the following solvents will best dissolve wax, which is a complex mixture of compounds that mostly are CH3 - CH2 - CH2 -CH2 -CH2 ¬ ?
H
H H
H 3C
C
C
H H
H
H H
C
C
H
CH3
C
C
C
C
C
C
(b) Benzene
O
Cl
C
H
H
H H
(a) Hexane
H3C
H
Cl
CH3
(c) Acetone
C
Cl
Cl
(d) Carbon tetrachloride
O
H
H
(e) Water
Practice Exercise 2
Arrange the following substances in order of increasing solubility in water:
(a)
H
H
(b)
H
H HH HH HH H
(c)H
H
H HH HH HH H
H
C HC CC CC CC CH C HO
H
CHO C CC CC CC COHC
H
H HH HH HH H
H
H
H HH HH HH H
H
H
H HH HH HH H
H
(d)H
H HH HH HH H
H
C HC CC CC CC COHC
H
H HH HH HH H
OH
H C HC CC CC CC CCl C
H
H H HH HH HH H
OH
Cl
H
Pressure Effects
The solubilities of solids and liquids are not appreciably affected by pressure,
whereas the solubility of a gas in any solvent is increased as the partial pressure of
the gas above the solvent increases. We can understand the effect of pressure on
gas solubility by considering Figure 13.12 , which shows carbon dioxide gas distributed between the gas and solution phases. When equilibrium is established,
the rate at which gas molecules enter the solution equals the rate at which solute
molecules escape from the solution to enter the gas phase. The equal number of
up and down arrows in the left container in Figure 13.12 represent these opposing
processes.
Now suppose we exert greater pressure on the piston and compress the gas
above the solution, as shown in the middle container in Figure 13.12. If we reduce
the gas volume to half its original value, the pressure of the gas increases to about
twice its original value. As a result of this pressure increase, the rate at which
gas molecules strike the liquid surface and enter the solution phase increases.
Thus, the solubility of the gas in the solution increases until equilibrium is again
established; that is, solubility increases until the rate at which gas molecules
enter the solution equals the rate at which they escape from the solution. Thus,
541
542
chapter 13 Properties of Solutions
Go Figure
If the partial pressure of a gas over a solution is doubled, how has the concentration of gas in the solution changed after
equilibrium is restored?
Equilibrium
Pressure is increased.
More CO2 dissolves
Equilibrium restored
▲ Figure 13.12 Effect of pressure on gas solubility.
Go Figure
How do the slopes of the lines vary
with the molecular weight of the gas?
Explain the trend.
O2
CO
Solubility (mM)
1.00
N2
0.50
He
0
0.50
1.00
Partial pressure (atm)
▲ Figure 13.13 The solubility of a gas in
water is directly proportional to the partial
pressure of the gas. The solubilities are in
millimoles of gas per liter of solution.
the solubility of a gas in a liquid solvent increases in direct proportion to the partial
pressure of the gas above the ­solution (◀ Figure 13.13).
The relationship between pressure and gas solubility is expressed by Henry’s law:
Sg = kPg[13.4]
Here, Sg is the solubility of the gas in the solvent (usually expressed as molarity), Pg
is the partial pressure of the gas over the solution, and k is a proportionality constant
known as the Henry’s law constant. The value of this constant depends on the solute,
solvent, and temperature. As an example, the solubility of N2 gas in water at 25 °C and
0.78 atm pressure is 4.75 * 10-4 M. The Henry’s law constant for N2 in 25 °C water is
thus 14.75 * 10-4 mol>L2> 0.78 atm = 6.1 * 10-4 mol>L@atm. If the partial pressure
of N2 is doubled, Henry’s law predicts that the solubility in water at 25 °C also doubles
to 9.50 * 10-4 M.
Bottlers use the effect of pressure on solubility in producing carbonated beverages, which are bottled under a carbon dioxide pressure greater than 1 atm. When the
bottles are opened to the air, the partial pressure of CO2 above the solution decreases.
Hence, the solubility of CO2 decreases, and CO21g2 escapes from the solution as bubbles (▶ Figure 13.14).
S a mpl e
Exercise 13.2 A Henry’s Law Calculation
Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2
of 4.0 atm over the liquid at 25 °C. The Henry’s law constant for CO2 in water at this temperature is 3.4 * 10-2 mol>L@atm.
Solution
Analyze We are given the partial pressure of CO2, PCO2, and the Henry’s law constant, k, and
asked to calculate the concentration of CO2 in the solution.
Plan With the information given, we can use Henry’s law, Equation 13.4, to calculate the
­solubility, S CO2.
section 13.3 Factors Affecting Solubility
543
Solve SCO2 = kPCO2 = 13.4 * 10-2 mol>[email protected] atm2 = 0.14 mol>L = 0.14 M
Check The units are correct for solubility, and the answer has two significant figures consistent
with both the partial pressure of CO2 and the value of Henry’s constant.
Practice Exercise 1
You double the partial pressure of a gas over a liquid at constant temperature. Which of these
statements is then true?
(a)The Henry’s law constant is doubled.
(b)The Henry’s law constant is decreased by half.
(c)There are half as many gas molecules in the liquid.
(d)There are twice as many gas molecules in the liquid.
(e)There is no change in the number of gas molecules in the liquid.
Practice Exercise 2
Calculate the concentration of CO2 in a soft drink after the bottle is opened and the solution
equilibrates at 25 °C under a CO2 partial pressure of 3.0 * 10-4 atm.
Temperature Effects
The solubility of most solid solutes in water increases as the solution temperature
­increases, as ▼ Figure 13.15 shows. There are exceptions to this rule, however, as seen
for Ce21SO423, whose solubility curve slopes downward with increasing temperature.
In contrast to solid solutes, the solubility of gases in water decreases with increasing
temperature (▼ Figure 13.16). If a glass of cold tap water is warmed, you can see bubbles on the inside of the glass because some of the dissolved air comes out of solution.
Similarly, as carbonated beverages are allowed to warm, the solubility of CO2
decreases, and CO21g2 escapes from the solution.
Give It Some Thought
Why do bubbles form on the inside wall of a cooking pot when water is heated on the
stove, even though the water temperature is well below the boiling point of water?
Go Figure
Go Figure
How does the solubility of KCl at 80 °C compare with
that of NaCl at the same temperature?
Where would you expect N2 to fit on this graph?
CH4
100
2.0
O3
aN
O2
80
Pb
50
40
2C
)2
O3
(N
K
C
60
3
l2
aC
r2 O
7
70
KCl
Solubility (mM)
N
KN
O
Solubility (g of salt in 100 g H2O)
90
CO
1.0
NaCl
30
lO 3
KC
20
10
0
▲ Figure 13.14 Gas solubility decreases
as pressure decreases. CO2 bubbles out
of solution when a carbonated beverage is
opened because the CO2 partial pressure
above the solution is reduced.
He
Ce2(SO4)3
0
10
20
30 40 50 60 70
Temperature (°C)
80
90 100
▲ Figure 13.15 Solubilities of some ionic compounds in water
as a function of temperature.
0
10
20
30
Temperature (°C)
40
50
▲ Figure 13.16 Solubilities of four gases in water as a function
of temperature. The solubilities are in millimoles per liter of
solution, for a constant total pressure of 1 atm in the gas phase.
544
chapter 13 Properties of Solutions
Chemistry and Life
Blood Gases and Deep-Sea Diving
Because gas solubility increases with increasing pressure, divers who
breathe compressed air (▶ Figure 13.17) must be concerned about
the solubility of gases in their blood. Although the gases are not very
soluble at sea level, their solubilities can be appreciable at deep levels where their partial pressures are greater. Thus, divers must ascend
slowly to prevent dissolved gases from being released rapidly from solution and forming bubbles in the blood and other fluids in the body.
These bubbles affect nerve impulses and cause decompression sickness, or “the bends,” which is a painful and potentially fatal condition.
Nitrogen is the main problem because it is the most abundant gas in
air and because it can be removed from the body only through the
respiratory system. Oxygen, in contrast, is consumed in metabolism.
Deep-sea divers sometimes substitute helium for nitrogen in the
air they breathe because helium has a much lower solubility in biological fluids than N2. For example, divers working at a depth of 100 ft
experience a pressure of about 4 atm. At this pressure, a mixture of
95% helium and 5% oxygen gives an oxygen partial pressure of about
0.2 atm, which is the partial pressure of oxygen in normal air at 1 atm.
▲ Figure 13.17 Gas solubility increases as pressure increases. Divers
who use compressed gases must be concerned about the solubility of
the gases in their blood.
If the oxygen partial pressure becomes too great, the urge to breathe
is reduced, CO2 is not removed from the body, and CO2 poisoning
­occurs. At excessive concentrations in the body, carbon dioxide acts
as a neurotoxin, interfering with nerve conduction and transmission.
Related Exercises: 13.59, 13.60, 13.107
13.4 | Expressing Solution
Concentration
The concentration of a solution can be expressed either qualitatively or quantitatively.
The terms dilute and concentrated are used to describe a solution qualitatively. A solution with a relatively small concentration of solute is said to be dilute; one with a large
concentration is said to be concentrated. Chemists use various ways to express concentration quantitatively, and we examine several of these next.
Mass Percentage, ppm, and ppb
One of the simplest quantitative expressions of concentration is the mass percentage of
a component in a solution, given by
Mass % of component =
mass of component in soln
* 100[13.5]
total mass of soln
Because percent means “per hundred,” a solution of hydrochloric acid that is 36% HCl
by mass contains 36 g of HCl for each 100 g of solution.
We often express the concentration of very dilute solutions in parts per million
(ppm) or parts per billion (ppb). These quantities are similar to mass percentage
but use 106 (a million) or 109 (a billion), respectively, in place of 100, as a multiplier
for the ratio of the mass of solute to the mass of solution. Thus, parts per million is
defined as
ppm of component =
mass of component in soln
* 106[13.6]
total mass of soln
A solution whose solute concentration is 1 ppm contains 1 g of solute for each ­million
11062 grams of solution or, equivalently, 1 mg of solute per kilogram of solution.
­Because the density of water is 1 g>mL, 1 kg of a dilute aqueous solution has a volume
section 13.4 Expressing Solution Concentration
very close to 1 L. Thus, 1 ppm also corresponds to 1 mg of solute per liter of aqueous
solution.
The acceptable maximum concentrations of toxic or carcinogenic substances in the
environment are often expressed in ppm or ppb. For example, the maximum allowable
concentration of arsenic in drinking water in the United States is 0.010 ppm; that is,
0.010 mg of arsenic per liter of water. This concentration corresponds to 10 ppb.
Give It Some Thought
A solution of SO2 in water contains 0.00023 g of SO2 per liter of solution. What
is the concentration of SO2 in ppm? In ppb?
S a mpl e
Exercise 13.3 Calculation of Mass-Related Concentrations
(a) A solution is made by dissolving 13.5 g of glucose 1C6H12O62 in 0.100 kg of water. What is
the mass percentage of solute in this solution?
(b) A 2.5-g sample of groundwater was found to contain 5.4 mg of Zn2+. What is the concentration of
Zn2+ in parts per million?
Solution
(a) Analyze We are given the number of grams of solute (13.5 g) and the number of grams of
solvent 10.100 kg = 100 g2. From this, we must calculate the mass percentage of solute.
Plan We can calculate the mass percentage by using Equation 13.5. The mass of the solution is
the sum of the mass of solute (glucose) and the mass of solvent (water).
Solve Mass % of glucose =
mass glucose
mass soln
* 100 =
13.5 g
13.5 g + 100 g
* 100 = 11.9%
Comment The mass percentage of water in this solution is 1100 - 11.92% = 88.1%.
(b) Analyze In this case we are given the number of micrograms of solute. Because 1 mg is
1 * 10-6 g, 5.4 mg = 5.4 * 10-6 g.
Plan We calculate the parts per million using Equation 13.6.
Solve ppm =
5.4 * 10-6 g
mass of solute
* 106 =
* 106 = 2.2 ppm
mass of soln
2.5 g
Practice Exercise 1
Calculate the mass percentage of NaCl in a solution containing 1.50 g of NaCl in 50.0 g of water.
(a) 0.0291%, (b) 0.0300%, (c) 0.0513%, (d) 2.91%, (e) 3.00%.
Practice Exercise 2
A commercial bleaching solution contains 3.62% by mass of sodium hypochlorite, NaOCl.
What is the mass of NaOCl in a bottle containing 2.50 kg of bleaching solution?
Mole Fraction, Molarity, and Molality
Concentration expressions are often based on the number of moles of one or more
components of the solution. Recall from Section 10.6 that the mole fraction of a component of a solution is given by
Mole fraction of component =
moles of component
[13.7]
total moles of all components
The symbol X is commonly used for mole fraction, with a subscript to indicate the component of interest. For example, the mole fraction of HCl in a hydrochloric acid solution is represented as XHCl. Thus, if a solution contains 1.00 mol of HCl (36.5 g) and 8.00 mol of water
(144 g), the mole fraction of HCl is XHCl = 11.00 mol2>11.00 mol + 8.00 mol2 = 0.111.
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chapter 13 Properties of Solutions
Mole fractions have no units because the units in the numerator and the denominator cancel. The sum of the mole fractions of all components of a solution must equal 1. Thus, in
the aqueous HCl solution, XH2O = 1.000 - 0.111 = 0.889. Mole fractions are very useful when dealing with gases, as we saw in Section 10.6, but have limited use when dealing
with liquid solutions.
Recall from Section 4.5 that the molarity (M) of a solute in a solution is defined as
Molarity =
moles of solute
[13.8]
liters of soln
For example, if you dissolve 0.500 mol of Na2CO3 in enough water to form 0.250 L
of solution, the molarity of Na2CO3 in the solution is 10.500 mol2>10.250 L2 = 2.00 M.
Molarity is especially useful for relating the volume of a solution to the quantity
of solute contained in that volume, as we saw in our discussions of titrations.
(Section 4.6)
The molality of a solution, denoted m, is a concentration unit that is also based
on moles of solute. Molality equals the number of moles of solute per kilogram of
solvent:
Molality =
moles of solute
[13.9]
kilograms of solvent
Thus, if you form a solution by mixing 0.200 mol of NaOH (8.00 g) and 0.500 kg of
water (500 g), the concentration of the solution is 10.200 mol2>10.500 kg2 = 0.400 m
(that is, 0.400 molal) in NaOH.
The definitions of molarity and molality are similar enough that they can be easily
confused. Molarity depends on the volume of solution, whereas molality depends on the
mass of solvent. When water is the solvent, the molality and molarity of dilute solutions
are numerically about the same because 1 kg of solvent is nearly the same as 1 kg of
solution, and 1 kg of the solution has a volume of about 1 L.
The molality of a given solution does not vary with temperature because masses do
not vary with temperature. The molarity of the solution does change with temperature,
however, because the volume of the solution expands or contracts with temperature.
Thus, molality is often the concentration unit of choice when a solution is to be used
over a range of temperatures.
Give It Some Thought
If an aqueous solution is very dilute, will its molality be greater than its molarity,
nearly the same as its molarity, or smaller than its molarity?
S a mpl e
Exercise 13.4 Calculation of Molality
A solution is made by dissolving 4.35 g of glucose 1C6H12O62 in 25.0 mL of water at 25 °C.
­Calculate the molality of glucose in the solution. ­Water has a density of 1.00 g>mL.
Solution
Analyze We are asked to calculate a solution concentration in units
of molality. To do this, we must determine the number of moles of
solute (glucose) and the number of kilograms of solvent (water).
Solve Use the molar mass of glucose, 180.2 g>mol, to
convert grams to moles:
Because water has a density of 1.00 g>mL, the mass of
the solvent is
Finally, use Equation 13.9 to obtain the molality:
Plan We use the molar mass of C6H12O6 to convert grams of glucose
to moles. We use the density of water to convert milliliters of water
to kilograms. The molality equals the number of moles of solute
­(glucose) divided by the number of kilograms of solvent (water).
Mol C6H12O6 = 14.35 g C6H12O62a
1 mol C6H12O6
b = 0.0241 mol C6H12O6
180.2 g C6H12O6
125.0 mL211.00 g>mL2 = 25.0 g = 0.0250 kg
0.0241 mol C6H12O6
Molality of C6H12O6 =
= 0.964 m
0.0250 kg H2O
section 13.4 Expressing Solution Concentration
Practice Exercise 1
Suppose you take a solution and add more solvent, so that the
original mass of solvent is doubled. You take this new solution and
add more solute, so that the original mass of the solute is doubled.
What happens to the molality of the final solution, compared to
the original molality?
(a)It is doubled.
(b)It is decreased by half.
547
(c)It is unchanged.
(d)It will increase or decrease depending on the molar mass of the
solute.
(e) There is no way to tell without knowing the molar mass of the
solute.
Practice Exercise 2
What is the molality of a solution made by dissolving 36.5 g of
naphthalene 1C10H82 in 425 g of toluene 1C7H82?
Converting Concentration Units
If you follow the dimensional analysis techniques you learned in Chapter 1, you can
convert between concentration units, as shown in Sample Exercise 13.5. To convert
­between molality and molarity, the density of the solution will be needed, as in Sample
Exercise 13.6.
S a mpl e
Exercise 13.5 Calculation of Mole Fraction and Molality
An aqueous solution of hydrochloric acid contains 36% HCl by mass. (a) Calculate the mole
f­ raction of HCl in the solution. (b) Calculate the molality of HCl in the solution.
Solution
Analyze We are asked to calculate the concentration of the solute,
HCl, in two related concentration units, given only the percentage by
mass of the solute in the solution.
Solve (a) To calculate the mole fraction of HCl, we convert the
masses of HCl and H2O to moles and then use Equation 13.7:
(b) To calculate the molality of HCl in the solution, we use
Equation 13.9. We calculated the number of moles of HCl
in part (a), and the mass of solvent is 64 g = 0.064 kg:
Plan In converting concentration units based on the mass or moles
of solute and solvent (mass percentage, mole fraction, and molality),
it is useful to assume a certain total mass of solution. Let’s assume
that there is exactly 100 g of solution. Because the solution is 36%
HCl, it contains 36 g of HCl and 1100 - 362 g = 64 g of H2O.
We must convert grams of solute (HCl) to moles to calculate either
mole fraction or molality. We must convert grams of solvent 1H2O2
to moles to calculate mole fractions and to kilograms to calculate
molality.
1 mol HCl
b = 0.99 mol HCl
36.5 g HCl
1 mol H2O
b = 3.6 mol H2O
Moles H2O = 164 g H2O2a
18 g H2O
moles HCl
0.99
0.99
=
=
= 0.22
XHCl =
moles H2O + moles HCl
3.6 + 0.99
4.6
Moles HCl = 136 g HCl2a
Molality of HCl =
0.99 mol HCl
= 15 m
0.064 kg H2O
Notice that we can’t readily calculate the molarity of the solution because we don’t know the
­volume of the 100 g of solution.
Practice Exercise 1
The solubility of oxygen gas in water at 40 °C is 1.0 mmol per liter
of solution. What is this concentration in units of mole fraction?
(a)1.00 * 10-6, (b) 1.80 * 10-5, (c) 1.00 * 10-2,
(d) 1.80 * 10-2, (e) 5.55 * 10-2.
Practice Exercise 2
A commercial bleach solution contains 3.62% by mass of NaOCl
in water. Calculate (a) the mole fraction and (b) the molality of
NaOCl in the solution.
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chapter 13 Properties of Solutions
S a mpl e
Exercise 13.6 Calculation of Molarity Using the Density of the Solution
A solution with a density of 0.876 g>mL contains 5.0 g of toluene 1C7H82 and 225 g of benzene.
Calculate the molarity of the solution.
Solution
Analyze Our goal is to calculate the molarity of a solution, given the
masses of solute (5.0 g) and solvent (225 g) and the density of the solution 10.876 g>mL2.
Solve The number of moles of solute is:
The density of the solution is used to convert the mass
of the solution to its volume:
Molarity is moles of solute per liter of solution:
Plan The molarity of a solution is the number of moles of solute di-
vided by the number of liters of solution (Equation 13.8). The number
of moles of solute 1C7H82 is calculated from the number of grams
of solute and its molar mass. The volume of the solution is obtained
from the mass of the solution (mass of solution = mass of solute +
mass of solvent = 5.0 g + 225 g = 230 g) and its density.
Moles C7H8 = 15.0 g C7H82a
Milliliters soln = 1230 g2a
Molarity = a
1 mol C7H8
b = 0.054 mol
92 g C7H8
1 mL
b = 263 mL
0.876 g
moles C7H8
0.054 mol C7H8 1000 mL soln
b = a
ba
b = 0.21 M
liter soln
263 mL soln
1 L soln
Check The magnitude of our answer is reasonable. Rounding moles to 0.05 and liters to 0.25
gives a ­molarity of 10.05 mol2>10.25 L2 = 0.2 M.
The units for our answer 1mol>L2 are correct, and the answer, 0.21, has two significant figures,
corresponding to the number of significant figures in the mass of solute (2).
Comment Because the mass of the solvent (0.225 kg) and the volume of the solution (0.263)
are similar in magnitude, the molarity and molality are also similar in magnitude:
10.054 mol C7H82>10.225 kg solvent2 = 0.24 m.
Practice Exercise 1
Maple syrup has a density of 1.325 g>mL, and 100.00 g of maple
syrup contains 67 mg of calcium in the form of Ca2+ ions. What is
the molarity of calcium in maple syrup?
(a) 0.017 M, (b) 0.022 M, (c) 0.89 M, (d) 12.6 M, (e) 45.4 M
Practice Exercise 2
A solution containing equal masses of glycerol 1C3H8O32 and
­water has a density of 1.10 g>mL. Calculate (a) the molality of
glycerol, (b) the mole fraction of glycerol, (c) the molarity of
­glycerol in the solution.
13.5 | Colligative Properties
Some physical properties of solutions differ in important ways from those of the pure
solvent. For example, pure water freezes at 0 °C, but aqueous solutions freeze at lower
temperatures. We apply this behavior when we add ethylene glycol antifreeze to a car’s
radiator to lower the freezing point of the solution. The added solute also raises the
boiling point of the solution above that of pure water, making it possible to operate the
engine at a higher temperature.
Lowering of the freezing point and raising of the boiling point are physical properties of solutions that depend on the quantity (concentration) but not on the kind
or identity of the solute particles. Such properties are called colligative properties.
(Colligative means “depending on the collection”; colligative properties depend on
the collective effect of the number of solute particles.)
In addition to freezing-point lowering and boiling-point raising, vapor-pressure
lowering and osmotic pressure are also colligative properties. As we examine each one,
notice how solute concentration quantitatively affects the property.
Vapor-Pressure Lowering
(Section 11.5)
A liquid in a closed container establishes equilibrium with its vapor.
The vapor pressure is the pressure exerted by the vapor when it is at equilibrium with
the liquid (that is, when the rate of vaporization equals the rate of condensation).
A substance that has no measurable vapor pressure is nonvolatile, whereas one that
exhibits a vapor pressure is volatile.
A solution consisting of a volatile liquid solvent and a nonvolatile solute forms
spontaneously because of the increase in entropy that accompanies their mixing. In
section 13.5 Colligative Properties
Volatile solvent particles
Nonvolatile solute particles
Add
nonvolatile
solute
Equilibrium
Rate of vaporization
reduced by presence
of nonvolatile solute
Equilibrium reestablished
with fewer molecules in
gas phase
▲ Figure 13.18 Vapor-pressure lowering. The presence of nonvolatile solute particles in a
liquid solvent results in a reduction of the vapor pressure above the liquid.
effect, the solvent molecules are stabilized in their liquid state by this process and thus
have a lower tendency to escape into the vapor state. Therefore, when a nonvolatile solute is present, the vapor pressure of the solvent is lower than the vapor pressure of the
pure solvent, as illustrated in ▲ Figure 13.18.
Ideally, the vapor pressure of a volatile solvent above a solution containing a nonvolatile solute is proportional to the solvent’s concentration in the solution. This relationship is expressed quantitatively by Raoult’s law, which states that the partial pressure
exerted by solvent vapor above the solution, Psolution, equals the product of the mole frac°
tion of the solvent, Xsolvent, times the vapor pressure of the pure solvent, P solvent
:
[13.10]
Psolution = Xsolvent P°
solvent
For example, the vapor pressure of pure water at 20 °C is P°H2O = 17.5 torr. Imagine holding the temperature constant while adding glucose 1C6H12O62 to the water so
that the mole fractions in the resulting solution are XH2O = 0.800 and XC6H12O6 = 0.200.
According to Equation 13.10, the vapor pressure of the water above this solution is 80.0%
of that of pure water:
Psolution = 10.8002117.5 torr2 = 14.0 torr
The presence of the nonvolatile solute lowers the vapor pressure of the volatile solvent
by 17.5 torr - 14.0 torr = 3.5 torr.
The vapor-pressure lowering, ∆P, is directly proportional to the mole fraction of
the solute, Xsolute :
∆P = Xsolute P°solvent[13.11]
Thus, for the example of the solution of glucose in water, we have
∆P = XC6H12O6 P°H2O = 10.2002117.5 torr2 = 3.50 torr
The vapor-pressure lowering caused by adding a nonvolatile solute depends on the
total concentration of solute particles, regardless of whether they are molecules or ions.
Remember that vapor-pressure lowering is a colligative property, so its value for any solution depends on the concentration of solute particles and not on their kind or identity.
Give It Some Thought
Adding 1 mol of NaCl to 1 kg of water lowers the vapor pressure of water more
than adding 1 mol of C6H12O6. Explain.
S a mpl e
Exercise 13.7 Calculation of Vapor Pressure of a Solution
Glycerin 1C3H8O32 is a nonvolatile nonelectrolyte with a density of 1.26 g>mL at 25 °C. Calculate
the vapor pressure at 25 °C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water.
The vapor pressure of pure water at 25 °C is 23.8 torr (Appendix B), and its density is 1.00 g>mL.
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chapter 13 Properties of Solutions
Solution
Analyze Our goal is to calculate the vapor pressure of a solution,
given the volumes of solute and solvent and the density of the
solute.
Solve To calculate the mole fraction of water in the
solution, we must determine the number of moles of
C3H8O3 and H2O:
pressure of a solution. The mole fraction of the solvent in the solution,
Xsolvent, is the ratio of the number of moles of solvent 1H2O2to total
moles of solution 1moles C3H8O3 + moles H2O2.
Moles C3H8O3 = 150.0 mL C3H8O32a
Moles H2O = 1500.0 mL H2O2a
XH2O =
We now use Raoult’s law to calculate the vapor pressure of water for the solution:
Plan We can use Raoult’s law (Equation 13.10) to calculate the vapor
1.26 g C3H8O3
1 mL C3H8O3
1.00 g H2O
1 mL H2O
ba
ba
1 mol C3H8O3
b = 0.684 mol
92.1 g C3H8O3
1 mol H2O
b = 27.8 mol
18.0 g H2O
mol H2O
27.8
=
= 0.976
mol H2O + mol C3H8O3
27.8 + 0.684
PH2O = XH2O P °H2O = 10.9762123.8 torr2 = 23.2 torr
Comment The vapor pressure of the solution has been lowered by
23.8 torr - 23.2 torr = 0.6 torr relative to that of pure water. The
vapor-pressure lowering can be calculated directly using Equation 13.11 together with the mole fraction of the solute, C3H8O3:
Practice Exercise 1
The vapor pressure of benzene, C6H6, is 100.0 torr at 26.1 °C. Assuming Raoult’s law is obeyed, how many moles of a nonvolatile solute
must be added to 100.0 mL of benzene to decrease its vapor pressure
by 10.0% at 26.1 °C? The density of benzene is 0.8765 g>cm3.
(a) 0.011237, (b) 0.11237, (c) 0.1248, (d) 0.1282, (e) 8.765.
∆P = XC3H8O3P °H2O = 10.0242123.8 torr2 = 0.57 torr. Notice that the
use of Equation 13.11 gives one more significant figure than the number obtained by subtracting the vapor pressure of the solution from
that of the pure solvent.
Practice Exercise 2
The vapor pressure of pure water at 110 °C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm
at 110 °C. Assuming that Raoult’s law is obeyed, what is the mole
fraction of ethylene glycol in the solution?
A Closer Look
Ideal Solutions with Two or More
Volatile Components
Solutions sometimes have two or more volatile components. Gasoline,
for example, is a solution of several volatile liquids. To gain some understanding of such mixtures, consider an ideal solution of two volatile liquids, A and B. (For our purposes here, it does not matter which
we call the solute and which the solvent.) The partial pressures above
the solution are given by Raoult’s law:
PA = XAPA° and PB = XBPB°
and the total vapor pressure above the solution is
Ptotal = PA + PB = XAPA° + XB PB°
Consider a mixture of 1.0 mol of benzene 1C6H62 and 2.0 mol of
toluene 1C7H82 1Xben = 0.33, Xtol = 0.672. At 20 °C, the vapor pressures of the pure substances are P °ben = 75 torr and P °tol = 22 torr.
Thus, the partial pressures above the solution are
Pben = 10.332175 torr2 = 25 torr
Ptol = 10.672122 torr2 = 15 torr
and the total vapor pressure above the liquid is
Ptotal = Pben + Ptol = 25 torr + 15 torr = 40 torr
Note that the vapor is richer in benzene, the more volatile component.
The mole fraction of benzene in the vapor is given by the
­r atio of its vapor pressure to the total pressure (Equations 10.14
and 10.15):
Xben in vapor =
▲ Figure 13.19 The volatile components of organic mixtures can be
separated on an industrial scale in these distillation towers.
Pben
25 torr
=
= 0.63
Ptol
40 torr
Although benzene constitutes only 33% of the molecules in the solution, it makes up 63% of the molecules in the vapor.
When an ideal liquid solution containing two volatile
­components is in equilibrium with its vapor, the more volatile component will be relatively richer in the vapor. This fact forms the
basis of distillation, a technique used to separate (or partially sepa (Section 1.3)
rate) mixtures containing volatile components.
Distillation is a way of purifying liquids, and is the procedure by
which petrochemical plants achieve the separation of crude petroleum into gasoline, diesel fuel, lubricating oil, and other products
( ◀ Figure 13.19 ). Distillation is also used routinely on a small
scale in the laboratory.
Related Exercises: 13.67, 13.68
section 13.5 Colligative Properties
(Section 10.4),
An ideal gas is defined as one that obeys the ideal-gas equation
and an ideal solution is defined as one that obeys Raoult’s law. Whereas ideality for
a gas arises from a complete lack of intermolecular interaction, ideality for a solution
implies total uniformity of interaction. The molecules in an ideal solution all influence
one another in the same way—in other words, solute–solute, solvent–solvent, and solute–solvent interactions are indistinguishable from one another. Real solutions best
approximate ideal behavior when the solute concentration is low and solute and solvent
have similar molecular sizes and take part in similar types of intermolecular attractions.
Many solutions do not obey Raoult’s law exactly and so are not ideal. If, for
instance, the solvent–solute interactions in a solution are weaker than either the solvent–solvent or solute–solute interactions, the vapor pressure tends to be greater than
that predicted by Raoult’s law. When the solute–solvent interactions in a solution are
exceptionally strong, as might be the case when hydrogen bonding exists, the vapor
pressure is lower than that predicted by Raoult’s law. Although you should be aware
that these departures from ideality occur, we will ignore them for the remainder of this
chapter.
Boiling-Point Elevation
In Sections 11.5 and 11.6, we examined the vapor pressures of pure substances and how
to use them to construct phase diagrams. How does the phase diagram of a solution and,
hence, its boiling and freezing points differ from that of the pure solvent? The addition
of a nonvolatile solute lowers the vapor pressure of the solution. Thus, in ▼ Figure 13.20
the vapor-pressure curve of the solution is shifted downward relative to the vapor-pressure curve of the pure solvent.
Recall from Section 11.5 that the normal boiling point of a liquid is the temperature at which its vapor pressure equals 1 atm. Because the solution has a lower vapor
pressure than the pure solvent, a higher temperature is required for the solution to
achieve a vapor pressure of 1 atm. As a result, the boiling point of the solution is higher
than that of the pure solvent. This effect is seen in Figure 13.20. We find the normal
boiling point of the pure solvent on the graph by locating the point where the 1-atm
pressure horizontal line intersects the black vapor-pressure curve and then tracing this
point down to the temperature axis. For the solution, the 1-atm line intersects the blue
vapor-pressure curve at a higher temperature, indicating that the solution has a higher
boiling point than the pure solvent.
The increase in the boiling point of a solution, relative to the pure solvent, depends
on the molality of the solute. But it is important to remember that boiling-point
Pressure
1 atm
Solid
Liquid
Vapor pressure of
pure liquid solvent
Vapor
pressure
of solution
Gas
Pure solvent
boiling point
Temperature
Solution
boiling point
∆Tb
▲ Figure 13.20 Phase diagram illustrating boiling-point elevation. The black lines show the
pure solvent’s phase equilibria curves, and the blue lines show the solution’s phase equilibria
curves.
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chapter 13 Properties of Solutions
elevation is proportional to the total concentration of solute particles, regardless of
whether the particles are molecules or ions. When NaCl dissolves in water, 2 mol of
solute particles (1 mol of Na+ and 1 mol of Cl- ) are formed for each mole of NaCl that
dissolves. We take this into account by defining i, the van’t Hoff factor, as the number
of fragments that a solute breaks up into for a particular solvent. The change in boiling
point for a solution compared to the pure solvent is:
∆Tb = Tb(solution) - Tb(solvent) = iKbm[13.12]
In this equation, Tb1solution2 is the boiling point of the solution, Tb1solvent2 is the
boiling point of the pure solvent, m is the molality of the solute, Kb is the molal boilingpoint-elevation constant for the solvent (which is a proportionality constant that is
experimentally determined for each solvent), and i is the van’t Hoff factor. For a nonelectrolyte, we can always assume i = 1; for an electrolyte, i will depend on how the substance ionizes in that solvent. For instance, i = 2 for NaCl in water, assuming complete
dissociation of ions. As a result, we expect the boiling-point elevation of a 1 m aqueous
solution of NaCl to be twice as large as the boiling-point elevation of a 1 m solution of a
nonelectrolyte such as sucrose. Thus, to properly predict the effect of a particular solute
on boiling-point elevation (or any other colligative property), it is important to know
(Sections 4.1 and 4.3)
whether the solute is an electrolyte or a nonelectrolyte.
Give It Some Thought
A solute dissolved in water causes the boiling point to increase by 0.51 °C. Does
this necessarily mean that the concentration of the solute is 1.0 m (▶ Table 13.3)?
Freezing-Point Depression
The vapor-pressure curves for the liquid and solid phases meet at the triple point.
(Section 11.6) In ▼ Figure 13.21 we see that the triple-point temperature of the
solution is lower than the triple-point temperature of pure liquid because the solution
has a lower vapor pressure than the pure liquid.
The freezing point of a solution is the temperature at which the first crystals of
pure solvent form in equilibrium with the solution. Recall from Section 11.6 that the
line representing the solid–liquid equilibrium rises nearly vertically from the triple
point. It is easy to see in Figure 13.21 that the triple-point temperature of the solution is
lower than that of the pure liquid, but it is also true for all points along the solid–liquid
equilibrium curve: the freezing point of the solution is lower than that of the pure liquid.
1 atm
Liquid
Pressure
552
Solid
Vapor pressure
of pure solid
solvent
Vapor pressure
of pure liquid
solvent
Pure solvent
triple point
Vapor pressure
of solution
Gas
Solution freezing
point
Pure solvent
freezing point
Solution triple
point
∆Tf
Temperature
▲ Figure 13.21 Phase diagram illustrating freezing-point depression. The black lines show the
pure solvent’s phase equilibria curves, and the blue lines show the solution’s phase equilibria
curves.
section 13.5 Colligative Properties
Table 13.3 Molal Boiling-Point-Elevation and Freezing-Point-Depression Constants
Normal Boiling
Point 1°C2
100.0
Kb1°C>m2
Benzene, C6H6
80.1
Ethanol, C2H5OH
78.4
Carbon tetrachloride, CCl4
Chloroform, CHCl3
Solvent
Water, H2O
Normal Freezing
Point 1°C2
0.0
Kf 1°C>m2
2.53
5.5
5.12
1.22
-114.6
1.99
76.8
5.02
- 22.3
29.8
61.2
3.63
- 63.5
4.68
0.51
Like the boiling-point elevation, the change in freezing point ∆Tf is directly proportional to solute molality, taking into account the van’t Hoff factor i:
∆Tf = Tf 1solution2 - Tf 1solvent2 = -iKf m[13.13]
The proportionality constant Kf is the molal freezing-point-depression constant,
analogous to Kb for boiling point elevation. Note that because the solution freezes at a
lower temperature than does the pure solvent, the value of ∆Tf is negative.
Some typical values of Kb and Kf for several common solvents are given in
▲ Table 13.3 . For water, the table shows Kb = 0.51 °C>m, which means that the
boiling point of any aqueous solution that is 1 m in nonvolatile solute particles is
0.51 °C higher than the boiling point of pure water. Because solutions generally do
not behave ideally, the constants listed in Table 13.3 serve well only for solutions
that are rather dilute.
For water, Kf is 1.86 °C>m. Therefore, any aqueous solution that is 1 m in nonvolatile solute particles (such as 1 m C6H12O6 or 0.5 m NaCl) freezes at the temperature that
is 1.86 °C lower than the freezing point of pure water.
The freezing-point depression caused by solutes has useful applications: it is why
antifreeze works in car cooling systems, and why calcium chloride 1CaCl22 promotes
the melting of ice on roads during winter.
S a mpl e
Exercise 13.8 Calculation of Boiling-Point Elevation
and Freezing-Point Depression
Automotive antifreeze contains ethylene glycol, CH21OH2CH21OH2, a nonvolatile nonelectrolyte, in water. Calculate the boiling point and freezing point of a 25.0% by mass solution of
ethylene glycol in water.
Solution
Analyze We are given that a solution contains 25.0% by mass of a nonvolatile, nonelectrolyte solute and asked to calculate the boiling and freezing points of the solution. To do this, we need to
calculate the boiling-point elevation and freezing-point depression.
Plan To calculate the boiling-point elevation and the freezing-point depression using Equations
13.12 and 13.13, we must express the concentration of the solution as molality. Let’s assume for
convenience that we have 1000 g of solution. Because the solution is 25.0% by mass ethylene
glycol, the masses of ethylene glycol and water in the solution are 250 and 750 g, respectively.
Using these quantities, we can calculate the molality of the solution, which we use with the molal
boiling-point-elevation and freezing-point-depression constants (Table 13.3) to calculate ∆Tb
and ∆Tf. We add ∆Tb to the boiling point and ∆Tf to the freezing point of the solvent to obtain
the boiling point and freezing point of the solution.
Solve The molality of the solution is calculated as follows:
Molality =
250 g C2H6O2
1000 g H2O
moles C2H6O2
1 mol C2H6O2
= a
ba
ba
b
kilograms H2O
750 g H2O
62.1 g C2H6O2
1 kg H2O
= 5.37 m
1.86
553
554
chapter 13 Properties of Solutions
We can now use Equations 13.12 and 13.13 to calculate the changes in the boiling and freezing
points:
∆Tb = iKbm = 11210.51 °C>m215.37 m2 = 2.7 °C
∆Tf = -iKfm = -11211.86 °C>m215.37 m2 = - 10.0 °C
Hence, the boiling and freezing points of the solution are readily calculated:
∆Tb = Tb1solution2 - Tb1solvent2
2.7 °C = Tb1solution2 - 100.0 °C
Tb1solution2 = 102.7 °C
∆Tf = Tf 1solution2 - Tf 1solvent2
- 10.0 °C = Tf 1solution2 - 0.0 °C
Tf 1solution2 = -10.0 °C
Comment Notice that the solution is a liquid over a larger temperature range than the pure
solvent.
Practice Exercise 1
Which aqueous solution will have the lowest freezing point? (a) 0.050 m CaCl2, (b) 0.15 m
NaCl, (c) 0.10 m HCl, (d) 0.050 m CH3COOH, (e) 0.20 m C12H22O11.
Practice Exercise 2
Referring to Table 13.3, calculate the freezing point of a solution containing 0.600 kg of
CHCl3 and 42.0 g of eucalyptol 1C10H18O2, a fragrant substance found in the leaves of eucalyptus trees.
Osmosis
Certain materials, including many membranes in biological systems and synthetic substances such as cellophane, are semipermeable. When in contact with a solution, these
materials allow only ions or small molecules—water molecules, for instance—to pass
through their network of tiny pores.
Consider a situation in which only solvent molecules are able to pass through a
semipermeable membrane placed between two solutions of different concentrations.
The rate at which the solvent molecules pass from the less concentrated solution (lower
solute concentration but higher solvent concentration) to the more concentrated solution (higher solute concentration but lower solvent concentration) is greater than the
rate in the opposite direction. Thus, there is a net movement of solvent molecules from
the solution with a lower solute concentration into the one with a higher solute concentration. In this process, called osmosis, the net movement of solvent is always toward
the solution with the lower solvent (higher solute) concentration, as if the solutions were
driven to attain equal concentrations.
▶ Figure 13.22 shows the osmosis that occurs between an aqueous solution and
pure water, separated by a semipermeable membrane. The U-tube contains water on
the left and an aqueous solution on the right. Initially, there is a net movement of water
through the membrane from left to right, leading to unequal liquid levels in the two
arms of the U-tube. Eventually, at equilibrium (middle panel of Figure 13.22), the pressure difference resulting from the unequal liquid heights becomes so large that the net
flow of water ceases. This pressure, which stops osmosis, is the osmotic pressure, Π,
of the solution. If an external pressure equal to the osmotic pressure is applied to the
solution, the liquid levels in the two arms can be equalized, as shown in the right panel
of Figure 13.22.
The osmotic pressure obeys a law similar in form to the ideal-gas law, ΠV = inRT
where Π is the osmotic pressure, V is the volume of the solution, i is the van’t Hoff
­factor, n is the number of moles of solute, R is the ideal-gas constant, and T is the
­absolute temperature. From this equation, we can write
n
Π = i a b RT = iMRT[13.14]
V
section 13.5 Colligative Properties
555
Go Figure
If the pure water in the left arm of the U-tube is replaced by a solution more concentrated than the one in the right arm, what
will happen?
Initially
Pure
solvent
Solution
Solvent
particle
(H2O)
Solute
particle
At equilibrium
Semipermeable
membrane
Net movement of H2O
is from pure water side
to solution side
At equilibrium, the flow of H2O is
the same in both directions, so there
is no net movement of H2O
▲ Figure 13.22 Osmosis is the process of a solvent moving from one compartment to another,
across a semipermeable membrane, toward higher solute concentration. Osmotic pressure is
generated at equilibrium due to the different heights of liquid on either side of the membrane
and is equivalent to the pressure needed to equalize the fluid levels across the membrane.
where M is the molarity of the solution. Because the osmotic pressure for any
solution depends on the solution concentration, osmotic pressure is a colligative
property.
If two solutions of identical osmotic pressure are separated by a semipermeable
membrane, no osmosis will occur. The two solutions are isotonic with respect to each
other. If one solution is of lower osmotic pressure, it is hypotonic with respect to the
more concentrated solution. The more concentrated solution is hypertonic with respect
to the dilute solution.
Give It Some Thought
Of two KBr solutions, one 0.50 m and the other 0.20 m, which is hypotonic with
respect to the other?
Osmosis plays an important role in living systems. The membranes of red
blood cells, for example, are semipermeable. Placing a red blood cell in a solution
that is hypertonic relative to the intracellular solution (the solution inside the cells)
causes water to move out of the cell ( Figure 13.23). This causes the cell to shrivel,
a process called crenation. Placing the cell in a solution that is hypotonic relative to
the intracellular fluid causes water to move into the cell, which may cause the cell
to rupture, a process called hemolysis. People who need body fluids or nutrients
replaced but cannot be fed orally are given solutions by intravenous (IV) infusion,
which feeds nutrients directly into the veins. To prevent crenation or hemolysis of
red blood cells, the IV solutions must be isotonic with the intracellular fluids of the
blood cells.
Pressure applied
to equalize fluid
levels in both arms
equals osmotic
pressure, ∏
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chapter 13 Properties of Solutions
Go Figure
If the fluid surrounding a patient’s red blood cells is depleted in electrolytes, is crenation or hemolysis more likely to occur?
The arrows represent the net movement of water molecules.
Low solute
concentration
High solute
concentration
Low solute
concentration
Red blood cell in isotonic medium
neither swells nor shrinks.
Crenation of red blood cell placed
in hypertonic environment
High solute
concentration
Hemolysis of red blood cell placed
in hypotonic environment
▲ Figure 13.23 Osmosis through red blood cell walls. If water moves out of the red blood
cell, it shrivels (crenation); if water moves into the red blood cell, it will swell and may burst
(hemolysis).
S a mpl e
Exercise 13.9 Osmotic Pressure Calculations
The average osmotic pressure of blood is 7.7 atm at 25 °C. What molarity of glucose 1C6H12O62
will be isotonic with blood?
Solution
Analyze We are asked to calculate the concentration of glucose in water that would be isotonic
with blood, given that the osmotic pressure of blood at 25 °C is 7.7 atm.
Plan Because we are given the osmotic pressure and temperature, we can solve for the concentra-
tion, using Equation 13.14. Because glucose is a nonelectrolyte, i = 1.
Solve
Π = iMRT
M =
Π
=
iRT
17.7 atm2
L@atm
112a0.0821
b1298 K2
mol@K
= 0.31 M
Comment In clinical situations, the concentrations of solutions are generally expressed as mass
percentages. The mass percentage of a 0.31 M solution of glucose is 5.3%. The concentration of
NaCl that is isotonic with blood is 0.16 M, because i = 2 for NaCl in water (a 0.155 M solution
of NaCl is 0.310 M in particles). A 0.16 M solution of NaCl is 0.9% mass in NaCl. This kind of
solution is known as a physiological saline solution.
Practice Exercise 1
Which of the following actions will raise the osmotic pressure of a solution? (a) decreasing the
solute concentration, (b) decreasing the temperature, (c) adding more solvent, (d) increasing
the temperature, (e) none of the above.
Practice Exercise 2
What is the osmotic pressure, in atm, of a 0.0020 M sucrose 1C12H22O112 solution at 20.0 °C?
section 13.5 Colligative Properties
557
There are many interesting biological examples of osmosis. A cucumber placed in
concentrated brine loses water via osmosis and shrivels into a pickle. People who eat a
lot of salty food retain water in tissue cells and intercellular space because of osmosis.
The resultant swelling or puffiness is called edema. Water moves from soil into plant
roots partly because of osmosis. Bacteria on salted meat or candied fruit lose water
through osmosis, shrivel, and die—thus preserving the food.
Give It Some Thought
Is the osmotic pressure of a 0.10 M solution of NaCl greater than, less than, or
equal to that of a 0.10 M solution of KBr?
Determination of Molar Mass from Colligative Properties
The colligative properties of solutions provide a useful means of determining the molar
mass of solutes. Any of the four colligative properties can be used, as shown in Sample
Exercises 13.10 and 13.11.
S a mpl e
Exercise 13.10 Molar Mass from Freezing-Point Depression
A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g of the
substance in 40.0 g of CCl4. The boiling point of the resultant solution was 0.357 °C higher than
that of the pure solvent. Calculate the molar mass of the solute.
Solution
Analyze Our goal is to calculate the molar mass of a solute based
on knowledge of the boiling-point elevation of its solution in CCl4,
∆Tb = 0.357 °C, and the masses of solute and solvent. Table 13.3
gives Kb for the solvent 1CCl42, Kb = 5.02 °C>m.
Plan We can use Equation 13.12, ∆Tb = iKbm, to calculate the molality of the solution. Because the solute is a nonelectrolyte, i = 1.
Then we can use molality and the quantity of solvent (40.0 g CCl4)
to calculate the number of moles of solute. Finally, the molar mass
of the solute equals the number of grams per mole, so we divide the
number of grams of solute (0.250 g) by the number of moles we have
just calculated.
Solve From Equation 13.12, we have
Molality =
∆Tb
0.357 °C
=
= 0.0711 m
iKb
1125.02 °C>m
Thus, the solution contains 0.0711 mol of solute per kilogram of solvent. The solution was prepared using 40.0 g = 0.0400 kg of solvent
1CCl42. The number of moles of solute in the solution is therefore
10.0400 kg CCl42a0.0711
mol solute
b = 2.84 * 10-3 mol solute
kg CCl4
The molar mass of the solute is the number of grams per mole of the
substance:
Molar mass =
0.250 g
2.84 * 10-3 mol
= 88.0 g>mol
Practice Exercise 1
A mysterious white powder could be powdered sugar
1C12H22O112, cocaine 1C17H21NO42, codeine 1C18H21NO32, norfenefrine 1C8H11NO22, or fructose 1C6H12O62. When 80 mg of
the powder is dissolved in 1.50 mL of ethanol (d = 0.789 g>cm3,
normal freezing point -114.6 °C, Kf = 1.99 °C>m), the freezing
point is lowered to -115.5 °C. What is the identity of the white
powder? (a) powdered sugar, (b) cocaine, (c) codeine, (d) norfenefrine, (e) fructose.
Practice Exercise 2
Camphor 1C10H16O2 melts at 179.8 °C, and it has a particularly
large freezing-point-depression constant, Kf = 40.0 °C>m. When
0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 °C. What is the molar mass of the solute?
S a mpl e
Exercise 13.11 Molar Mass from Osmotic Pressure
The osmotic pressure of an aqueous solution of a certain protein was measured to determine
the protein’s molar mass. The solution contained 3.50 mg of protein dissolved in sufficient
water to form 5.00 mL of solution. The osmotic pressure of the solution at 25 °C was found to
be 1.54 torr. Treating the protein as a nonelectrolyte, calculate its molar mass.
Solution
Analyze Our goal is to calculate the molar mass of a high-molecular-
mass protein, based on its osmotic pressure and a knowledge of the
mass of protein and solution volume. Since the protein will be considered as a nonelectrolyte, i = 1.
Plan The temperature 1T = 25 °C2 and osmotic pressure
1Π = 1.54 torr2 are given, and we know the value of R so we can use
Equation 13.14 to calculate the molarity of the solution, M. In doing
so, we must convert temperature from °C to K and the osmotic pressure from torr to atm. We then use the molarity and the volume of
558
chapter 13 Properties of Solutions
the solution (5.00 mL) to determine the number of moles of solute.
Finally, we obtain the molar mass by dividing the mass of the solute
(3.50 mg) by the number of moles of solute.
Solve Solving Equation 13.14 for molarity gives
1 atm
11.54 torr2a
b
760 torr
mol
Π
=
= 8.28 * 10-5
Molarity =
iRT
L
L@atm
112a0.0821
b1298 K2
mol@K
Because the volume of the solution is 5.00 mL = 5.00 * 10-3 L, the
number of moles of protein must be
Moles = 18.28 * 10-5 mol>L215.00 * 10-3 L2 = 4.14 * 10-7 mol
The molar mass is the number of grams per mole of the substance.
Because we know the sample has a mass of 3.50 mg = 3.50 * 10-3 g,
we can calculate the molar mass by dividing the number of grams in
the sample by the number of moles we just calculated:
Molar mass =
grams
moles
=
3.50 * 10-3 g
4.14 * 10-7 mol
= 8.45 * 103 g>mol
Comment Because small pressures can be measured easily and accurately, osmotic pressure measurements provide a useful way to determine the molar masses of large molecules.
Practice Exercise 1
Proteins frequently form complexes in which 2, 3, 4, or even more
individual proteins (“monomers”) interact specifically with each other
via hydrogen bonds or electrostatic interactions. The entire assembly
of proteins can act as one unit in solution, and this assembly is called
the “quaternary structure” of the protein. Suppose you discover a new
protein whose monomer molar mass is 25,000 g>mol. You measure
an osmotic pressure of 0.0916 atm at 37 °C for 7.20 g of the protein in
10.00 mL of an aqueous solution. How many protein monomers form
the quaternary protein structure in solution? Treat the protein as a
nonelectrolyte. (a) 1, (b) 2, (c) 3, (d) 4 , (e) 8.
Practice Exercise 2
A sample of 2.05 g of polystyrene of uniform polymer chain length
was dissolved in enough toluene to form 0.100 L of solution. The
osmotic pressure of this solution was found to be 1.21 kPa at
25 °C. Calculate the molar mass of the polystyrene.
A Closer Look
The van’t Hoff Factor
The colligative properties of solutions depend on the total concentration
of solute particles, regardless of whether the particles are ions or molecules. Thus, we expect a 0.100 m solution of NaCl to have a freezingpoint ­depression of 12210.100 m211.86 °C>m2 = 0.372 °C because it is
0.100 m in Na+1aq2 and 0.100 m in Cl-1aq2. The measured freezingpoint depression is only 0.348 °C, however, and the situation is similar
for other strong electrolytes. A 0.100 m solution of KCl, for example,
freezes at -0.344 °C.
The difference between expected and observed colligative
properties for strong electrolytes is due to electrostatic attractions
between ions. As the ions move about in solution, ions of opposite charge collide and “stick together” for brief moments. While
they are together, they behave as a single particle called an ion pair
(▶ Figure 13.24). The number of independent particles is thereby
reduced, causing a reduction in the freezing-point depression (as
well as in boiling-point elevation, vapor-pressure reduction, and
­osmotic pressure).
We have been assuming that the van’t Hoff factor, i, is equal to
the number of ions per formula unit of the electrolyte. The true (measured) value of this factor, however, is given by the ratio of the measured value of a colligative property to the value calculated when the
substance is assumed to be a nonelectrolyte. Using the freezing-point
depression, for example, we have
i =
∆Tf 1measured2
∆Tf 1calculated for nonelectrolyte2
one SO42- per formula unit. In the absence of any information about
the actual value of i for a solution, we will use the limiting value in
calculations.
Two trends are evident in ▶ Table 13.4 , which gives
­m easured van’t Hoff factors for several substances at different
Na+
Cl−
[13.15]
The limiting value of i can be determined for a salt from the
number of ions per formula unit. For NaCl, for example, the limiting
van’t Hoff factor is 2 because NaCl consists of one Na+ and one Cl- per
formula unit; for K2SO4 it is 3 because K2SO4 consists of two K+ and
One ion pair
▲ Figure 13.24 Ion pairing and colligative properties. A
solution of NaCl contains not only separated Na+ 1aq2 and
Cl- 1aq2 ions but ion pairs as well.
559
section 13.6 Colloids
­ ilutions. First, dilution affects the value of i for electrolytes; the
d
more dilute the solution, the more closely i approaches the expected value based on the number of ions in the formula unit.
Thus, we conclude that the extent of ion pairing in electrolyte solutions decreases upon dilution. Second, the lower the charges on the
ions, the less i departs from the expected value because the extent
of ion pairing decreases as the ionic charges decrease. Both trends
are consistent with simple electrostatics: The force of interaction
between charged particles decreases as their separation increases
and as their charges decrease.
Table 13.4 Measured and Expected van’t Hoff Factors
for Several Substances at 25 °C
Concentration
0.100 m
0.0100 m
0.00100 m
Expected
Value
Sucrose
1.00
1.00
1.00
1.00
NaCl
1.87
1.94
1.97
2.00
K2SO4
2.32
2.70
2.84
3.00
MgSO4
1.21
1.53
1.82
2.00
Compound
Related Exercises: 13.83, 13.84, 13.103, 13.105
13.6 | Colloids
Some substances appear to initially dissolve in a solvent, but over time, the substance
separates from the pure solvent. For example, finely divided clay particles dispersed in
water eventually settle out because of gravity. Gravity affects the clay particles because
they are much larger than most molecules, consisting of thousands or even millions of
atoms. In contrast, the dispersed particles in a true solution (ions in a salt solution or
glucose molecules in a sugar solution) are small. Between these extremes lie dispersed
particles that are larger than typical molecules but not so large that the components of
the mixture separate under the influence of gravity. These intermediate types of dispersions are called either colloidal dispersions or simply colloids. Colloids form the dividing line between solutions and heterogeneous mixtures. Like solutions, colloids can
be gases, liquids, or solids. Examples of each are listed in ▼ Table 13.5.
Particle size can be used to classify a mixture as colloid or solution. Colloid particles range in diameter from 5 to 1000 nm; solute particles are smaller than 5 nm in
(Section 12.9), when dispersed in
diameter. The nanomaterials we saw in Chapter 12
a liquid, are colloids. A colloid particle may even consist of a single giant molecule. The
hemoglobin molecule, for example, which carries oxygen in your blood, has molecular
dimensions of 6.5 * 5.5 * 5.0 nm and a molar mass of 64,500 g>mol.
Although colloid particles may be so small that the dispersion appears uniform
even under a microscope, they are large enough to scatter light. Consequently, most
colloids appear cloudy or opaque unless they are very dilute. (For example, homogenized milk is a colloid of fat and protein molecules dispersed in water.) Furthermore,
because they scatter light, a light beam can be seen as it passes through a colloidal
Table 13.5 Types of Colloids
Phase of
Colloid
Dispersing (solvent-like)
Substance
Dispersed (solute-like)
Substance
Colloid Type
Example
Gas
Gas
Gas
—
None (all are solutions)
Gas
Gas
Liquid
Aerosol
Fog
Gas
Gas
Solid
Aerosol
Smoke
Liquid
Liquid
Gas
Foam
Whipped cream
Liquid
Liquid
Liquid
Emulsion
Milk
Liquid
Liquid
Solid
Sol
Paint
Solid
Solid
Gas
Solid foam
Marshmallow
Solid
Solid
Liquid
Solid emulsion
Butter
Solid
Solid
Solid
Solid sol
Ruby glass
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chapter 13 Properties of Solutions
▲ Figure 13.25 Tyndall effect in the laboratory. The glass on the right contains a colloidal
dispersion; that on the left contains a solution.
dispersion (▲ Figure 13.25). This scattering of light by colloidal particles, known as
the Tyndall effect, makes it possible to see the light beam of an automobile on a dusty
dirt road, or the sunlight streaming through trees or clouds. Not all wavelengths are
scattered to the same extent. Colors at the blue end of the visible spectrum are scattered
more than those at the red end by the molecules and small dust particles in the atmosphere. As a result, our sky appears blue. At sunset, light from the sun travels through
more of the atmosphere; blue light is scattered even more, allowing the reds and yellows to pass through and be seen.
Go Figure
Hydrophilic and Hydrophobic Colloids
What is the chemical composition
of the groups that carry a negative
charge?
Hydrophilic polar and charged groups on
molecule surface help molecule remain
dispersed in water and other polar solvents
−
N
H
O C
The most important colloids are those in which the dispersing medium is water.
These colloids may be hydrophilic (“water loving”) or hydrophobic (“water fearing”). Hydrophilic colloids are most like the solutions that we have previously
examined. In the human body, the extremely large protein molecules such as enzymes and antibodies are kept in suspension by interaction with surrounding water
molecules. A hydrophilic molecule folds in such a way that its hydrophobic groups
are away from the water molecules, on the inside of the folded molecule, while its
­hydrophilic, polar groups are on the surface, interacting with the water molecules.
The hydrophilic groups generally contain oxygen or nitrogen and often carry a
charge (◀ Figure 13.26).
Give it Some Thought
−
+
▲ Figure 13.26 Hydrophilic colloidal
particle. Examples of the hydrophilic
groups that help to keep a giant molecule
(macromolecule) suspended in water.
Some proteins reside in the hydrophobic lipid bilayer of the cell membrane.
Would hydrophilic groups of these proteins still be facing the lipid “solvent”?
Hydrophobic colloids can be dispersed in water only if they are stabilized in
some way. Otherwise, their natural lack of affinity for water causes them to separate
from the water. One method of stabilization involves adsorbing ions on the surface
of the hydrophobic particles ( ▶ Figure 13.27 ). (Adsorption means to adhere to a
surface. It differs from absorption, which means to pass into the interior, as when a
section 13.6 Colloids
Like charges keep particles from colliding,
sticking together, and growing large
enough to settle out of solution
−
+
Adsorbed anions
can interact
with water
+
−
+
Cations in
solution
−
Hydrophobic
particle
−
−
−
−
+
−
+
Repulsion
−
Water
+
−
Hydrophobic
particle
−
−
+
+
−
−
−
−
−
+
+
−
+
+
−
▲ Figure 13.27 Hydrophobic colloids stabilized in water by adsorbed anions.
sponge absorbs water.) The adsorbed ions can interact with water, thereby stabilizing the colloid. At the same time, the electrostatic repulsion between adsorbed ions
on neighboring colloid particles keeps the particles from sticking together rather
than dispersing in the water.
Hydrophobic colloids can also be stabilized by hydrophilic groups on their surfaces. Oil drops are hydrophobic, for example, and they do not remain suspended in
water. Instead, they aggregate, forming an oil slick on the water surface. Sodium stearate (▼ Figure 13.28), or any similar substance having one end that is hydrophilic
(either polar or charged) and one end that is hydrophobic (nonpolar), will stabilize a
suspension of oil in water. Stabilization results from the interaction of the hydrophobic
ends of the stearate ions with the oil drops and the hydrophilic ends with the water.
Give It Some Thought
Why don’t oil drops stabilized by sodium stearate coagulate to form larger oil drops?
Go Figure
Which kind of intermolecular force attracts the stearate ion to the oil drop?
+
−
Na+
−
Stearate
ion
+
−
+
+
Hydrophobic
end
−
−
−
Hydrophilic
end
+
−
−
+
+ −
−
+
Oil
droplet
−
−
+
Water
+
▲ Figure 13.28 Stabilization of an emulsion of oil in water by stearate ions.
561
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chapter 13 Properties of Solutions
Colloid stabilization has an interesting application in the human digestive system.
When fats in our diet reach the small intestine, a hormone causes the gallbladder to
excrete a fluid called bile. Among the components of bile are compounds that have
chemical structures similar to sodium stearate; that is, they have a hydrophilic (polar)
end and a hydrophobic (nonpolar) end. These compounds emulsify the fats in the intestine and thus permit digestion and absorption of fat-soluble vitamins through the intestinal wall. The term emulsify means “to form an emulsion,” a suspension of one liquid in
another, with milk being one example (Table 13.5). A substance that aids in the formation of an emulsion is called an emulsifying agent. If you read the labels on foods and
other materials, you will find that a variety of chemicals are used as emulsifying agents.
These chemicals typically have a hydrophilic end and a hydrophobic end.
Chemistry and Life
Sickle-Cell Anemia
Our blood contains the complex protein hemoglobin, which carries
oxygen from the lungs to other parts of the body. In the genetic disease sickle-cell anemia, hemoglobin molecules are abnormal and have
a lower solubility in water, especially in their unoxygenated form.
Consequently, as much as 85% of the hemoglobin in red blood cells
crystallizes out of solution.
The cause of the insolubility is a structural change in one part of
an amino acid. Normal hemoglobin molecules contain an amino acid
that has a ¬ CH2CH2COOH group:
This change leads to the aggregation of the defective form of
hemoglobin into particles too large to remain suspended in biological fluids. It also causes the cells to distort into the sickle shape
shown in ▼ Figure 13.29 . The sickled cells tend to clog capillaries, causing severe pain, weakness, and the gradual deterioration
of vital organs. The disease is hereditary, and if both parents carry
the defective genes, it is likely that their children will possess only
­abnormal hemoglobin.
You might wonder how it is that a life-threatening disease such
as sickle-cell anemia has persisted in humans through evolutionary
time. The answer in part is that people with the disease are far less
susceptible to malaria. Thus, in tropical climates rife with malaria,
those with sickle-cell disease have lower incidence of this debilitating
disease.
O
CH2
CH2
C
Normal
OH
Normal
The polarity of the ¬ COOH group contributes to the solubility
of the hemoglobin molecule in water. In the hemoglobin molecules of
sickle-cell anemia patients, the ¬ CH2CH2COOH chain is absent and
in its place is the nonpolar (hydrophobic) ¬ CH1CH322 group:
CH
▲ Figure 13.29 A scanning electron micrograph of normal (round)
and sickle (crescent-shaped) red blood cells. Normal red blood cells are
about 6 * 10-3 mm in diameter.
CH3
CH3
Abnormal
Abnormal
Colloidal Motion in Liquids
We learned in Chapter 10 that gas molecules move at some average speed that
­depends inversely on their molar mass, in a straight line, until they collide with something. The mean free path is the average distance molecules travel between collisions.
(Section 10.8) Recall also that the kinetic-molecular theory of gases assumes that
(Section 10.7)
gas molecules are in continuous, random motion.
section 13.6 Colloids
Colloidal particles in a solution undergo random motion as a result of collisions
with solvent molecules. Because the colloidal particles are massive in comparison
with solvent molecules, their movements from any one collision are very tiny. However, there are many such collisions, and they cause a random motion of the entire
colloidal particle, called Brownian motion. In 1905, Einstein developed an equation
for the average square of the displacement of a colloidal particle, a historically very
important development. As you might expect, the larger the colloidal particle, the
shorter its mean free path in a given liquid (▼ Table 13.6). Today, the understanding
of Brownian motion is applied to diverse problems in everything from cheese-making
to medical imaging.
Table 13.6 Calculated Mean Free Path, after One Hour,
for Uncharged Colloidal Spheres in Water at 20 °C
Radius of sphere, nm
Mean Free Path, mm
1
1.23
10
0.390
100
0.123
1000
0.039
S a mpl e
Integrative Exercise Putting Concepts Together
A 0.100-L solution is made by dissolving 0.441 g of CaCl21s2 in water. (a) Calculate the osmotic
pressure of this solution at 27 °C, assuming that it is completely dissociated into its component
ions. (b) The measured osmotic pressure of this solution is 2.56 atm at 27 °C. Explain why it is
less than the value calculated in (a), and calculate the van’t Hoff factor, i, for the solute in this
solution. (c) The enthalpy of solution for CaCl2 is ∆H = - 81.3 kJ>mol. If the final temperature
of the solution is 27 °C, what was its initial temperature? (Assume that the density of the solution is 1.00 g>mL, that its specific heat is 4.18 J>g@K, and that the solution loses no heat to its
surroundings.)
Solution
(a) The osmotic pressure is given by Equation 13.14, Π = iMRT. We know the temperature,
T = 27 °C = 300 K, and the gas constant, R = 0.0821 L@atm/mol@K. We can calculate the
molarity of the solution from the mass of CaCl2 and the volume of the solution:
Molarity = a
0.441 g CaCl2
0.100 L
ba
1 mol CaCl2
b = 0.0397 mol CaCl2 >L
110 g CaCl2
(Sections 4.1 and 4.3) Thus, CaCl2
Soluble ionic compounds are strong electrolytes.
consists of metal cations 1Ca2+2 and nonmetal anions 1Cl-2. When completely dissociated,
each CaCl2 unit forms three ions (one Ca2+ and two Cl-). Hence, the calculated osmotic
pressure is
Π = iMRT = 13210.0397 mol>L210.0821 L@atm>mol@K21300 K2 = 2.93 atm
(b) The actual values of colligative properties of electrolytes are less than those calculated because the electrostatic interactions between ions limit their independent movements. In this
case, the van’t Hoff factor, which measures the extent to which electrolytes actually dissociate into ions, is given by
i =
=
Π1measured2
Π1calculated for nonelectrolyte2
2.56 atm
= 2.62
10.0397 mol>L210.0821 L@atm>mol@K21300 K2
Thus, the solution behaves as if the CaCl2 has dissociated into 2.62 particles instead of the
ideal 3.
(c) If the solution is 0.0397 M in CaCl2 and has a total volume of 0.100 L, the number of moles
of solute is 10.100 L210.0397 mol>L2 = 0.00397 mol. Hence, the quantity of heat generated in forming the solution is 10.00397 mol21-81.3 kJ>mol2 = -0.323 kJ. The solution
563
564
chapter 13 Properties of Solutions
absorbs this heat, causing its temperature to increase. The relationship between temperature
change and heat is given by Equation 5.22:
q = 1specific heat21grams21∆T2
The heat absorbed by the solution is q = + 0.323 kJ = 323 J. The mass of the 0.100 L of
solution is 1100 mL211.00 g>mL2 = 100 g (to three significant figures). Thus, the temperature change is
∆T =
=
q
1specific heat of solution21grams of solution2
323 J
= 0.773 K
14.18 J>g@K21100 g2
(Section 1.4) Because the solution temA kelvin has the same size as a degree Celsius.
perature increases by 0.773 °C, the initial temperature was 27.0 °C - 0.773 °C = 26.2 °C.
Chapter Summary and Key Terms
The Solution Process (Section 13.1) Solutions form when
one substance disperses uniformly throughout another. The attractive
interaction of solvent molecules with solute is called solvation. When
the solvent is water, the interaction is called hydration. The dissolution
of ionic substances in water is promoted by hydration of the separated
ions by the polar water molecules. The overall enthalpy change upon
solution formation may be either positive or negative. Solution formation is favored both by a positive entropy change, corresponding to an
increased dispersal of the components of the solution, and by a negative enthalpy change, indicating an exothermic process.
Saturated Solutions and Solubility (Section 13.2) The
equilibrium between a saturated solution and undissolved solute is
dynamic; the process of solution and the reverse process, crystallization,
occur simultaneously. In a solution in equilibrium with undissolved solute, the two processes occur at equal rates, giving a saturated solution.
If there is less solute present than is needed to saturate the solution, the
solution is unsaturated. When solute concentration is greater than the
equilibrium concentration value, the solution is supersaturated. This is an
unstable condition, and separation of some solute from the solution will
occur if the process is initiated with a solute seed crystal. The amount of
solute needed to form a saturated solution at any particular temperature
is the solubility of that solute at that temperature.
Factors Affecting Solubility (Section 13.3) The solubility
of one substance in another depends on the tendency of systems to
become more random, by becoming more dispersed in space, and on
the relative intermolecular solute–solute and solvent–solvent energies
compared with solute–solvent interactions. Polar and ionic solutes
tend to dissolve in polar solvents, and nonpolar solutes tend to dissolve in nonpolar solvents (“like dissolves like”). Liquids that mix in
all proportions are miscible; those that do not dissolve significantly in
one another are immiscible. Hydrogen-bonding interactions between
solute and solvent often play an important role in determining solubility; for example, ethanol and water, whose molecules form hydrogen bonds with each other, are miscible. The solubilities of gases in a
liquid are generally proportional to the pressure of the gas over the
solution, as expressed by Henry’s law : Sg = kPg. The solubilities of
most solid solutes in water increase as the temperature of the solution increases. In contrast, the solubilities of gases in water generally
decrease with increasing temperature.
Expressing Solution Concentrations (Section 13.4) Concentrations of solutions can be expressed quantitatively by several different measures, including mass percentage [(mass solute/mass
solution) * 100] parts per million (ppm), parts per billion (ppb), and
mole fraction. Molarity, M, is defined as moles of solute per liter of
solution; molality, m, is defined as moles of solute per kilogram of solvent. Molarity can be converted to these other concentration units if
the density of the solution is known.
Colligative Properties (Section 13.5) A physical property
of a solution that depends on the concentration of solute particles
present, regardless of the nature of the solute, is a colligative property.
Colligative properties include vapor-pressure lowering, freezingpoint lowering, b
­ oiling-point elevation, and osmotic pressure. Raoult’s
law expresses the lowering of vapor pressure. An ideal solution obeys
Raoult’s law. Differences in solvent–solute as compared with solvent–
solvent and solute–solute intermolecular forces cause many solutions
to depart from ideal behavior.
A solution containing a nonvolatile solute possesses a higher
boiling point than the pure solvent. The molal boiling-point-elevation
constant, Kb, represents the increase in boiling point for a 1 m solution of solute particles as compared with the pure solvent. Similarly,
the molal freezing-point-depression constant, Kf , measures the lowering
of the freezing point of a solution for a 1 m solution of solute particles.
The temperature changes are given by the equations ∆Tb = iKbm and
∆Tf = -iKf m where i is the van’t Hoff factor, which represents how
many particles the solute breaks up into in the solvent. When NaCl
dissolves in water, two moles of solute particles are formed for each
mole of dissolved salt. The boiling point or freezing point is thus elevated or depressed, respectively, approximately twice as much as that
of a nonelectrolyte solution of the same concentration. Similar considerations apply to other strong electrolytes.
Osmosis is the movement of solvent molecules through a semipermeable membrane from a less concentrated to a more concentrated solution.
This net movement of solvent generates an osmotic pressure, Π, which
can be measured in units of gas pressure, such as atm. The osmotic pressure of a solution is proportional to the solution molarity: Π = iMRT.
Osmosis is a very important process in living systems, in which cell walls
act as semipermeable membranes, permitting the passage of water but
restricting the passage of ionic and macromolecular components.
Key Equations
Colloids (Section 13.6) Particles that are large on the molecular
scale but still small enough to remain suspended indefinitely in a solvent system form colloids, or colloidal dispersions. Colloids, which are
intermediate between solutions and heterogeneous mixtures, have many
practical applications. One useful physical property of colloids, the scattering of visible light, is referred to as the Tyndall effect. Aqueous colloids are classified as hydrophilic or hydrophobic. Hydrophilic colloids
565
are common in living organisms, in which large molecular aggregates
(enzymes, antibodies) remain suspended because they have many polar,
or charged, atomic groups on their surfaces that interact with water.
Hydrophobic colloids, such as small droplets of oil, may remain in suspension through adsorption of charged particles on their surfaces.
Colloids undergo Brownian motion in liquids, analogous to the
random three-dimensional motion of gas molecules.
Learning Outcomes After studying this chapter, you should be able to:
• Describe how enthalpy and entropy changes affect solution
­formation. (Section 13.1)
• Describe the relationship between intermolecular forces and s­ olubility,
including use of the “like dissolves like” rule. (Sections 13.1 and 13.3)
• Describe the role of equilibrium in the solution process and its
­relationship to the solubility of a solute. (Section 13.2)
• Describe the effect of temperature on the solubility of solids and
gases in liquids. (Section 13.3)
• Describe the relationship between the partial pressure of a gas and
its solubility. (Section 13.3)
• Calculate the concentration of a solution in terms of ­molarity, molality,
mole fraction, percent composition, and parts per ­million and be able
to interconvert between them. (Section 13.4)
• Describe what a colligative property is and explain the difference
between the effects of nonelectrolytes and electrolytes on colligative properties. (Section 13.5)
• Calculate the vapor pressure of a solvent over a solution.
(Section 13.5)
• Calculate the boiling-point elevation and freezing-point depression
of a solution. (Section 13.5)
• Calculate the osmotic pressure of a solution. (Section 13.5)
• Explain the difference between a solution and a colloid.
(Section 13.6)
• Describe the similarities between the motions of gas molecules
and the motions of colloids in a liquid. (Section 13.6)
Key Equations
• Sg = kPg
• Mass % of component =
• ppm of component =
• Mole fraction of component =
• Molarity =
moles of solute
liters of soln
[13.8]
Concentration in terms of molarity
• Molality =
moles of solute
kilograms of solvent
[13.9]
Concentration in terms of molality
°
• Psolution = Xsolvent Psolvent
[13.10] Raoult’s law, calculating vapor pressure of solvent above a solution
• ∆Tb = iKbm [13.12] Calculating the boiling-point elevation of a solution
• ∆Tf = - iKfm
[13.13] Calculating the freezing-point depression of a solution
• Π = ia bRT = iMRT
n
V
mass of component in soln
total mass of soln
mass of component in soln
total mass of soln
* 100
* 106
moles of component
total moles of all components
[13.4]
Henry’s law, which relates gas solubility to partial pressure
[13.5]
Concentration in terms of mass percent
[13.6]
Concentration in terms of parts per million (ppm)
[13.7]
Concentration in terms of mole fraction
[13.14] Calculating the osmotic pressure of a solution
566
chapter 13 Properties of Solutions
Exercises
Visualizing Concepts
13.1 Rank the contents of the following containers in order of
­increasing entropy: [Section 13.1]
13.6The solubility of Xe in water at 1 atm pressure and 20 °C is
­approximately 5 * 10-3 M. (a) Compare this with the solubilities of Ar and Kr in water (Table 13.1). (b) What properties
of the rare gas atoms account for the variation in solubility?
[Section 13.3]
13.7The structures of vitamins E and B6 are shown below. Predict
which is more water soluble and which is more fat soluble.
Explain. [Section 13.3]
(a)
(b)
(c)
13.2This figure shows the interaction of a cation with surrounding water molecules.
+
Vitamin B6
(a) Which atom of water is associated with the cation? Explain.
(b) Which of the following explanations accounts for the
fact that the ion-solvent interaction is greater for Li+
than for K+?
a. Li+ is of lower mass than K+.
b. The ionization energy of Li is higher than that for K.
c. Li+ has a smaller ionic radius than K+.
d. Li has a lower density than K.
e. Li reacts with water more slowly than K. [Section 13.1]
13.3Consider two ionic solids, both composed of singly-charged
ions, that have different lattice energies. (a) Will the solids have
the same solubility in water? (b) If not, which solid will be more
soluble in water, the one with the larger lattice energy or the
one with the smaller lattice energy? Assume that solute-solvent
interactions are the same for both solids. [Section 13.1]
Vitamin E
13.8You take a sample of water that is at room temperature and in
contact with air and put it under a vacuum. Right away, you
see bubbles leave the water, but after a little while, the bubbles
stop. As you keep applying the vacuum, more bubbles appear.
A friend tells you that the first bubbles were water vapor, and
the low pressure had reduced the boiling point of water, causing the water to boil. Another friend tells you that the first
bubbles were gas molecules from the air (oxygen, nitrogen,
and so forth) that were dissolved in the water. Which friend is
mostly likely to be correct? What, then, is responsible for the
second batch of bubbles? [Section 13.4]
13.9The figure shows two identical volumetric flasks containing
the same solution at two temperatures.
(a) Does the molarity of the solution change with the change
in temperature? Explain.
(b) Does the molality of the solution change with the change
in temperature? Explain. [Section 13.4]
13.4Are gases always miscible with each other? Explain. [Section 13.1]
13.5 Which of the following is the best representation of a
­saturated solution? Explain your reasoning. [Section 13.2]
(a)
(b)
(c)
25 °C
55 °C
Exercises
13.10This portion of a phase diagram shows the vapor-pressure
curves of a volatile solvent and of a solution of that solvent
containing a nonvolatile solute. (a) Which line represents the
solution? (b) What are the normal boiling points of the solvent and the solution? [Section 13.5]
1.0
567
(b) In making a solution, the enthalpy of mixing is always a
positive number.
(c) An increase in entropy favors mixing.
13.14Indicate whether each statement is true or false: (a) NaCl dissolves in water but not in benzene 1C6H62 because benzene is
denser than water. (b) NaCl dissolves in water but not in benzene because water has a large dipole moment and benzene
has zero dipole moment. (c) NaCl dissolves in water but not
in benzene because the water–ion interactions are stronger
than benzene–ion interactions.
P (atm)
13.15Indicate the type of solute–solvent interaction (Section 11.2)
that should be most important in each of the following solutions: (a) CCl4 in benzene 1C6H62, (b) methanol 1CH3OH2 in
water, (c) KBr in water, (d) HCl in acetonitrile 1CH3CN2.
40
50
60
T (°C)
13.16Indicate the principal type of solute–solvent interaction in
each of the following solutions and rank the solutions from
weakest to strongest solute–solvent interaction: (a) KCl in water, (b) CH2Cl2 in benzene 1C6H62, (c) methanol 1CH3OH2 in
water.
70
13.11Suppose you had a balloon made of some highly flexible semipermeable membrane. The balloon is filled completely with
a 0.2 M solution of some solute and is submerged in a 0.1 M
solution of the same solute:
13.17An ionic compound has a very negative ∆Hsoln in water.
(a) Would you expect it to be very soluble or nearly insoluble
in water? (b) Which term would you expect to be the largest
negative number: ∆Hsolvent, ∆Hsolute, or ∆Hmix?
13.18When ammonium chloride dissolves in water, the solution
becomes colder. (a) Is the solution process exothermic or endothermic? (b) Why does the solution form?
13.19(a) In Equation 13.1, which of the enthalpy terms for dissolving an ionic solid would correspond to the lattice energy?
(b) Which energy term in this equation is always exothermic?
0.1 M
0.2 M
Initially, the volume of solution in the balloon is 0.25 L.
­Assuming the volume outside the semipermeable membrane
is large, as the illustration shows, what would you expect for
the solution volume inside the balloon once the system has
come to equilibrium through osmosis? [Section 13.5]
13.12The molecule n-octylglucoside, shown here, is widely used in
biochemical research as a nonionic detergent for “solubilizing”
large hydrophobic protein molecules. What characteristics of
this molecule are important for its use in this way? [Section 13.6]
13.20For the dissolution of LiCl in water, ∆Hsoln = - 37 kJ>mol.
Which term would you expect to be the largest negative number: ∆Hsolvent, ∆Hsolute, or ∆Hmix?
13.21Two nonpolar organic liquids, hexane 1C6H142 and heptane
1C7H162, are mixed. (a) Do you expect ∆Hsoln to be a large
positive number, a large negative number, or close to zero?
­Explain. (b) Hexane and heptane are miscible with each other
in all proportions. In making a solution of them, is the entropy
of the system increased, decreased, or close to zero, compared
to the separate pure liquids?
13.22The enthalpy of solution of KBr in water is about
+198 kJ>mol. Nevertheless, the solubility of KBr in water
is relatively high. Why does the solution process occur even
though it is endothermic?
Saturated Solutions; Factors Affecting Solubility
(Sections 13.2 and 13.3)
13.23The solubility of Cr1NO323 # 9 H2O in water is 208 g per 100 g
of water at 15 °C. A solution of Cr1NO323 # 9 H2O in water at
35 °C is formed by dissolving 324 g in 100 g of water. When
this solution is slowly cooled to 15 °C, no precipitate forms.
(a) What term describes this solution? (b) What action might
you take to initiate crystallization? Use molecular-level processes to explain how your suggested procedure works.
The Solution Process (Section 13.1)
13.13Indicate whether each statement is true or false:
(a) A solute will dissolve in a solvent if solute–solute interactions are stronger than solute-solvent interactions.
13.24The solubility of MnSO4 # H2O in water at 20 °C is 70 g per
100 mL of water. (a) Is a 1.22 M solution of MnSO4 # H2O
in water at 20 °C saturated, supersaturated, or unsaturated?
(b) Given a solution of MnSO4 # H2O of unknown concentration, what experiment could you perform to determine
whether the new solution is saturated, supersaturated, or
unsaturated?
568
chapter 13 Properties of Solutions
13.25By referring to Figure 13.15, determine whether the addition
of 40.0 g of each of the following ionic solids to 100 g of water
at 40 °C will lead to a saturated solution: (a) NaNO3, (b) KCl,
(c) K2Cr2O7, (d) Pb1NO322.
13.26By referring to Figure 13.15, determine the mass of each of the
following salts required to form a saturated solution in 250 g
of water at 30 °C: (a) KClO3, (b) Pb1NO322, (c) Ce21SO423.
13.27Consider water and glycerol, CH21OH2CH1OH2CH2OH.
(a) Would you expect them to be miscible in all proportions?
Explain. (b) List the intermolecular attractions that occur between a water molecule and a glycerol molecule.
13.28Oil and water are immiscible. Which is the most likely reason?
(a) Oil molecules are denser than water. (b) Oil molecules are
composed mostly of carbon and hydrogen. (c) Oil molecules
have higher molar masses than water. (d) Oil molecules have
higher vapor pressures than water. (e) Oil molecules have
higher boiling points than water.
13.29Common laboratory solvents include acetone 1CH3COCH32,
methanol 1CH3OH2, toluene 1C6H5CH32, and water. Which
of these is the best solvent for nonpolar solutes?
13.30Would you expect alanine (an amino acid) to be more soluble
in water or in hexane? Explain.
1C6H62 or glycerol, CH21OH2CH1OH2CH2OH, (c) octanoic
acid, CH3CH2CH2CH2CH2CH2CH2COOH, or acetic acid,
CH3COOH? Explain your answer in each case.
13.34Which of the following in each pair is likely to be more soluble in water: (a) cyclohexane 1C6H122 or glucose 1C6H12O62,
(b) propionic acid 1CH3CH2COOH2 or sodium propionate
1CH3CH2COONa2, (c) HCl or ethyl chloride 1CH3CH2Cl2?
Explain in each case.
13.35(a) Explain why carbonated beverages must be stored in
sealed containers. (b) Once the beverage has been opened,
why does it maintain more carbonation when refrigerated
than at room temperature?
13.36Explain why pressure substantially affects the solubility of O2
in water but has little effect on the solubility of NaCl in water.
13.37The Henry’s law constant for helium gas in water at 30 °C
is 3.7 * 10-4 M>atm and the constant for N2 at 30 °C is
6.0 * 10-4 M>atm. If the two gases are each present at 1.5
atm pressure, calculate the solubility of each gas.
13.38The partial pressure of O2 in air at sea level is 0.21 atm. Using
the data in Table 13.1, together with Henry’s law, calculate the
molar concentration of O2 in the surface water of a mountain
lake saturated with air at 20 °C and an atmospheric pressure
of 650 torr.
Concentrations of Solutions (Section 13.4)
13.39(a) Calculate the mass percentage of Na2SO4 in a solution
containing 10.6 g of Na2SO4 in 483 g of water. (b) An ore contains 2.86 g of silver per ton of ore. What is the concentration
of silver in ppm?
Alanine
13.31(a) Would you expect stearic acid, CH31CH2216COOH, to be
more soluble in water or in carbon tetrachloride? Explain.
(b) Which would you expect to be more soluble in water,
cyclohexane or dioxane? Explain.
CH2
O
H2C
H2C
CH2
H2C
CH2
H2C
CH2
CH2
O
CH2
Dioxane
Cyclohexane
13.32Ibuprofen, widely used as a pain reliever, has a limited solubility in water, less than 1 mg>mL. Which part of the molecule’s
structure (gray, white, red) contributes to its water solubility?
Which part of the molecule (gray, white, red) contributes to
its water insolubility?
13.40(a) What is the mass percentage of iodine in a solution containing 0.035 mol I2 in 125 g of CCl4? (b) Seawater ­contains
0.0079 g of Sr2+ per kilogram of water. What is the concentration of Sr2+ in ppm?
13.41A solution is made containing 14.6 g of CH3OH in 184 g of
H2O. Calculate (a) the mole fraction of CH3OH, (b) the mass
percent of CH3OH, (c) the molality of CH3OH.
13.42A solution is made containing 20.8 g of phenol 1C6H5OH2 in
425 g of ethanol 1CH3CH2OH2. Calculate (a) the mole fraction of phenol, (b) the mass percent of phenol, (c) the molality of phenol.
13.43Calculate the molarity of the following aqueous solutions:
(a) 0.540 g of Mg1NO322 in 250.0 mL of solution, (b) 22.4 g of
LiClO4 # 3 H2O in 125 mL of solution, (c) 25.0 mL of 3.50 M
HNO3 diluted to 0.250 L.
13.44What is the molarity of each of the following solutions:
(a) 15.0 g of Al21SO423 in 0.250 mL solution, (b) 5.25 g of
Mn1NO322 # 2 H2O in 175 mL of solution, (c) 35.0 mL of
9.00 M H2SO4 diluted to 0.500 L?
13.45Calculate the molality of each of the following solutions:
(a) 8.66 g of benzene 1C6H62 dissolved in 23.6 g of carbon tetrachloride 1CCl42, (b) 4.80 g of NaCl dissolved in 0.350 L of
water.
Ibuprofen
13.33Which of the following in each pair is likely to be more
soluble in hexane, C6H14: (a) CCl4 or CaCl2, (b) benzene
13.46(a) What is the molality of a solution formed by dissolving
1.12 mol of KCl in 16.0 mol of water? (b) How many grams
of sulfur 1S82 must be dissolved in 100.0 g of naphthalene
1C10H82 to make a 0.12 m solution?
13.47A sulfuric acid solution containing 571.6 g of H2SO4 per
liter of solution has a density of 1.329 g>cm3. Calculate
Exercises
569
(a) the mass percentage, (b) the mole fraction, (c) the molality,
(d) the molarity of H2SO4 in this solution.
13.48Ascorbic acid 1vitamin C, C6H8O62 is a water-soluble vitamin. A solution containing 80.5 g of ascorbic acid dissolved in
210 g of water has a density of 1.22 g>mL at 55 °C. Calculate
(a) the mass percentage, (b) the mole fraction, (c) the molality, (d) the molarity of ascorbic acid in this solution.
13.49The density of acetonitrile 1CH3CN2 is 0.786 g>mL and the
density of methanol 1CH3OH2 is 0.791 g>mL. A solution is
made by dissolving 22.5 mL of CH3OH in 98.7 mL of CH3CN.
(a) What is the mole fraction of methanol in the solution?
(b) What is the molality of the solution? (c) Assuming that
the volumes are additive, what is the molarity of CH3OH in
the solution?
13.50The density of toluene 1C7H82 is 0.867 g>mL, and the density of thiophene 1C4H4S2 is 1.065 g>mL. A solution is made
by ­dissolving 8.10 g of thiophene in 250.0 mL of toluene.
(a) Calculate the mole fraction of thiophene in the solution. (b) Calculate the molality of thiophene in the solution.
(c) Assuming that the volumes of the solute and solvent are
additive, what is the ­molarity of thiophene in the solution?
13.51Calculate the number of moles of solute present in each of
the following aqueous solutions: (a) 600 mL of 0.250 M SrBr2,
(b) 86.4 g of 0.180 m KCl, (c) 124.0 g of a solution that is 6.45%
glucose 1C6H12O62 by mass.
13.52Calculate the number of moles of solute present in each of the following solutions: (a) 255 mL of 1.50 M HNO31aq2, (b) 50.0 mg
of an aqueous solution that is 1.50 m NaCl, (c) 75.0 g of an
aqueous solution that is 1.50% sucrose 1C12H22O112 by mass.
13.53Describe how you would prepare each of the following
aqueous solutions, starting with solid KBr: (a) 0.75 L of
1.5 * 10-2 M KBr, (b) 125 g of 0.180 m KBr, (c) 1.85 L of a
solution that is 12.0% KBr by mass (the density of the solution is 1.10 g>mL), (d) a 0.150 M solution of KBr that contains just enough KBr to precipitate 16.0 g of AgBr from a
solution containing 0.480 mol of AgNO3.
13.54Describe how you would prepare each of the following aqueous solutions: (a) 1.50 L of 0.110 M 1NH422SO4 solution,
starting with solid 1NH422SO4; (b) 225 g of a solution that is
0.65 m in Na2CO3, starting with the solid solute; (c) 1.20 L of
a solution that is 15.0% Pb1NO322 by mass (the density of the
solution is 1.16 g>mL), starting with solid solute; (d) a 0.50 M
solution of HCl that would just neutralize 5.5 g of Ba1OH22
starting with 6.0 M HCl.
13.55Commercial aqueous nitric acid has a density of 1.42 g>mL
and is 16 M. Calculate the percent HNO3 by mass in the
solution.
13.56Commercial concentrated aqueous ammonia is 28% NH3 by
mass and has a density of 0.90 g>mL. What is the molarity of
this solution?
13.57Brass is a substitutional alloy consisting of a solution of copper and zinc. A particular sample of red brass consisting of
80.0% Cu and 20.0% Zn by mass has a density of 8750 kg>m3.
(a) What is the molality of Zn in the solid solution? (b) What
is the molarity of Zn in the solution?
13.58Caffeine 1C8H10N4O22 is a stimulant found in coffee and tea.
If a solution of caffeine in the solvent chloroform 1CHCl32
has a concentration of 0.0500 m, calculate (a) the percentage
of caffeine by mass, (b) the mole fraction of caffeine in the
solution.
Caffeine
13.59During a person’s typical breathing cycle, the CO2 concentration in the expired air rises to a peak of 4.6% by volume.
(a) Calculate the partial pressure of the CO2 in the expired
air at its peak, assuming 1 atm pressure and a body temperature of 37 °C. (b) What is the molarity of the CO2 in
the ­e xpired air at its peak, assuming a body temperature
of 37 °C?
13.60Breathing air that contains 4.0% by volume CO2 over time
causes rapid breathing, throbbing headache, and nausea,
among other symptoms. What is the concentration of CO2 in
such air in terms of (a) mol percentage, (b) molarity, assuming 1 atm pressure and a body temperature of 37 °C?
Colligative Properties (Section 13.5)
13.61You make a solution of a nonvolatile solute with a liquid solvent. Indicate whether each of the following statements is true
or false. (a) The freezing point of the solution is higher than
that of the pure solvent. (b) The freezing point of the solution
is lower than that of the pure solvent. (c) The boiling point of
the solution is higher than that of the pure solvent. (d) The
boiling point of the solution is lower than that of the pure
solvent.
13.62You make a solution of a nonvolatile solute with a liquid solvent. Indicate if each of the following statements is true or
false. (a) The freezing point of the solution is unchanged by
addition of the solvent. (b) The solid that forms as the solution freezes is nearly pure solute. (c) The freezing point of the
solution is independent of the concentration of the solute.
(d) The boiling point of the solution increases in proportion
to the concentration of the solute. (e) At any temperature, the
vapor pressure of the solvent over the solution is lower than
what it would be for the pure solvent.
13.63Consider two solutions, one formed by adding 10 g of glucose
1C6H12O62 to 1 L of water and the other formed by adding 10 g
of sucrose 1C12H22O112 to 1 L of water. Calculate the vapor
pressure for each solution at 20 °C; the vapor pressure of pure
water at this temperature is 17.5 torr.
13.64(a) What is an ideal solution? (b) The vapor pressure of pure
water at 60 °C is 149 torr. The vapor pressure of water over a
solution at 60 °C containing equal numbers of moles of water
and ethylene glycol (a nonvolatile solute) is 67 torr. Is the solution ideal according to Raoult’s law? Explain.
13.65(a) Calculate the vapor pressure of water above a solution
prepared by adding 22.5 g of lactose 1C12H22O112 to 200.0 g
of water at 338 K. (Vapor-pressure data for water are given
in Appendix B.) (b) Calculate the mass of propylene glycol
1C3H8O22 that must be added to 0.340 kg of water to reduce
the vapor pressure by 2.88 torr at 40 °C.
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chapter 13 Properties of Solutions
13.66(a) Calculate the vapor pressure of water above a solution
prepared by dissolving 28.5 g of glycerin 1C3H8O32 in 125 g
of water at 343 K. (The vapor pressure of water is given in
Appendix B.) (b) Calculate the mass of ethylene glycol
1C2H6O22 that must be added to 1.00 kg of ethanol 1C2H5OH2
to reduce its vapor pressure by 10.0 torr at 35 °C. The vapor
pressure of pure ethanol at 35 °C is 1.00 * 102 torr.
[13.67] At 63.5 °C, the vapor pressure of H2O is 175 torr, and that of
ethanol 1C2H5OH2 is 400 torr. A solution is made by mixing
equal masses of H2O and C2H5OH. (a) What is the mole fraction of ethanol in the solution? (b) Assuming ideal-solution
behavior, what is the vapor pressure of the solution at 63.5 °C?
(c) What is the mole fraction of ethanol in the vapor above the
solution?
[13.68] At 20 °C, the vapor pressure of benzene 1C6H62 is 75 torr,
and that of toluene 1C7H82 is 22 torr. Assume that benzene
and toluene form an ideal solution. (a) What is the composition in mole fraction of a solution that has a vapor pressure of
35 torr at 20 °C? (b) What is the mole fraction of benzene in
the vapor above the solution described in part (a)?
13.69(a) Does a 0.10 m aqueous solution of NaCl have a higher
boiling point, a lower boiling point, or the same boiling point
as a 0.10 m aqueous solution of C6H12O6? (b) The experimental boiling point of the NaCl solution is lower than that
calculated assuming that NaCl is completely dissociated in
­solution. Why is this the case?
13.70Arrange the following aqueous solutions, each 10% by
mass in solute, in order of increasing boiling point: glucose
1C6H12O62, sucrose 1C12H22O112, sodium nitrate 1NaNO32.
13.71List the following aqueous solutions in order of increasing boiling point: 0.120 m glucose, 0.050 m LiBr, 0.050 m
Zn1NO322.
13.72List the following aqueous solutions in order of decreasing
freezing point: 0.040 m glycerin 1C3H8O32, 0.020 m KBr,
0.030 m phenol 1C6H5OH2.
13.73Using data from Table 13.3, calculate the freezing and boiling
points of each of the following solutions: (a) 0.22 m glycerol
1C3H8O32 in ethanol, (b) 0.240 mol of naphthalene 1C10H82
in 2.45 mol of chloroform, (c) 1.50 g NaCl in 0.250 kg of water,
(d) 2.04 g KBr and 4.82 g glucose 1C6H12O62 in 188 g of water.
13.74Using data from Table 13.3, calculate the freezing and boiling
points of each of the following solutions: (a) 0.25 m glucose in
ethanol; (b) 20.0 g of decane, C10H22, in 50.0 g CHCl3; (c) 3.50 g
NaOH in 175 g of water, (d) 0.45 mol ethylene glycol and
0.15 mol KBr in 150 g H2O.
13.75How many grams of ethylene glycol 1C2H6O22 must be
added to 1.00 kg of water to produce a solution that freezes at
-5.00 °C?
point by 0.49 °C. Calculate the approximate molar mass of
adrenaline from this data.
Adrenaline
13.80Lauryl alcohol is obtained from coconut oil and is used to
make detergents. A solution of 5.00 g of lauryl alcohol in
0.100 kg of benzene freezes at 4.1 °C. What is the molar mass
of lauryl alcohol from this data?
13.81Lysozyme is an enzyme that breaks bacterial cell walls. A solution containing 0.150 g of this enzyme in 210 mL of solution has an osmotic pressure of 0.953 torr at 25 °C. What is the
molar mass of lysozyme?
13.82A dilute aqueous solution of an organic compound soluble
in water is formed by dissolving 2.35 g of the compound in
water to form 0.250 L of solution. The resulting solution has
an osmotic pressure of 0.605 atm at 25 °C. Assuming that the
organic compound is a nonelectrolyte, what is its molar mass?
[13.83] The osmotic pressure of a 0.010 M aqueous solution of CaCl2
is found to be 0.674 atm at 25 °C. (a) Calculate the van’t Hoff
factor, i, for the solution. (b) How would you expect the value
of i to change as the solution becomes more concentrated?
Explain.
[13.84] Based on the data given in Table 13.4, which solution would
give the larger freezing-point lowering, a 0.030 m solution of
NaCl or a 0.020 m solution of K2SO4? How do you explain the
departure from ideal behavior and the differences observed
between the two salts?
Colloids (Section 13.6)
13.85(a) Do colloids made only of gases exist? Why or why not?
(b) In the 1850’s, Michael Faraday prepared ruby-red colloids of gold nanoparticles in water that are still stable today.
These brightly colored colloids look like solutions. What
experiment(s) could you do to determine whether a given colored preparation is a solution or colloid?
13.86Choose the best answer: A colloidal dispersion of one liquid in another is called (a) a gel, (b) an emulsion, (c) a foam,
(d) an aerosol.
13.77What is the osmotic pressure formed by dissolving 44.2 mg of
aspirin 1C9H8O42 in 0.358 L of water at 25 °C?
13.87An “emulsifying agent” is a compound that helps stabilize a hydrophobic colloid in a hydrophilic solvent (or a hydrophilic colloid in
a hydrophobic solvent). Which of the following choices is the best
emulsifying agent? (a) CH3COOH, (b) CH3CH2CH2COOH,
(c) CH31CH2211COOH, (d) CH31CH2211COONa.
13.79Adrenaline is the hormone that triggers the release of extra
glucose molecules in times of stress or emergency. A solution
of 0.64 g of adrenaline in 36.0 g of CCl4 elevates the boiling
[13.89] Proteins can be precipitated out of aqueous solution by the
addition of an electrolyte; this process is called “salting out”
13.76What is the freezing point of an aqueous solution that boils at
105.0 °C?
13.78Seawater contains 3.4 g of salts for every liter of solution. Assuming that the solute consists entirely of NaCl (in fact, over
90% of the salt is indeed NaCl), calculate the osmotic pressure
of seawater at 20 °C.
13.88Aerosols are important components of the atmosphere.
Does the presence of aerosols in the atmosphere increase or
decrease the amount of sunlight that arrives at the Earth’s
surface, compared to an “aerosol-free” atmosphere? Explain
your reasoning.
Additional Exercises
the protein. (a) Do you think that all proteins would be precipitated out to the same extent by the same concentration of
the same electrolyte? (b) If a protein has been salted out, are
the protein–protein interactions stronger or weaker than they
were before the electrolyte was added? (c) A friend of yours
who is taking a biochemistry class says that salting out works
because the waters of hydration that surround the protein
prefer to surround the electrolyte as the electrolyte is added;
therefore, the protein’s hydration shell is stripped away, leading to protein precipitation. Another friend of yours in the
571
same biochemistry class says that salting out works because
the incoming ions adsorb tightly to the protein, making ion
pairs on the protein surface, which end up giving the protein a
zero net charge in water and therefore leading to precipitation.
Discuss these two hypotheses. What kind of measurements
would you need to make to distinguish between these two
hypotheses?
13.90Explain how (a) a soap such as sodium stearate stabilizes a
colloidal dispersion of oil droplets in water; (b) milk curdles
upon addition of an acid.
Additional Exercises
13.91Butylated hydroxytoluene (BHT) has the following molecular
structure:
13.97 The maximum allowable concentration of lead in drinking
water is 9.0 ppb. (a) Calculate the molarity of lead in a 9.0ppb solution. (b) How many grams of lead are in a swimming
pool containing 9.0 ppb lead in 60 m3 of water?
CH3
H3C
CH3
CH3
C
C
CH3
OH
that seawater contains 13 ppt of gold, calculate the number of
grams of gold contained in 1.0 * 103 gal of seawater.
CH3
CH3
BHT
It is widely used as a preservative in a variety of foods, including dried cereals. Based on its structure, would you expect BHT to be more soluble in water or in hexane 1C6H142?
Explain.
13.92A saturated solution of sucrose 1C12H22O112 is made by dissolving excess table sugar in a flask of water. There are 50 g
of undissolved sucrose crystals at the bottom of the flask in
contact with the saturated solution. The flask is stoppered and
set aside. A year later a single large crystal of mass 50 g is at
the bottom of the flask. Explain how this experiment provides
evidence for a dynamic equilibrium between the saturated solution and the undissolved solute.
13.93 Most fish need at least 4 ppm dissolved O2 in water for survival. (a) What is this concentration in mol>L? (b) What partial pressure of O2 above water is needed to obtain 4 ppm O2
in water at 10 °C? (The Henry’s law constant for O2 at this
temperature is 1.71 * 10-3 mol>L@atm.)
13.98 Acetonitrile 1CH3CN2 is a polar organic solvent that dissolves
a wide range of solutes, including many salts. The density of
a 1.80 M LiBr solution in acetonitrile is 0.826 g>cm3. Calculate the concentration of the solution in (a) molality, (b) mole
fraction of LiBr, (c) mass percentage of CH3CN.
13.99 A “canned heat” product used to warm buffet dishes consists
of a homogeneous mixture of ethanol 1C2H5OH2 and paraffin, which has an average formula of C24H50. What mass of
C2H5OH should be added to 620 kg of the paraffin to produce
8 torr of ethanol vapor pressure at 35 °C? The vapor pressure
of pure ethanol at 35 °C is 100 torr.
13.100 A solution contains 0.115 mol H2O and an unknown number
of moles of sodium chloride. The vapor pressure of the solution at 30 °C is 25.7 torr. The vapor pressure of pure water at
this temperature is 31.8 torr. Calculate the number of grams
of sodium chloride in the solution. (Hint: Remember that
­sodium chloride is a strong electrolyte.)
[13.101]Two beakers are placed in a sealed box at 25 °C. One beaker
contains 30.0 mL of a 0.050 M aqueous solution of a nonvolatile nonelectrolyte. The other beaker contains 30.0 mL of
a 0.035 M aqueous solution of NaCl. The water vapor from
the two solutions reaches equilibrium. (a) In which beaker
does the solution level rise, and in which one does it fall?
(b) What are the volumes in the two beakers when equilibrium is ­attained, assuming ideal behavior?
13.94 The presence of the radioactive gas radon (Rn) in well water presents a possible health hazard in parts of the United
States. (a) Assuming that the solubility of radon in water
with 1 atm pressure of the gas over the water at 30 °C is
7.27 * 10-3 M, what is the Henry’s law constant for radon
in water at this temperature? (b) A sample consisting of
various gases contains 3.5 * 10-6 mole fraction of radon.
This gas at a total pressure of 32 atm is shaken with ­water
at 30 °C. Calculate the molar concentration of radon in
the water.
13.102 A car owner who knows no chemistry has to put antifreeze in
his car’s radiator. The instructions recommend a mixture of
30% ethylene glycol and 70% water. Thinking he will improve
his protection he uses pure ethylene glycol, which is a liquid
at room temperature. He is saddened to find that the solution does not provide as much protection as he hoped. The
pure ethylene glycol freezes solid in his radiator on a very cold
day, while his neighbor, who did use the 30/70 mixture, has
no problem. Suggest an explanation.
13.95 Glucose makes up about 0.10% by mass of human blood.
­C alculate this concentration in (a) ppm, (b) molality.
(c) What further information would you need to determine
the molarity of the solution?
13.103 Calculate the freezing point of a 0.100 m aqueous solution of
K2SO4, (a) ignoring interionic attractions, and (b) taking interionic attractions into consideration by using the van’t Hoff
factor (Table 13.4).
13.96The concentration of gold in seawater has been reported to
be between 5 ppt (parts per trillion) and 50 ppt. Assuming
13.104 Carbon disulfide 1CS22 boils at 46.30 °C and has a density of
1.261 g>mL. (a) When 0.250 mol of a nondissociating solute
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chapter 13 Properties of Solutions
is dissolved in 400.0 mL of CS2, the solution boils at 47.46 °C.
What is the molal boiling-point-elevation constant for CS2?
(b) When 5.39 g of a nondissociating unknown is dissolved
in 50.0 mL of CS2, the solution boils at 47.08 °C. What is the
molecular weight of the unknown?
[13.105]A lithium salt used in lubricating grease has the formula
LiCnH2n + 1O2. The salt is soluble in water to the extent of
0.036 g per 100 g of water at 25 °C. The osmotic pressure of
this solution is found to be 57.1 torr. Assuming that molality and molarity in such a dilute solution are the same and
that the lithium salt is completely dissociated in the solution, determine an appropriate value of n in the formula for
the salt.
Integrative Exercises
13.106 Fluorocarbons (compounds that contain both carbon and fluorine) were, until recently, used as refrigerants. The compounds
listed in the following table are all gases at 25 °C, and their solubilities in water at 25 °C and 1 atm fluorocarbon pressure are
given as mass percentages. (a) For each fluorocarbon, calculate
the molality of a saturated solution. (b) Explain why the molarity
of each of the solutions should be very close numerically to the
molality. (c) Based on their molecular structures, account for the
differences in solubility of the four fluorocarbons. (d) Calculate
the Henry’s law constant at 25 °C for CHClF2, and compare its
magnitude to that for N2 16.8 * 10-4 mol>L@atm2. Suggest a
reason for the difference in magnitude.
Fluorocarbon
Solubility (mass %)
CF4
0.0015
CClF3
0.009
CCl2F2
0.028
CHClF2
0.30
[13.107]At ordinary body temperature 137 °C2, the solubility of N2 in
­water at ordinary atmospheric pressure (1.0 atm) is 0.015 g>L.
Air is approximately 78 mol % N2. (a) Calculate the number of
moles of N2 dissolved per liter of blood, assuming blood is a simple aqueous solution. (b) At a depth of 100 ft in water, the external
pressure is 4.0 atm. What is the solubility of N2 from air in blood
at this pressure? (c) If a scuba diver suddenly surfaces from this
depth, how many milliliters of N2 gas, in the form of tiny bubbles,
are released into the bloodstream from each liter of blood?
[13.108] Consider the following values for enthalpy of vaporization
1kJ>mol2 of several organic substances:
O
CH3C
H
30.4
Acetaldehyde
O
H2C
28.5
CH2
Ethylene oxide
O
CH3CCH3
32.0
Acetone
CH2
H2C
CH2
Cyclopropane
24.7
(a) Account for the variations in heats of vaporization for
these substances, considering their relative intermolecular
forces. (b) How would you expect the solubilities of these substances to vary in hexane as solvent? In ethanol? Use intermolecular forces, including hydrogen-bonding interactions
where applicable, to explain your responses.
[13.109]A textbook on chemical thermodynamics states, “The heat
of solution represents the difference between the lattice energy of the crystalline solid and the solvation energy of the
gaseous ions.” (a) Draw a simple energy diagram to illustrate
this statement. (b) A salt such as NaBr is insoluble in most
polar nonaqueous solvents such as acetonitrile 1CH3CN2 or
nitromethane 1CH3NO22, but salts of large cations, such as
tetramethylammonium bromide 31CH324NBr4, are generally more soluble. Use the thermochemical cycle you drew
in part (a) and the factors that determine the lattice energy
(Section 8.2) to explain this fact.
13.110 (a) A sample of hydrogen gas is generated in a closed container by reacting 2.050 g of zinc metal with 15.0 mL of
1.00 M sulfuric acid. Write the balanced equation for the
reaction, and calculate the number of moles of hydrogen
formed, assuming that the reaction is complete. (b) The
volume over the solution in the container is 122 mL. Calculate the partial pressure of the hydrogen gas in this volume
at 25 °C, ignoring any solubility of the gas in the solution.
(c) The Henry’s law constant for hydrogen in water at 25 °C
is 7.8 * 10-4 mol>L@atm. Estimate the number of moles of
hydrogen gas that remain dissolved in the solution. What
fraction of the gas molecules in the system is dissolved in
the solution? Was it reasonable to ignore any dissolved hydrogen in part (b)?
[13.111]The following table presents the solubilities of several
gases in water at 25 °C under a total pressure of gas and
water vapor of 1 atm. (a) What volume of CH41g2 under
standard conditions of temperature and pressure is contained in 4.0 L of a saturated solution at 25 °C? (b) Explain the variation in solubility among the hydrocarbons
listed (the first three compounds), based on their molecular structures and intermolecular forces. (c) Compare the
solubilities of O2, N2, and NO, and account for the variations based on molecular structures and intermolecular
forces. (d) Account for the much larger values observed
for H2S and SO2 as compared with the other gases listed.
(e) Find several pairs of substances with the same or nearly
the same molecular masses (for example, C2H4 and N2),
and use intermolecular interactions to explain the
­
differences in their solubilities.
Design an Experiment
Gas
Solubility (mM)
CH4 1methane2
1.3
C2H6 1ethane2
1.8
N2
0.6
O2
1.2
4.7
C2H4 1ethylene2
NO
1.9
H2S
99
SO2
1476
13.112 A small cube of lithium 1density = 0.535 g/cm32 measuring
1.0 mm on each edge is added to 0.500 L of water. The following reaction occurs:
573
bonds between acetone and chloroform molecules to explain
the deviation from ideal behavior. (c) Based on the behavior
of the solution, predict whether the mixing of acetone and
chloroform is an exothermic 1∆Hsoln 6 02 or endothermic
1∆Hsoln 7 02 process. (d) Would you expect the same vaporpressure behavior for acetone and chloromethane (CH3Cl)?
Explain.
13.114 Compounds like sodium stearate, called “surfactants” in general, can form structures known as micelles in water, once the
solution concentration reaches the value known as the critical
micelle concentration (cmc). Micelles contain dozens to hundreds of molecules. The cmc depends on the substance, the
solvent, and the temperature.
Surfactant
tail
Surfactant
head
2 Li1s2 + 2 H2O1l2 ¡ 2 LiOH1aq2 + H21g2
cmc
What is the freezing point of the resultant solution, assuming
that the reaction goes to completion?
[13.113]At 35 °C the vapor pressure of acetone, 1CH322CO, is 360
torr, and that of chloroform, CHCl3, is 300 torr. Acetone and
chloroform can form very weak hydrogen bonds between one
another; the chlorines on the carbon give the carbon a sufficient partial positive charge to enable this behavior:
Cl
Cl
C
H
O
C
Cl
CH3
CH3
A solution composed of an equal number of moles of acetone
and chloroform has a vapor pressure of 250 torr at 35 °C.
(a) What would be the vapor pressure of the solution if it
exhibited ideal behavior? (b) Use the existence of hydrogen
Surfactant monomers
Micelle
At and above the cmc, the properties of the solution vary
drastically.
(a) The turbidity (the amount of light scattering) of solutions increases dramatically at the cmc. Suggest an explanation. (b) The ionic conductivity of the solution dramatically
changes at the cmc. Suggest an explanation. (c) Chemists
have developed fluorescent dyes that glow brightly only
when the dye molecules are in a hydrophobic environment. Predict how the intensity of such fluorescence would
­relate to the concentration of sodium stearate as the sodium
­stearate concentration approaches and then increases past
the cmc.
Design an Experiment
Based on Figure 13.18, you might think that the reason volatile solvent molecules in a solution
are less likely to escape to the gas phase, compared to the pure solvent, is because the solute
­molecules are physically blocking the solvent molecules from leaving at the surface. This is a common ­misconception. Design an experiment to test the hypothesis that solute blocking of solvent
­vaporization is not the reason that solutions have lower vapor pressures than pure solvents.
14
Chemical Kinetics
Chemical reactions take time to occur. Some reactions, such as
the rusting of iron or the changing of color in leaves, occur relatively
slowly, requiring days, months, or years to complete. Others, such as
the combustion reaction that generates the thrust for a rocket, as in the
chapter-opening photograph, happen much more rapidly. As chemists,
we need to be concerned about the speed with which chemical
reactions occur as well as the products of the reactions. For example,
the chemical reactions that govern the metabolism of food, the transport of
essential nutrients, and your body’s ability to adjust to temperature changes (see
the Chemistry and Life box on the regulation of body temperature in Section 5.5)
all require that reactions occur with the appropriate speed. Indeed, considerations
of the speeds of reactions are among the most important aspects of designing new
chemistry and chemical processes. The area of chemistry concerned with the speeds, or
rates, of reactions is chemical kinetics.
So far, we have focused on the beginning and end of chemical reactions: We start
with certain reactants and see what products they yield. This view is useful but does not
tell us what happens in the middle—that is, which chemical bonds are broken, which
are formed, and in what order these events occur. The speed at which a chemical reaction occurs is called the reaction rate. Reaction rates can occur over very different time
What’s
Ahead
14.1 Factors that Affect Reaction Rates We see that
four variables affect reaction rates: concentration, physical states
of reactants, temperature, and presence of catalysts. These factors
can be understood in terms of the collisions among reactant
molecules that lead to reaction.
14.2 Reaction Rates We examine how to express reaction
rates and how reactant disappearance rates and product
appearance rates are related to the reaction stoichiometry.
▶ Launch of the Juno Spacecraft at
Cape Canaveral, Florida, in August 2011.
The spacecraft is mounted to an Atlas V
rocket, which at launch uses the very rapid
combustion of kerosene and liquid oxygen
to generate its thrust.
14.3 Concentration and Rate Laws We show that the
effect of concentration on rate is expressed quantitatively by rate
laws and show how rate laws and rate constants are determined
experimentally.
14.4 The Change of Concentration with Time We
learn that rate equations can be written to express how
concentrations change with time and look at several classifications
of rate equations: zero-order, first-order, and second-order
reactions.
14.5 Temperature and Rate We explore the effect of
temperature on rate. In order to occur, most reactions require a
minimum input of energy called the activation energy.
14.6 Reaction Mechanisms We look more closely at
reaction mechanisms, the step-by-step molecular pathways
leading from reactants to products.
14.7 Catalysis We end the chapter with a discussion of
how catalysts increase reaction rates, including a discussion of
biological catalysts called enzymes.
576
Energy
chapter 14 Chemical Kinetics
10−15 s
109 s
(30 years)
1s
1015 s
(30 million years)
Time scale
Go Figure
If a heated steel nail were placed in
pure O2, would you expect it to burn as
readily as the steel wool does?
Steel wool heated in air
(about 20% O2) glows red-hot
but oxidizes to Fe2O3 slowly
▲ Figure 14.1 Reaction rates span an enormous range of time scales. The absorption of light
by an atom or a molecule is complete within one femtosecond; explosions occur within seconds;
corrosion can occur over years; and the weathering of rocks can occur over millions of years.
scales (▲ Figure 14.1). To investigate how reactions happen, we must examine the reaction rates and the factors that influence them. Experimental information on the rate of
a given reaction provides important evidence that helps us formulate a reaction mechanism, which is a step-by-step, molecular-level view of the pathway from reactants to
products.
Our goal in this chapter is to understand how to determine reaction rates and to
consider the factors that control these rates. What factors determine how rapidly food
spoils, for instance? How does one design an automotive airbag that fills extremely rapidly following a car crash? What determines the rate at which steel rusts? How can we
remove hazardous pollutants in automobile exhaust before the exhaust leaves the tailpipe? Although we will not address these specific questions, we will see that the rates of
all chemical reactions are subject to the same principles.
14.1 | Factors that Affect
Reaction Rates
Four factors affect the rate at which any particular reaction occurs:
Red-hot steel wool in 100%
O2 burns vigorously, forming
Fe2O3 quickly
▲ Figure 14.2 Effect of concentration on
reaction rate. The difference in behavior is
due to the different concentrations of O2 in
the two environments.
1.Physical state of the reactants. Reactants must come together to react. The more
readily reactant molecules collide with one another, the more rapidly they react.
Reactions may broadly classified as being either homogeneous, involving either all
gases or all liquids, or as heterogeneous, in which reactants are in different phases.
Under heterogeneous conditions, a reaction is limited by the area of contact of the
reactants. Thus, heterogeneous reactions that involve solids tend to proceed more
rapidly if the surface area of the solid is increased. For example, a medicine in the
form of a fine powder dissolves in the stomach and enters the blood more quickly
than the same medicine in the form of a tablet.
2.Reactant concentrations. Most chemical reactions proceed more quickly if the concentration of one or more reactants is increased. For example, steel wool burns
only slowly in air, which contains 20% O2, but bursts into flame in pure oxygen
(◀ Figure 14.2). As reactant concentration increases, the frequency with which the
reactant molecules collide increases, leading to increased rates.
3.Reaction temperature. Reaction rates generally increase as temperature is increased. The bacterial reactions that spoil milk, for instance, proceed more rapidly
at room temperature than at the lower temperature of a refrigerator. Increasing
(Section 10.7) As moltemperature increases the kinetic energies of molecules.
ecules move more rapidly, they collide more frequently and with higher energy,
leading to increased reaction rates.
section 14.2 Reaction Rates
4.The presence of a catalyst. Catalysts are agents that increase reaction rates without
themselves being used up. They affect the kinds of collisions (and therefore alter
the mechanism) that lead to reaction. Catalysts play many crucial roles in living
organisms, including ourselves.
On a molecular level, reaction rates depend on the frequency of collisions between
molecules. The greater the frequency of collisions, the higher the reaction rate. For a collision to lead to a reaction, however, it must occur with sufficient energy to break bonds
and with suitable orientation for new bonds to form in the proper locations. We will
consider these factors as we proceed through this chapter.
Give It Some Thought
In a reaction involving reactants in the gas state, how does increasing the partial
pressures of the gases affect the reaction rate?
14.2 | Reaction Rates
The speed of an event is defined as the change that occurs in a given time interval, which
means that whenever we talk about speed, we necessarily bring in the notion of time.
For example, the speed of a car is expressed as the change in the car’s position over a
certain time interval. In the United States, the speed of cars is usually measured in units
of miles per hour—that is, the quantity that is changing (position measured in miles)
divided by a time interval (measured in hours).
Similarly, the speed of a chemical reaction—its reaction rate—is the change in the
concentration of reactants or products per unit of time. The units for reaction rate are
usually molarity per second 1M>s2—that is, the change in concentration measured in
molarity divided by a time interval measured in seconds.
Let’s consider the hypothetical reaction A ¡ B, depicted in ▼ Figure 14.3.
Each red sphere represents 0.01 mol of A, each blue sphere represents 0.01 mol of B,
and the container has a volume of 1.00 L. At the beginning of the reaction, there is 1.00
mol A, so the concentration is 1.00 mol >L = 1.00 M. After 20 s, the concentration of A
has fallen to 0.54 M and the concentration of B has risen to 0.46 M. The sum of the concentrations is still 1.00 M because 1 mol of B is produced for each mole of A that reacts.
After 40 s, the concentration of A is 0.30 M and that of B is 0.70 M.
Go Figure
Estimate the number of moles of A in the mixture after 30 s.
0s
20 s
1.00 mol A
0 mol B
0.54 mol A
0.46 mol B
40 s
0.30 mol A
0.70 mol B
▲ Figure 14.3 Progress of a hypothetical reaction a ¡ B. The volume of the flask is 1.0 L.
577
578
chapter 14 Chemical Kinetics
The rate of this reaction can be expressed either as the rate of disappearance of
reactant A or as the rate of appearance of product B. The average rate of appearance of
B over a particular time interval is given by the change in concentration of B divided by
the change in time:
Average rate of appearance of B =
=
change in concentration of B
change in time
3B4 at t2 - 3B4 at t1
∆3B4
=
t2 - t1
∆t
[14.1]
We use brackets around a chemical formula, as in [B], to indicate molarity. The
Greek letter delta, ∆, is read “change in” and is always equal to a final value minus an
initial value.
(Equation 5.4, Section 5.2) The average rate of appearance of B over
the 20-s interval from the beginning of the reaction 1t1 = 0 s to t2 = 20 s2 is
0.46 M - 0.00 M
Average rate =
= 2.3 * 10-2 M>s
20 s - 0 s
We could equally well express the reaction rate in term of the reactant, A. In this
case, we would be describing the rate of disappearance of A, which we express as
Average rate of disappearance of A = -
change in concentration of A
change in time
∆3A4
[14.2]
∆t
Notice the minus sign in this equation, which we use to indicate that the concentration of A decreases. By convention, rates are always expressed as positive quantities.
Because [A] decreases, ∆3A4 is a negative number. The minus sign we put in the equation converts the negative ∆3A4 to a positive rate of disappearance.
Because one molecule of A is consumed for every molecule of B that forms, the
average rate of disappearance of A equals the average rate of appearance of B:
= -
Average rate = -
∆3A4
0.54 M - 1.00 M
= = 2.3 * 10-2 M>s
∆t
20 s - 0 s
Sample
Exercise 14.1 Calculating an Average Rate of Reaction
From the data in Figure 14.3, calculate the average rate at which A disappears over the time
interval from 20 s to 40 s.
Solution
Analyze We are given the concentration of A at 20 s 10.54 M2 and at 40 s 10.30 M2 and asked to
calculate the average rate of reaction over this time interval.
Plan The average rate is given by the change in concentration, ∆3A4, divided by the change in time, ∆t.
Because A is a reactant, a minus sign is used in the calculation to make the rate a positive quantity.
Solve
Average rate = -
∆3A4
∆t
= -
0.30 M - 0.54 M
= 1.2 * 10-2 M>s
40 s - 20 s
Practice Exercise 1
If the experiment in Figure 14.3 is run for 60 s, 0.16 mol A remain. Which of the f­ ollowing
statements is or are true?
(i) After 60 s there are 0.84 mol B in the flask.
(ii) The ­decrease in the number of moles of A from t1 = 0 s to t2 = 20 s is greater than that
from t1 = 40 to t2 = 60 s.
(iii) The average rate for the reaction from t1 = 40 s to t2 = 60 s is 7.0 * 10-3 M>s.
(a) Only one of the statements is true.
(b) Statements (i) and (ii) are true.
(c) Statements (i) and (iii) are true.
(d) Statements (ii) and (iii) are true.
(e) All three statements are true.
Practice Exercise 2
Use the data in Figure 14.3 to calculate the average rate of appearance of B over the time
­interval from 0 s to 40 s.
section 14.2 Reaction Rates
579
Table 14.1 Rate Data for Reaction of C 4h 9Cl with Water
Time, t 1s 2
0.0
50.0
100.0
150.0
200.0
300.0
400.0
500.0
800.0
10,000
3 C4h 9Cl4 1M 2
0.1000
0.0905
0.0820
0.0741
0.0671
0.0549
0.0448
0.0368
0.0200
0
Average Rate 1M , s 2
1.9
1.7
1.6
1.4
1.22
1.01
0.80
0.560
*
*
*
*
*
*
*
*
10-4
10-4
10-4
10-4
10-4
10-4
10-4
10-4
Change of Rate with Time
Now let’s consider the reaction between butyl chloride 1C4H9Cl2 and water to form
butyl alcohol 1C4H9OH2 and hydrochloric acid:
C4H9Cl1aq2 + H2O1l2 ¡ C4H9OH1aq2 + HCl1aq2[14.3]
Suppose we prepare a 0.1000@M aqueous solution of C4H9Cl and then measure the
concentration of C4H9Cl at various times after time zero (which is the instant at which the
reactants are mixed, thereby initiating the reaction). We can use the resulting data, shown in the first two columns of ▲ Table 14.1, to calculate the
Go Figure
average rate of disappearance of C4H9Cl over various time intervals; these
rates are given in the third column. Notice that the average rate decreases How does the instantaneous rate of reaction change as
over each 50-s interval for the first several measurements and continues to the reaction proceeds?
decrease over even larger intervals through the remaining measurements.
C4H9Cl(aq) + H2O(l)
C4H9OH(aq) + HCl(aq)
It is typical for rates to decrease as a reaction proceeds because the concen0.100
tration of reactants decreases. The change in rate as the reaction proceeds
Instantaneous
is also seen in a graph of 3C4H9Cl4 versus time (▶ Figure 14.4). Notice
0.090
rate at t = 0 s
how the steepness of the curve decreases with time, indicating a decreas(initial rate)
ing reaction rate.
0.080
Instantaneous Rate
∆3C4H9Cl4
10.017 - 0.0422 M
= ∆t
1800 - 4002 s
-5
= 6.3 * 10 M>s
Instantaneous rate = -
0.060
0.050
0.040
∆ [C4H9Cl]
[C4H9Cl] (M)
Graphs such as Figure 14.4 that show how the concentration of a reactant or product changes with time allow us to evaluate the instantaneous rate of a reaction, which is the rate at a particular instant during
the reaction. The instantaneous rate is determined from the slope of the
curve at a particular point in time. We have drawn two tangent lines
in Figure 14.4, a dashed line running through the point at t = 0 s
and a solid line running through the point at t = 600 s. The slopes
of these tangent lines give the instantaneous rates at these two time
points.* To determine the instantaneous rate at 600 s, for example, we
construct horizontal and vertical lines to form the blue right triangle
in Figure 14.4. The slope of the tangent line is the ratio of the height of
the vertical side to the length of the horizontal side:
Instantaneous rate at
time t = slope of
tangent to the line at
time t.
0.070
0.030
0.020
∆t
0.010
0
Instantaneous
rate at t = 600 s
100 200 300 400 500 600 700 800 900
Time (s)
▲ Figure 14.4 Concentration of butyl chloride 1c4h9cl2 as a
function of time.
*You may wish to review graphical determination of slopes in Appendix A. If you are familiar with calculus, you
may recognize that the average rate approaches the instantaneous rate as the time interval approaches zero. This
limit, in the notation of calculus, is the negative of the derivative of the curve at time t, - d3C4H9Cl4>dt.
580
chapter 14 Chemical Kinetics
In discussions that follow, the term rate means instantaneous rate unless indicated otherwise. The instantaneous rate at t = 0 is called the initial rate of the reaction.
To understand the difference between average and instantaneous rates, imagine you
have just driven 98 mi in 2.0 h. Your average speed for the trip is 49 mi>hr, but your
instantaneous speed at any moment during the trip is the speedometer reading at that
moment.
Sample
Exercise 14.2 Calculating an Instantaneous Rate of Reaction
Using Figure 14.4, calculate the instantaneous rate of disappearance of C4H9Cl at t = 0 s (the
initial rate).
Solution
Analyze We are asked to determine an instantaneous rate from a graph of reactant concentration
versus time.
Plan To obtain the instantaneous rate at t = 0 s, we must determine the slope of the curve at
t = 0. The tangent is drawn on the graph as the hypotenuse of the tan triangle. The slope of this
straight line equals the change in the vertical axis divided by the corresponding change in the
horizontal axis (which, in the case of this example, is the change in molarity over change in time).
Solve The tangent line falls from 3C4H9Cl4 = 0.100 M to 0.060 M in the time change from 0 s
to 210 s. Thus, the initial rate is
Rate = -
∆3C4H9Cl4
∆t
= -
10.060 - 0.1002 M
1210 - 02 s
= 1.9 * 10-4 M>s
Practice Exercise 1
Which of the following could be the instantaneous rate of the reaction in Figure 14.4 at t = 1000 s?
(a)1.2 * 10-4 M>s, (b) 8.8 * 10-5 M>s, (c) 6.3 * 10-5 M>s, (d) 2.7 * 10-5 M>s,
(e)More than one of these.
Practice Exercise 2
Using Figure 14.4, determine the instantaneous rate of disappearance of C4H9Cl at t = 300 s.
Give It Some Thought
In Figure 14.4, order the following three rates from fastest to slowest: (i) The
average rate of the reaction between 0 s and 600 s, (ii) the instantaneous rate at
t = 0 s, and (iii) the instantaneous rate at t = 600 s. You should not have to do
any calculations.
Reaction Rates and Stoichiometry
During our discussion of the hypothetical reaction A ¡ B, we saw that the stoichiometry requires that the rate of disappearance of A equal the rate of appearance
of B. Likewise, the stoichiometry of Equation 14.3 indicates that 1 mol of C4H9OH
is produced for each mole of C4H9Cl consumed. Therefore, the rate of appearance of
C4H9OH equals the rate of disappearance of C4H9Cl:
Rate = -
∆3C4H9Cl4
∆3C4H9OH4
=
∆t
∆t
What happens when the stoichiometric relationships are not one-to-one? For
example, consider the reaction 2 HI1g2 ¡ H21g2 + I21g2. We can measure either
the rate of disappearance of HI or the rate of appearance of either H2 or I2. Because 2
mol of HI disappears for each mole of H2 or I2 that forms, the rate of disappearance of
HI is twice the rate of appearance of either H2 or I2. How do we decide which number
to use for the rate of the reaction? Depending on whether we monitor HI, I2, or H2,
the rates can differ by a factor of 2. To fix this problem, we need to take into account
the reaction stoichiometry. To arrive at a number for the reaction rate that does not
section 14.3 Concentration and Rate Laws
depend on which component we measured, we must divide the rate of disappearance of
HI by 2 (its coefficient in the balanced chemical equation):
Rate = -
∆3H24
∆3I24
1 ∆3HI4
=
=
2 ∆t
∆t
∆t
In general, for the reaction
aA + bB ¡ cC + dD
the rate is given by
Rate = -
1 ∆3A4
1 ∆3B4
1 ∆3C4
1 ∆3D4
= =
=
[14.4]
a ∆t
c ∆t
b ∆t
d ∆t
When we speak of the rate of a reaction without specifying a particular reactant or
product, we utilize the definition in Equation 14.4.*
Sample
Exercise 14.3 Relating Rates at Which Products Appear and Reactants Disappear
(a) How is the rate at which ozone disappears related to the rate at which oxygen appears in the
reaction 2 O31g2 ¡ 3 O21g2?
(b) If the rate at which O2 appears, ∆3O24> ∆t, is 6.0 * 10-5 M>s at a particular instant, at what
rate is O3 disappearing at this same time, - ∆3O34> ∆t?
Solution
Analyze We are given a balanced chemical equation and asked to relate
the rate of appearance of the product to the rate of disappearance of the
reactant.
Plan We can use the coefficients in the chemical equation as shown in
Equation 14.4 to express the relative rates of reactions.
Solve
(a) Using the coefficients in the balanced equation and the
relationship given by Equation 14.4, we have:
1 ∆3O34
1 ∆3O24
Rate = =
2 ∆t
3 ∆t
(b) Solving the equation from part (a) for the rate at which
O3 disappears, - ∆3O34> ∆t, we have:
-
∆3O34
∆t
=
2 ∆3O24
2
= 16.0 * 10-5M>s2 = 4.0 * 10-5M>s
3 ∆t
3
Practice Exercise 1
At a certain time in a reaction, substance A is disappearing at
a rate of 4.0 * 10-2 M>s, substance B is appearing at a rate
of 2.0 * 10-2 M>s, and substance C is appearing at a rate of
6.0 * 10-2 M>s. Which of the following could be the
stoichiometry for the reaction being studied?
(a) 2A + B ¡ 3C (b) A ¡ 2B + 3C (c) 2A ¡ B + 3C (d) 4A ¡ 2B + 3C
(e)A + 2B ¡ 3C
Practice Exercise 2
If the rate of decomposition of N2O5 in the reaction
2 N2O51g2 ¡ 4 NO21g2 + O21g2 at a particular instant is
4.2 * 10 - 7M>s, what is the rate of appearance of (a) NO2 and
(b) O2 at that instant?
Check We can apply a stoichiometric factor to convert the
O2 formation rate to the O3 disappearance rate:
-
∆3O34
∆t
= a6.0 * 10-5
mol O2 >L
= 4.0 * 10-5M>s
s
ba
mol O3 >L
2 mol O3
b = 4.0 * 10-5
s
3 mol O2
14.3 | Concentration and Rate Laws
One way of studying the effect of concentration on reaction rate is to determine the
way in which the initial rate of a reaction depends on the initial concentrations. For
example, we might study the rate of the reaction
NH4+ 1aq2 + NO2- 1aq2 ¡ N21g2 + 2 H2O1l2
*Equation 14.4 does not hold true if substances other than C and D are formed in significant amounts. For
example, sometimes intermediate substances build in concentration before forming the final products. In
that case, the relationship between the rate of disappearance of reactants and the rate of appearance of products is not given by Equation 14.4. All reactions whose rates we consider in this chapter obey Equation 14.4.
581
582
chapter 14 Chemical Kinetics
Table 14.2 Rate Data for the Reaction of Ammonium and Nitrite Ions in
Water at 25 °C
Experiment
Number
Initial nh 4+
Concentration (M)
Initial no2−
Concentration (M)
1
0.0100
0.200
2
0.0200
0.200
10.8 * 10 - 7
3
0.0400
0.200
21.5 * 10 - 7
4
0.200
0.0202
10.8 * 10 - 7
5
0.200
0.0404
21.6 * 10 - 7
6
0.200
0.0808
43.3 * 10 - 7
Observed Initial
Rate 1M , s 2
5.4 * 10 - 7
by measuring the concentration of NH4+ or NO2- as a function of time or by measuring the volume of N2 collected as a function of time. Because the stoichiometric coefficients on NH4 + , NO2 - , and N2 are the same, all of these rates are the same.
▲ Table 14.2 shows that changing the initial concentration of either reactant
changes the initial reaction rate. If we double 3NH4+ 4 while holding 3NO2 - 4 constant,
the rate doubles (compare experiments 1 and 2). If we increase 3NH4+4 by a factor of 4
but leave 3NO2- 4 unchanged (experiments 1 and 3), the rate changes by a factor of 4, and
so forth. These results indicate that the initial reaction rate is proportional to 3NH4+ 4.
When 3NO2- 4 is similarly varied while 3NH4+ 4 is held constant, the rate is affected in the
same manner. Thus, the rate is also directly proportional to the concentration of 3NO2- 4.
A Closer Look
Using Spectroscopic Methods to
Measure Reaction Rates: Beer’s Law
A = ebc[14.5]
In this equation, A is the measured absorbance, e is the extinction coefficient (a characteristic of the substance being monitored at a
given wavelength of light), b is the path length through which the light
0.4
Spectrometer measures
intensity of purple color as I2
concentration increases
Absorbance
A variety of techniques can be used to monitor reactant and product
concentration during a reaction, including spectroscopic methods,
which rely on the ability of substances to absorb (or emit) light. Spectroscopic kinetic studies are often performed with the reaction mixture in the sample compartment of a spectrometer, an instrument that
measures the amount of light transmitted or absorbed by a sample at
different wavelengths. For kinetic studies, the spectrometer is set to
measure the light absorbed at a wavelength characteristic of one of the
reactants or products. In the decomposition of HI(g) into H21g2 and
I21g2, for example, both HI and H2 are colorless, whereas I2 is violet.
During the reaction, the violet color of the reaction mixture gets more
intense as I2 forms. Thus, visible light of appropriate wavelength can
be used to monitor the reaction (▶ Figure 14.5).
▶ Figure 14.6 shows the components of a spectrometer. The
spectrometer measures the amount of light absorbed by the sample by
comparing the intensity of the light emitted from the light source with
the intensity of the light transmitted through the sample, for various
wavelengths. As the concentration of I2 increases and its color becomes
more intense, the amount of light absorbed by the reaction mixture increases, as Figure 14.5 shows, causing less light to reach the detector.
How can we relate the amount of light detected by the spectrometer
to the concentration of a species? A relationship called Beer’s law gives us
a direct route to the information we seek. Beer’s law connects the amount
of light absorbed to the concentration of the absorbing substance:
100 mg/L
70 mg/L
40 mg/L
10 mg/L
1 mg/L
0.2
0.0
400
450
500
550
Wavelength (nm)
600
▲ Figure 14.5 Visible spectra of l2 at different concentrations.
passes, and c is the molar concentration of the absorbing substance.
Thus, the concentration is directly proportional to absorbance. Many
chemical and pharmaceutical companies routinely use Beer’s law to
calculate the concentration of purified solutions of the compounds
that they make. In the laboratory portion of your course, you may very
well perform one or more experiments in which you use Beer’s law to
relate absorption of light to concentration.
Related Exercises: 14.101, 14.102, Design an Experiment
650
section 14.3 Concentration and Rate Laws
Source
Lenses/slits/
collimators
Monochromator
(selects wavelength)
Sample
Detector
▲ Figure 14.6 Components of a spectrometer.
We express the way in which the rate depends on the reactant concentrations by
the equation
Rate = k3NH4+ 43NO2- 4[14.6]
An equation such as Equation 14.6, which shows how the rate depends on reactant
concentrations, is called a rate law. For the general reaction
aA + bB ¡ cC + dD
the rate law generally has the form
Rate = k3A4m3B4n[14.7]
Notice that only the concentrations of the reactants generally appear in the rate law. The
constant k is called the rate constant. The magnitude of k changes with temperature
and therefore determines how temperature affects rate, as we will see in Section 14.5.
The exponents m and n are typically small whole numbers. As we will learn shortly, if
we know m and n for a reaction, we can gain great insight into the individual steps that
occur during the reaction.
Give It Some Thought
How do reaction rate, rate law, and rate constant differ?
Once we know the rate law for a reaction and the reaction rate for a set of reactant concentrations, we can calculate the value of k. For example, using the values for
experiment 1 in Table 14.2, we can substitute into Equation 14.6:
5.4 * 10-7 M>s = k10.0100 M210.200 M2
k =
5.4 * 10-7 M>s
10.0100 M210.200 M2
= 2.7 * 10-4 M -1s-1
You should verify that this same value of k is obtained using any of the other experimental results in Table 14.2.
Once we have both the rate law and the k value for a reaction, we can calculate the reaction rate for any set of concentrations. For example, using Equation
14.7 with k = 2.7 * 10-4 M -1 s-1, m = 1, and n = 1, we can calculate the rate for
3NH4+ 4 = 0.100 M and 3NO2- 4 = 0.100 M:
Rate = 12.7 * 10-4M -1 s-1210.100 M210.100 M2 = 2.7 * 10-6M>s
Give It Some Thought
Does the rate constant have the same units as the rate?
Computer
583
584
chapter 14 Chemical Kinetics
Reaction Orders: The Exponents in the Rate Law
The rate law for most reactions has the form
Rate = k3reactant 14m3reactant 24n c [14.8]
The exponents m and n are called reaction orders. For example, consider again the
rate law for the reaction of NH4+ with NO2- :
Rate = k3NH4+ 43NO2- 4
Because the exponent of 3NH4+ 4 is 1, the rate is first order in NH4+ . The rate is also
first order in NO2- . (The exponent 1 is not shown in rate laws.) The overall reaction
order is the sum of the orders with respect to each reactant represented in the rate
law. Thus, for the NH4+ - NO2- reaction, the rate law has an overall reaction order of
1 + 1 = 2, and the reaction is second order overall.
The exponents in a rate law indicate how the rate is affected by each reactant concentration. Because the rate at which NH4+ reacts with NO2- depends on 3NH4+ 4 raised
to the first power, the rate doubles when 3NH4+ 4 doubles, triples when 3NH4+ 4 triples,
and so forth. Doubling or tripling 3NO2- 4 likewise doubles or triples the rate. If a rate
law is second order with respect to a reactant, 3A42, then doubling the concentration of
that substance causes the reaction rate to quadruple because 3242 = 4, whereas tripling
the concentration causes the rate to increase ninefold: 3342 = 9.
The following are some additional examples of experimentally determined
rate laws:
2 N2O51g2 ¡ 4 NO21g2 + O21g2 Rate = k3N2O54[14.9]
H21g2 + I21g2 ¡ 2 HI1g2
Rate = k3H243I24[14.10]
CHCl31g2 + Cl21g2 ¡ CCl41g2 + HCl1g2 Rate = k3CHCl343Cl241>2[14.11]
Although the exponents in a rate law are sometimes the same as the coefficients in the
balanced equation, this is not necessarily the case, as Equations 14.9 and 14.11 show. For
any reaction, the rate law must be determined experimentally. In most rate laws, reaction orders are 0, 1, or 2. However, we also occasionally encounter rate laws in which
the reaction order is fractional (as is the case with Equation 14.11) or even negative.
Give It Some Thought
The experimentally determined rate law for the reaction
2 NO1g2 + 2 H21g2 ¡ N21g2 + 2 H2O1g2 is rate = k3NO423H24.
(a) What are the reaction orders in this rate law?
(b) Would the reaction rate increase more if we doubled the concentration of NO
or the concentration of H2?
Sample
Exercise 14.4 Relating a Rate Law to the Effect of Concentration on Rate
Consider a reaction A + B ¡ C for which rate = k3A43B42. Each of the following boxes represents a reaction mixture in which A is shown as red spheres and B as purple ones. Rank these mixtures in order of increasing rate of reaction.
(1)
(2)
(3)
section 14.3 Concentration and Rate Laws
585
Solution
Analyze We are given three boxes containing different numbers of
spheres representing mixtures containing different reactant concentrations. We are asked to use the given rate law and the compositions of the boxes to rank the mixtures in order of increasing
reaction rates.
Plan Because all three boxes have the same volume, we can put the
number of spheres of each kind into the rate law and calculate the rate
for each box.
Solve Box 1 contains 5 red spheres and 5 purple spheres, giving the
following rate:
Box 1: Rate = k1521522 = 125k
Box 2 contains 7 red spheres and 3 purple spheres:
Box 2: Rate = k1721322 = 63k
Box 3 contains 3 red spheres and 7 purple spheres:
Box 3: Rate = k1321722 = 147k
The slowest rate is 63k (Box 2), and the highest is 147k (Box 3). Thus,
the rates vary in the order 2 6 1 6 3.
Check Each box contains 10 spheres. The rate law indicates that in
this case [B] has a greater influence on rate than [A] because B has a
larger reaction order. Hence, the mixture with the highest concentration of B (most purple spheres) should react fastest. This analysis confirms the order 2 6 1 6 3.
Practice Exercise 1
Suppose the rate law for the reaction in this Sample Exercise were
rate = k3A423B4. What would be the ordering of the rates for the
three mixtures shown above, from slowest to fastest?
(a)1 6 2 6 3 (b) 1 6 3 6 2 (c) 3 6 2 6 1
(d) 2 6 1 6 3 (e) 3 6 1 6 2
Practice Exercise 2
Assuming that rate = k3A43B4, rank the mixtures represented in
this Sample Exercise in order of increasing rate.
Magnitudes and Units of Rate Constants
If chemists want to compare reactions to evaluate which ones are relatively fast and
which ones are relatively slow, the quantity of interest is the rate constant. A good general rule is that a large value of k (∼109 or higher) means a fast reaction and a small
value of k (10 or lower) means a slow reaction.
Give It Some Thought
Suppose the reactions A ¡ B and X ¡ Y have the same value of k. When
3A4 = 3X4, will the two reactions necessarily have the same rate?
The units of the rate constant depend on the overall reaction order of the rate law.
In a reaction that is second order overall, for example, the units of the rate constant
must satisfy the equation:
Units of rate = 1units of rate constant21units of concentration22
Hence, in our usual units of molarity for concentration and seconds for time,
we have
Units of rate constant =
M>s
units of rate
=
= M -1 s-1
1units of concentration22
M2
Sample
Exercise 14.5 Determining Reaction Orders and Units for Rate Constants
(a) What are the overall reaction orders for the reactions described in Equations 14.9 and 14.11?
(b) What are the units of the rate constant for the rate law in Equation 14.9?
Solution
order in CHCl3 and one-half order in Cl2. The overall reaction
order is three halves.
Analyze We are given two rate laws and asked to express (a) the overall
reaction order for each and (b) the units for the rate constant for the first
reaction.
Plan The overall reaction order is the sum of the exponents in the rate
(b) For the rate law for Equation 14.9, we have
Units of rate = 1units of rate constant21units of concentration2
law. The units for the rate constant, k, are found by using the normal
units for rate 1M>s2 and concentration (M) in the rate law and applying algebra to solve for k.
so
(a) The rate of the reaction in Equation 14.9 is first order in N2O5
and first order overall. The reaction in Equation 14.11 is first
Notice that the units of the rate constant change as the overall order of
the reaction changes.
Solve
Units of rate constant =
M>s
units of rate
=
= s-1
units of concentration
M
586
chapter 14 Chemical Kinetics
Practice Exercise 1
Which of the following are the units of the rate constant for
Equation 14.11?
-1
-1
-1
-1
-1
1
-3
-1
-3 -1
(a) M 2 s (b) M 2 s 2 (c) M 2 s (d) M 2 s (e) M 2s 2
Practice Exercise 2
(a) What is the reaction order of the reactant H2 in Equation
14.10? (b) What are the units of the rate constant for Equation
14.10?
Using Initial Rates to Determine Rate Laws
We have seen that the rate law for most reactions has the general form
Rate = k3reactant 14m3reactant 24n c
Thus, the task of determining the rate law becomes one of determining the reaction orders, m and n. In most reactions, the reaction orders are 0, 1, or 2. As noted
earlier in this section, we can use the response of the reaction rate to a change in initial
concentration to determine the reaction order.
In working with rate laws, it is important to realize that the rate of a reaction
depends on concentration but the rate constant does not. As we will see later in this
chapter, the rate constants (and hence the reaction rate) are affected by temperature
and by the presence of a catalyst.
Sample
Exercise 14.6 Determining a Rate Law from Initial Rate Data
The initial rate of a reaction A + B ¡ C was measured for several different starting concentrations of
A and B, and the results are as follows:
Experiment
Number
[A] (M)
[B] (M)
1
0.100
0.100
2
0.100
0.200
4.0 * 10-5
3
0.200
0.100
16.0 * 10-5
Initial Rate 1M , s 2
4.0 * 10-5
Using these data, determine (a) the rate law for the reaction, (b) the rate constant, (c) the rate of the
reaction when 3A4 = 0.050M and 3B4 = 0.100 M.
Solution
Analyze We are given a table of data that relates concentrations of reactants with initial rates of reaction and
asked to determine (a) the rate law, (b) the rate constant, and (c) the rate of reaction for a set of concentrations not listed in the table.
Plan (a) We assume that the rate law has the following form:
Rate = k3A4m3B4n. We will use the given data to deduce the reaction orders m and n by determining how changes in the concentration change the rate. (b) Once we know m and n, we can use the
rate law and one of the sets of data to determine the rate constant
k. (c) Upon determining both the rate constant and the reaction
orders, we can use the rate law with the given concentrations to
calculate rate.
Solve
(a) If we compare experiments 1 and 2, we see that [A] is held constant
and [B] is doubled. Thus, this pair of experiments shows how [B]
affects the rate, allowing us to deduce the order of the rate law with
In experiments 1 and 3, [B] is held constant, so these data show how
[A] affects rate. Holding [B] constant while doubling [A] increases
the rate fourfold. This result indicates that rate is proportional to 3A42
(that is, the reaction is second order in A). Hence, the rate law is
(b) Using the rate law and the data from experiment 1, we have
(c) Using the rate law from part (a) and the rate constant from
part (b), we have
respect to B. Because the rate remains the same when [B] is doubled,
the concentration of B has no effect on the reaction rate. The rate
law is therefore zero order in B (that is, n = 0).
Rate = k3A423B40 = k3A42
k =
4.0 * 10-5 M>s
rate
=
= 4.0 * 10-3 M -1 s-1
2
3A4
10.100 M22
Rate = k3A42 = 14.0 * 10-3 M -1s-1210.050 M22 = 1.0 * 10-5 M>s
Because [B] is not part of the rate law, it is irrelevant to the rate if there is at least some B present to react with A.
section 14.4 The Change of Concentration with Time
Check A good way to check our rate law is to use the concentrations in experiment 2 or 3 and see
if we can correctly calculate the rate. Using data from experiment 3, we have
Rate = k3A42 = 14.0 * 10-3 M -1 s-1210.200 M22 = 1.6 * 10-4 M>s
Thus, the rate law correctly reproduces the data, giving both the correct number and the correct
units for the rate.
Practice Exercise 1
A certain reaction X + Y ¡ Z is described as being first order in [X] and third order
overall. Which of the following statements is or are true?:
(i) The rate law for the reaction is: Rate = k3X43Y42.
(ii) If the concentration of X is increased by a factor of 1.5, the rate will i­ ncrease by a factor
of 2.25.
(iii) If the concentration of Y is increased by a factor of 1.5, the rate will increase by a factor of 2.25.
(a) Only one of the statements is true.
(c) Statements (i) and (iii) are true.
(e) All three statements are true.
(b) Statements (i) and (ii) are true.
(d) Statements (ii) and (iii) are true.
Practice Exercise 2
The following data were measured for the reaction of nitric oxide with hydrogen:
2 NO1g2 + 2 H21g2 ¡ N21g2 + 2 H2O1g2
Experiment
Number
[NO] (M)
1
0.10
3 h 2 4 (M)
0.10
Initial Rate 1M , s 2
2
0.10
0.20
2.46 * 10-3
3
0.20
0.10
4.92 * 10-3
1.23 * 10-3
(a)Determine the rate law for this reaction. (b) Calculate the rate constant.
(c)Calculate the rate when 3NO4 = 0.050 M and 3H24 = 0.150 M.
14.4 | The Change of Concentration
with Time
The rate laws we have examined so far enable us to calculate the rate of a reaction from
the rate constant and reactant concentrations. In this section, we will show that rate
laws can also be converted into equations that show the relationship between concentrations of reactants or products and time. The mathematics required to accomplish
this conversion involves calculus. We do not expect you to be able to perform the calculus operations, but you should be able to use the resulting equations. We will apply this
conversion to three of the simplest rate laws: those that are first order overall, those that
are second order overall, and those that are zero order overall.
First-Order Reactions
A first-order reaction is one whose rate depends on the concentration of a single reactant raised to the first power. If a reaction of the type A ¡ products is first order,
the rate law is:
Rate = -
∆3A4
= k3A4
∆t
This form of a rate law, which expresses how rate depends on concentration, is called
the differential rate law. Using the operation from calculus called integration, this relationship can be transformed into an equation known as the integrated rate law for a
587
588
chapter 14 Chemical Kinetics
first-order reaction that relates the initial concentration of A, 3A40, to its concentration
at any other time t, 3A4t :
ln3A4t - ln3A40 = -kt or ln
3A4t
= -kt[14.12]
3A40
The function “ln” in Equation 14.12 is the natural logarithm (Appendix A.2).
Equation 14.12 can also be rearranged to
ln3A4t = -kt + ln3A40[14.13]
Equations 14.12 and 14.13 can be used with any concentration units as long as the
units are the same for both 3A4t and 3A40.
For a first-order reaction, Equation 14.12 or 14.13 can be used in several ways.
Given any three of the following quantities, we can solve for the fourth: k, t, 3A40, and
3A4t. Thus, you can use these equations to determine (1) the concentration of a reactant
remaining at any time after the reaction has started, (2) the time interval required for a
given fraction of a sample to react, or (3) the time interval required for a reactant concentration to fall to a certain level.
Sample
Exercise 14.7 Using the Integrated First-Order Rate Law
The decomposition of a certain insecticide in water at 12 °C follows
first-order kinetics with a rate constant of 1.45 yr-1. A quantity of this
insecticide is washed into a lake on June 1, leading to a concentration of 5.0 * 10-7 g>cm3. Assume that the temperature of the lake is
constant (so that there are no effects of temperature variation on the
rate). (a) What is the concentration of the insecticide on June 1 of the
following year? (b) How long will it take for the insecticide concentration to decrease to 3.0 * 10 - 7 g>cm3?
Solution
Plan
Analyze We are given the rate constant for a reaction that obeys first-
(a) We are given k = 1.45 yr-1, t = 1.00 yr, and
3insecticide40 = 5.0 * 10-7 g>cm3, and so Equation 14.13 can be
solved for 3insecticide4t.
order kinetics, as well as information about concentrations and times,
and asked to calculate how much reactant (insecticide) remains after
1 yr. We must also determine the time interval needed to reach a particular insecticide concentration. Because the exercise gives time in
(a) and asks for time in (b), we will find it most useful to use the integrated rate law, Equation 14.13.
(b) We have k = 1.45 yr-1, 3insecticide40 = 5.0 * 10-7 g>cm3, and
3insecticide4t = 3.0 * 10-7 g>cm3, and so we can solve Equation
14.13 for time, t.
Solve
(a) Substituting the known quantities into Equation 14.13, we have
We use the ln function on a calculator to evaluate the second term on
the right [that is, ln15.0 * 10-724, giving
To obtain 3insecticide4t = 1 yr, we use the inverse natural logarithm, or
ex, function on the calculator:
Note that the concentration units for 3A4t and 3A40 must be the same.
(b) Again substituting into Equation 14.13, with
3insecticide4t = 3.0 * 10-7 g>cm3, gives
Solving for t gives
ln3insecticide4t = 1 yr = -11.45 yr-1211.00 yr2 + ln15.0 * 10-72
ln3insecticide4t = 1 yr = -1.45 + 1-14.512 = - 15.96
3insecticide4t = 1 yr = e - 15.96 = 1.2 * 10 - 7g>cm3
ln13.0 * 10-72 = -11.45 yr-121t2 + ln15.0 * 10-72
t = -3ln13.0 * 10-72 - ln15.0 * 10-724>1.45 yr-1
= -1-15.02 + 14.512>1.45 yr-1 = 0.35 yr
Check In part (a) the concentration remaining after 1.00 yr (that
is, 1.2 * 10 - 7 g>cm3) is less than the original concentration
15.0 * 10 - 7 g>cm32, as it should be. In (b) the given concentration
13.0 * 10-7 g>cm32 is greater than that remaining after 1.00 yr, indicating that the time must be less than a year. Thus, t = 0.35 yr is a reasonable answer.
Practice Exercise 1
At 25 °C, the decomposition of dinitrogen pentoxide, N2O51g2, into NO21g2 and O21g2 follows
first-order kinetics with k = 3.4 * 10-5 s-1. A sample of N2O5 with an initial pressure of
760 torr decomposes at 25 °C until its partial pressure is 650 torr. How much time (in seconds) has elapsed?
(a) 5.3 * 10-6 (b) 2000 (c) 4600 (d) 34,000 (e) 190,000
Practice Exercise 2
The decomposition of dimethyl ether, 1CH322O, at 510 °C is a first-order process with a rate constant of 6.8 * 10-4 s-1:
1CH322O1g2 ¡ CH41g2 + H21g2 + CO1g2
If the initial pressure of 1CH322O is 135 torr, what is its pressure after 1420 s?
section 14.4 The Change of Concentration with Time
589
Equation 14.13 can be used to verify whether a reaction is first order and to
determine its rate constant. This equation has the form of the general equation for
a straight line, y = mx + b, in which m is the slope and b is the y-intercept of the
line (Appendix A.4):
ln [A]t = −kt
y
Methyl isonitrile
+ ln [A]0
= mx +
b
For a first-order reaction, therefore, a graph of ln3A4t versus time gives a straight line
with a slope of -k and a y-intercept of ln3A40. A reaction that is not first order will not
yield a straight line.
As an example, consider the conversion of methyl isonitrile 1CH3NC2 to its isomer
acetonitrile 1CH3CN2 (▶ Figure 14.7). Because experiments show that the reaction is
first order, we can write the rate equation:
ln3CH3NC4t = -kt + ln3CH3NC40
We run the reaction at a temperature at which methyl isonitrile is a gas 1199 °C2, and
▼ Figure 14.8 (a) shows how the pressure of this gas varies with time. We can use
pressure as a unit of concentration for a gas because we know from the ideal-gas
law the pressure is directly proportional to the number of moles per unit volume.
Figure 14.8(b) shows that a plot of the natural logarithm of the pressure versus time
is a straight line. The slope of this line is -5.1 * 10-5 s-1. (You should verify this for
yourself, remembering that your result may vary slightly from ours because of inaccuracies associated with reading the graph.) Because the slope of the line equals -k, the
rate constant for this reaction equals 5.1 * 10-5 s-1.
Second-Order Reactions
A second-order reaction is one for which the rate depends either on a reactant concentration raised to the second power or on the concentrations of two reactants each raised
to the first power. For simplicity, let’s consider reactions of the type A ¡ products
or A + B ¡ products that are second order in just one reactant, A:
Rate = -
∆3A4
= k3A42
∆t
Go Figure
160
140
120
100
80
60
40
20
0
ln pressure, CH3NC
Pressure, CH3NC (torr)
What can you conclude given that the plot of ln P versus t is linear?
0
10,000
20,000
Time (s)
(a)
30,000
5.2
5.0
4.8
4.6
4.4
4.2
4.0
3.8
3.6
3.4
0
10,000
20,000
Time (s)
(b)
▲ Figure 14.8 Kinetic data for conversion of methyl isonitrile into acetonitrile.
30,000
Acetonitrile
▲ Figure 14.7 The first-order reaction of
ch3nc conversion into ch3cn.
590
chapter 14 Chemical Kinetics
With the use of calculus, this differential rate law can be used to derive the integrated
rate law for second-order reactions:
1
1
= kt +
[14.14]
3A4t
3A40
This equation, like Equation 14.13, has four variables, k, t, 3A40, and 3A4t, and any one
of these can be calculated knowing the other three. Equation 14.14 also has the form
of a straight line 1y = mx + b2. If the reaction is second order, a plot of 1>3A4t versus t yields a straight line with slope k and y-intercept 1>3A40. One way to distinguish
between first- and second-order rate laws is to graph both ln3A4t and 1>3A4t against
t. If the ln3A4t plot is linear, the reaction is first order; if the 1>3A4t plot is linear, the
reaction is second order.
Sample
Exercise 14.8 Determining Reaction Order from the Integrated Rate Law
The following data were obtained for the gas-phase decomposition
of nitrogen dioxide at 300 °C, NO21g2 ¡ NO1g2 + 12 O21g2. Is the
reaction first or second order in NO2?
Time (s)
0.0
3no2 4 1M 2
50.0
0.00787
100.0
0.00649
200.0
0.00481
300.0
0.00380
0.01000
Solution
Analyze We are given the concentrations of a reactant at various times
during a reaction and asked to determine whether the reaction is first
or second order.
Plan We can plot ln3NO24 and 1>3NO24 against time. If one plot or the
other is linear, we will know the reaction is either first or second order.
Solve To graph ln3NO24 and 1>3NO24 against time, we first make the
following calculations from the data given:
0.01000
ln 3 no2 4
-4.605
1 , 3no2 4 11 , M 2
50.0
0.00787
-4.845
127
100.0
0.00649
-5.037
154
200.0
0.00481
-5.337
208
300.0
0.00380
-5.573
263
ln[NO2]
–4.8
–5.2
–5.4
–5.8
1/[NO2] (1/M)
–4.6
As ▶ Figure 14.9 shows, only the plot of 1>3NO24 versus
–4.8
time is linear. Thus, the reaction obeys a second-order rate law:
2
Rate = k3NO24 . From the slope of this–5.0
straight-line graph,
we determine that k = 0.543 M -1 s-1 for the disappearance
–5.2
of NO2.
ln[NO2]
–5.0
–5.6
100
–5.4
Practice Exercise 1
–5.6
For a certain reaction A ¡ products, a plot of ln[A] versus
-2 -1
time produces a straight line with a slope
–5.8 of - 3.0 * 10 s .
0
100
200
Which of the following statements is or are true?:
Time (s)
(i) The ­reaction follows first-order kinetics.
(ii) The rate constant for the reaction is 3.0 * 10-2 s-1.
(iii) The initial concentration of [A] was 1.0 M.
1/[NO2] (1/M)
0.0
3 no2 4 1M 2
Time (s)
–4.6
300
0
100
200
Time (s)
300
250
150
50
0
100
200
Time (s)
300
▲ Figure 14.9 Kinetic data for decomposition
of no2.
250
150
50
0
100
200
Time (s)
section 14.4 The Change of Concentration with Time
(a) Only one of the statements is true.
(b) Statements (i) and (ii) are true.
(c) Statements (i) and (iii) are true.
(d)Statements (ii) and (iii) are true.
(e) All three statements are true.
591
Practice Exercise 2
The decomposition of NO2 discussed in the Sample Exercise is
second order in NO2 with k = 0.543 M -1 s-1. If the initial
concentration of NO2 in a closed vessel is 0.0500 M, what is the
concentration of this reactant after 0.500 h?
Zero-Order Reactions
We have seen that in a first-order reaction the concentration of a reactant A decreases
nonlinearly, as shown by the red curve in ▶ Figure 14.10. As [A] declines, the rate at
which it disappears declines in proportion. A zero-order reaction is one in which the
rate of disappearance of A is independent of [A]. The rate law for a zero-order reaction is
Rate =
- ∆3A4
= k
∆t
Go Figure
At which times during the reaction
would you have trouble distinguishing
a zero-order reaction from a first-order
reaction?
A0
The integrated rate law for a zero-order reaction is
where 3A4t is the concentration of A at time t and 3A40 is the initial concentration.
This is the equation for a straight line with vertical intercept 3A40 and slope -kt, as
shown in the blue curve in Figure 14.10.
The most common type of zero-order reaction occurs when a gas undergoes
decomposition on the surface of a solid. If the surface is completely covered by
decomposing molecules, the rate of reaction is constant because the number of reacting surface molecules is constant, so long as there is some gas-phase substance left.
[A]
3A4t = -kt + 3A40
First-order reaction
Slope = −kt
Half-Life
The half-life of a reaction, t1>2, is the time required for the concentration of a reactant to reach half its initial value, 3A4t1>2 = 123A40. Half-life is a convenient way to describe how fast a reaction occurs, especially if it is a first-order process. A fast reaction
has a short half-life.
We can determine the half-life of a first-order reaction by substituting 3A4t1>2 = 123A40
for 3A4t and t1>2 for t in Equation 14.12:
ln
1
2 3A40
3A40
Zero-order reaction
Time
▲ Figure 14.10 Comparison of first-order
and zero-order reactions for the disappearance
of reactant A with time.
= -kt1>2
ln 12 = -kt1>2
t1>2 = -
ln 12
k
=
0.693
k
150
[14.15]
Pressure, CH3NC (torr)
From Equation 14.15, we see that t1>2 for a first-order rate law does
not depend on the initial concentration of any reactant. Consequently,
the half-life remains constant throughout the reaction. If, for example,
the concentration of a reactant is 0.120 M at some instant in the reac75
tion, it will be 1210.120 M2 = 0.060 M after one half-life. After one more
half-life passes, the concentration will drop to 0.030 M, and so on. Equation 14.15 also indicates that, for a first-order reaction, we can calculate
t1>2 if we know k and calculate k if we know t1>2.
37.5
The change in concentration over time for the first-order rearrangement of gaseous methyl isonitrile at 199 °C is graphed in ▶ Figure 14.11.
Because the concentration of this gas is directly proportional to its pres0
sure during the reaction, we have chosen to plot pressure rather than
concentration in this graph. The first half-life occurs at 13,600 s (3.78
h). At a time 13,600 s later, the methyl isonitrile pressure (and therefore,
concentration) has decreased to half of one-half, or one-fourth, of the initial value. In a
first-order reaction, the concentration of the reactant decreases by one-half in each of a
series of regularly spaced time intervals, each interval equal to t1>2.
t1/2
t1/2
10,000
20,000
Time (s)
30,000
▲ Figure 14.11 Kinetic data for the
rearrangement of methyl isonitrile to
acetonitrile at 199 °c, showing the half-life of
the reaction.
592
chapter 14 Chemical Kinetics
Give It Some Thought
If a solution containing 10.0 g of a substance reacts by first-order kinetics, how
many grams remain after three half-lives?
Chemistry Put to Work
Methyl Bromide in
the Atmosphere
The compounds known as chlorofluorocarbons (CFCs) are well-known
agents responsible for the destruction of Earth’s protective ozone layer.
Another simple molecule that has the potential to destroy the stratospheric ozone layer is methyl bromide, CH3Br (▼ Figure 14.12).
Because this substance has a wide range of uses, including antifungal treatment of plant seeds, it has been produced in large quantities
in the past (about 150 million pounds per year worldwide in 1997, at
the height of its production). In the stratosphere, the C ¬ Br bond is
Stratosphere
Diffusion to
stratosphere
50% decomposes in 0.8 years
Troposphere
Lower atmosphere
Methyl bromide
applied as antifungal
treatment
▲ Figure 14.12 Distribution and fate of methyl bromide in Earth’s
atmosphere.
broken through absorption of short-wavelength radiation. The resultant
Br atoms then catalyze decomposition of O3.
Methyl bromide is removed from the lower atmosphere by a variety of mechanisms, including a slow reaction with ocean water:
CH3Br1g2 + H2O1l2 ¡ CH3OH1aq2 + HBr1aq2[14.16]
To determine the potential importance of CH3Br in destruction
of the ozone layer, it is important to know how rapidly the reaction in
Equation 14.16 and all other reactions remove CH3Br from the lower
atmosphere before it can diffuse into the stratosphere.
The average lifetime of CH3Br in Earth’s lower atmosphere is
difficult to measure because the conditions that exist in the atmosphere are too complex to be simulated in the laboratory. Instead,
scientists analyzed nearly 4000 atmospheric samples collected
above the Pacific Ocean for the presence of several trace organic
substances, including methyl bromide. From these measurements,
it was possible to estimate the atmospheric residence time for CH3Br.
The atmospheric residence time is related to the half-life
for CH3Br in the lower atmosphere, assuming CH3Br decomposes by a first-order process. From the experimental data,
the half-life for methyl bromide in the lower atmosphere is
estimated to be 0.8 { 0.1 yr. That is, a collection of CH3Br
molecules present at any given time will, on average, be 50%
decomposed after 0.8 yr, 75% decomposed after 1.6 yr, and so
on. A half-life of 0.8 yr, while comparatively short, is still sufficiently long so that CH3Br contributes significantly to the destruction of the ozone layer.
In 1997 an international agreement was reached to phase
out use of methyl bromide in developed countries by 2005. However, in recent years exemptions for critical agricultural use have
been requested and granted. Nevertheless, authorized worldwide
production was down to 26 million pounds in 2012, three-fourths
of which is used in the United States.
Related Exercise: 14.122
Sample
Exercise 14.9 Determining the Half-Life of a First-Order Reaction
The reaction of C4H9Cl with water is a first-order reaction. (a) Use Figure 14.4 to estimate the halflife for this reaction. (b) Use the half-life from (a) to calculate the rate constant.
Solution
Analyze We are asked to estimate the half-life of a reaction from a
graph of concentration versus time and then to use the half-life to calculate the rate constant for the reaction.
(b) Equation 14.15 is used to calculate the rate constant from the
half-life.
Plan
(a) From the graph, we see that the initial value of 3C4H9Cl4 is 0.100 M.
The half-life for this first-order reaction is the time required
for 3C4H9Cl4 to decrease to 0.050 M, which we can read off the
graph. This point occurs at approximately 340 s.
(a) To estimate a half-life, we can select a concentration and then
determine the time required for the concentration to decrease to
half of that value.
Solve
section 14.5 Temperature and Rate
(b) Solving Equation 14.15 for k, we have
0.693
0.693
k =
=
= 2.0 * 10-3 s-1
t1>2
340 s
Check At the end of the second half-life, which should occur at 680 s,
the concentration should have decreased by yet another factor of 2, to
0.025 M. Inspection of the graph shows that this is indeed the case.
Practice Exercise 1
We noted in an earlier Practice Exercise that at 25 °C the decomposition of N2O51g2 into NO21g2 and O21g2 follows first-order
593
kinetics with k = 3.4 * 10-5 s-1. How long will it take for a
sample originally containing 2.0 atm of N2O5 to reach a partial
pressure of 380 torr?
(a) 5.7 h (b) 8.2 h (c) 11 h (d) 16 h (e) 32 h
Practice Exercise 2
(a)Using Equation 14.15, calculate t1>2 for the decomposition of
the insecticide described in Sample Exercise 14.7.
(b)How long does it take for the concentration of the insecticide
to reach one-quarter of the initial value?
The half-life for second-order and other reactions depends on reactant concentrations and therefore changes as the reaction progresses. We obtained Equation 14.15 for
the half-life for a first-order reaction by substituting 3A4t1>2 = 123A40 for 3A4t and t1>2
for t in Equation 14.12. We find the half-life of a second-order reaction by making the
same substitutions into Equation 14.14:
1
1
= kt1>2 +
1
3A40
2 3A40
2
1
= kt1>2
3A40
3A40
t1>2 =
1
[14.17]
k3A40
In this case, the half-life depends on the initial concentration of reactant—the lower the
initial concentration, the longer the half-life.
Give It Some Thought
Why can we report the half-life for a first-order reaction without knowing the
initial concentration, but not for a second-order reaction?
14.5 | Temperature and Rate
The rates of most chemical reactions increase as the temperature rises. For example,
dough rises faster at room temperature than when refrigerated, and plants grow more
rapidly in warm weather than in cold. We can see the effect of temperature on reaction
rate by observing a chemiluminescence reaction (one that produces light), such as that
in Cyalume® light sticks (▶ Figure 14.13).
How is this experimentally observed temperature effect reflected in the rate law?
The faster rate at higher temperature is due to an increase in the rate constant with
increasing temperature. For example, let’s reconsider the first-order reaction we saw in
Figure 14.7, namely CH3NC ¡ CH3CN. Figure 14.14 shows the rate constant for
this reaction as a function of temperature. The rate constant and, hence, the rate of the
reaction increase rapidly with temperature, approximately doubling for each 10 °C rise.
Go Figure
Why does the light stick glow with less
light in cold water than in hot water?
The Collision Model
Reaction rates are affected both by reactant concentrations and by temperature. The
(Section 10.7), accounts for
collision model, based on the kinetic-molecular theory
both of these effects at the molecular level. The central idea of the collision model is
that molecules must collide to react. The greater the number of collisions per second,
the greater the reaction rate. As reactant concentration increases, therefore, the number
of collisions increases, leading to an increase in reaction rate. According to the kineticmolecular theory of gases, increasing the temperature increases molecular speeds. As
molecules move faster, they collide more forcefully (with more energy) and more frequently, increasing reaction rates.
Hot water
Cold water
▲ Figure 14.13 Temperature affects the
rate of the chemiluminescence reaction in
light sticks: The chemiluminescent reaction
occurs more rapidly in hot water, and more
light is produced.
594
chapter 14 Chemical Kinetics
Go Figure
Would you expect this curve to
eventually go back down to lower
values? Why or why not?
3 × 10−3
k (s–1)
CH3NC
For a reaction to occur, though, more is required than simply a collision—it
must be the right kind of collision. For most reactions, in fact, only a tiny fraction of
collisions leads to a reaction. For example, in a mixture of H2 and I2 at ordinary temperatures and pressures, each molecule undergoes about 1010 collisions per second. If
every collision between H2 and I2 resulted in the formation of HI, the reaction would
be over in much less than a second. Instead, at room temperature the
reaction proceeds very slowly because only about one in every 1013 collisions produces a reaction. What keeps the reaction from occurring
more rapidly?
CH3CN
The Orientation Factor
In most reactions, collisions between molecules result in a chemical reaction only if the molecules are oriented in a certain way when they collide.
The relative orientations of the molecules during collision determine
whether the atoms are suitably positioned to form new bonds. For example, consider the reaction
2 × 10−3
1 × 10−3
Cl + NOCl ¡ NO + Cl2
180 190 200 210 220 230 240 250
Temperature (°C)
▲ Figure 14.14 Temperature dependence
of the rate constant for methyl isonitrile
conversion to acetonitrile. The four points
indicated are used in Sample Exercise 14.11.
which takes place if the collision brings Cl atoms together to form Cl2,
as shown in the top panel of ▼ Figure 14.15. In contrast, in the collision
shown in the lower panel, the two Cl atoms are not colliding directly with
one another, and no products are formed.
Activation Energy
Molecular orientation is not the only factor influencing whether a molecular collision
will produce a reaction. In 1888 the Swedish chemist Svante Arrhenius suggested that
molecules must possess a certain minimum amount of energy to react. According to
the collision model, this energy comes from the kinetic energies of the colliding molecules. Upon collision, the kinetic energy of the molecules can be used to stretch, bend,
and ultimately break bonds, leading to chemical reactions. That is, the kinetic energy
is used to change the potential energy of the molecule. If molecules are moving too
slowly—in other words, with too little kinetic energy—they merely bounce off one another without changing. The minimum energy required to initiate a chemical reaction
is called the activation energy, Ea, and its value varies from reaction to reaction.
Effective collision,
reaction occurs,
Cl2 forms
Before collision
Collision
After collision
Ineffective collision,
no reaction possible,
no Cl2
Before collision
Collision
▲ Figure 14.15 Molecular collisions may or may not lead to a chemical reaction between Cl and NOCl.
After collision
section 14.5 Temperature and Rate
595
Go Figure
If the barrier were lower than as shown in the figure, would the golfer have to hit the ball as hard?
▲ Figure 14.16 Energy is needed to overcome a barrier between initial and final states.
The situation during reactions is analogous to that shown in ▲ Figure 14.16.
The golfer hits the ball to make it move over the hill in the direction of the cup.
The hill is a barrier between ball and cup. To reach the cup, the player must impart
enough kinetic energy with the putter to move the ball to the top of the barrier. If
he does not impart enough energy, the ball will roll partway up the hill and then
back down toward him. In the same way, molecules require a certain minimum
energy to break existing bonds during a chemical reaction. We can think of this
minimum energy as an energy barrier. In the rearrangement of methyl isonitrile
to acetonitrile, for example, we might imagine the reaction passing through an
intermediate state in which the N ‚ C portion of the methyl isonitrile molecule is
sideways:
H3C
N
C
H3C
C
N
H3C
C
N
Go Figure
shows that energy must be supplied to stretch the bond between the How does the energy needed to
H3C group and the N ‚ C group to allow the N ‚ C group to rotate. After the N ‚ C overcome the energy barrier compare
group has twisted sufficiently, the C ¬ C bond begins to form, and the energy of the with the overall change in energy for
molecule drops. Thus, the barrier to formation of acetonitrile represents the energy this reaction?
necessary to force the molecule through the relatively unstable interActivated complex forms
mediate state, analogous to forcing the ball in Figure 14.16 over
Molecule bends,
the hill. The difference between the energy of the starting molecule
C
C — N bond
H3C. . .
and the highest energy along the reaction pathway is the activation
begins to break
N
energy, Ea. The molecule having the arrangement of atoms shown
C
at the top of the barrier is called either the activated complex or the
C — C bond forms
H3C N
transition state.
Ea
The conversion of H3C ¬ N ‚ C to H3C ¬ C ‚ N is exotherH 3C C
mic. Figure 14.17 therefore shows the product as having a lower
N
energy than the reactant. The energy change for the reaction, ∆E,
has no effect on reaction rate, however. The rate depends on the
H3C N C
magnitude of Ea; generally, the lower the value of Ea is, the faster the
∆E
reaction.
Notice that the reverse reaction is endothermic. The activation
H3C C N
energy for the reverse reaction is equal to the energy that must be
Reaction progress
overcome if approaching the barrier from the right: ∆E + Ea. Thus,
to reach the activated complex for the reverse reaction requires more energy than for ▲ Figure 14.17 Energy profile for
the forward reaction—for this reaction, there is a larger barrier to overcome going from conversion of methyl isonitrile 1h3cnc2 to its
isomer acetonitrile 1h3ccn2.
right to left than from left to right.
Potential energy
▶ Figure 14.17
596
chapter 14 Chemical Kinetics
Give It Some Thought
Suppose you could measure the rates for both the forward and reverse
reactions of the process in Figure 14.17. In which direction would the rate
be larger? Why?
Any particular methyl isonitrile molecule acquires sufficient energy to overcome
the energy barrier through collisions with other molecules. Recall from the kineticmolecular theory of gases that, at any instant, gas molecules are distributed in energy
(Section 10.7) ▼ Figure 14.18 shows the distribution of kinetic
over a wide range.
energies for two temperatures, comparing them with the minimum energy needed for
reaction, Ea. At the higher temperature a much greater fraction of the molecules have
kinetic energy greater than Ea, which leads to a greater rate of reaction.
Give It Some Thought
Suppose we have two reactions, A ¡ B and B ¡ C. You can isolate B, and
it is stable. Is B the transition state for the reaction A ¡ C?
For a collection of molecules in the gas phase, the fraction of molecules that have kinetic
energy equal to or greater than Ea is given by the expression
Go Figure
Fraction of molecules
What would the curve look like for a
temperature higher than that for the
red curve in the figure?
Minimum energy
needed for reaction, Ea
Lower T
Higher T
Larger fraction of molecules
reacts at higher temperature
Kinetic energy
▲ Figure 14.18 The effect of temperature
on the distribution of kinetic energies of
molecules in a sample.
f = e -Ea>RT [14.18]
In this equation, R is the gas constant 18.314 J>mol@K2 and T is the
absolute temperature. To get an idea of the magnitude of f, let’s suppose that Ea is 100 kJ>mol, a value typical of many reactions, and that
T is 300 K. The calculated value of f is 3.9 * 10-18, an extremely small
number! At 320 K, f = 4.7 * 10-17. Thus, only a 20° increase in temperature produces a more than tenfold increase in the fraction of molecules possessing at least 100 kJ>mol of energy.
The Arrhenius Equation
Arrhenius noted that for most reactions the increase in rate with
increasing temperature is nonlinear (Figure 14.14). He found that
most reaction-rate data obeyed an equation based on (a) the fraction of molecules possessing energy Ea or greater, (b) the number of collisions per
second, and (c) the fraction of collisions that have the appropriate orientation.
These three factors are incorporated into the Arrhenius equation:
k = Ae -Ea>RT [14.19]
In this equation, k is the rate constant, Ea is the activation energy, R is the gas constant 18.314 J>mol@K2, and T is the absolute temperature. The frequency factor, A, is
constant, or nearly so, as temperature is varied. This factor is related to the frequency
of collisions and the probability that the collisions are favorably oriented for reaction.* As the magnitude of Ea increases, k decreases because the fraction of molecules
that possess the required energy is smaller. Thus, at fixed values of T and A, reaction
rates decrease as Ea increases.
*Because collision frequency increases with temperature, A also has some temperature dependence, but this
dependence is much smaller than the exponential term. Therefore, A is considered approximately constant.
section 14.5 Temperature and Rate
597
Sample
Exercise 14.10 Activation Energies and Speeds of Reaction
Consider a series of reactions having these energy profiles:
2
Potential energy
1
3
25 kJ/mol
20 kJ/mol
15 kJ/mol
−10 kJ/mol
Reaction progress
5 kJ/mol
−15 kJ/mol
Reaction progress
Reaction progress
Rank the reactions from slowest to fastest assuming that they have nearly the same value for the frequency factor A.
Solution
The lower the activation energy, the faster the reaction. The value of
∆E does not affect the rate. Hence, the order from slowest reaction to
fastest is 2 6 3 6 1.
Practice Exercise 1
Which of the following statements is or are true?
(i) The activation energies for the forward and reverse directions
of a reaction can be different.
(ii) Assuming that A is constant, if both Ea and T increase, then k
will increase.
(iii) For two different reactions, the one with the smaller value of
Ea will necessarily have the larger value for k.
(a) Only one of the statements is true.
(b) Statements (i) and (ii) are true.
(c) Statements (i) and (iii) are true.
(d) Statements (ii) and (iii) are true.
(e) All three statements are true.
Practice Exercise 2
Rank the reverse reactions from slowest to fastest.
Determining the Activation Energy
We can calculate the activation energy for a reaction by manipulating the Arrhenius
equation. Taking the natural log of both sides of Equation 14.19, we obtain
Ea
+ ln A
ln k = − –––
RT
=
[14.20]
y
mx + b
which has the form of the equation for a straight line. A graph of ln k versus 1>T is a
line with a slope equal to -Ea >R and a y-intercept equal to ln A. Thus, the activation
energy can be determined by measuring k at a series of temperatures, graphing ln k
versus 1>T, and calculating Ea from the slope of the resultant line.
We can also use Equation 14.20 to evaluate Ea in a nongraphical way if we know
the rate constant of a reaction at two or more temperatures. For example, suppose that
at two different temperatures T1 and T2 a reaction has rate constants k1 and k2. For each
condition, we have
ln k1 = -
Ea
Ea
+ ln A and ln k2 = + ln A
RT1
RT2
Subtracting ln k2 from ln k1 gives
ln k1 - ln k2 = a-
Ea
Ea
+ ln A b - a+ ln A b
RT1
RT2
598
chapter 14 Chemical Kinetics
Simplifying this equation and rearranging gives
ln
Ea 1
k1
1
=
a - b [14.21]
k2
R T2
T1
Equation 14.21 provides a convenient way to calculate a rate constant k1 at some temperature T1 when we know the activation energy and the rate constant k2 at some other
temperature T2.
Sample
Exercise 14.11 Determining the Activation Energy
The following table shows the rate constants for the rearrangement of methyl isonitrile at various
temperatures (these are the data points in Figure 14.14):
Temperature 1 °C 2
189.7
2.52 * 10-5
198.9
5.25 * 10-5
230.3
6.30 * 10-4
251.2
3.16 * 10-3
k 1s−1 2
(a) From these data, calculate the activation energy for the reaction. (b) What is the value of the
rate constant at 430.0 K?
Solution
Analyze We are given rate constants, k, measured at several temperatures and asked to determine the
activation energy, Ea, and the rate constant, k, at a particular temperature.
Plan We can obtain Ea from the slope of a graph of ln k versus 1>T. Once we know Ea, we can use
Equation 14.21 together with the given rate data to calculate the rate constant at 430.0 K.
Solve
(a) We must first convert the temperatures from degrees
Celsius to kelvins. We then take the inverse of each temperature, 1>T, and the natural log of each rate constant,
ln k. This gives us the table shown at the right:
A graph of ln k versus 1>T is a straight line (▼ Figure 14.19).
The slope of the line is obtained by choosing any two
well-separated points and using the coordinates of each:
∆y
∆x
=
-6.6 - 1-10.42
0.00195 - 0.00215
462.9
472.1
1 , T 1K −1 2
ln k
2.160 * 10
-3
-10.589
-9.855
2.118 * 10
-3
503.5
1.986 * 10
-3
-7.370
524.4
1.907 * 10-3
-5.757
= -1.9 * 104
Because logarithms have no units, the numerator in this equation is dimensionless. The denominator has the units of 1>T,
namely, K - 1. Thus, the overall units for the slope are K. The slope
equals -Ea >R. We use the value for the gas constant R in units of
J>mol@K (Table 10.2). We thus obtain
Ea
Slope = R
Ea = - 1slope21R2 = - 1- 1.9 * 104 K2a8.314
= 1.6 * 102 kJ>mol = 160 kJ>mol
−5
−6
−7
ln k
Slope =
T 1K 2
Slope =
−Ea
R
−8
−9
1 kJ
J
ba
b −10
mol@K 1000 J
−11
0.0019
We report the activation energy to only two significant figures
because we are limited by the precision with which we can
read the graph in Figure 14.19.
0.0020
0.0021
1/T (K−1)
0.0022
▲ Figure 14.19 Graphical determination of activation energy Ea.
section 14.6 Reaction Mechanisms
(b) To determine the rate constant, k1, at T1 = 430.0 K, we can
use Equation 14.21 with Ea = 160 kJ>mol and one of the
rate constants and temperatures from the given data, such as
k2 = 2.52 * 10-5 s-1 and T2 = 462.9 K:
Practice Exercise 1
Using the data in Sample Exercise 14.11, which of the following is the
rate constant for the rearrangement of methyl isonitrile at 320 °C?
(a)8.1 * 10-15 s-1 (b) 2.2 * 10-13 s-1 (c) 2.7 * 10-9 s-1 (d) 2.3 * 10-1 s-1 (e) 9.2 * 103 s-1
k1
b =
2.52 * 10 - 5 s - 1
160 kJ>mol
1000 J
1
1
ba
ba
b = -3.18
a
8.314 J>mol@K 462.9 K
430.0 K
1 kJ
ln a
Practice Exercise 2
To one significant figure, what is the value for the frequency factor
A for the data presented in Sample Exercise 14.11.
Thus,
k1
= e - 3.18 = 4.15 * 10 - 2
2.52 * 10 - 5 s - 1
k1 = 14.15 * 10 - 2212.52 * 10 - 5 s - 12 = 1.0 * 10 - 6 s - 1
Note that the units of k1 are the same as those of k2.
14.6 | Reaction Mechanisms
A balanced equation for a chemical reaction indicates the substances present at the start
of the reaction and those present at the end of the reaction. It provides no information, however, about the detailed steps that occur at the molecular level as the reactants
are turned into products. The steps by which a reaction occurs is called the reaction
mechanism. At the most sophisticated level, a reaction mechanism describes the order
in which bonds are broken and formed and the changes in relative positions of the atoms in the course of the reaction.
Elementary Reactions
We have seen that reactions take place because of collisions between reacting molecules. For example, the collisions between molecules of methyl isonitrile 1CH3NC2 can
provide the energy to allow the CH3NC to rearrange to acetonitrile:
H3C
N
C
H3C
C
N
H3C
C
N
Similarly, the reaction of NO and O3 to form NO2 and O2 appears to occur as a result
of a single collision involving suitably oriented and sufficiently energetic NO and O3
molecules:
599
NO1g2 + O31g2 ¡ NO21g2 + O21g2[14.22]
Both reactions occur in a single event or step and are called elementary reactions.
The number of molecules that participate as reactants in an elementary reaction
defines the molecularity of the reaction. If a single molecule is involved, the reaction is
unimolecular. The rearrangement of methyl isonitrile is a unimolecular process. Elementary reactions involving the collision of two reactant molecules are bimolecular.
The reaction between NO and O3 is bimolecular. Elementary reactions involving the
simultaneous collision of three molecules are termolecular. Termolecular reactions are
far less probable than unimolecular or bimolecular processes and are extremely rare.
The chance that four or more molecules will collide simultaneously with any regularity
is even more remote; consequently, such collisions are never proposed as part of a reaction mechanism. Thus, nearly all reaction mechanisms contain only unimolecular and
bimolecular elementary reactions.
600
chapter 14 Chemical Kinetics
Give It Some Thought
What is the molecularity of the elementary reaction?
NO1g2 + Cl21g2 ¡ NOCl1g2 + Cl1g2
Multistep Mechanisms
The net change represented by a balanced chemical equation often occurs by a multistep mechanism consisting of a sequence of elementary reactions. For example, below
225 °C, the reaction
Go Figure
For this profile, is it easier for a
molecule of the intermediate to
convert to reactants or products?
appears to proceed in two elementary reactions (or two elementary steps), each of which
is bimolecular. First, two NO2 molecules collide, and an oxygen atom is transferred
from one to the other. The resultant NO3 then collides with a CO molecule and transfers an oxygen atom to it:
NO21g2 + NO21g2 ¡ NO31g2 + NO1g2
NO31g2 + CO1g2 ¡ NO21g2 + CO21g2
Potential energy
Transition
state
Transition
state
Thus, we say that the reaction occurs by a two-step mechanism.
The chemical equations for the elementary reactions in a multistep mechanism must always add to give the chemical equation of the overall process. In
the present example, the sum of the two elementary reactions is
Intermediate
Products
Reactants
Reaction progress
▲ Figure 14.20 The energy profile of a
reaction, showing transition states and an
intermediate.
NO21g2 + CO1g2 ¡ NO1g2 + CO21g2[14.23]
2 NO21g2 + NO31g2 + CO1g2 ¡ NO21g2 + NO31g2 + NO1g2 + CO21g2
Simplifying this equation by eliminating substances that appear on both sides
gives Equation 14.23, the net equation for the process.
Because NO3 is neither a reactant nor a product of the reaction—it is
formed in one elementary reaction and consumed in the next—it is called
an intermediate. Multistep mechanisms involve one or more intermediates.
Intermediates are not the same as transition states, as shown in ◀ Figure 14.20.
Intermediates can be stable and can therefore sometimes be identified and even
isolated. Transition states, on the other hand, are always inherently unstable and as such
can never be isolated. Nevertheless, the use of advanced “ultrafast” techniques sometimes
allows us to characterize them.
Sample
Exercise 14.12 Determining Molecularity and Identifying Intermediates
It has been proposed that the conversion of ozone into O2 proceeds by a two-step mechanism:
O31g2 ¡ O21g2 + O1g2
O31g2 + O1g2 ¡ 2 O21g2
(a) Describe the molecularity of each elementary reaction in this mechanism.
(b) Write the equation for the overall reaction.
(c) Identify the intermediate(s).
Solution
Analyze We are given a two-step mechanism and asked for (a) the
molecularities of each of the two elementary reactions, (b) the equation for the overall process, and (c) the intermediate.
Plan The molecularity of each elementary reaction depends on the
number of reactant molecules in the equation for that reaction. The
overall equation is the sum of the equations for the elementary reactions. The intermediate is a substance formed in one step of the
mechanism and used in another and therefore not part of the equation for the overall reaction.
Solve
(a) The first elementary reaction involves a single reactant and is
consequently unimolecular. The second reaction, which involves
two reactant molecules, is bimolecular.
section 14.6 Reaction Mechanisms
(b) Adding the two elementary reactions gives
2 O31g2 + O1g2 ¡ 3 O21g2 + O1g2
Because O(g) appears in equal amounts on both sides of the equation,
it can be eliminated to give the net equation for the chemical process:
2 O31g2 ¡ 3 O21g2
(c) The intermediate is O(g). It is neither an original reactant nor a
final product but is formed in the first step of the mechanism and
consumed in the second.
Practice Exercise 1
Consider the following two-step reaction mechanism:
A1g2 + B1g2 ¡ X1g2 + Y1g2
(iii) The substance X(g) is an intermediate in this mechanism.
(a) Only one of the statements is true.
(b) Statements (i) and (ii) are true.
(c) Statements (i) and (iii) are true.
(d) Statements (ii) and (iii) are true.
(e) All three statements are true.
Practice Exercise 2
For the reaction
Mo1CO26 + P1CH323 ¡ Mo1CO25P1CH323 + CO
the proposed mechanism is
Mo1CO26 ¡ Mo1CO25 + CO
Mo1CO25 + P1CH323 ¡ Mo1CO25P1CH323
X1g2 + C1g2 ¡ Y1g2 + Z1g2
Which of the following statements about this mechanism is or are true?
(i) Both of the steps in this mechanism are bimolecular.
(ii) The overall reaction is A1g2 + B1g2 + C1g2 ¡ Y1g2 + Z1g2.
(a)Is the proposed mechanism consistent with the equation for
the overall reaction? (b) What is the molecularity of each step of
the mechanism? (c) Identify the intermediate(s).
Rate Laws for Elementary Reactions
In Section 14.3, we stressed that rate laws must be determined experimentally; they cannot
be predicted from the coefficients of balanced chemical equations. We are now in a position
to understand why this is so. Every reaction is made up of a series of one or more elementary steps, and the rate laws and relative speeds of these steps dictate the overall rate law for
the reaction. Indeed, the rate law for a reaction can be determined from its mechanism, as
we will see shortly, and compared with the experimental rate law. Thus, our next challenge
in kinetics is to arrive at reaction mechanisms that lead to rate laws consistent with those
observed experimentally. We start by examining the rate laws of elementary reactions.
Elementary reactions are significant in a very important way: If a reaction is elementary, its rate law is based directly on its molecularity. For example, consider the unimolecular reaction
A ¡ products
As the number of A molecules increases, the number that reacts in a given time interval
increases proportionally. Thus, the rate of a unimolecular process is first order:
Rate = k3A4
For bimolecular elementary steps, the rate law is second order, as in the reaction
A + B ¡ products
601
Rate = k3A43B4
The second-order rate law follows directly from collision theory. If we double the concentration of A, the number of collisions between the molecules of A and B doubles;
likewise, if we double [B], the number of collisions between A and B doubles. Therefore, the rate law is first order in both [A] and [B] and second order overall.
The rate laws for all feasible elementary reactions are given in ▼ Table 14.3. Notice
how each rate law follows directly from the molecularity of the reaction. It is important
Table 14.3 Elementary Reactions and Their Rate Laws
Molecularity
Elementary Reaction
Rate Law
Unimolecular
A ¡ products
Rate = k3A4
Bimolecular
A + A ¡ products
Rate = k3A42
Bimolecular
A + B ¡ products
Rate = k3A43B4
Termolecular
A + A + A ¡ products
Rate = k3A43
Termolecular
A + A + B ¡ products
Termolecular
A + B + C ¡ products
Rate = k3A423B4
Rate = k3A43B43C4
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chapter 14 Chemical Kinetics
to remember, however, that we cannot tell by merely looking at a balanced, overall
chemical equation whether the reaction involves one or several elementary steps.
Sample
Exercise 14.13 Predicting the Rate Law for an Elementary Reaction
If the following reaction occurs in a single elementary reaction, predict its rate law:
Solution
H21g2 + Br21g2 ¡ 2 HBr1g2
Analyze We are given the equation and asked for its rate law, assuming that it is an elementary process.
Plan Because we are assuming that the reaction occurs as a single
elementary reaction, we are able to write the rate law using the coefficients for the reactants in the equation as the reaction orders.
Solve The reaction is bimolecular, involving one molecule of H2 and
one molecule of Br2. Thus, the rate law is first order in each reactant
and second order overall:
Rate = k3H243Br24
Comment Experimental studies of this reaction show that the reaction
actually has a very different rate law:
Rate = k3H243Br241>2
Because the experimental rate law differs from the one obtained by assuming a single elementary reaction, we can conclude that the mechanism cannot occur by a single elementary step. It must, therefore,
involve two or more elementary steps.
Practice Exercise 1
Consider the following reaction: 2 A + B ¡ X + 2 Y. You are
told that the first step in the mechanism of this reaction has the
following rate law: Rate = k3A43B4. Which of the following could
be the first step in the reaction mechanism (note that substance Z
is an intermediate)?
(a)A + A ¡ Y + Z
(b)A ¡ X + Z
(c)A + A + B ¡ X + Y + Y
(d)B ¡ X + Y
(e)A + B ¡ X + Z
Practice Exercise 2
Consider the following reaction: 2 NO1g2 + Br21g2 ¡ 2 NOBr1g2.
(a) Write the rate law for the reaction, assuming it involves a single
elementary reaction. (b) Is a single-step mechanism likely for this
reaction?
The Rate-Determining Step for a Multistep
Mechanism
As with the reaction in Sample Exercise 14.13, most reactions occur by mechanisms
that involve two or more elementary reactions. Each step of the mechanism has its own
rate constant and activation energy. Often one step is much slower than the others, and
the overall rate of a reaction cannot exceed the rate of the slowest elementary step. Because the slow step limits the overall reaction rate, it is called the rate-determining step
(or rate-limiting step).
To understand the concept of the rate-determining step for a reaction, consider
a toll road with two toll plazas ( ▶ Figure 14.21). Cars enter the toll road at point 1
and pass through toll plaza A. They then pass an intermediate point 2 before passing
through toll plaza B and arriving at point 3. We can envision this trip along the toll
road as occurring in two elementary steps:
Step 1: Point 1 ¡ Point 2 1through toll plaza A2
Step 2: Point 2 ¡ Point 3 1through toll plaza B2
Overall: Point 1 ¡ Point 3 1through both toll plazas2
Now suppose that one or more gates at toll plaza A are malfunctioning, so that
traffic backs up behind the gates, as depicted in Figure 14.21(a). The rate at which cars
can get to point 3 is limited by the rate at which they can get through the traffic jam
at plaza A. Thus, step 1 is the rate-determining step of the journey along the toll road.
If, however, all gates at A are functioning but one or more at B are not, traffic flows
quickly through A but gets backed up at B, as depicted in Figure 14.21(b). In this case,
step 2 is the rate-determining step.
In the same way, the slowest step in a multistep reaction determines the overall rate.
By analogy to Figure 14.21(a), the rate of a fast step following the rate-determining step
does not speed up the overall rate. If the slow step is not the first one, as is the case in
section 14.6 Reaction Mechanisms
Go Figure
For which of the two scenarios in the figure will one get from point 1 to point 3 most rapidly?
Toll plaza A
1
Toll plaza B
2
3
(a) Cars slowed at toll plaza A, rate-determining step is passage through A
Toll plaza A
1
Toll plaza B
2
3
(b) Cars slowed at toll plaza B, rate-determining step is passage through B
▲ Figure 14.21 Rate-determining steps in traffic flow on a toll road.
Figure 14.21(b), the faster preceding steps produce intermediate products that accumulate before being consumed in the slow step. In either case, the rate-determining step
governs the rate law for the overall reaction.
Give It Some Thought
Why can’t the rate law for a reaction generally be deduced from the balanced
equation for the reaction?
Mechanisms with a Slow Initial Step
We can most easily see the relationship between the slow step in a mechanism and
the rate law for the overall reaction by considering an example in which the first step
in a multistep mechanism is the rate-determining step. Consider the reaction of NO2
and CO to produce NO and CO2 (Equation 14.23). Below 225 °C, it is found experimentally that the rate law for this reaction is second order in NO2 and zero order in
CO: Rate = k3NO242. Can we propose a reaction mechanism consistent with this rate
law? Consider the two-step mechanism:*
k1
Step 1: NO21g2 + NO21g2 ¡ NO31g2 + NO1g2 1slow2
Step 2:
k2
NO31g2 + CO1g2 ¡ NO21g2 + CO21g2 1fast2
Overall: NO21g2 + CO1g2 ¡ NO1g2 + CO21g2
Step 2 is much faster than step 1; that is, k2 77 k1, telling us that the intermediate
NO31g2 is slowly produced in step 1 and immediately consumed in step 2.
Because step 1 is slow and step 2 is fast, step 1 is the rate-determining step. Thus,
the rate of the overall reaction depends on the rate of step 1, and the rate law of the
*Note the rate constants k1 and k2 written above the reaction arrows. The subscript on each rate constant
identifies the elementary step involved. Thus, k1 is the rate constant for step 1, and k2 is the rate constant for
step 2. A negative subscript refers to the rate constant for the reverse of an elementary step. For example, k - 1
is the rate constant for the reverse of the first step.
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chapter 14 Chemical Kinetics
overall reaction equals the rate law of step 1. Step 1 is a bimolecular process that has the
rate law
Rate = k13NO242
Thus, the rate law predicted by this mechanism agrees with the one observed experimentally. The reactant CO is absent from the rate law because it reacts in a step that
follows the rate-determining step.
A scientist would not, at this point, say that we have “proved” that this mechanism is correct. All we can say is that the rate law predicted by the mechanism is
consistent with experiment. We can often envision a different sequence of steps that
leads to the same rate law. If, however, the predicted rate law of the proposed mechanism disagrees with experiment, we know for certain that the mechanism cannot be
correct.
Sample
Exercise 14.14 Determining the Rate Law for a Multistep Mechanism
The decomposition of nitrous oxide, N2O, is believed to occur by a two-step mechanism:
N2O1g2 ¡ N21g2 + O1g2 1slow2
N2O1g2 + O1g2 ¡ N21g2 + O21g2 1fast2
(a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction.
Solution
Analyze Given a multistep mechanism with the relative speeds of the
steps, we are asked to write the overall reaction and the rate law for
that overall reaction.
Plan (a) Find the overall reaction by adding the elementary steps and
eliminating the intermediates. (b) The rate law for the overall reaction
will be that of the slow, rate-determining step.
Solve
(a) Adding the two elementary reactions gives
are told that the rate of this reaction is second order overall and
second order in [C]. Could any of the following be a rate-determining first step in a reaction mechanism that is consistent with
the observed rate law for the reaction (note that substance Z is an
intermediate)?
(a) C + C ¡ K + Z (b) C ¡ J + Z (c) C + D ¡ J + Z
(d) D ¡ J + K (e) None of these are consistent with the
observed rate law.
2 N2O1g2 + O1g2 ¡ 2 N21g2 + 2 O21g2 + O1g2
Practice Exercise 2
Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen:
2 N2O1g2 ¡ 2 N21g2 + O21g2
O31g2 + 2 NO21g2 ¡ N2O51g2 + O21g2
Omitting the intermediate, O(g), which occurs on both sides of the
equation, gives the overall reaction:
(b) The rate law for the overall reaction is just the rate law for the
slow, rate-determining elementary reaction. Because that slow step
is a unimolecular elementary reaction, the rate law is first order:
Rate = k3N2O4
Practice Exercise 1
Let’s consider a hypothetical reaction similar to that in Practice
Exercise 1 of Sample Exercise 14.13: 2 C + D ¡ J + 2 K. You
The reaction is believed to occur in two steps:
O31g2 + NO21g2 ¡ NO31g2 + O21g2
NO31g2 + NO21g2 ¡ N2O51g2
The experimental rate law is rate = k3O343NO24. What
can you say about the relative rates of the two steps of the
mechanism?
Mechanisms with a Fast Initial Step
It is possible, but not particularly straightforward, to derive the rate law for a mechanism in which an intermediate is a reactant in the rate-determining step. This situation
arises in multistep mechanisms when the first step is fast and therefore not the ratedetermining step. Let’s consider one example: the gas-phase reaction of nitric oxide
(NO) with bromine 1Br22:
2 NO1g2 + Br21g2 ¡ 2 NOBr1g2[14.24]
section 14.6 Reaction Mechanisms
The experimentally determined rate law for this reaction is second order in NO and
first order in Br2:
Rate = k3NO423Br24[14.25]
We seek a reaction mechanism that is consistent with this rate law. One possibility is
that the reaction occurs in a single termolecular step:
NO1g2 + NO1g2 + Br21g2 ¡ 2 NOBr1g2 Rate = k3NO423Br24[14.26]
As noted in Practice Exercise 2 of Exercise 14.13, this does not seem likely because termolecular processes are so rare.
Give It Some Thought
Why are termolecular elementary steps rare in gas-phase reactions?
Let’s consider an alternative mechanism that does not involve a termolecular step:
Step 1:
Step 2:
k1
NO1g2 + Br21g2 ∆
NOBr21g2
k
-1
k2
1fast2
NOBr21g2 + NO1g2 ¡ 2 NOBr1g2 1slow2[14.27]
In this mechanism, step 1 involves two processes: a forward reaction and its reverse.
Because step 2 is the rate-determining step, the rate law for that step governs the
rate of the overall reaction:
Rate = k23NOBr243NO4[14.28]
Note that NOBr2 is an intermediate generated in the forward reaction of step 1. Intermediates are usually unstable and have a low, unknown concentration. Thus, the rate
law of Equation 14.28 depends on the unknown concentration of an intermediate,
which isn’t desirable. We want instead to express the rate law for a reaction in terms of
the reactants, or the products if necessary, of the reaction.
With the aid of some assumptions, we can express the concentration of the intermediate NOBr2 in terms of the concentrations of the starting reactants NO and Br2. We
first assume that NOBr2 is unstable and does not accumulate to any significant extent
in the reaction mixture. Once formed, NOBr2 can be consumed either by reacting with
NO to form NOBr or by falling back apart into NO and Br2. The first of these possibilities is step 2 of our alternative mechanism, a slow process. The second is the reverse of
step 1, a unimolecular process:
k -1
NOBr21g2 ¡ NO1g2 + Br21g2
Because step 2 is slow, we assume that most of the NOBr2 falls apart according to this
reaction. Thus, we have both the forward and reverse reactions of step 1 occurring
much faster than step 2. Because they occur rapidly relative to step 2, the forward and
reverse reactions of step 1 establish an equilibrium. As in any other dynamic equilibrium, the rate of the forward reaction equals that of the reverse reaction:
k3NO43Br24
Rate of forward reaction
Solving for 3NOBr24, we have
3NOBr24 =
=
k-13NOBr24
Rate of reverse reaction
k1
3NO43Br24
k-1
Substituting this relationship into Equation 14.28, we have
Rate = k2
k1
3NO43Br243NO4 = k3NO423Br24
k-1
where the experimental rate constant k equals k2k1 >k-1. This expression is consistent
with the experimental rate law (Equation 14.25). Thus, our alternative mechanism
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chapter 14 Chemical Kinetics
(Equation 14.27), which involves two steps but only unimolecular and bimolecular processes, is far more probable than the single-step termolecular mechanism of
Equation 14.26.
In general, whenever a fast step precedes a slow one, we can solve for the concentration of an intermediate by assuming that an equilibrium is established in the fast step.
Sample
Exercise 14.15 Deriving the Rate Law for a Mechanism with a Fast Initial Step
Show that the following mechanism for Equation 14.24 also produces a rate law consistent with the
experimentally observed one:
Step 1:
Step 2:
Solution
k1
NO1g2 + NO1g2 ∆
N2O21g2 1fast, equilibrium2
k
-1
k2
N2O21g2 + Br21g2 ¡ 2 NOBr1g2 1slow2
Analyze We are given a mechanism with a fast initial step and asked
to write the rate law for the overall reaction.
Plan The rate law of the slow elementary step in a mechanism deter-
mines the rate law for the overall reaction. Thus, we first write the rate
law based on the molecularity of the slow step. In this case, the slow step
involves the intermediate N2O2 as a reactant. Experimental rate laws,
however, do not contain the concentrations of intermediates; instead
they are expressed in terms of the concentrations of starting substances.
Thus, we must relate the concentration of N2O2 to the concentration of
NO by assuming that an equilibrium is established in the first step.
Solve The second step is rate determining, so the overall rate is
Rate = k23N2O243Br24
We solve for the concentration of the intermediate N2O2 by assuming
that an equilibrium is established in step 1; thus, the rates of the forward and reverse reactions in step 1 are equal:
k13NO42 = k-13N2O24
Solving for the concentration of the intermediate, N2O2, gives
k1
3N2O24 =
3NO42
k-1
Substituting this expression into the rate expression gives
k1
Rate = k2 3NO423Br24 = k3NO423Br24
k-1
Thus, this mechanism also yields a rate law consistent with the experimental one. Remember: There may be more than one mechanism that
leads to an observed experimental rate law!
Practice Exercise 1
Consider the following hypothetical reaction:
2 P + Q ¡ 2 R + S. The following mechanism is proposed
for this reaction:
P + P ∆ T (fast)
Q + T ¡ R + U
1slow2
U ¡ R + S 1fast2
Substances T and U are unstable intermediates. What rate law is
predicted by this mechanism?
(a) Rate = k3P42 (b) Rate = k3P43Q4 (c) Rate = k3P423Q4
2
(d) Rate = k3P43Q4 (e) Rate = k3U4
Practice Exercise 2
The first step of a mechanism involving the reaction of bromine is
k1
Br21g2 ∆
2 Br1g2 1fast, equilibrium2
k
-1
What is the expression relating the concentration of Br(g) to that
of Br21g2?
So far we have considered only three reaction mechanisms: one for a reaction that
occurs in a single elementary step and two for simple multistep reactions where there is
one rate-determining step. There are other more complex mechanisms, however. If you
take a biochemistry class, for example, you will learn about cases in which the concentration of an intermediate cannot be neglected in deriving the rate law. Furthermore,
some mechanisms require a large number of steps, sometimes 35 or more, to arrive at a
rate law that agrees with experimental data!
14.7 | Catalysis
A catalyst is a substance that changes the speed of a chemical reaction without undergoing a permanent chemical change itself. Most reactions in the body, the atmosphere,
and the oceans occur with the help of catalysts. Much industrial chemical research is
section 14.7 Catalysis
607
devoted to the search for more effective catalysts for reactions of commercial importance. Extensive research efforts also are devoted to finding means of inhibiting or removing certain catalysts that promote undesirable reactions, such as those that corrode
metals, age our bodies, and cause tooth decay.
Homogeneous Catalysis
A catalyst that is present in the same phase as the reactants in a reaction mixture is
called a homogeneous catalyst. Examples abound both in solution and in the gas phase.
Consider, for example, the decomposition of aqueous hydrogen peroxide, H2O21aq2,
into water and oxygen:
2 H2O21aq2 ¡ 2 H2O1l2 + O21g2[14.29]
In the absence of a catalyst, this reaction occurs extremely slowly. Many substances are capable of catalyzing the reaction, however, including bromide ion, which reacts with hydrogen
peroxide in acidic solution, forming aqueous bromine and water (▼ Figure 14.22).
2 Br- 1aq2 + H2O21aq2 + 2 H+ ¡ Br21aq2 + 2 H2O1l2[14.30]
Go Figure
Why does the solution in the middle cylinder have a brownish color?
NaBr catalyst about
to be added to
reaction mixture
H2O2(aq) in
acidic solution
2 Br−(aq) + H2O2(aq) + 2 H+(aq)
Br2(aq) + H2O2(aq)
Br2(aq) + 2 H2O(l)
2 Br−(aq) + 2 H+(aq) + O2(g)
brown
colorless
▲ Figure 14.22 Homogeneous catalysis. Effect of catalyst on the speed of hydrogen peroxide
decomposition to water and oxygen gas.
bubbles
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chapter 14 Chemical Kinetics
If this were the complete reaction, bromide ion would not be a catalyst because it
undergoes chemical change during the reaction. However, hydrogen peroxide also
reacts with the Br21aq2 generated in Equation 14.30:
Br21aq2 + H2O21aq2 ¡ 2 Br - 1aq2 + 2 H + 1aq2 + O21g2[14.31]
Potential energy
The sum of Equations 14.30 and 14.31 is just Equation 14.29, a result which you can
check for yourself.
When the H2O2 has been completely decomposed, we are left with a colorless
solution of Br - 1aq2, which means that this ion is indeed a catalyst of the reaction because it speeds up the reaction without itself undergoing any net change. In
contrast, Br2 is an intermediate because it is first formed (Equation 14.30) and then
consumed (Equation 14.31). Neither the catalyst nor the intermediate appears in
the equation for the overall reaction. Notice, however, that the catalyst is there at
the start of the reaction, whereas the intermediate is formed during the course of the
Go Figure
reaction.
Where are the intermediates and
How does a catalyst work? If we think about the general
transition states in this diagram?
form of rate laws (Equation 14.7, rate = k3A4m3B4n), we must
conclude that the catalyst must affect the numerical value of
k, the rate constant. On the basis of the Arrhenius equation
Uncatalyzed reaction
(Equation 14.19, k = Ae -Ea>RT ), k is determined by the activaCatalyzed
reaction
tion energy 1Ea2 and the frequency factor (A). A catalyst may
affect the rate of reaction by altering the value of either Ea or A.
We can envision this happening in two ways: The catalyst could
2 H2O2
provide a new mechanism for the reaction that has an Ea value
lower than the Ea value for the uncatalyzed reaction, or the cat2 H2O2 + 2 Br− + 2 H+
alyst could assist in the orientation of reactants and so increase
A. The most dramatic catalytic effects come from lowering Ea.
2 H2O + O2
As a general rule, a catalyst lowers the overall activation energy
2 H2O + O2 +
for a chemical reaction.
−
+
2 Br + 2 H
A catalyst can lower the activation energy for a reaction
by providing a different mechanism for the reaction. In the
Reaction progress
decomposition of hydrogen peroxide, for example, two succes▲ Figure 14.23 Energy profiles for
sive reactions of H2O2, first with bromide and then with bromine, take place. Because
the uncatalyzed and bromide-catalyzed
these two reactions together serve as a catalytic pathway for hydrogen peroxide decomdecomposition of h2o2.
position, both of them must have significantly lower activation energies than the uncatalyzed decomposition (◀ Figure 14.23).
Heterogeneous Catalysis
A heterogeneous catalyst is one that exists in a phase different from the phase of the
reactant molecules, usually as a solid in contact with either gaseous reactants or reactants in a liquid solution. Many industrially important reactions are catalyzed by the
surfaces of solids. For example, raw petroleum is transformed into smaller hydrocarbon molecules by using what are called “cracking” catalysts. Heterogeneous catalysts
are often composed of metals or metal oxides.
The initial step in heterogeneous catalysis is usually adsorption of reactants.
Adsorption refers to the binding of molecules to a surface, whereas absorption refers
(Section 13.6) Adsorpto the uptake of molecules into the interior of a substance.
tion occurs because the atoms or ions at the surface of a solid are extremely reactive.
Because the catalyzed reaction occurs on the surface, special methods are often used
to prepare catalysts so that they have very large surface areas. Unlike their counterparts in the interior of the substance, surface atoms and ions have unused bonding
capacity that can be used to bond molecules from the gas or solution phase to the
surface of the solid.
section 14.7 Catalysis
Hydrogen
Carbon
1
3
H2 and C2H4 adsorbed
on metal surface
2
After H—H bond breaks, H atoms
migrate along metal surface
One free H has attached
to C2H4 to form C2H5
(ethyl group) intermediate
4
Second free H about to attach to
C2H5 intermediate to form C2H6
5
Ethane, C2H6, desorbs from
metal surface
▲ Figure 14.24 Heterogeneous catalysis. Mechanism for reaction of ethylene with hydrogen on
a catalytic surface.
The reaction of hydrogen gas with ethylene gas to form ethane gas provides an
example of heterogeneous catalysis:
C2H41g2 + H21g2 ¡ C2H61g2
Ethylene ∆H° = -137 kJ>mol[14.32]
Ethane
Even though this reaction is exothermic, it occurs very slowly in the absence of a catalyst. In the presence of a finely powdered metal, however, such as nickel, palladium,
or platinum, the reaction occurs easily at room temperature via the mechanism
diagrammed in ▲ Figure 14.24. Both ethylene and hydrogen are adsorbed on the
metal surface. Upon adsorption, the H ¬ H bond of H2 breaks, leaving two H atoms
initially bonded to the metal surface but relatively free to move. When a hydrogen
encounters an adsorbed ethylene molecule, it can form a s bond to one of the carbon
atoms, effectively destroying the C ¬ C p bond and leaving an ethyl group 1C2H52
bonded to the surface via a metal-to-carbon s bond. This s bond is relatively weak,
so when the other carbon atom also encounters a hydrogen atom, a sixth C ¬ H s
bond is readily formed, and an ethane molecule 1C2H62 is released from the metal
surface.
Give It Some Thought
How does a homogeneous catalyst compare with a heterogeneous one regarding
the ease of recovery of the catalyst from the reaction mixture?
Enzymes
The human body is characterized by an extremely complex system of interrelated
chemical reactions, all of which must occur at carefully controlled rates to maintain
life. A large number of marvelously efficient biological catalysts known as enzymes
are necessary for many of these reactions to occur at suitable rates. Most enzymes
609
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chapter 14 Chemical Kinetics
Chemistry Put to Work
Catalytic Converters
Heterogeneous catalysis plays a major role in the fight against urban air
pollution. Two components of automobile exhausts that help form photochemical smog are nitrogen oxides and unburned hydrocarbons. In addition, automobile exhaust may contain considerable quantities of carbon
monoxide. Even with the most careful attention to engine design, it is impossible under normal driving conditions to reduce the quantity of these
pollutants to an acceptable level in the exhaust gases. It is therefore necessary to remove them from the exhaust before they are vented to the air.
This removal is accomplished in the catalytic converter.
The catalytic converter, which is part of an automobile’s exhaust
system, must perform two functions: (1) oxidation of CO and unburned hydrocarbons 1CxHy2 to carbon dioxide and water, and (2)
reduction of nitrogen oxides to nitrogen gas:
O2
CO, CxHy ¡ CO2 + H2O
NO, NO2 ¡ N2
These two functions require different catalysts, so the development
of a successful catalyst system is a difficult challenge. The catalysts must be
effective over a wide range of operating temperatures. They must continue
to be active despite the fact that various components of the exhaust can
block the active sites of the catalyst. And the catalysts must be sufficiently
rugged to withstand exhaust gas turbulence and the mechanical shocks of
driving under various conditions for thousands of miles.
Catalysts that promote the combustion of CO and hydrocarbons are,
in general, the transition-metal oxides and the noble metals. These materials are supported on a structure (▶ Figure 14.25) that allows the best
possible contact between the flowing exhaust gas and the catalyst surface.
A honeycomb structure made from alumina 1Al2O32 and impregnated
with the catalyst is employed. Such catalysts operate by first adsorbing oxygen gas present in the exhaust gas. This adsorption weakens the O ¬ O
bond in O2, so that oxygen atoms are available for reaction with adsorbed
CO to form CO2. Hydrocarbon oxidation probably proceeds somewhat
similarly, with the hydrocarbons first being adsorbed followed by rupture
of a C ¬ H bond.
Go Figure
Why is the reaction faster when the liver is ground up?
Catalase present in beef liver rapidly
converts H2O2 to water and O2
O2 gas
Ground beef liver
▲ Figure 14.26 Enzymes speed up reactions.
H2O2
and
H2O
Transition-metal oxides and noble metals are also the most effective
catalysts for reduction of NO to N2 and O2. The catalysts that are most
effective in one reaction, however, are usually much less effective in the
other. It is therefore necessary to have two catalytic components.
Catalytic converters contain remarkably efficient heterogeneous
catalysts. The automotive exhaust gases are in contact with the catalyst
for only 100 to 400 ms, but in this very short time, 96% of the hydrocarbons and CO is converted to CO2 and H2O, and the emission of
nitrogen oxides is reduced by 76%.
There are costs as well as benefits associated with the use of catalytic converters, one being that some of the metals are very expensive.
Catalytic converters currently account for about 35% of the platinum,
65% of the palladium, and 95% of the rhodium used annually. All of
these metals, which come mainly from Russia and South Africa, can
be far more expensive than gold.
Related Exercises: 14.62, 14.81, 14.82, 14.124
Metal catalyst
impregnated on high
surface area alumina
“honeycomb”
▲ Figure 14.25 Cross section of a catalytic converter.
are large protein molecules with molecular weights ranging from about
10,000 to about 1 million amu. They are very selective in the reactions
they catalyze, and some are absolutely specific, operating for only one substance in only one reaction. The decomposition of hydrogen peroxide, for
example, is an important biological process. Because hydrogen peroxide is
strongly oxidizing, it can be physiologically harmful. For this reason, the
blood and liver of mammals contain an enzyme, catalase, that catalyzes
the decomposition of hydrogen peroxide into water and oxygen (Equation
14.29). ◀ Figure 14.26 shows the dramatic acceleration of this chemical
reaction by the catalase in beef liver.
The reaction any given enzyme catalyzes takes place at a specific location in the enzyme called the active site. The substances that react at this
site are called substrates. The lock-and-key model provides a simple explanation for the specificity of an enzyme (▶ Figure 14.27). The substrate is
pictured as fitting neatly into the active site, much like a key fits into a lock.
Lysozyme is an enzyme that is important to the functioning of our
immune system because it accelerates reactions that damage (or “lyse”)
bacterial cell walls. ▶ Figure 14.28 shows a model of the enzyme lysozyme without and with a bound substrate molecule.
section 14.7 Catalysis
611
Go Figure
Which molecules must bind more tightly to the active site, substrates or products?
Substrate
Enzyme
Products
Enzyme–substrate
complex
Enzyme
▲ Figure 14.27 Lock-and-key model for enzyme action.
The combination of enzyme and substrate is called the enzyme–substrate complex.
Although Figure 14.27 shows both the active site and its substrate as having a fixed
shape, the active site is often fairly flexible and so may change shape as it binds the substrate. The binding between substrate and active site involves dipole–dipole attractions,
(Section 11.2)
hydrogen bonds, and dispersion forces.
As substrate molecules enter the active site, they are somehow activated so that
they are capable of reacting rapidly. This activation process may occur, for example,
by the withdrawal or donation of electron density from a particular bond or group of
atoms in the enzyme’s active site. In addition, the substrate may become distorted in
the process of fitting into the active site and made more reactive. Once the reaction
occurs, the products depart from the active site, allowing another substrate molecule
to enter.
The activity of an enzyme is destroyed if some molecule other than the substrate
specific to that enzyme binds to the active site and blocks entry of the substrate. Such
substances are called enzyme inhibitors. Nerve poisons and certain toxic metal ions,
such as lead and mercury, are believed to act in this way to inhibit enzyme activity.
Some other poisons act by attaching elsewhere on the enzyme, thereby distorting the
active site so that the substrate no longer fits.
Enzymes are enormously more efficient than nonbiochemical catalysts. The number of individual catalyzed reaction events occurring at a particular active site, called
the turnover number, is generally in the range of 103 to 107 per second. Such large turnover numbers correspond to very low activation energies. Compared with a simple
chemical catalyst, enzymes can increase the rate constant for a given reaction by a millionfold or more.
Give It Some Thought
Is it reasonable to say that enzymes lower the energy of the transition state for a
reaction?
▲ Figure 14.28 Lysozyme was one of the
first enzymes for which a structure–function
relationship was described. This model shows
how the substrate (yellow) “fits” into the
active site of the enzyme.
612
chapter 14 Chemical Kinetics
Chemistry and Life
Nitrogen Fixation
and Nitrogenase
Nitrogen is one of the most essential elements in living organisms, found in
many compounds vital to life, including proteins, nucleic acids, vitamins,
and hormones. Nitrogen is continually cycling through the biosphere in
Nitrogen in
atmosphere
(N2)
Ammonia
(NH3)
Groundwater
Plant and
animal waste
Nitrate (NO3−)
▲ Figure 14.29 Simplified picture of the nitrogen cycle.
S
Fe
various forms, as shown in ◀ Figure 14.29. For example, certain microorganisms convert the nitrogen in animal waste and dead plants and
animals into N21g2, which then returns to the atmosphere. For the food
chain to be sustained, there must be a means of converting atmospheric
N21g2 into a form plants can use. For this reason, if a chemist were
asked to name the most important chemical reaction in the world,
she might easily say nitrogen fixation, the process by which atmospheric N21g2 is converted into compounds suitable for
plant use. Some fixed nitrogen results from the action of
lightning on the atmosphere, and some is produced
industrially using a process we will discuss in
Chapter 15. About 60% of fixed nitrogen,
however, is a consequence of the action of the
Nitrogen
remarkable and complex enzyme nitrogenase.
fixation
caused by
This enzyme is not present in humans or other
lightning
animals; rather, it is found in bacteria that live
in the root nodules of certain plants, such as
the legumes clover and alfalfa.
Nitrogenase converts N2 into NH3, a
process that, in the absence of a catalyst, has
a very large activation energy. This process is
a reduction reaction in which the oxidation
state of N is reduced from 0 in N2 to - 3 in
NH3. The mechanism by which nitrogenase
reduces N2 is not fully understood. Like many
other enzymes, including catalase, the active
site of nitrogenase contains transition-metal
Nitrite
atoms; such enzymes are called metalloen(NO2−)
zymes. Because transition metals can readily
change oxidation state, metalloenzymes are
especially useful for effecting transformations in
which substrates are either oxidized or reduced.
It has been known for nearly 40 years that a portion of
nitrogenase contains iron and molybdenum atoms. This portion,
called the FeMo-cofactor, is thought to serve as the active site of the
enzyme. The FeMo-cofactor of nitrogenase is a cluster of seven Fe atoms
and one Mo atom, all linked by sulfur atoms (▼ Figure 14.30).
It is one of the wonders of life that simple bacteria can contain beautifully complex
and vitally important enzymes such as nitrogenase. Because of this enzyme, nitrogen is
continually cycled between its comparatively
inert role in the atmosphere and its critical
role in living organisms. Without nitrogenase,
life as we know it could not exist on Earth.
Mo
Related Exercises: 14.86, 14.115, 14.116
◀ Figure 14.30 The FeMo-cofactor of
nitrogenase. Nitrogenase is found in nodules
in the roots of certain plants, such as the
white clover roots shown at the left. The
cofactor, which is thought to be the active
site of the enzyme, contains seven Fe atoms
and one Mo atom, linked by sulfur atoms.
The molecules on the outside of the cofactor
connect it to the rest of the protein.
613
section 14.7 Catalysis
Formic acid (HCOOH) decomposes in the gas phase at elevated temperatures as follows:
HCOOH1g2 ¡ CO21g2 + H21g2
The uncatalyzed decomposition reaction is determined to be first order. A graph of the partial
pressure of HCOOH versus time for decomposition at 838 K is shown as the red curve in
▶ Figure 14.31. When a small amount of solid ZnO is added to the reaction chamber, the
partial pressure of acid versus time varies as shown by the blue curve in Figure 14.31.
(a) Estimate the half-life and first-order rate constant for formic acid decomposition.
(b) What can you conclude from the effect of added ZnO on the decomposition of formic acid?
(c) The progress of the reaction was followed by measuring the partial pressure of formic acid
vapor at selected times. Suppose that, instead, we had plotted the concentration of formic
acid in units of mol>L. What effect would this have had on the calculated value of k?
(d) The pressure of formic acid vapor at the start of the reaction is 3.00 * 102 torr. Assuming
constant temperature and ideal-gas behavior, what is the pressure in the system at the end
of the reaction? If the volume of the reaction chamber is 436 cm3, how many moles of gas
occupy the reaction chamber at the end of the reaction?
(e) The standard heat of formation of formic acid vapor is ∆Hf° = -378.6 kJ>mol. Calculate ∆H°
for the overall reaction. If the activation energy 1Ea2 for the reaction is 184 kJ>mol, sketch an
approximate energy profile for the reaction, and label Ea, ∆H °, and the transition state.
Solution
(a) The initial pressure of HCOOH is 3.00 * 102 torr. On the graph we move to the level at which the
partial pressure of HCOOH is 1.50 * 102 torr, half the initial value. This corresponds to a time of
about 6.60 * 102 s, which is therefore the half-life. The first-order rate constant is given by Equation 14.15: k = 0.693>t1>2 = 0.693>660 s = 1.05 * 10-3 s-1.
(b) The reaction proceeds much more rapidly in the presence of solid ZnO, so the surface of the
oxide must be acting as a catalyst for the decomposition of the acid. This is an example of
heterogeneous catalysis.
(c) If we had graphed the concentration of formic acid in units of moles per liter, we would still
have determined that the half-life for decomposition is 660 s, and we would have computed
the same value for k. Because the units for k are s-1, the value for k is independent of the
units used for concentration.
(d) According to the stoichiometry of the reaction, two moles of product are formed for each mole
of reactant. When reaction is completed, therefore, the pressure will be 600 torr, just twice the
initial pressure, assuming ideal-gas behavior. (Because we are working at quite high temperature and fairly low gas pressure, assuming ideal-gas behavior is reasonable.) The number of
(Section 10.4):
moles of gas present can be calculated using the ideal-gas equation
n =
1600>760 atm210.436 L2
PV
=
= 5.00 * 10-3 mol
RT
10.08206 L@atm>mol@K21838 K2
(e) We first calculate the overall change in energy, ∆H°
(Section 5.7 and Appendix C), as in
∆H° = ∆Hf°1CO21g22 + ∆Hf°1H21g22 - ∆Hf°1HCOOH1g22
= -393.5 kJ>mol + 0 - 1-378.6 kJ>mol2
= -14.9 kJ>mol
From this and the given value for Ea, we can draw an approximate energy profile for the reaction,
in analogy to Figure 14.17.
Potential energy
Transition state
Ea
∆H°
Reaction progress
Pressure, HCOOH (torr)
Sample
Integrative Exercise Putting Concepts Together
300
225
HCOOH(g) alone
150
75
HCOOH(g)
with ZnO(s)
0
500
1000
Time (s)
1500
▲ Figure 14.31 Variation in pressure of
HCOOH(g) as a function of time at 838 K.
614
chapter 14 Chemical Kinetics
Chapter Summary and Key Terms
Introduction to Kinetics (Introduction and Section 14.1) Chemical kinetics is the area of chemistry in which reaction rates are
studied. Factors that affect reaction rate are the physical state of the
reactants; concentration; temperature; and the presence of catalysts.
Reaction Rates (Section 14.2) Reaction rates are usually expressed
as changes in concentration per unit time: Typically, for reactions in solution, rates are given in units of molarity per second 1M>s2. For most reactions, a plot of molarity versus time shows that the rate slows down as
the reaction proceeds. The instantaneous rate is the slope of a line drawn
tangent to the concentration-versus-time curve at a specific time. Rates
can be written in terms of the appearance of products or the disappearance of reactants; the stoichiometry of the reaction dictates the relationship
between rates of appearance and disappearance.
Rates and Concentration (Section 14.3) The quantitative
relationship between rate and concentration is expressed by a rate law,
which usually has the following form:
Rate = k3reactant 14m3reactant 24n c
The constant k in the rate law is called the rate constant; the exponents
m, n, and so forth are called reaction orders for the reactants. The sum
of the reaction orders gives the overall reaction order. Reaction orders
must be determined experimentally. The units of the rate constant
depend on the overall reaction order. For a reaction in which the overall reaction order is 1, k has units of s-1; for one in which the overall
reaction order is 2, k has units of M -1 s-1.
Spectroscopy is one technique that can be used to monitor the
course of a reaction. According to Beer’s law, the absorption of electromagnetic radiation by a substance at a particular wavelength is
directly proportional to its concentration.
Concentration and Time (Section 14.4) Rate laws can be used
to determine the concentrations of reactants or products at any time during a reaction. In a first-order reaction the rate is proportional to the concentration of a single reactant raised to the first power: Rate = k3A4. In
such cases the integrated form of the rate law is ln3A4t = - kt + ln3A40,
where 3A4t is the concentration of reactant A at time t, k is the rate constant, and 3A40 is the initial concentration of A. Thus, for a first-order
reaction, a graph of ln[A] versus time yields a straight line of slope -k.
A second-order reaction is one for which the overall reaction order
is 2. If a second-order rate law depends on the concentration of only
one reactant, then rate = k3A42, and the time dependence of [A] is
given by the integrated form of the rate law: 1>3A4t = 1>3A40 + kt.
In this case, a graph of 1>3A4t versus time yields a straight line. A zeroorder reaction is one for which the overall reaction order is 0. Rate = k
if the reaction is zero order.
The half-life of a reaction, t1>2, is the time required for the concentration of a reactant to drop to one-half of its original value. For
a first-order reaction, the half-life depends only on the rate constant
and not on the initial concentration: t1>2 = 0.693>k. The half-life of
a second-order reaction depends on both the rate constant and the
initial concentration of A: t1>2 = 1>k3A40.
The Effect of Temperature on Rates (Section 14.5) The
collision model, which assumes that reactions occur as a result of collisions
between molecules, helps explain why the magnitudes of rate constants
increase with increasing temperature. The greater the kinetic energy of the
colliding molecules, the greater is the energy of collision. The minimum
energy required for a reaction to occur is called the activation energy, Ea.
A collision with energy Ea or greater can cause the atoms of the colliding
molecules to reach the activated complex (or transition state), which is the
highest energy arrangement in the pathway from reactants to products.
Even if a collision is energetic enough, it may not lead to reaction; the reactants must also be correctly oriented relative to one another in order for a
collision to be effective.
Because the kinetic energy of molecules depends on temperature, the rate constant of a reaction is very dependent on temperature.
The relationship between k and temperature is given by the Arrhenius equation: k = Ae -Ea>RT. The term A is called the frequency factor;
it relates to the number of collisions that are favorably oriented for
reaction. The Arrhenius equation is often used in logarithmic form:
ln k = ln A - Ea >RT. Thus, a graph of ln k versus 1>T yields a
straight line with slope - Ea >R.
Reaction Mechanisms (Section 14.6) A reaction mechanism
details the individual steps that occur in the course of a reaction. Each of
these steps, called elementary reactions, has a well-defined rate law that
depends on the number of molecules (the molecularity) of the step. Elementary reactions are defined as either unimolecular, bimolecular, or termolecular,
depending on whether one, two, or three reactant molecules are involved,
respectively. Termolecular elementary reactions are very rare. Unimolecular, bimolecular, and termolecular reactions follow rate laws that are first
order overall, second order overall, and third order overall, respectively.
Many reactions occur by a multistep mechanism, involving two
or more elementary reactions, or steps. An intermediate is produced
in one elementary step, is consumed in a later elementary step, and
therefore does not appear in the overall equation for the reaction.
When a mechanism has several elementary steps, the overall rate is
limited by the slowest elementary step, called the rate-determining step.
A fast elementary step that follows the rate-determining step will have
no effect on the rate law of the reaction. A fast step that precedes the
rate-determining step often creates an equilibrium that involves an
intermediate. For a mechanism to be valid, the rate law predicted by
the mechanism must be the same as that observed experimentally.
Catalysts (Section 14.7) A catalyst is a substance that increases
the rate of a reaction without undergoing a net chemical change itself.
It does so by providing a different mechanism for the reaction, one that
has a lower activation energy. A homogeneous catalyst is one that is in
the same phase as the reactants. A heterogeneous catalyst has a different
phase from the reactants. Finely divided metals are often used as heterogeneous catalysts for solution- and gas-phase reactions. Reacting molecules can undergo binding, or adsorption, at the surface of the catalyst.
The adsorption of a reactant at specific sites on the surface makes bond
breaking easier, lowering the activation energy. Catalysis in living organisms is achieved by enzymes, large protein molecules that usually catalyze a very specific reaction. The specific reactant molecules involved in
an enzymatic reaction are called substrates. The site of the enzyme where
the catalysis occurs is called the active site. In the lock-and-key model for
enzyme catalysis, substrate molecules bind very specifically to the active
site of the enzyme, after which they can undergo reaction.
Learning Outcomes After studying this chapter, you should be able to:
• List the factors that affect the rate of chemical reactions. (Section 14.1)
• Determine the rate of a reaction given time and concentration.
(Section 14.2)
• Relate the rate of formation of products and the rate of disappearance of reactants given the balanced chemical equation for the
reaction. (Section 14.2)
Exercises
615
• Explain the form and meaning of a rate law, including the ideas of
• Apply the relationship between the rate constant of a first-order
• Determine the rate law and rate constant for a reaction from a se-
• Explain how the activation energy affects a rate and be able to use
reaction order and rate constant. (Section 14.3)
ries of experiments given the measured rates for various concentrations of reactants. (Section 14.3)
• Apply the integrated form of a rate law to determine the concentration of a reactant at a given time. (Section 14.4)
reaction and its half-life. (Section 14.4)
the Arrhenius equation. (Section 14.5)
• Predict a rate law for a reaction having a multistep mechanism
given the individual steps in the mechanism. (Section 14.6)
• Explain the principles underlying catalysis. (Section 14.7)
Key Equations
Rate = -
1 ∆3A4
1 ∆3B4
1 ∆3C4
1 ∆3D4
= =
=
a ∆t
c ∆t
b ∆t
d ∆t
Rate = k3A4m3B4n
ln3A4t - ln3A40 = - kt or ln
1
1
= kt +
3A4t
3A40
[14.4]Definition of reaction rate in terms of the components of the balanced chemical equation a A + b B ¡ c C + d D
[14.7]General form of a rate law for the reaction A + B ¡ products
3A4t
3A40
= -kt
[14.12]The integrated form of a first-order rate law for the reaction
A ¡ products
[14.14]The integrated form of the second-order rate law for the reaction
A ¡ products
0.693
k
[14.15]Relating the half-life and rate constant for a first-order reaction
k = Ae -Ea>RT [14.19]The Arrhenius equation, which expresses how the rate constant
depends on temperature
t1>2 =
ln k = -
Ea
+ ln A
RT
[14.20]
Logarithmic form of the Arrhenius equation
Exercises
14.1 An automotive fuel injector dispenses a fine spray of gasoline
into the automobile cylinder, as shown in the bottom drawing here. When an injector gets clogged, as shown in the top
drawing, the spray is not as fine or even and the performance
of the car declines. How is this observation related to chemical kinetics? [Section 14.1]
14.2Consider the following graph of the concentration of a substance X over time. Is each of the following statements true
or false? (a) X is a product of the reaction. (b) The rate of
the reaction remains the same as time progresses. (c) The
average rate between points 1 and 2 is greater than the average rate between points 1 and 3. (d) As time progresses,
the curve will eventually turn downward toward the x-axis.
[Section 14.2]
3
[X]
Visualizing Concepts
2
1
Time
616
chapter 14 Chemical Kinetics
14.3 You study the rate of a reaction, measuring both the concentration of the reactant and the concentration of the product as
a function of time, and obtain the following results:
of the following mixtures will have the fastest initial rate?
[Section 14.3]
Concentration
A
(1)
B
Time
(a) Which chemical equation is consistent with these data:
(i) A ¡ B, (ii) B ¡ A, (iii) A ¡ 2 B,
(iv) B ¡ 2 A?
(b) Write equivalent expressions for the rate of the reaction
in terms of the appearance or disappearance of the two
substances. [Section 14.2]
(2)
(3)
14.6A friend studies a first-order reaction and obtains the following three graphs for experiments done at two different temperatures. (a) Which two graphs represent experiments done
at the same temperature? What accounts for the difference in
these two graphs? In what way are they the same? (b) Which
two graphs represent experiments done with the same starting concentration but at different temperatures? Which graph
probably represents the lower temperature? How do you
know? [Section 14.4]
14.4Suppose that for the reaction K + L ¡ M, you monitor
the production of M over time, and then plot the following
graph from your data:
ln [A]
3
[M]
2
1
Time
0
5
10
15
20
t, min
25
30
(a) Is the reaction occurring at a constant rate from t = 0 to
t = 15 min? (b) Is the reaction completed at t = 15 min?
(c) Suppose the reaction as plotted here were started with
0.20 mol K and 0.40 mol L. After 30 min, an additional
0.20 mol K are added to the reaction mixture. Which of
the following correctly describes how the plot would look
from t = 30 min to t = 60 min? (i) [M] would remain
at the same constant value it has at t = 30 min, (ii) [M]
would increase with the same slope as t = 0 to 15 min, until t = 45 min at which point the plot becomes horizontal
again, or (iii) [M] decreases and reaches 0 at t = 45 min.
[Section 14.2]
14.5The following diagrams represent mixtures of NO(g) and
O21g2. These two substances react as follows:
2 NO1g2 + O21g2 ¡ 2 NO21g2
It has been determined experimentally that the rate is second
order in NO and first order in O2. Based on this fact, which
14.7(a) Given the following diagrams at t = 0 min and t = 30
min, what is the half-life of the reaction if it follows first-order
kinetics?
t = 0 min
t = 30 min
(b) After four half-life periods for a first-order reaction, what
fraction of reactant remains? [Section 14.4]
14.8
Which of the following linear plots do you expect for a reaction
A ¡ products if the kinetics are (a) zero order, (b) first
­order, or (c) second order? [Section 14.4]
617
Exercises
ln[A]
(ii)
ln[A]
(i)
Time
Time
Time
Time
(vi)
Reaction progress
[A]
[A]
(v)
Potential energy
1/[A]
(iv)
1/[A]
(iii)
14.11The following graph shows two different reaction pathways
for the same overall reaction at the same temperature. Is each
of the following statements true or false? (a) The rate is faster
for the red path than for the blue path. (b) For both paths, the
rate of the reverse reaction is slower than the rate of the forward reaction. (c) The energy change ∆E is the same for both
paths. [Section 14.6]
Time
Time
14.9 The following diagram shows a reaction profile. Label the
components indicated by the boxes. [Section 14.5]
14.12 Consider the diagram that follows, which represents
two steps in an overall reaction. The red spheres are oxygen, the blue ones nitrogen, and the green ones fluorine.
(a) Write the chemical equation for each step in the reaction.
(b) Write the equation for the overall reaction. (c) Identify
the intermediate in the mechanism. (d) Write the rate law
for the overall reaction if the first step is the slow, rate-determining step. [Section 14.6]
Energy
(2)
+
+
(1)
(3)
+
(4)
Reaction progress
14.13Based on the following reaction profile, how many intermediates are formed in the reaction A ¡ C? How many transition states are there? Which step, A ¡ B or B ¡ C,
is the faster? For the reaction A ¡ C, is ∆E positive,
­negative, or zero? [Section 14.6]
ln k
Potential energy
14.10The accompanying graph shows plots of ln k versus 1>T for
two different reactions. The plots have been extrapolated to
the y-intercepts. Which reaction (red or blue) has (a) the
larger value for Ea, and (b) the larger value for the frequency
factor, A? [Section 14.5]
B
A
C
1/T
Reaction progress
618
chapter 14 Chemical Kinetics
14.14Draw a possible transition state for the bimolecular reaction depicted here. (The blue spheres are nitrogen atoms, and the red
ones are oxygen atoms.) Use dashed lines to represent the bonds
that are in the process of being broken or made in the transition
state. [Section 14.6]
+
+
14.15The following diagram represents an imaginary two-step
mechanism. Let the red spheres represent element A, the green
ones element B, and the blue ones element C. (a) Write the
equation for the net reaction that is occurring. (b) Identify the
intermediate. (c) Identify the catalyst. [Sections 14.6 and 14.7]
14.16 Draw a graph showing the reaction pathway for an overall
exothermic reaction with two intermediates that are produced at different rates. On your graph indicate the reactants,
products, intermediates, transition states, and activation energies. [Sections 14.6 and 14.7]
Reaction Rates (Sections 14.1 and 14.2)
14.17 (a) What is meant by the term reaction rate? (b) Name three
factors that can affect the rate of a chemical reaction. (c) Is the
rate of disappearance of reactants always the same as the rate
of appearance of products?
14.18(a) What are the units usually used to express the rates of reactions occurring in solution? (b) From your everyday experience, give two examples of the effects of temperature on the
rates of reactions. (c) What is the difference between average
rate and instantaneous rate?
14.19 Consider the following hypothetical aqueous reaction:
A1aq2 S B1aq2. A flask is charged with 0.065 mol of A in a
total volume of 100.0 mL. The following data are collected:
Time (min)
0
10
20
30
40
Moles of A
0.065
0.051
0.042
0.036
0.031
(a) Calculate the number of moles of B at each time in the table,
assuming that there are no molecules of B at time zero, and that
A cleanly converts to B with no intermediates. (b) Calculate the
average rate of disappearance of A for each 10-min interval in
units of M>s. (c) Between t = 10 min and t = 30 min, what
is the average rate of appearance of B in units of M>s? Assume
that the volume of the solution is constant.
14.20A flask is charged with 0.100 mol of A and allowed to react
to form B according to the hypothetical gas-phase reaction
A1g2 ¡ B1g2. The following data are collected:
Time (s)
Moles of A
0
40
80
120
160
0.100
0.067
0.045
0.030
0.020
(a) Calculate the number of moles of B at each time in the
table, assuming that A is cleanly converted to B with no intermediates. (b) Calculate the average rate of disappearance
of A for each 40-s interval in units of mol>s. (c) Which of
the following would be needed to calculate the rate in units
of concentration per time: (i) the pressure of the gas at each
time, (ii) the volume of the reaction flask, (iii) the temperature, or (iv) the molecular weight of A?
14.21 The isomerization of methyl isonitrile 1CH3NC2 to acetonitrile 1CH3CN2 was studied in the gas phase at 215 °C, and the
following data were obtained:
Time (s)
0
3Ch 3nC 4 1M 2
0.0165
2000
0.0110
5000
0.00591
8000
0.00314
12,000
0.00137
15,000
0.00074
(a) Calculate the average rate of reaction, in M>s, for the
time interval between each measurement. (b) Calculate
the average rate of reaction over the entire time of the
data from t = 0 to t = 15,000 s. (c) Which is greater,
the average rate between t = 2000 and t = 12,000 s,
or between t = 8000 and t = 15,000 s? (d) Graph
3CH3NC4 versus time and determine the instantaneous
rates in M>s at t = 5000 s and t = 8000 s.
14.22The rate of disappearance of HCl was measured for the following reaction:
CH3OH1aq2 + HCl1aq2 ¡ CH3Cl1aq2 + H2O1l2
The following data were collected:
Time (min)
[HCl] (M)
0.0
1.85
54.0
1.58
107.0
1.36
215.0
1.02
430.0
0.580
(a) Calculate the average rate of reaction, in M>s, for the
time interval between each measurement. (b) Calculate the
average rate of reaction for the entire time for the data from
t = 0.0 min to t = 430.0 min. (c) Which is greater, the average rate between t = 54.0 and t = 215.0 min, or between
t = 107.0 and t = 430.0 min? (d) Graph [HCl] versus time
and determine the instantaneous rates in M>min and M>s at
t = 75.0 min and t = 250 min.
14.23 For each of the following gas-phase reactions, indicate how
the rate of disappearance of each reactant is related to the rate
of appearance of each product:
(a) H2O21g2 ¡ H21g2 + O21g2
(b) 2 N2O1g2 ¡ 2 N21g2 + O21g2
(c) N21g2 + 3 H21g2 ¡ 2 NH31g2
(d) C2H5NH21g2 ¡ C2H41g2 + NH31g2
Exercises
14.24For each of the following gas-phase reactions, write the rate
expression in terms of the appearance of each product and
disappearance of each reactant:
(a) 2 H2O1g2 ¡ 2 H21g2 + O21g2
(b) 2 SO21g2 + O21g2 ¡ 2 SO31g2
(c) 2 NO1g2 + 2 H21g2 ¡ N21g2 + 2 H2O1g2
(d) N21g2 + 2 H21g2 ¡ N2H41g2
14.25 ( a ) C o n s i d e r t h e c o m b u s t i o n o f H21g2: 2 H21g2 +
O21g2 ¡ 2 H2O1g2. If hydrogen is burning at the rate
of 0.48 mol>s, what is the rate of consumption of oxygen?
What is the rate of formation of water vapor? (b) The reaction
2 NO1g2 + Cl21g2 ¡ 2 NOCl1g2 is carried out in a closed
vessel. If the partial pressure of NO is decreasing at the rate of
56 torr/min, what is the rate of change of the total pressure of
the vessel?
14.26(a) Consider the combustion of ethylene, C2H41g2 +
3 O21g2 ¡ 2 CO21g2 + 2 H2O1g2. If the concentration of
C2H4 is decreasing at the rate of 0.036 M>s, what are the rates
of change in the concentrations of CO2 and H2O? (b) The rate
of decrease in N2H4 partial pressure in a closed reaction vessel from the reaction N2H41g2 + H21g2 ¡ 2 NH31g2 is
74 torr per hour. What are the rates of change of NH3 partial
pressure and total pressure in the vessel?
14.31 Consider the following reaction:
CH3Br1aq2 + OH -1aq2 ¡ CH3OH1aq2 + Br-1aq2
The rate law for this reaction is first order in CH3Br and first
order in OH - . When 3CH3Br4 is 5.0 * 10-3 M and 3OH - 4
is 0.050 M, the reaction rate at 298 K is 0.0432 M>s. (a) What
is the value of the rate constant? (b) What are the units of the
rate constant? (c) What would happen to the rate if the concentration of OH - were tripled? (d) What would happen to
the rate if the concentration of both reactants were tripled?
14.32The reaction between ethyl bromide 1C2H5Br2 and hydroxide
ion in ethyl alcohol at 330 K, C2H5Br1alc2 + OH - 1alc2 ¡
C2H5OH1l2 + Br - 1alc2, is first order each in ethyl bromide
and hydroxide ion. When 3C2H5Br4 is 0.0477 M and 3OH - 4
is 0.100 M, the rate of disappearance of ethyl bromide is
1.7 * 10-7 M>s. (a) What is the value of the rate constant?
(b) What are the units of the rate constant? (c) How would
the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl
alcohol to the solution?
14.33 The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way:
OCl - + I - ¡ OI - + Cl - . This rapid reaction gives the
following rate data:
Rate Laws (Section 14.3)
14.27 A reaction A + B ¡ C obeys the following rate law:
Rate = k3B42. (a) If [A] is doubled, how will the rate change?
Will the rate constant change? (b) What are the reaction orders for A and B? What is the overall reaction order? (c) What
are the units of the rate constant?
14.28Consider a hypothetical reaction between A, B, and C that is
first order in A, zero order in B, and second order in C. (a)
Write the rate law for the reaction. (b) How does the rate
change when [A] is doubled and the other reactant concentrations are held constant? (c) How does the rate change when [B]
is tripled and the other reactant concentrations are held constant? (d) How does the rate change when [C] is tripled and the
other reactant concentrations are held constant? (e) By what
factor does the rate change when the concentrations of all three
reactants are tripled? (f) By what factor does the rate change
when the concentrations of all three reactants are cut in half?
14.29 The decomposition reaction of N2O5 in carbon tetrachloride is
2 N2O5 ¡ 4 NO2 + O2. The rate law is first order in N2O5.
At 64 °C the rate constant is 4.82 * 10-3 s-1. (a) Write the
rate law for the reaction. (b) What is the rate of reaction when
3N2O54 = 0.0240 M? (c) What happens to the rate when the
concentration of N2O5 is doubled to 0.0480 M? (d) What happens
to the rate when the concentration of N2O5 is halved to 0.0120 M?
14.30Consider the following reaction:
2 NO1g2 + 2 H21g2 ¡ N21g2 + 2 H2O1g2
(a) The rate law for this reaction is first order in H2 and second
order in NO. Write the rate law. (b) If the rate constant for
this reaction at 1000 K is 6.0 * 104 M -2 s-1, what is the reaction rate when 3NO4 = 0.035 M and 3H24 = 0.015 M?
(c) What is the reaction rate at 1000 K when the concentration
of NO is increased to 0.10 M, while the concentration of H2
is 0.010 M? (d) What is the reaction rate at 1000 K if [NO] is
decreased to 0.010 M and 3H24 is increased to 0.030 M?
619
3oCl− 4 1M 2
3i− 4 1M 2
1.5 * 10
initial Rate 1M , s2
3.0 * 10-3
1.5 * 10-3
2.72 * 10-4
1.5 * 10-3
3.0 * 10-3
2.72 * 10-4
1.5 * 10
-3
-3
1.36 * 10-4
(a) Write the rate law for this reaction. (b) Calculate the
rate constant with proper units. (c) Calculate the rate when
3OCl-4 = 2.0 * 10-3 M and 3I - 4 = 5.0 * 10 - 4 M.
14.34The reaction 2 ClO21aq2 + 2 OH - 1aq2 ¡ ClO3- 1aq2 +
ClO2- 1aq2 + H2O1l2 was studied with the following results:
1
1Clo2 2 1M 2
0.060
3oh− 4 1M 2
0.030
initial Rate 1M , s2
2
0.020
0.030
0.00276
3
0.020
0.090
0.00828
Experiment
0.0248
(a) Determine the rate law for the reaction. (b) Calculate the
rate constant with proper units. (c) Calculate the rate when
3ClO24 = 0.100 M and 3OH - 4 = 0.050 M.
14.35 The following data were measured for the reaction
BF31g2 + NH31g2 ¡ F3BNH31g2:
Experiment
1
3BF3 4 1m 2
0.250
3 nh 3 4 1M 2
0.250
initial Rate 1M , s2
0.2130
2
0.250
0.125
0.1065
3
0.200
0.100
0.0682
4
0.350
0.100
0.1193
5
0.175
0.100
0.0596
(a) What is the rate law for the reaction? (b) What is
the overall order of the reaction? (c) Calculate the rate
constant with proper units? (d) What is the rate when
3BF34 = 0.100 M and 3NH34 = 0.500 M?
620
chapter 14 Chemical Kinetics
14.36The following data were collected for the rate of disappearance
of NO in the reaction 2 NO1g2 + O21g2 ¡ 2 NO21g2:
1
3 no 4 1M 2
0.0126
3 o2 4 1M 2
0.0125
initial Rate 1M , s 2
2
0.0252
0.0125
5.64 * 10-2
3
0.0252
0.0250
1.13 * 10-1
Experiment
1.41 * 10-2
(a) What is the rate law for the reaction? (b) What are the
units of the rate constant? (c) What is the average value of the
rate constant calculated from the three data sets? (d) What
is the rate of disappearance of NO when 3NO4 = 0.0750 M
and 3O24 = 0.0100 M ? (e) What is the rate of disappearance
of O2 at the concentrations given in part (d)?
[14.37] Consider the gas-phase reaction between nitric oxide and
bromine at 273 °C: 2 NO1g2 + Br21g2 ¡ 2 NOBr1g2. The
following data for the initial rate of appearance of NOBr were
obtained:
Experiment
initial Rate 1M , s 2
1
3 no 4 1m2
0.10
3 Br2 4 1M 2
2
0.25
0.20
150
3
0.10
0.50
60
4
0.35
0.50
735
0.20
24
(a) Determine the rate law. (b) Calculate the average value
of the rate constant for the appearance of NOBr from the
four data sets. (c) How is the rate of appearance of NOBr
related to the rate of disappearance of Br2? (d) What is the
rate of disappearance of Br2 when 3NO4 = 0.075 M and
3Br24 = 0.25 M ?
[14.38] Consider the reaction of peroxydisulfate ion 1S2O82-2 with
iodide ion 1I - 2 in aqueous solution:
S2O82 - 1aq2 + 3 I - 1aq2 ¡ 2 SO42 - 1aq2 + I3- 1aq2
At a particular temperature the initial rate of disappearance
of S2O82 - varies with reactant concentrations in the following
manner:
1
3 s2o82− 4 1M 2
0.018
3 i 4 1M 2
0.036
initial Rate 1M , s 2
2
0.027
0.036
3.9 * 10-6
3
0.036
0.054
7.8 * 10-6
4
0.050
0.072
-5
Experiment
−
2.6 * 10-6
1.4 * 10
(a) Determine the rate law for the reaction and state the units
of the rate constant. (b) What is the average value of the rate
constant for the disappearance of S2O82 - based on the four
sets of data? (c) How is the rate of disappearance of S2O82 related to the rate of disappearance of I - ? (d) What is the
rate of disappearance of I - when 3S2O82 - 4 = 0.025 M and
3I - 4 = 0.050 M ?
Change of Concentration
with Time (Section 14.4)
14.39 (a) Define the following symbols that are encountered in rate
equations for the generic reaction A S B: 3A40, t1>2 3A4t, k.
(b) What quantity, when graphed versus time, will yield a
straight line for a first-order reaction? (c) How can you calculate the rate constant for a first-order reaction from the graph
you made in part (b)?
14.40(a) For a generic second-order reaction A ¡ B, what
quantity, when graphed versus time, will yield a straight line?
(b) What is the slope of the straight line from part (a)?
(c) How do the half-lives of first-order and second-order
­reactions differ?
14.41 (a) The gas-phase decomposition of SO2Cl2, SO2Cl21g2 ¡
SO21g2 + Cl21g2, is first order in SO2Cl2. At 600 K the halflife for this process is 2.3 * 105 s. What is the rate constant at this temperature? (b) At 320 °C the rate constant is
2.2 * 10-5 s-1. What is the half-life at this temperature?
14.42Molecular iodine, I21g2, dissociates into iodine atoms at
625 K with a first-order rate constant of 0.271 s-1. (a) What is
the half-life for this reaction? (b) If you start with 0.050 M I2
at this temperature, how much will remain after 5.12 s assuming that the iodine atoms do not recombine to form I2?
14.43 As described in Exercise 14.41, the decomposition of sulfuryl
chloride 1SO2Cl22 is a first-order process. The rate constant for
the decomposition at 660 K is 4.5 * 10-2 s-1. (a) If we begin
with an initial SO2Cl2 pressure of 450 torr, what is the partial
pressure of this substance after 60 s? (b) At what time will the
partial pressure of SO2Cl2 decline to one-tenth its initial value?
14.44The first-order rate constant for the decomposition of
N2O5, 2 N2O51g2 ¡ 4 NO21g2 + O21g2, at 70 °C is 6.82 *
10-3 s-1. Suppose we start with 0.0250 mol of N2O51g2 in a
­volume of 2.0 L. (a) How many moles of N2O5 will remain after
5.0 min? (b) How many minutes will it take for the quantity of
N2O5 to drop to 0.010 mol? (c) What is the half-life of N2O5
at 70 °C ?
14.45 The reaction SO2Cl21g2 ¡ SO21g2 + Cl21g2 is first order
in SO2Cl2. Using the following kinetic data, determine the
magnitude and units of the first-order rate constant:
Time (s)
Pressure so2Cl 2 (atm)
0
1.000
2500
0.947
5000
0.895
7500
0.848
10,000
0.803
14.46From the following data for the first-order gas-phase isomerization of CH3NC at 215 °C, calculate the first-order rate constant and half-life for the reaction:
Time (s)
Pressure Ch 3nC (torr)
0
502
2000
335
5000
180
8000
95.5
12,000
41.7
15,000
22.4
14.47Consider the data presented in Exercise 14.19. (a) By using
appropriate graphs, determine whether the reaction is first
order or second order. (b) What is the rate constant for the
reaction? (c) What is the half-life for the reaction?
Exercises
14.48Consider the data presented in Exercise 14.20. (a) Determine
whether the reaction is first order or second order. (b) What
is the rate constant? (c) What is the half-life?
14.49 The gas-phase decomposition of NO2, 2 NO21g2 ¡
2 NO1g2 + O21g2, is studied at 383 °C, giving the following
data:
Time (s)
0.0
3 no2 4 (M)
5.0
0.017
10.0
0.0090
15.0
0.0062
20.0
0.0047
0.100
(a) Is the reaction first order or second order with respect
to the concentration of NO2? (b) What is the rate constant?
(c) Predict the reaction rates at the beginning of the reaction for initial concentrations of 0.200 M, 0.100 M, and
0.050 M NO2.
14.50Sucrose 1C12H22O112, commonly known as table sugar, reacts
in dilute acid solutions to form two simpler sugars, glucose
and fructose, both of which have the formula C6H22O6. At
23 °C and in 0.5 M HCl, the following data were obtained for
the disappearance of sucrose:
Time (min)
0
3 C12h 22o11 4 1M 2
39
0.274
80
0.238
140
0.190
210
0.146
0.316
(a) Is the reaction first order or second order with respect to
3C12H22O114? (b) What is the rate constant? (c) Using this rate
constant, calculate the concentration of sucrose at 39, 80, 140,
and 210 min if the initial sucrose concentration was 0.316 M
and the reaction were zero order in sucrose.
Temperature and Rate (Section 14.5)
14.51 (a) What factors determine whether a collision between two
molecules will lead to a chemical reaction? (b) According to
the collision model, why does temperature affect the value
of the rate constant? (c) Does the rate constant for a reaction
generally increase or decrease with an increase in reaction
temperature?
14.52(a) In which of the following reactions would you expect the
orientation factor to be least important in leading to reaction:
NO + O ¡ NO2 or H + Cl ¡ HCl? (b) How does
the kinetic-molecular theory help us understand the temperature dependence of chemical reactions?
14.53 Calculate the fraction of atoms in a sample of argon gas at
400 K that has an energy of 10.0 kJ or greater.
14.54(a) The activation energy for the isomerization of methyl isonitrile (Figure 14.7) is 160 kJ>mol. Calculate the fraction of
methyl isonitrile molecules that has an energy of 160.0 kJ or
greater at 500 K. (b) Calculate this fraction for a temperature
of 520 K. What is the ratio of the fraction at 520 K to that at
500 K?
621
14.55 The gas-phase reaction Cl1g2 + HBr1g2 ¡ HCl1g2 + Br1g2
has an overall energy change of - 66 kJ. The activation energy
for the reaction is 7 kJ. (a) Sketch the energy profile for the reaction, and label Ea and ∆E. (b) What is the activation energy
for the reverse reaction?
14.56For the elementary process N2O51g2 ¡ NO21g2 + NO31g2
the activation energy 1Ea2 and overall ∆E are 154 kJ>mol and
136 kJ>mol, respectively. (a) Sketch the energy profile for this
reaction, and label Ea and ∆E. (b) What is the activation energy for the reverse reaction?
14.57 Indicate whether each statement is true or false.
(a) If you compare two reactions with similar collision factors,
the one with the larger activation energy will be faster.
(b) A reaction that has a small rate constant must have a
small frequency factor.
(c) Increasing the reaction temperature increases the fraction of successful collisions between reactants.
14.58Indicate whether each statement is true or false.
(a) If you measure the rate constant for a reaction at different temperatures, you can calculate the overall enthalpy
change for the reaction.
(b) Exothermic reactions are faster than endothermic
reactions.
(c) If you double the temperature for a reaction, you cut the
activation energy in half.
14.59 Based on their activation energies and energy changes and
assuming that all collision factors are the same, which of the
following reactions would be fastest and which would be
slowest?
(a) Ea = 45 kJ>mol; ∆E = - 25 kJ>mol
(b) Ea = 35 kJ>mol; ∆E = - 10 kJ>mol
(c) Ea = 55 kJ>mol; ∆E = 10 kJ>mol
14.60Which of the reactions in Exercise 14.61 will be fastest in the
reverse direction? Which will be slowest?
14.61 (a) A certain first-order reaction has a rate constant of
2.75 * 10-2 s-1 at 20 °C. What is the value of k at 60 °C if
Ea = 75.5 kJ>mol ? (b) Another first-order reaction also has a
rate constant of 2.75 * 10-2 s-1 at 20 °C What is the value of
k at 60 °C if Ea = 125 kJ>mol ? (c) What assumptions do you
need to make in order to calculate answers for parts (a) and (b)?
14.62Understanding the high-temperature behavior of nitrogen
oxides is essential for controlling pollution generated in automobile engines. The decomposition of nitric oxide (NO) to N2
and O2 is second order with a rate constant of 0.0796 M -1s-1
at 737 °C and 0.0815 M -1s-1 at 947 °C. Calculate the activation energy for the reaction.
14.63 The rate of the reaction
CH3COOC2H51aq2 + OH - 1aq2 ¡ CH3COO - 1aq2 + C2H5OH1aq2
was measured at several temperatures, and the following data
were collected:
Temperature 1 °C 2
15
k 1M −1 s−1 2
25
0.101
35
0.184
45
0.332
0.0521
622
chapter 14 Chemical Kinetics
14.64The temperature dependence of the rate constant for a reaction is tabulated as follows:
600
k 1M −1 s−1 2
0.028
650
0.22
700
1.3
750
6.0
800
Reaction progess
(a) How many elementary reactions are in the reaction
mechanism? (b) How many intermediates are formed in
the reaction? (c) Which step is rate limiting? (d) For the
overall reaction, is ∆E positive, negative, or zero?
23
Calculate Ea and A.
14.71 The following mechanism has been proposed for the gasphase reaction of H2 with ICl:
Reaction Mechanisms (Section 14.6)
14.65 (a) What is meant by the term elementary reaction?
(b) What is the difference between a unimolecular and a
bimolecular elementary reaction? (c) What is a reaction
mechanism? (d) What is meant by the term rate-determining step?
14.66(a) What is meant by the term molecularity? (b) Why are
termolecular elementary reactions so rare? (c) What is an intermediate in a mechanism? (d) What are the differences between an intermediate and a transition state?
14.67 What is the molecularity of each of the following elementary
reactions? Write the rate law for each.
(a) Cl21g2 ¡ 2 Cl1g2
(b) OCl - 1aq2 + H2O1l2 ¡ HOCl1aq2 + OH - 1aq2
(c) NO1g2 + Cl21g2 ¡ NOCl21g2
14.68What is the molecularity of each of the following elementary
reactions? Write the rate law for each.
(a) 2 NO1g2 ¡ N2O21g2
HI1g2 + ICl1g2 ¡ I21g2 + HCl1g2
(a) Write the balanced equation for the overall reaction.
(b) Identify any intermediates in the mechanism. (c) If the
first step is slow and the second one is fast, which rate law do
you expect to be observed for the overall reaction?
14.72The decomposition of hydrogen peroxide is catalyzed by iodide ion. The catalyzed reaction is thought to proceed by a
two-step mechanism:
H2O21aq2 + I - 1aq2 ¡ H2O1l2 + IO - 1aq2 1slow2
IO - 1aq2 + H2O21aq2 ¡ H2O1l2 + O21g2 + I - 1aq2 1fast2
(a) Write the chemical equation for the overall process. (b)
Identify the intermediate, if any, in the mechanism. (c) Assuming that the first step of the mechanism is rate determining, predict the rate law for the overall process.
[14.73] The reaction 2 NO1g2 + Cl21g2 ¡ 2 NOCl1g2 was performed and the following data obtained under conditions of
constant 3Cl24:
CH2
H21g2 + ICl1g2 ¡ HI1g2 + HCl1g2
(b) H2C ¬ CH21g2 ¡ CH2 “ CH ¬ CH31g2
(c) SO31g2 ¡ SO21g2 + O1g2
14.69 (a) Based on the following reaction profile, how many intermediates are formed in the reaction A ¡ D? (b) How
many transition states are there? (c) Which step is the fastest? (d) For the reaction A ¡ D, is ∆E positive, negative, or zero?
1/[NO]
Temperature (K)
14.70Consider the following energy profile.
Potential energy
Calculate the value of Ea by constructing an appropriate
graph.
(under conditions
of constant [Cl2])
Potential energy
Time
(a) Is the following mechanism consistent with the data?
C
A
B
D
NO1g2 + Cl21g2 ¡ NOCl21g2
NOCl21g2 + NO1g2 ¡ 2 NOCl1g2
1fast2
1slow2
(b) Does the linear plot guarantee that the overall rate law is
second order?
14.74You have studied the gas-phase oxidation of HBr by O2:
Reaction progress
4 HBr1g2 + O21g2 ¡ 2 H2O1g2 + 2 Br21g2
Exercises
You find the reaction to be first order with respect to HBr
and first order with respect to O2. You propose the following
mechanism:
HBr1g2 + O21g2 ¡ HOOBr1g2
HOOBr1g2 + HBr1g2 ¡ 2 HOBr1g2
HOBr1g2 + HBr1g2 ¡ H2O1g2 + Br21g2
(a) Confirm that the elementary reactions add to give the
overall reaction. (b) Based on the experimentally determined
rate law, which step is rate determining? (c) What are the
intermediates in this mechanism? (d) If you are unable to detect HOBr or HOOBr among the products, does this disprove
your mechanism?
Catalysis (Section 14.7)
14.75 (a) What is a catalyst? (b) What is the difference between a
homogeneous and a heterogeneous catalyst? (c) Do catalysts
affect the overall enthalpy change for a reaction, the activation energy, or both?
14.76(a) Most commercial heterogeneous catalysts are extremely
finely divided solid materials. Why is particle size important?
(b) What role does adsorption play in the action of a heterogeneous catalyst?
14.77 In Figure 14.22, we saw that Br - 1aq2 catalyzes the decomposition of H2O21aq2 into H2O1l2 and O21g2. Suppose that some
KBr1s2 is added to an aqueous solution of hydrogen peroxide.
Make a sketch of 3Br - 1aq24 versus time from the addition of
the solid to the end of the reaction.
14.78In solution, chemical species as simple as H+ and OH - can
serve as catalysts for reactions. Imagine you could measure the
3H+4 of a solution containing an acid-catalyzed reaction as it
occurs. Assume the reactants and products themselves are neither acids nor bases. Sketch the 3H+4 concentration profile you
would measure as a function of time for the reaction, assuming
t = 0 is when you add a drop of acid to the reaction.
14.79 The oxidation of SO2 to SO3 is accelerated by NO2. The reaction proceeds according to:
NO21g2 + SO21g2 ¡ NO1g2 + SO31g2
2 NO1g2 + O21g2 ¡ 2 NO21g2
(a) Show that, with appropriate coefficients, the two reactions
can be summed to give the overall oxidation of SO2 by O2 to
give SO3. (b) Do we consider NO2 a catalyst or an intermediate in this reaction? (c) Is this an example of homogeneous
catalysis or heterogeneous catalysis?
14.80The addition of NO accelerates the decomposition of N2O,
possibly by the following mechanism:
623
14.81 Many metallic catalysts, particularly the precious-metal ones,
are often deposited as very thin films on a substance of high
surface area per unit mass, such as alumina 1Al2O32 or silica
1SiO22. (a) Why is this an effective way of utilizing the catalyst material compared to having powdered metals? (b) How
does the surface area affect the rate of reaction?
14.82(a) If you were going to build a system to check the effectiveness
of automobile catalytic converters on cars, what substances would
you want to look for in the car exhaust? (b) Automobile catalytic
converters have to work at high temperatures, as hot exhaust
gases stream through them. In what ways could this be an advantage? In what ways a disadvantage? (c) Why is the rate of flow of
exhaust gases over a catalytic converter important?
14.83 When D2 reacts with ethylene 1C2H42 in the presence
of a finely divided catalyst, ethane with two deuteriums,
CH2D ¬ CH2D, is formed. (Deuterium, D, is an isotope of hydrogen of mass 2). Very little ethane forms in which two deuteriums are bound to one carbon (for example, CH3 ¬ CHD2).
Use the sequence of steps involved in the reaction (Figure
14.24) to explain why this is so.
14.84Heterogeneous catalysts that perform hydrogenation reactions, as illustrated in Figure 14.24, are subject to “poisoning,”
which shuts down their catalytic ability. Compounds of sulfur
are often poisons. Suggest a mechanism by which such compounds might act as poisons.
[14.85] The enzyme carbonic anhydrase catalyzes the reaction
CO21g2 + H2O1l2 ¡ HCO3 - 1aq2 + H+1aq2. In water,
without the enzyme, the reaction proceeds with a rate constant
of 0.039 s-1 at 25 °C. In the presence of the enzyme in water,
the reaction proceeds with a rate constant of 1.0 * 106 s-1 at
25 °C. Assuming the collision factor is the same for both situations, calculate the difference in activation energies for the
uncatalyzed versus enzyme-catalyzed reaction.
[14.86] The enzyme urease catalyzes the reaction of urea, 1NH2CONH22,
with water to produce carbon dioxide and ammonia. In water,
without the enzyme, the reaction proceeds with a first-order rate
constant of 4.15 * 10-5 s-1 at 100 °C. In the presence of the
enzyme in water, the reaction proceeds with a rate constant of
3.4 * 104 s-1 at 21 °C. (a) Write out the balanced equation for
the reaction catalyzed by urease. (b) If the rate of the catalyzed
reaction were the same at 100 °C as it is at 21 °C, what would
be the difference in the activation energy between the catalyzed
and uncatalyzed reactions? (c) In actuality, what would you expect for the rate of the catalyzed reaction at 100 °C as compared
to that at 21 °C? (d) On the basis of parts (c) and (d), what can
you conclude about the difference in activation energies for the
catalyzed and uncatalyzed reactions?
NO1g2 + N2O1g2 ¡ N21g2 + NO21g2
[14.87] The activation energy of an uncatalyzed reaction is 95 kJ>mol.
The addition of a catalyst lowers the activation energy to
55 kJ>mol. Assuming that the collision factor remains the
same, by what factor will the catalyst increase the rate of the
reaction at (a) 25 °C, (b) 125 °C?
(a) What is the chemical equation for the overall reaction?
Show how the two steps can be added to give the overall equation. (b) Is NO serving as a catalyst or an intermediate in this
reaction? (c) If experiments show that during the decomposition of N2O, NO2 does not accumulate in measurable quantities, does this rule out the proposed mechanism?
[14.88] Suppose that a certain biologically important reaction is quite
slow at physiological temperature 137 °C2 in the absence of a
catalyst. Assuming that the collision factor remains the same,
by how much must an enzyme lower the activation energy of
the reaction to achieve a 1 * 105@fold increase in the reaction rate?
2 NO21g2 ¡ 2 NO1g2 + O21g2
624
chapter 14 Chemical Kinetics
Additional Exercises
14.89Consider the reaction A + B ¡ C + D. Is each of the following statements true or false? (a) The rate law for the reaction
must be Rate = k3A43B4. (b) If the reaction is an elementary
reaction, the rate law is second order. (c) If the reaction is an
elementary reaction, the rate law of the reverse reaction is first
order. (d) The activation energy for the reverse reaction must
be greater than that for the forward reaction.
14.90Hydrogen sulfide 1H2S2 is a common and troublesome pollutant in industrial wastewaters. One way to remove H2S is
to treat the water with chlorine, in which case the following
reaction occurs:
H2S1aq2 + Cl21aq2 ¡ S1s2 + 2 H+1aq2 + 2 Cl - 1aq2
The rate of this reaction is first order in each reactant.
The rate constant for the disappearance of H2S at 28 °C is
3.5 * 10-2 M -1 s-1. If at a given time the concentration of
H2S is 2.0 * 10 - 4 M and that of Cl2 s 0.025 M, what is the
rate of formation of Cl - ?
14.91 The reaction 2 NO1g2 + O21g2 ¡ 2 NO21g2 is second order in NO and first order in O2. When 3NO4 = 0.040 M, and
3O24 = 0.035 M, the observed rate of disappearance of NO is
9.3 * 10-5 M>s. (a) What is the rate of disappearance of O2 at
this moment? (b) What is the value of the rate constant? (c) What
are the units of the rate constant? (d) What would happen to the
rate if the concentration of NO were increased by a factor of 1.8?
14.92You perform a series of experiments for the reaction
A ¡ B + C and find that the rate law has the form
rate = k3A4x. Determine the value of x in each of the following cases: (a) There is no rate change when 3A40 is tripled.
(b) The rate increases by a factor of 9 when 3A40 is tripled.
(c) When 3A40 is doubled, the rate increases by a factor of 8.
14.93Consider the following reaction between mercury(II) chloride
and oxalate ion:
(a) What is the rate law for this reaction? (b) What is the
value of the rate constant with proper units? (c) What is the
reaction rate when the initial concentration of X is 0.75 M
and that of Z is 1.25 M?
14.95 The reaction 2 NO2 ¡ 2 NO + O2 has the rate constant
k = 0.63 M - 1s - 1. (a) Based on the units for k, is the reaction first or second order in NO2? (b) If the initial concentration of NO2 is 0.100 M, how would you determine
how long it would take for the concentration to decrease to
0.025 M?
14.96Consider two reactions. Reaction (1) has a constant half-life,
whereas reaction (2) has a half-life that gets longer as the reaction proceeds. What can you conclude about the rate laws of
these reactions from these observations?
[14.97] A first-order reaction A ¡ B has the rate constant
k = 3.2 * 10-3 s-1. If the initial concentration of A is
2.5 * 10-2 M, what is the rate of the reaction at t = 660 s?
14.98(a) The reaction H2O21aq2 ¡ H2O1l2 + 12O21g2 is first
order. Near room temperature, the rate constant equals
7.0 * 10-4 s-1. Calculate the half-life at this temperature. (b) At 415 °C, 1CH222O decomposes in the gas phase,
1CH222O1g2 ¡ CH41g2 + CO1g2. If the reaction is first
order with a half-life of 56.3 min at this temperature, calculate
the rate constant in s-1.
14.99 Americium-241 is used in smoke detectors. It has a first–order
rate constant for radioactive decay of k = 1.6 * 10-3 yr-1.
By contrast, iodine-125, which is used to test for thyroid
functioning, has a rate constant for radioactive decay of
k = 0.011 day -1. (a) What are the half-lives of these two isotopes? (b) Which one decays at a faster rate? (c) How much of
a 1.00-mg sample of each isotope remains after 3 half-lives?
(d) How much of a 1.00-mg sample of each isotope remains
after 4 days?
2 HgCl21aq2 + C2O42 - 1aq2 ¡ 2 Cl - 1aq2 + 2 CO21g2 + Hg2Cl21s2 14.100 Urea 1NH2CONH22 is the end product in protein metabolism
in animals. The decomposition of urea in 0.1 M HCl occurs
according to the reaction
The initial rate of this reaction was determined for several
2concentrations of HgCl2 and C2O4 , and the following rate
NH2CONH21aq2 + H +1aq2 + 2 H2O1l2 ¡ 2 NH4+1aq2 + HCO3- 1aq2
data were obtained for the rate of disappearance of C2O42 - :
The reaction is first order in urea and first order overall.
Experiment
3 hgCl 2 4 1M 2
Rate 1m , s 2
3 C2o42 − 4 1M 2
When 3NH2CONH24 = 0.200 M, the rate at 61.05 °C is
-5
8.56 * 10-5 M>s. (a) What is the rate constant, k? (b) What
1
0.164
0.15
3.2 * 10
is the concentration of urea in this solution after 4.00 * 103 s
2
0.164
0.45
2.9 * 10-4
if the starting concentration is 0.500 M? (c) What is the half3
0.082
0.45
life for this reaction at 61.05 °C ?
1.4 * 10-4
[14.101] The rate of a first-order reaction is followed by spectroscopy,
4
0.246
0.15
4.8 * 10-5
monitoring the absorbance of a colored reactant at 520 nm.
(a) What is the rate law for this reaction? (b) What is the
The reaction occurs in a 1.00-cm sample cell, and the only
value of the rate constant with proper units? (c) What is
colored species in the reaction has an extinction coefficient of
the reaction rate when the initial concentration of HgCl2
5.60 * 103 M -1 cm-1 at 520 nm. (a) Calculate the initial conis 0.100 M and that of 1C2O4 2-2 is 0.25 M if the temperacentration of the colored reactant if the absorbance is 0.605
ture is the same as that used to obtain the data shown?
at the beginning of the reaction. (b) The absorbance falls to
0.250 at 30.0 min. Calculate the rate constant in units of s - 1.
14.94The following kinetic data are collected for the initial rates of
(c) Calculate the half-life of the reaction. (d) How long does it
a reaction 2 X + Z ¡ products:
take for the absorbance to fall to 0.100?
Experiment
1
3 x 4 0 1M 2
0.25
0.25
Rate 1M , s2
3 Z4 0 1M 2
4.0 * 101
2
0.50
0.50
3.2 * 102
3
0.50
0.75
7.2 * 102
[14.102] A colored dye compound decomposes to give a colorless
product. The original dye absorbs at 608 nm and has an extinction coefficient of 4.7 * 104 M -1cm-1 at that wavelength.
You perform the decomposition reaction in a 1-cm cuvette in
a spectrometer and obtain the following data:
Additional Exercises
Time (min)
Absorbance at 608 nm
0
1.254
30
0.941
60
0.752
90
0.672
120
0.545
From these data, determine the rate law for the reaction
“dye ¡ product” and determine the rate constant.
14.103 Cyclopentadiene 1C5H62 reacts with itself to form dicyclopentadiene 1C10H122. A 0.0400 M solution of C5H6 was monitored
as a function of time as the reaction 2 C5H6 ¡ C10H12
proceeded. The following data were collected:
Time (s)
0.0
3 C5h 6 4 1M 2
50.0
0.0300
100.0
0.0240
150.0
0.0240
200.0
0.0174
0.0400
Plot 3C5H64 versus time, ln3C5H64 versus time, and 1>3C5H64
versus time. (a) What is the order of the reaction? (b) What is
the value of the rate constant?
14.104 The first-order rate constant for reaction of a particular organic compound with water varies with temperature as
follows:
300
Rate Constant 1s−1 2
320
1.0 * 10-9
340
3.0 * 10-8
355
2.4 * 10-7
Temperature (K)
3.2 * 10-11
From these data, calculate the activation energy in units of
kJ>mol.
14.105 At 28 °C, raw milk sours in 4.0 h but takes 48 h to sour in a
refrigerator at 5 °C. Estimate the activation energy in kJ>mol
for the reaction that leads to the souring of milk?
[14.106] The following is a quote from an article in the August 18,
1998, issue of The New York Times about the breakdown
of cellulose and starch: “A drop of 18 degrees Fahrenheit
[from 77 °F to 59 °F4 lowers the reaction rate six times; a
36-degree drop [from 77 °F to 41 °F] produces a fortyfold
decrease in the rate.” (a) Calculate activation energies for
the breakdown process based on the two estimates of the
effect of temperature on rate. Are the values consistent?
(b) Assuming the value of Ea calculated from the 36° drop
and that the rate of breakdown is first order with a half-life
at 25 °C of 2.7 yr, calculate the half-life for breakdown at a
temperature of -15 °C.
14.107 The following mechanism has been proposed for the reaction
of NO with H2 to form N2O and H2O:
NO1g2 + NO1g2 ¡ N2O21g2
N2O21g2 + H21g2 ¡ N2O1g2 + H2O1g2
625
(a) Show that the elementary reactions of the proposed
mechanism add to provide a balanced equation for the reaction. (b) Write a rate law for each elementary reaction in the
mechanism. (c) Identify any intermediates in the mechanism.
(d) The observed rate law is rate = k3NO423H24. If the proposed mechanism is correct, what can we conclude about the
relative speeds of the first and second reactions?
14.108 Ozone in the upper atmosphere can be destroyed by the following two-step mechanism:
Cl1g2 + O31g2 ¡ ClO1g2 + O21g2
ClO1g2 + O1g2 ¡ Cl1g2 + O21g2
(a) What is the overall equation for this process? (b) What is
the catalyst in the reaction? (c) What is the intermediate in
the reaction?
14.109 Using Figure 14.23 as your basis, draw the energy profile for
the bromide-catalyzed decomposition of hydrogen peroxide.
(a) Label the curve with the activation energies for reactions
[14.30] and [14.31]. (b) Notice from Figure 14.22 that when
Br - 1aq2 is first added, Br2 accumulates to some extent during the reaction and the solution turns brown. What does this
tell us about the relative rates of the reactions represented by
Equations 14.30 and 14.31?
[14.110]The following mechanism has been proposed for the gasphase reaction of chloroform 1CHCl32 and chlorine:
k1
2 Cl1g2 1fast2
Step 1: Cl21g2 ∆
k
-1
k2
Step 2: Cl1g2 + CHCl31g2 ¡ HCl1g2 + CCl31g2 1slow2
k3
Step 3: Cl1g2 + CCl31g2 ¡ CCl4 1fast2
(a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity
of each of the elementary reactions? (d) What is the ratedetermining step? (e) What is the rate law predicted by
this mechanism? (Hint: The overall reaction order is not
an integer.)
[14.111] Consider the hypothetical reaction 2 A + B ¡ 2 C + D.
The following two-step mechanism is proposed for the
reaction:
Step 1: A + B ¡ C + X
Step 2: A + X ¡ C + D
X is an unstable intermediate. (a) What is the predicted rate
law expression if Step 1 is rate determining? (b) What is the
predicted rate law expression if Step 2 is rate determining?
(c) Your result for part (b) might be considered surprising
for which of the following reasons: (i) The concentration of a
product is in the rate law. (ii) There is a negative reaction order in the rate law. (iii) Both reasons (i) and (ii). (iv) Neither
reasons (i) nor (ii).
[14.112] In a hydrocarbon solution, the gold compound 1CH323AuPH3
decomposes into ethane 1C2H62 and a different gold compound, 1CH32AuPH3. The following mechanism has been
proposed for the decomposition of 1CH323AuPH3:
k1
Step 1: 1CH323 AuPH3 ∆
1CH323Au + PH3 1fast2
k
-1
k2
Step 2: 1CH323 Au ¡ C2H6 + 1CH32Au 1slow2
k3
Step 3: 1CH32Au + PH3 ¡ 1CH32AuPH3 1fast2
626
chapter 14 Chemical Kinetics
(a) What is the overall reaction? (b) What are the intermediates in the mechanism? (c) What is the molecularity of each
of the elementary steps? (d) What is the rate-determining
step? (e) What is the rate law predicted by this mechanism?
(f) What would be the effect on the reaction rate of adding
PH3 to the solution of 1CH323AuPH3?
[14.113] Platinum nanoparticles of diameter ∼ 2 nm are important
catalysts in carbon monoxide oxidation to carbon dioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of 3.924 Å. (a) Estimate how many
platinum atoms would fit into a 2.0-nm sphere; the volume
of a sphere is 14>32pr3. Recall that 1 A° = 1 * 10-10 m and
1 nm = 1 * 10-9 m. (b) Estimate how many platinum atoms are on the surface of a 2.0-nm Pt sphere, using the surface area of a sphere 14pr22 and assuming that the “footprint”
of one Pt atom can be estimated from its atomic diameter of
2.8 A° . (c) Using your results from (a) and (b), calculate the
percentage of Pt atoms that are on the surface of a 2.0-nm
nanoparticle. (d) Repeat these calculations for a 5.0-nm platinum nanoparticle. (e) Which size of nanoparticle would you
expect to be more catalytically active and why?
14.114 One of the many remarkable enzymes in the human body is
carbonic anhydrase, which catalyzes the interconversion of
carbon dioxide and water with bicarbonate ion and protons.
If it were not for this enzyme, the body could not rid itself
rapidly enough of the CO2 accumulated by cell metabolism.
The enzyme catalyzes the dehydration (release to air) of up
to 107 CO2 molecules per second. Which components of this
description correspond to the terms enzyme, substrate, and
turnover number?
14.115 Suppose that, in the absence of a catalyst, a certain biochemical reaction occurs x times per second at normal body temperature 137 °C2. In order to be physiologically useful, the
reaction needs to occur 5000 times faster than when it is uncatalyzed. By how many kJ>mol must an enzyme lower the
activation energy of the reaction to make it useful?
14.116 Enzymes are often described as following the two-step
mechanism:
E + S ∆ ES 1fast2
ES ¡ E + P 1slow2
where E = enzyme, S = substrate,
ES = enzyme9substrate complex, and P = product.
(a) If an enzyme follows this mechanism, what rate law is
expected for the reaction? (b) Molecules that can bind to the
active site of an enzyme but are not converted into product
are called enzyme inhibitors. Write an additional elementary
step to add into the preceding mechanism to account for the
reaction of E with I, an inhibitor.
Integrative Exercises
[14.117]Dinitrogen pentoxide 1N2O52 decomposes in chloroform as a solvent to yield NO2 and O2. The decomposition is first order with
a rate constant at 45 °C of 1.0 * 10-5 s-1. Calculate the partial
pressure of O2 produced from 1.00 L of 0.600 M N2O5 solution at
45 °C over a period of 20.0 h if the gas is collected in a 10.0-L container. (Assume that the products do not dissolve in chloroform.)
[14.118] The reaction between ethyl iodide and hydroxide ion in ethanol 1C2H5OH2 solution, C2H5I1alc2 + OH - 1alc2 ¡
C2H5OH1l2 + I - 1alc2, has an activation energy of 86.8 kJ>mol
and a frequency factor of 2.10 * 1011 M -1 s-1. (a) Predict the
rate constant for the reaction at 35 °C. (b) A solution of KOH
in ethanol is made up by dissolving 0.335 g KOH in ethanol
to form 250.0 mL of solution. Similarly, 1.453 g of C2H5I is
dissolved in ethanol to form 250.0 mL of solution. Equal volumes of the two solutions are mixed. Assuming the reaction
is first order in each reactant, what is the initial rate at 35 °C?
(c) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? (d) Assuming the frequency
factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at 50 °C.
ln k
[14.119] You obtain kinetic data for a reaction at a set of different temperatures. You plot ln k versus 1>T and obtain the following graph:
Suggest a molecular-level interpretation of these unusual
data.
14.120 The gas-phase reaction of NO with F2 to form NOF and F has
an activation energy of Ea = 6.3 kJ>mol. and a frequency factor of A = 6.0 * 108 M -1 s-1. The reaction is believed to be
bimolecular:
NO1g2 + F21g2 ¡ NOF1g2 + F1g2
(a) Calculate the rate constant at 100 °C. (b) Draw the Lewis
structures for the NO and the NOF molecules, given that
the chemical formula for NOF is misleading because the
nitrogen atom is actually the central atom in the molecule.
(c) Predict the shape for the NOF molecule. (d) Draw a possible transition state for the formation of NOF, using dashed
lines to indicate the weak bonds that are beginning to form.
(e) Suggest a reason for the low activation energy for the
reaction.
14.121 The mechanism for the oxidation of HBr by O2 to form
2 H2O and Br2 is shown in Exercise 14.74. (a) Calculate the
overall standard enthalpy change for the reaction process.
(b) HBr does not react with O2 at a measurable rate at room
temperature under ordinary conditions. What can you infer
from this about the magnitude of the activation energy for the
rate-determining step? (c) Draw a plausible Lewis structure
for the intermediate HOOBr. To what familiar compound of
hydrogen and oxygen does it appear similar?
[14.122] The rates of many atmospheric reactions are accelerated by
the absorption of light by one of the reactants. For example,
consider the reaction between methane and chlorine to produce methyl chloride and hydrogen chloride:
1/T
Reaction 1: CH41g2 + Cl21g2 ¡ CH3Cl1g2 + HCl1g2
Design an Experiment
This reaction is very slow in the absence of light. However,
Cl21g2 can absorb light to form Cl atoms:
Reaction 2: Cl21g2 + hv ¡ 2 Cl1g2
Once the Cl atoms are generated, they can catalyze the reaction of CH4 and Cl2, according to the following proposed
mechanism:
Reaction 3: CH41g2 + Cl1g2 ¡ CH31g2 + HCl1g2
Reaction 4: CH31g2 + Cl21g2 ¡ CH3Cl1g2 + Cl1g2
The enthalpy changes and activation energies for these two
reactions are tabulated as follows:
3
∆h° 1kJ , mol2
4
- 109
Reaction
+4
Ea 1kJ , mol2
17
4
(a) By using the bond enthalpy for Cl2 (Table 8.4), determine
the longest wavelength of light that is energetic enough
to cause reaction 2 to occur. In which portion of the electromagnetic spectrum is this light found? (b) By using the
data tabulated here, sketch a quantitative energy profile
for the catalyzed reaction represented by reactions 3 and 4.
(c) By using bond enthalpies, estimate where the reactants,
CH41g2 + Cl21g2, should be placed on your diagram in part
(b). Use this result to estimate the value of Ea for the reaction
CH41g2 + Cl21g2 ¡ CH31g2 + HCl1g2 + Cl1g2.
(d) The species Cl(g) and CH31g2 in reactions 3 and 4 are
radicals, that is, atoms or molecules with unpaired
electrons. Draw a Lewis structure of CH3, and verify that it is
a radical. (e) The sequence of reactions 3 and 4 comprises a
627
radical chain mechanism. Why do you think this is called a
“chain reaction”? Propose a reaction that will terminate the
chain reaction.
[14.123]Many primary amines, RNH2, where R is a carbon-containing
fragment such as CH3, CH3CH2, and so on, undergo reactions
where the transition state is tetrahedral. (a) Draw a hybrid orbital picture to visualize the bonding at the nitrogen in a primary
amine (just use a C atom for “R”). (b) What kind of reactant with
a primary amine can produce a tetrahedral intermediate?
[14.124] The NOx waste stream from automobile exhaust includes species such as NO and NO2. Catalysts that convert these species to N2. are desirable to reduce air pollution. (a) Draw the
Lewis dot and VSEPR structures of NO, NO2, and N2. (b)
Using a resource such as Table 8.4, look up the energies of
the bonds in these molecules. In what region of the electromagnetic spectrum are these energies? (c) Design a spectroscopic experiment to monitor the conversion of NOx into N2,
describing what wavelengths of light need to be monitored as
a function of time.
[14.125] As shown in Figure 14.24, the first step in the heterogeneous hydrogenation of ethylene is adsorption of the
ethylene molecule on a metal surface. One proposed explanation for the “sticking” of ethylene to a metal surface is
the interaction of the electrons in the C ¬ C p bond with
vacant orbitals on the metal surface. (a) If this notion is
correct, would ethane be expected to adsorb to a metal surface, and, if so, how strongly would ethane bind compared
to ethylene? (b) Based on its Lewis structure, would you
expect ammonia to adsorb to a metal surface using a similar explanation as for ethylene?
Design an Experiment
Let’s explore the chemical kinetics of our favorite hypothetical reaction: a A + b B ¡ c C + d D.
We shall assume that all the substances are soluble in water and that we carry out the reaction in aqueous solution. Substances A and C both absorb visible light, and the absorption maxima are 510 nm
for A and for 640 nm for C. Substances B and D are colorless. You are provided with pure samples
of all four substances and you know their chemical formulas. You are also provided appropriate instrumentation to obtain visible absorption spectra (see the Closer Look box on using spectroscopic
methods in Section 14.3). Let’s design an experiment to ascertain the kinetics of our reaction. (a) What
experiments could you design to determine the rate law and the rate constant for the reaction at room
temperature? Would you need to know the values of the stoichiometric constants a and c in order to
find the rate law? (b) Design an experiment to determine the activation energy for the reaction. What
challenges might you face in actually carrying out this experiment? (c) You now want to test whether
a particular water-soluble substance Q is a homogeneous catalyst for the reaction. What experiments
can you carry out to test this notion? (d) If Q does indeed catalyze the reaction, what follow-up experiments might you undertake to learn more about the reaction profile for the reaction?
15
Chemical Equilibrium
To be in equilibrium is to be in a state of balance. A tug of war in which the
two sides pull with equal force so that the rope does not move is an example
of a static equilibrium, one in which an object is at rest. Equilibria can also be
dynamic, whereby a forward process and the reverse process take place at the
same rate so that no net change occurs.
(Section 13.2) of an ionic compound in contact with undisA saturated solution
solved crystals of the same compound is a good example of dynamic equilibrium. The
rate at which ions leave the surface of the crystals and enter the solution (dissolution) is
equal to the rate at which ions leave the solution and become part of the solid (crystallization). Hence the concentration of ions in solution and the amount of undissolved
solid do not change with time.
If some of the solvent is lost due to evaporation, the solution becomes more concentrated, which increases the rate of crystallization. This change leads to a net migration of
ions from the solution into the solid until the solution concentration is reduced so that
the rate of crystallization and dissolution are once again equal and equilibrium is reestablished. A striking example of this effect is the formation of intricate salt formations
in the Dead Sea, as illustrated in the chapter-opening photograph. The Dead Sea, which
borders Jordan and Israel, is the lowest point on the surface of the Earth, and has a salt
concentration that is almost nine times higher than the ocean. An extended period of hot
weather leads to extensive evaporation of the Dead Sea, resulting in an increase in the
salt concentration and subsequent crystallization and growth of intricate salt formations.
A saturated solution is one of many instances of dynamic equilibrium that we have
(Section 11.5) is another example of dynamic
already encountered. Vapor pressure
equilibrium. The pressure of a vapor above a liquid in a closed container reaches equilibrium with the liquid phase, and therefore stops changing when the rate at which
molecules escape from the liquid into the gas phase equals the rate at which molecules
What’s
Ahead
15.1 The Concept of Equilibrium We begin by examining
reversible reactions and the concept of equilibrium.
15.2 The Equilibrium Constant We define the equilibrium
constant based on rates of forward and reverse reactions,
and learn how to write equilibrium-constant expressions for
homogeneous reactions.
▶ SALT PILLARS in the Dead Sea. These
pillars are formed in shallow bays where
evaporation can lead to salt concentrations
that exceed equilibrium values.
15.3 Understanding and Working with Equilibrium
Constants We learn to interpret the magnitude of an
equilibrium constant and how its value depends on the way the
corresponding chemical equation is expressed.
15.4 Heterogeneous Equilibria We learn how to write
equilibrium-constant expressions for heterogeneous reactions.
15.5 Calculating Equilibrium Constants We see
that the value of an equilibrium constant can be calculated from
equilibrium concentrations of reactants and products.
15.6 Applications of Equilibrium Constants We learn
that equilibrium constants can be used to predict equilibrium
concentrations of reactants and products and to determine the
direction in which a reaction mixture must proceed to achieve
equilibrium.
15.7 Le Châtelier’s Principle We discuss Le Châtelier’s
principle, which predicts how a system at equilibrium
responds to changes in concentration, volume, pressure,
and temperature.
630
chapter 15 Chemical Equilibrium
(Section 13.3), which govin the gas phase become part of the liquid. Henry’s law
erns the solubility of gases in a solvent, is yet another example of dynamic equilibrium.
In this chapter, we consider dynamic equilibria in chemical reactions. Chemical
equilibrium occurs when opposing reactions proceed at equal rates: The rate at which
the products form from the reactants equals the rate at which the reactants form from
the products. As a result, concentrations cease to change, making the reaction appear
to be stopped. In this and the next two chapters, we will explore chemical equilibrium
in some detail. Later, in Chapter 19, we will learn how to relate chemical equilibria to
thermodynamics. Here, we learn how to express the equilibrium state of a reaction in
quantitative terms and study the factors that determine the relative concentrations of
reactants and products in equilibrium mixtures.
15.1 | The Concept of Equilibrium
Let’s examine a simple chemical reaction to see how it reaches an equilibrium state—a
mixture of reactants and products whose concentrations no longer change with time. We
begin with N2O4, a colorless substance that dissociates to form brown NO2. ▼ Figure 15.1
shows a sample of frozen N2O4 inside a sealed tube. The solid N2O4 becomes a gas as it is
warmed above its boiling point 121.2 °C2, and the gas turns darker as the colorless N2O4
gas dissociates into brown NO2 gas. Eventually, even though there is still N2O4 in the tube,
the color stops getting darker because the system reaches equilibrium. We are left with an
Go Figure
If you were to let the tube on the right sit overnight and then take another picture would the brown color look darker, lighter,
or the same?
NO2
N2O4
Frozen N2O4 sample is
nearly colorless
On warming, the N2O4
becomes a gas and
partially dissociates to
form brown NO2(g)
▲ Figure 15.1 The equilibrium between NO2 and N2O4.
Colors stop changing, equilibrium
reached: rate of reaction
N2O4(g)
2 NO2(g) = rate of
reaction 2 NO2(g)
N2O4(g)
section 15.1 The Concept of Equilibrium
equilibrium mixture of N2O4 and NO2 in which the concentrations of the gases no longer
change as time passes. Because the reaction is in a closed system, where no gases can escape, equilibrium will eventually be reached.
The equilibrium mixture results because the reaction is reversible: N2O4 can
form NO2, and NO2 can form N2O4. Dynamic equilibrium is represented by writing the equation for the reaction with two half arrows pointing in opposite directions
(Section 4.1):
N2O41g2 ∆ 2 NO21g2[15.1]
Colorless
Brown
We can analyze this equilibrium using our knowledge of kinetics. Let’s call the
decomposition of N2O4 the forward reaction and the formation of N2O4 the reverse
reaction. In this case, both the forward reaction and the reverse reaction are elementary
reactions. As we learned in Section 14.6, the rate laws for elementary reactions can be
written from their chemical equations:
Forward reaction: N2O41g2 ¡ 2 NO21g2
Ratef = kf 3N2O44[15.2]
Rater = kr 3NO242[15.3]
Reverse reaction: 2 NO21g2 ¡ N2O41g2
At equilibrium, the rate at which NO2 forms in the forward reaction equals the rate
at which N2O4 forms in the reverse reaction:
kf 3N2O44 = kr 3NO242[15.4]
Forward reaction
Rearranging this equation gives
Reverse reaction
kf
3NO242
=
= a constant[15.5]
3N2O44
kr
From Equation 15.5, we see that the quotient of two rate constants is another constant.
We also see that, at equilibrium, the ratio of the concentration terms equals this same
constant. (We consider this constant, called the equilibrium constant, in Section 15.2.)
It makes no difference whether we start with N2O4 or with NO2, or even with some
mixture of the two. At equilibrium, at a given temperature, the ratio equals a specific
value. Thus, there is an important constraint on the proportions of N2O4 and NO2 at
equilibrium.
Once equilibrium is established, the concentrations of N2O4 and NO2 no longer
change, as shown in ▼ Figure 15.2(a). However, the fact that the composition of the
equilibrium mixture remains constant with time does not mean that N2O4 and NO2
stop reacting. On the contrary, the equilibrium is dynamic—which means some N2O4 is
always converting to NO2 and some NO2 is always converting to N2O4. At equilibrium,
however, the two processes occur at the same rate, as shown in Figure 15.2(b).
Go Figure
At equilibrium, is the ratio 3NO24>3N2O44 less than, greater to, or equal to 1?
N2O4
Equilibrium
achieved
Time
(a)
Rate
Concentration
0
kf [N2O4]
NO2
Equilibrium
achieved
(rates are equal)
kr[NO2]2
0
Time
(b)
▲ Figure 15.2 Achieving chemical equilibrium in the N2O4 1g 2 ∆ 2 NO2 1g 2 reaction.
Equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction.
631
chapter 15 Chemical Equilibrium
We learn several important lessons about equilibrium from this example:
• At equilibrium, the concentrations of reactants and products no longer change
with time.
• For equilibrium to occur, neither reactants nor products can escape from the
system.
• At equilibrium, a particular ratio of concentration terms equals a constant.
Give It Some Thought
(a) Which quantities are equal in a dynamic equilibrium?
(b) If the rate constant for the forward reaction in Equation 15.1 is larger than
the rate constant for the reverse reaction, will the constant in Equation 15.5
be greater than 1 or smaller than 1?
15.2 | The Equilibrium Constant
A reaction in which reactants convert to products and products convert to reactants in
the same reaction vessel naturally leads to an equilibrium, regardless of how complicated the reaction is and regardless of the nature of the kinetic processes for the forward
and reverse reactions. Consider the synthesis of ammonia from nitrogen and hydrogen:
N21g2 + 3 H21g2 ∆ 2 NH31g2[15.6]
Concentration
This reaction is the basis for the Haber process, which is critical for the production of
fertilizers and therefore critical to the world’s food supply.
theonly
Haber
process, N2 and
StartingIn
with
reactants
H2 react at high pressure and temperature in the presence of a catalyst to form ammonia.
In a closed system, however, the reaction does not lead to complete
consumption
of
Equilibrium
achieved
the N2 and H2. Rather, at some point the reaction appears to stop with all three components of the reaction mixture present at the same time.
How the concentrations of H2, N2, and NH3 vary with time is shown in ▼ Figure 15.3.
H2
Notice that an equilibrium mixture is obtained regardless of whether we begin
with
NH3
N2 and H2 or with NH3. The equilibrium condition is reached from either direction.N
2
Time
Starting with only products
Equilibrium achieved
H2
NH3
N2
Time
Concentration
Concentration
Starting with only reactants
Equilibrium achieved
H2
NH3
N2
Time
▲ Figure
15.3 equilibrium is reached whether we start with only reactants
Starting
withThe
onlysame
products
(N2 and H2) or with only product 1NH3 2.
Concentration
632
Equilibrium achieved
Give It Some Thought
How do we know when H
equilibrium has been reached in a chemical reaction?
2
NH3
An expression similar N
to2 Equation 15.5 governs the concentrations of N2, H2,
and NH3 at equilibrium.
If we were to systematically change the relative amounts of the
Time
three gases in the starting mixture and then analyze each equilibrium mixture, we could
determine the relationship among the equilibrium concentrations.
section 15.2 The Equilibrium Constant
633
Chemists carried out studies of this kind on other chemical systems in the nineteenth century before Haber’s work. In 1864, Cato Maximilian Guldberg (1836–1902)
and Peter Waage (1833–1900) postulated their law of mass action, which expresses,
for any reaction, the relationship between the concentrations of the reactants
and products present at equilibrium. Suppose we have the general equilibrium
equation
a A + b B ∆ d D + e E[15.7]
where A, B, D, and E are the chemical species involved and a, b, d, and e are their coefficients in the balanced chemical equation. According to the law of mass action, the
equilibrium condition is described by the expression
Kc =
3D4d3E4e — products
[15.8]
3A4a3B4b — reactants
We call this relationship the equilibrium-constant expression (or merely the equilibrium
expression) for the reaction. The constant Kc, the equilibrium constant, is the numerical
value obtained when we substitute molar equilibrium concentrations into the equilibriumconstant expression. The subscript c on the K indicates that concentrations expressed in
molarity are used to evaluate the constant.
Chemistry Put to Work
The Haber Process
The quantity of food required to feed the ever-increasing human
population far exceeds that provided by nitrogen-fixing plants.
(Section 14.7, “Nitrogen Fixation and Nitrogenase”) Therefore,
human agriculture requires substantial amounts of ammonia-based
fertilizers for croplands. Of all the chemical reactions that humans have
learned to control for their own purposes, the synthesis of ammonia
from hydrogen and atmospheric nitrogen is one of the most important.
In 1912, the German chemist Fritz Haber (1868–1934) developed
the Haber process (Equation 15.6). The process is sometimes also
called the Haber–Bosch process to honor Karl Bosch, the engineer who
developed the industrial process on a large scale (▶ Figure 15.4).
The engineering needed to implement the Haber process requires the
use of temperatures and pressures (approximately 500 °C and 200 to
600 atm) that were difficult to achieve at that time.
The Haber process provides a historically interesting example of the
complex impact of chemistry on our lives. At the start of World War I,
in 1914, Germany depended on nitrate deposits in Chile for the nitrogencontaining compounds needed to manufacture explosives. During the
war, the Allied naval blockade of South America cut off this supply. However, by using the Haber reaction to fix nitrogen from air, Germany was
able to continue to produce explosives. Experts have estimated that World
War I would have ended before 1918 had it not been for the Haber process.
From these unhappy beginnings as a major factor in international
warfare, the Haber process has become the world’s principal source
of fixed nitrogen. The same process that prolonged World War I has
enabled the manufacture of fertilizers that have increased crop yields,
thereby saving millions of people from starvation. About 40 billion
pounds of ammonia are manufactured annually in the United States,
mostly by the Haber process. The ammonia can be applied directly to
the soil, or it can be converted into ammonium salts that are also used
as fertilizers.
Haber was a patriotic German who gave enthusiastic support
to his nation’s war effort. He served as chief of Germany’s Chemical
Warfare Service during World War I and developed the use of chlorine as a poison-gas weapon. Consequently, the decision to award him
the Nobel Prize in Chemistry in 1918 was the subject of considerable
controversy and criticism. The ultimate irony, however, came in 1933
when Haber was expelled from Germany because he was Jewish.
Related Exercises: 15.44, 15.76, 15.92
▲ Figure 15.4 A high-pressure steel reactor used in
the Haber process is on display at Karlsruhe Institute
of Technology in Germany where the Haber process was
developed.
634
chapter 15 Chemical Equilibrium
The numerator of the equilibrium-constant expression is the product of the concentrations of all substances on the product side of the equilibrium equation, each raised
to a power equal to its coefficient in the balanced equation. The denominator is similarly
derived from the reactant side of the equilibrium equation. Thus, for the Haber process,
N21g2 + 3 H21g2 ∆ 2 NH31g2, the equilibrium-constant expression is
Kc =
3NH342
3N243H243
[15.9]
Once we know the balanced chemical equation for a reaction that reaches equilibrium, we can write the equilibrium-constant expression even if we do not know the
reaction mechanism. The equilibrium-constant expression depends only on the stoichiometry of the reaction, not on its mechanism.
The value of the equilibrium constant at any given temperature does not depend
on the initial amounts of reactants and products. It also does not matter whether other
substances are present, as long as they do not react with a reactant or a product. The
value of Kc depends only on the particular reaction and on the temperature.
Sample
Exercise 15.1 Writing Equilibrium-Constant Expressions
Write the equilibrium expression for Kc for the following reactions:
(a) 2 O31g2 ∆ 3 O21g2
(b) 2 NO1g2 + Cl21g2 ∆ 2 NOCl1g2
(c) Ag +1aq2 + 2 NH31aq2 ∆ Ag1NH322+1aq2
Solution
Analyze We are given three equations and are asked to write an
equilibrium-constant expression for each.
Plan Using the law of mass action, we write each expression as a quo-
tient having the product concentration terms in the numerator and
the reactant concentration terms in the denominator. Each concentration term is raised to the power of its coefficient in the balanced
chemical equation.
Practice Exercise 1
For the reaction 2 SO21g2 + O21g2 ∆ 2 SO31g2 which of the
following is the correct equilibrium-constant expression?
3SO2423O24
23SO243O24
(a) KC =
(b) KC =
2
23SO34
3SO34
(c) KC =
Solve
(a) Kc =
3O243
2
3O34
(b) Kc =
3NOCl42
2
3NO4 3Cl24
(c) Kc =
3Ag1NH322+4
+
2
3Ag 43NH34
3SO342
3SO2423O24
(d) KC =
23SO34
23SO243O24
Practice Exercise 2
Write the equilibrium-constant expression Kc for
(a) H21g2 + I21g2 ∆ 2 HI1g2,
(b) Cd2+1aq2 + 4 Br-1aq2 ∆ CdBr42-1aq2.
Evaluating Kc
We can illustrate how the law of mass action was discovered empirically and demonstrate that the equilibrium constant is independent of starting concentrations by examining a series of experiments involving dinitrogen tetroxide and nitrogen dioxide:
3NO242
N2O41g2 ∆ 2 NO21g2
Kc =
[15.10]
3N2O44
We start with several sealed tubes containing different concentrations of NO2 and N2O4.
The tubes are kept at 100 °C until equilibrium is reached. We then analyze the mixtures
and determine the equilibrium concentrations of NO2 and N2O4, which are shown in
▶ Table 15.1.
To evaluate Kc, we insert the equilibrium concentrations into the equilibriumconstant expression. For example, using Experiment 1 data, 3NO24 = 0.0172 M and
3N2O44 = 0.00140 M, we find
Kc =
3NO242
30.017242
=
= 0.211
3N2O44
0.00140
section 15.2 The Equilibrium Constant
635
Table 15.1 Initial and Equilibrium Concentrations of n2o4 1 g 2 and no2 1 g 2
at 100 °C
1
Initial
3 n2o4 4 1M 2
0.0
Initial
3 no2 4 1M 2
0.0200
Equilibrium
3n2o4 4 1M 2
0.00140
Equilibrium
3no2 4 1M 2
0.0172
0.211
2
0.0
0.0300
0.00280
0.0243
0.211
3
0.0
0.0400
0.00452
0.0310
0.213
4
0.0200
0.0
0.00452
0.0310
0.213
Kc
Proceeding in the same way, the values of Kc for the other samples are calculated.
Note from Table 15.1 that the value for Kc is constant (within the limits of experimental
error) even though the initial concentrations vary. Furthermore, Experiment 4 shows
that equilibrium can be achieved beginning with N2O4 rather than with NO2. That is,
equilibrium can be approached from either direction. ▶ Figure 15.5 shows how Experiments 3 and 4 result in the same equilibrium mixture even though the two experiments
start with very different NO2 concentrations.
Notice that no units are given for Kc either in Table 15.1 or in the calculation we
just did using Experiment 1 data. It is common practice to write equilibrium constants
without units for reasons that we address later in this section.
Give It Some Thought
How does the value of Kc in Equation 15.10 depend on the starting
concentrations of NO2 and N2O4?
Equilibrium Constants in Terms of Pressure, Kp
When the reactants and products in a chemical reaction are gases, we can formulate
the equilibrium-constant expression in terms of partial pressures. When partial pressures in atmospheres are used in the expression, we denote the equilibrium constant Kp
(where the subscript p stands for pressure). For the general reaction in Equation 15.7,
we have
1PD2d1PE2e
Kp =
[15.11]
1PA2a1PB2b
where PA is the partial pressure of A in atmospheres, PB is the partial pressure of B in
atmospheres, and so forth. For example, for our N2O4 >NO2 reaction, we have
Kp =
1PNO222
PN2O4
Give It Some Thought
What is the difference between the equilibrium constant Kc and the equilibrium
constant Kp?
For a given reaction, the numerical value of Kc is generally different from the
numerical value of Kp. We must therefore take care to indicate, via subscript c or p,
which constant we are using. It is possible, however, to calculate one from the other
(Section 10.4):
using the ideal-gas equation
n
PV = nRT, so P = RT[15.12]
V
The usual units for n>V are mol>L, which equals molarity, M. For substance A in our
generic reaction, we therefore see that
PA =
nA
RT = 3A4RT[15.13]
V
0.0400
[NO2] (M)
Experiment
Experiment 3
0.0300
0.0200
Experiment 4
0.0100
0
Time
▲ Figure 15.5 The same equilibrium
mixture is produced regardless of the initial
NO2 concentration. The concentration of
NO2 either increases or decreases until
equilibrium is reached.
636
chapter 15 Chemical Equilibrium
When we substitute Equation 15.13 and like expressions for the other gaseous components of the reaction into Equation 15.11, we obtain a general expression relating
Kp and Kc :
Kp = Kc 1RT2∆n[15.14]
The quantity ∆n is the change in the number of moles of gas in the balanced chemical
equation. It equals the sum of the coefficients of the gaseous products minus the sum of
the coefficients of the gaseous reactants:
∆n = 1moles of gaseous product2 - 1moles of gaseous reactant2[15.15]
For example, in the N2O41g2 ∆ 2 NO21g2 reaction, there are two moles of product
NO2 and one mole of reactant N2O4. Therefore, ∆n = 2 - 1 = 1, and Kp = Kc1RT2,
for this reaction.
Give It Some Thought
Is it possible to have a reaction where Kc = Kp? If so, under what conditions
would this relationship hold?
Sample
Exercise 15.2 Converting between Kc and Kp
For the Haber process,
N21g2 + 3 H21g2 ∆ 2 NH31g2
Kc = 9.60 at 300 °C. Calculate Kp for this reaction at this temperature.
Solution
Analyze We are given Kc for a reaction and asked to calculate Kp.
Plan The relationship between Kc and Kp is given by Equation 15.14. To apply that equation, we
must determine ∆n by comparing the number of moles of product with the number of moles of
reactants (Equation 15.15).
Solve With 2 mol of gaseous products 12 NH32 and 4 mol of gaseous reactants 11 N2 + 3 H22,
∆n = 2 - 4 = - 2. (Remember that ∆ functions are always based on products minus reactants.) The temperature is 273 + 300 = 573 K. The value for the ideal-gas constant, R, is
0.08206 L@atm>mol@K. Using Kc = 9.60, we therefore have
Kp = Kc 1RT2∆n = 19.60210.08206 * 5732-2 =
19.602
10.08206 * 57322
= 4.34 * 10-3
Practice Exercise 1
For which of the following reactions is the ratio Kp >Kc largest at 300 K?
(a) N21g2 + O21g2 ∆ 2 NO1g2 (b) CaCO31s2 ∆ CaO1s2 + CO21g2
(c) Ni1CO241g2 ∆ Ni1s2 + 4 CO1g2 (d) C1s2 + 2 H21g2 ∆ CH41g2
Practice Exercise 2
For the equilibrium 2 SO31g2 ∆ 2 SO21g2 + O21g2, Kc is 4.08 * 10-3 at 1000 K. Calculate the value for Kp.
Equilibrium Constants and Units
You may wonder why equilibrium constants are reported without units. The equilibrium constant is related to the kinetics of a reaction as well as to the thermodynamics.
(We explore this latter connection in Chapter 19.) Equilibrium constants derived from
thermodynamic measurements are defined in terms of activities rather than concentrations or partial pressures.
section 15.3 Understanding and Working with Equilibrium Constants
637
The activity of any substance in an ideal mixture is the ratio of the concentration
or pressure of the substance either to a reference concentration 11 M2 or to a reference pressure (1 atm). For example, if the concentration of a substance in an equilibrium mixture is 0.010 M, its activity is 0.010 M>1 M = 0.010. The units of such ratios
always cancel and, consequently, activities have no units. Furthermore, the numerical value of the activity equals the concentration. For pure solids and pure liquids,
the situation is even simpler because the activities then merely equal 1 (again with
no units).
In real systems, activities are also ratios that have no units. Even though these
activities may not be exactly numerically equal to concentrations, we will ignore
the differences. All we need to know at this point is that activities have no units.
As a result, the thermodynamic equilibrium constants derived from them also have
no units. It is therefore common practice to write all types of equilibrium constants
without units, a practice that we adhere to in this text. In more advanced chemistry courses, you may make more rigorous distinctions between concentrations and
activities.
Give It Some Thought
If the concentration of N2O4 in an equilibrium mixture is 0.00140 M, what is its
activity? (Assume the solution is ideal.)
15.3 | Understanding and Working
with Equilibrium Constants
Before doing calculations with equilibrium constants, it is valuable to understand what
the magnitude of an equilibrium constant can tell us about the relative concentrations
of reactants and products in an equilibrium mixture. It is also useful to consider how
the magnitude of any equilibrium constant depends on how the chemical equation is
expressed.
The Magnitude of Equilibrium Constants
The magnitude of the equilibrium constant for a reaction gives us important information about the composition of the equilibrium mixture. For example, consider the experimental data for the reaction of carbon monoxide gas and chlorine gas at 100 °C to
form phosgene 1COCl22, a toxic gas used in the manufacture of certain polymers and
insecticides:
CO1g2 + Cl21g2 ∆ COCl21g2
3COCl24
Kc =
= 4.56 * 109
3CO43Cl24
For the equilibrium constant to be so large, the numerator of the equilibrium-constant
expression must be approximately a billion 11092 times larger than the denominator.
Thus, the equilibrium concentration of COCl2 must be much greater than that of CO
or Cl2, and in fact, this is just what we find experimentally. We say that this equilibrium
lies to the right (that is, toward the product side). Likewise, a very small equilibrium
constant indicates that the equilibrium mixture contains mostly reactants. We then say
that the equilibrium lies to the left. In general,
If K W 1 (large K): Equilibrium lies to right, products predominate
If K V 1 (small K): Equilibrium lies to left, reactants predominate
These situations are summarized in ▶ Figure 15.6. Remember, it is forward and reverse
reaction rates, not reactant and product concentrations, that are equal at equilibrium.
Go Figure
What would this figure look like for a
reaction in which K ≈ 1?
Reactants
Products
K >> 1, equilibrium “lies to the right”
Reactants
Products
K << 1, equilibrium “lies to the left”
▲ Figure 15.6 Relationship between
magnitude of K and composition of an
equilibrium mixture.
638
chapter 15 Chemical Equilibrium
Sample
Exercise 15.3 Interpreting the Magnitude of an
Equilibrium Constant
The following diagrams represent three systems at equilibrium, all in the same-size containers.
(a) Without doing any calculations, rank the systems in order of increasing Kc. (b) If the volume
of the containers is 1.0 L and each sphere represents 0.10 mol, calculate Kc for each system.
(i)
(ii)
(iii)
Solution
Analyze We are asked to judge the relative magnitudes of three equilibrium constants and then
to calculate them.
Plan (a) The more product present at equilibrium, relative to reactant, the larger the equilibrium
constant. (b) The equilibrium constant is given by Equation 15.8.
Solve
(a) Each box contains 10 spheres. The amount of product in each varies as follows: (i) 6,
(ii) 1, (iii) 8. Therefore, the equilibrium constant varies in the order (ii) 6 (i) 6 (iii),
from smallest (most reactant) to largest (most products).
(b) In (i), we have 0.60 mol>L product and 0.40 mol>L reactant, giving Kc = 0.60>0.40 = 1.5.
(You will get the same result by merely dividing the number of spheres of each kind:
6 spheres>4 spheres = 1.5.) In (ii), we have 0.10 mol>L product and 0.90 mol>L
reactant, giving Kc = 0.10>0.90 = 0.11 (or 1 sphere>9 spheres = 0.11). In (iii),
we have 0.80 mol>L product and 0.20 mol>L reactant, giving Kc = 0.80>0.20 = 4.0
(or 8 spheres>2 spheres = 4.0). These calculations verify the order in (a).
Comment Imagine a drawing that represents a reaction with a very small or very large value of Kc. For
example, what would the drawing look like if Kc = 1 * 10-5? In that case there would need to be
100,000 reactant molecules for only 1 product molecule. But then, that would be impractical to draw.
Practice Exercise 1
The equilibrium constant for the reaction N2O41g2 ∆ 2 NO21g2 at 2 °C is Kc = 2.0. If
each yellow sphere represents 1 mol of N2O4 and each brown sphere 1 mol of NO2 which of the
following 1.0 L containers represents the equilibrium mixture at 2 °C?
2
(a)
(b)
(c)
(d)
(e)
Practice Exercise 2
For the reaction H21g2 + I21g2 ∆ 2 HI1g2, Kp = 794 at 298 K and Kp = 55 at 700 K. Is
the formation of HI favored more at the higher or lower temperature?
section 15.3 Understanding and Working with Equilibrium Constants
The Direction of the Chemical Equation and K
We have seen that we can represent the N2O4 >NO2 equilibrium as
N2O41g2 ∆ 2 NO21g2 Kc =
3NO242
= 0.212 1at 100 °C2[15.16]
3N2O44
We could equally well consider this equilibrium in terms of the reverse reaction:
2 NO21g2 ∆ N2O41g2
The equilibrium expression is then
Kc =
3N2O44
3NO242
=
1
= 4.72 1at 100 °C2[15.17]
0.212
Equation 15.17 is the reciprocal of the expression in Equation 15.16. The equilibrium-constant expression for a reaction written in one direction is the reciprocal of
the expression for the reaction written in the reverse direction. Consequently, the
numerical value of the equilibrium constant for the reaction written in one direction is the reciprocal of that for the reverse reaction. Both expressions are equally
valid, but it is meaningless to say that the equilibrium constant for the equilibrium
between NO2 and N2O4 is “0.212” or “4.72” unless we indicate how the equilibrium
reaction is written and specify the temperature. Therefore, whenever you are using
an equilibrium constant, you should always write the associated balanced chemical equation.
Give It Some Thought
For the reaction PCl51g2 ∆ PCl31g2 + Cl21g2, the equilibrium constant
KC = 1.1 * 10-2 at 400 K. What is the equilibrium constant for the reaction
PCl31g2 + Cl21g2 ∆ PCl51g2 at 400 K?
Relating Chemical Equation Stoichiometry
and Equilibrium Constants
There are many ways to write a balanced chemical equation for a given reaction. For
example, if we multiply Equation 15.1, N2O41g2 ∆ 2 NO21g2 by 2, we have
2 N2O41g2 ∆ 4 NO21g2
This chemical equation is balanced and might be written this way in some contexts.
Therefore, the equilibrium-constant expression for this equation is
Kc =
3NO244
3N2O442
which is the square of the equilibrium-constant expression given in Equation 15.10
for the reaction as written in Equation 15.1: 3NO242 >3N2O44. Because the new
equilibrium-constant expression equals the original expression squared, the new equilibrium constant Kc equals the original constant squared: 0.2122 = 0.0449 1at 100 °C2.
Once again, it is important to remember that you must relate each equilibrium constant you work with to a specific balanced chemical equation. The concentrations of the
substances in the equilibrium mixture will be the same no matter how you write the
chemical equation, but the value of Kc you calculate depends completely on how you
write the reaction.
Give It Some Thought
How does the magnitude of Kp for the reaction 2 HI1g2 ∆ H21g2 + I21g2
change if the equilibrium is written 6 HI1g2 ∆ 3 H21g2 + 3 I21g2?
639
640
chapter 15 Chemical Equilibrium
It is also possible to calculate the equilibrium constant for a reaction if we know
the equilibrium constants for other reactions that add up to give us the one we want,
(Section 5.6) For example, consider the following two reacsimilar to Hess’s law.
tions, their equilibrium-constant expressions, and their equilibrium constants at
100 °C:
1.2 NOBr1g2 ∆ 2 NO1g2 + Br21g2 Kc1 =
2.Br21g2 + Cl21g2 ∆ 2 BrCl1g2 Kc2 =
The net sum of these two equations is:
3NO423Br24
3NOBr42
= 0.014
3BrCl42
= 7.2
3Br243Cl24
3.2 NOBr1g2 + Cl21g2 ∆ 2 NO1g2 + 2 BrCl1g2
You can prove algebraically that the equilibrium-constant expression for reaction 3 is
the product of the expressions for reactions 1 and 2:
Kc3 =
Thus,
To summarize:
3NO423BrCl42
2
3NOBr4 3Cl24
=
3NO423Br24
3NOBr4
2
*
3BrCl42
3Br243Cl24
Kc3 = 1Kc121Kc22 = 10.014217.22 = 0.10
1.The equilibrium constant of a reaction in the reverse direction is the inverse (or
reciprocal) of the equilibrium constant of the reaction in the forward direction:
A + B ∆ C + D K1
C + D ∆ A + B K = 1>K1
2.The equilibrium constant of a reaction that has been multiplied by a number
is equal to the original equilibrium constant raised to a power equal to that
number.
A + B ∆ C + D
K1
nA + nB ∆ nC + nD K = K1n
3.The equilibrium constant for a net reaction made up of two or more reactions is the
product of the equilibrium constants for the individual reactions:
A + B ∆ C + D K1
C + F ∆ G + A K2
B + F ∆ D + G K3 = 1K121K22
Sample
Exercise 15.4 Combining Equilibrium Expressions
Given the reactions
HF1aq2 ∆ H+1aq2 + F-1aq2
+
H2C2O41aq2 ∆ 2 H 1aq2 +
C2O42 - 1aq2
Kc = 6.8 * 10-4
Kc = 3.8 * 10-6
determine the value of Kc for the reaction
2 HF1aq2 + C2O42 - 1aq2 ∆ 2 F-1aq2 + H2C2O41aq2
Solution
Analyze We are given two equilibrium equations and the corresponding equilibrium constants and are asked to determine the equilibrium
constant for a third equation, which is related to the first two.
Plan We cannot simply add the first two equations to get the third.
Instead, we need to determine how to manipulate these equations to
come up with equations that we can add to give us the desired equation.
section 15.4 Heterogeneous Equilibria
641
Solve If we multiply the first equation
by 2 and make the corresponding
change to its equilibrium constant
(raising to the power 2), we get
Reversing the second equation and
again making the corresponding change
to its equilibrium constant (taking the
reciprocal) gives
Now, we have two equations that sum
to give the net equation, and we can
multiply the individual Kc values to get
the desired equilibrium constant.
2 HF1aq2 ∆ 2 H+1aq2 + 2 F-1aq2
Kc = 16.8 * 10-422 = 4.6 * 10-7
2 H+1aq2 + C2O42 - 1aq2 ∆ H2C2O41aq2
Kc =
2 HF1aq2 ∆ 2 H+1aq2 + 2 F-1aq2
2 H+1aq2 + C2O42 - 1aq2 ∆ H2C2O41aq2
2 HF1aq2 + C2O42 - 1aq2 ∆ 2 F -1aq2 + H2C2O41aq2
Practice Exercise 1
Given the equilibrium constants for the following two reactions in aqueous solution at 25 °C,
HNO21aq2 ∆ H+1aq2 + NO2-1aq2 Kc = 4.5 * 10-4
H2SO31aq2 ∆ 2 H+1aq2 + SO3-1aq2 Kc = 1.1 * 10-9
what is the value of Kc for the reaction?
2 HNO21aq2 + SO32 - 1aq2 ∆ H2SO31aq2 + 2 NO2-1aq2
(a)4.9 * 10-13, (b) 4.1 * 105, (c) 8.2 * 105, (d) 1.8 * 102, (e) 5.4 * 10-3.
Practice Exercise 2
Given that, at 700 K, Kp = 54.0 for the reaction H21g2 + I21g2 ∆ 2 HI1g2 and Kp = 1.04 * 10-4
for the reaction N21g2 + 3 H21g2 ∆ 2 NH31g2, determine the value of Kp for the reaction
2 NH31g2 + 3 I21g2 ∆ 6 HI1g2 + N21g2 at 700 K.
15.4 | Heterogeneous Equilibria
Many equilibria involve substances that are all in the same phase. Such equilibria are
called homogeneous equilibria. In some cases, however, the substances in equilibrium
are in different phases, giving rise to heterogeneous equilibria. As an example of the latter, consider the equilibrium that occurs when solid lead(II) chloride dissolves in water
to form a saturated solution:
PbCl21s2 ∆ Pb2+ 1aq2 + 2 Cl- 1aq2[15.18]
This system consists of a solid in equilibrium with two aqueous species. If we want to write
the equilibrium-constant expression for this process, we encounter a problem we have not
encountered previously: How do we express the concentration of a solid? If we were to
carry out experiments starting with varying amounts of products and reactants we would
find that the equilibrium-constant expression for the reaction of Equation 15.18 is
Kc = 3Pb2+ 43Cl- 42[15.19]
Thus, our problem of how to express the concentration of a solid is not relevant in
the end, because PbCl21s2 does not show up in the equilibrium-constant expression.
More generally, we can state that whenever a pure solid or a pure liquid is involved in a
heterogeneous equilibrium, its concentration is not included in the equilibrium-constant
expression.
The fact that pure solids and pure liquids are excluded from equilibrium-constant
expressions can be explained in two ways. First, the concentration of a pure solid or liquid has a constant value. If the mass of a solid is doubled, its volume also doubles. Thus,
its concentration, which relates to the ratio of mass to volume, stays the same. Because
1
= 2.6 * 105
3.8 * 10-6
Kc = 4.6 * 10-7
Kc = 2.5 * 105
Kc = 14.6 * 10-7212.6 * 1052 = 0.12
642
chapter 15 Chemical Equilibrium
equilibrium-constant expressions include terms only for reactants and products whose
concentrations can change during a chemical reaction, the concentrations of pure solids and pure liquids are omitted.
The omission can also be rationalized in a second way. Recall from Section 15.2
that what is substituted into a thermodynamic equilibrium expression is the activity
of each substance, which is a ratio of the concentration to a reference value. For a pure
substance, the reference value is the concentration of the pure substance, so that the
activity of any pure solid or liquid is always 1.
Give It Some Thought
Write the equilibrium-constant expression for the evaporation of water,
H2O1l 2 ∆ H2O1g2, in terms of partial pressures.
Decomposition of calcium carbonate is another example of a heterogeneous
reaction:
CaCO31s2 ∆ CaO1s2 + CO21g2
Omitting the concentrations of the solids from the equilibrium-constant expression
gives
Kc = 3CO24 and Kp = PCO2
These equations tell us that at a given temperature, an equilibrium among CaCO3, CaO, and
CO2 always leads to the same CO2 partial pressure as long as all three components are present. As shown in ▼ Figure 15.7, we have the same CO2 pressure regardless of the relative
amounts of CaO and CaCO3.
Go Figure
If some of the CO21g2 were released from the upper bell jar and the seal then restored and the system allowed to return to
equilibrium, would the amount of CaCO31s2 increase, decrease or remain the same?
CaCO3(s)
CO2 (g)
CaCO3
CaO
Large amount of CaCO3 , small
amount of CaO, gas pressure P
CaO(s) + CO2(g)
CO2 (g)
CaCO3
CaO
Small amount of CaCO3 , large
amount of CaO, gas pressure still P
▲ Figure 15.7 At a given temperature, the equilibrium pressure of CO2 in the bell jars is the
same no matter how much of each solid is present.
section 15.4 Heterogeneous Equilibria
643
Sample
Exercise 15.5 Writing Equilibrium-Constant Expressions for Heterogeneous Reactions
Write the equilibrium-constant expression Kc for
(a) CO21g2 + H21g2 ∆ CO1g2 + H2O1l2
(b) SnO21s2 + 2 CO1g2 ∆ Sn1s2 + 2 CO21g2
Solution
Analyze We are given two chemical equations, both for heterogeneous
equilibria, and asked to write the corresponding equilibrium-constant
expressions.
Plan We use the law of mass action, remembering to omit any pure
solids and pure liquids from the expressions.
Solve
(a) The equilibrium-constant expression is
3CO4
Kc =
3CO243H24
Because H2O appears in the reaction as a liquid, its concentration
does not appear in the equilibrium-constant expression.
(b) The equilibrium-constant expression is
3CO242
Kc =
3CO42
Because SnO2 and Sn are pure solids, their concentrations do not appear in the equilibrium-constant expression.
Practice Exercise 1
Consider the equilibrium that is established in a saturated solution
of silver chloride, Ag +1aq2 + Cl-1aq2 ∆ AgCl1s2. If solid
AgCl is added to this solution, what will happen to the concentration of Ag + and Cl- ions in solution? (a) 3Ag +4 and 3Cl-4 will
both increase, (b) 3Ag +4 and 3Cl-4 will both decrease, (c) 3Ag +4
will increase and 3Cl-4 will decrease, (d) 3Ag +4 will decrease and
3Cl-4 will increase, (e) neither 3Ag +4 nor 3Cl-4 will change.
Practice Exercise 2
Write the following equilibrium-constant expressions:
(a)Kc for Cr1s2 + 3 Ag +1aq2 ∆ Cr3+1aq2 + 3 Ag1s2,
(b)Kp for 3 Fe1s2 + 4 H2O1g2 ∆ Fe3O41s2 + 4 H21g2.
Sample
Exercise 15.6 Analyzing a Heterogeneous Equilibrium
Each of these mixtures was placed in a closed container and allowed to stand:
(a) CaCO31s2
(b) CaO(s) and CO21g2 at a pressure greater than the value of Kp
(c) CaCO31s2 and CO21g2 at a pressure greater than the value of Kp
(d) CaCO31s2 and CaO1s2
Determine whether or not each mixture can attain the equilibrium
Solution
CaCO31s2 ∆ CaO1s2 + CO21g2
Analyze We are asked which of several combinations of species can
establish an equilibrium between calcium carbonate and its decomposition products, calcium oxide and carbon dioxide.
Plan For equilibrium to be achieved, it must be possible for both the
forward process and the reverse process to occur. For the forward
process to occur, some calcium carbonate must be present. For the
reverse process to occur, both calcium oxide and carbon dioxide
must be present. In both cases, either the necessary compounds may
be present initially or they may be formed by reaction of the other
species.
Solve Equilibrium can be reached in all cases except (c) as long as suf-
ficient quantities of solids are present. (a) CaCO3 simply decomposes,
forming CaO(s) and CO21g2 until the equilibrium pressure of CO2 is
attained. There must be enough CaCO3, however, to allow the CO2
pressure to reach equilibrium. (b) CO2 continues to combine with
CaO until the partial pressure of the CO2 decreases to the equilibrium
value. (c) Because there is no CaO present, equilibrium cannot be
attained; there is no way the CO2 pressure can decrease to its equilibrium value (which would require some CO2 to react with CaO).
(d) The situation is essentially the same as in (a): CaCO3 decomposes
until equilibrium is attained. The presence of CaO initially makes no
difference.
Practice Exercise 1
If 8.0 g of NH4HS1s2 is placed in a sealed vessel with
a volume of 1.0 L and heated to 200 °C the reaction
NH4HS1s2 ∆ NH31g2 + H2S1g2 will occur. When the system comes to equilibrium, some NH4HS1s2 is still present.
Which of the following changes will lead to a reduction in the
amount of NH4HS1s2 that is present, assuming in all cases that
equilibrium is re-established following the change? (a) Adding
more NH31g2 to the vessel, (b) Adding more H2S1g2 to the vessel, (c) Adding more NH4HS1s2 to the vessel, (d) Increasing the
volume of the vessel, (e) decreasing the volume of the vessel.
Practice Exercise 2
When added to Fe3O41s2 in a closed container, which one
of the following substances—H21g2, H2O1g2, O21g2—allows
equilibrium to be established in the reaction
3 Fe1s2 + 4 H2O1g2 ∆ Fe3O41s2 + 4 H21g2?
644
chapter 15 Chemical Equilibrium
When a solvent is a reactant or product in an equilibrium, its concentration is omitted from the equilibrium-constant expression, provided the concentrations of reactants
and products are low, so that the solvent is essentially a pure substance. Applying this
guideline to an equilibrium involving water as a solvent,
H2O1l2 + CO32- 1aq2 ∆ OH- 1aq2 + HCO3- 1aq2[15.20]
gives an equilibrium-constant expression that does not contain 3H2O4:
Kc =
3OH- 43HCO3- 4
3CO32- 4
[15.21]
Give It Some Thought
Write the equilibrium-constant expression for the reaction
NH31aq2 + H2O1l 2 ∆ NH4+1aq2 + OH-1aq2
15.5 | Calculating Equilibrium
Constants
If we can measure the equilibrium concentrations of all the reactants and products in
a chemical reaction, as we did with the data in Table 15.1, calculating the value of the
equilibrium constant is straightforward. We simply insert all the equilibrium concentrations into the equilibrium-constant expression for the reaction.
Sample
Exercise 15.7 Calculating K When All Equilibrium Concentrations Are Known
After a mixture of hydrogen and nitrogen gases in a reaction vessel is allowed to attain equilibrium
at 472 °C, it is found to contain 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. From these data,
calculate the equilibrium constant Kp for the reaction
N21g2 + 3 H21g2 ∆ 2 NH31g2
Solution
Analyze We are given a balanced equation and equilibrium partial
pressures and are asked to calculate the value of the equilibrium
constant.
Plan Using the balanced equation, we write the equilibrium-constant
expression. We then substitute the equilibrium partial pressures into
the expression and solve for Kp.
Solve
Kp =
1PNH322
3
PN21PH22
=
10.16622
12.46217.3823
= 2.79 * 10-5
Practice Exercise 1
A mixture of gaseous sulfur dioxide and oxygen are added to a
reaction vessel and heated to 1000 K where they react to form
SO31g2. If the vessel contains 0.669 atm SO21g2, 0.395 atm
O21g2, and 0.0851 atm SO31g2 after the system has reached
equilibrium, what is the equilibrium constant Kp for the reaction
2 SO21g2 + O21g2 ∆ 2 SO31g2? (a) 0.0410, (b) 0.322,
(c) 24.4, (d) 3.36, (e) 3.11.
Practice Exercise 2
An aqueous solution of acetic acid is found to have
the following equilibrium concentrations at 25 °C:
3CH3COOH4 = 1.65 * 10-2 M; 3H+4 = 5.44 * 10-4 M; and
3CH3COO-4 = 5.44 * 10-4 M. Calculate the equilibrium constant Kc for the ionization of acetic acid at 25 °C. The reaction is
CH3COOH1aq2 ∆ H+1aq2 + CH3COO-1aq2
Often, we do not know the equilibrium concentrations of all species in an equilibrium mixture. If we know the initial concentrations and the equilibrium concentration
of at least one species, however, we can generally use the stoichiometry of the reaction
section 15.5 Calculating Equilibrium Constants
645
to deduce the equilibrium concentrations of the others. The following steps outline the
procedure:
1.Tabulate all known initial and equilibrium concentrations of the species that appear in the equilibrium-constant expression.
2.For those species for which initial and equilibrium concentrations are known,
calculate the change in concentration that occurs as the system reaches
equilibrium.
3.Use the stoichiometry of the reaction (that is, the coefficients in the balanced
chemical equation) to calculate the changes in concentration for all other species
in the equilibrium-constant expression.
4.Use initial concentrations from step 1 and changes in concentration from step 3 to
calculate any equilibrium concentrations not tabulated in step 1.
5.Determine the value of the equilibrium constant.
The best way to illustrate how to do this type of calculation is by example, as we do in
Sample Exercise 15.8.
Sample
Exercise 15.8 Calculating K from Initial and Equilibrium Concentrations
A closed system initially containing 1.000 * 10-3 M H2 and 2.000 * 10-3 M I2 at 448 °C is allowed
to reach equilibrium, and at equilibrium the HI concentration is 1.87 * 10-3 M. Calculate Kc at
448 °C for the reaction taking place, which is
Solution
H21g2 + I21g2 ∆ 2 HI1g2
Analyze We are given the initial concentrations of H2 and I2 and the
equilibrium concentration of HI. We are asked to calculate the equilibrium constant Kc for H21g2 + I21g2 ∆ 2 HI1g2.
Plan We construct a table to find equilibrium concentrations of all
species and then use the equilibrium concentrations to calculate the
equilibrium constant.
Solve
(1) We tabulate the initial and equilibrium
concentrations of as many species as we
can. We also provide space in our table for
listing the changes in concentrations. As
shown, it is convenient to use the chemical equation as the heading for the table.
(2) We calculate the change in HI concentration, which is the difference between the
equilibrium and initial values:
(3) We use the coefficients in the balanced
equation to relate the change in [HI] to
the changes in 3H24 and 3I24:
(4) We calculate the equilibrium concentrations of H2 and I2, using initial concentrations and changes in concentration. The
equilibrium concentration equals the initial
concentration minus that consumed:
(5) Our table now is complete (with
equilibrium concentrations in blue
for emphasis):
Initial concentration (M)
H21g2
I21g2
+
1.000 * 10-3
2 HI1g2
∆
2.000 * 10-3
0
Change in concentration (M)
Equilibrium concentration (M)
1.87 * 10-3
Change in 3HI4 = 1.87 * 10-3 M - 0 = 1.87 * 10-3 M
a1.87 * 10-3
a1.87 * 10-3
mol H2
1 mol H2
mol HI
ba
b = 0.935 * 10-3
L
2 mol HI
L
mol I2
1 mol I2
mol HI
ba
b = 0.935 * 10-3
L
2 mol HI
L
3H24 = 1.000 * 10-3 M - 0.935 * 10-3 M = 0.065 * 10-3 M
3I24 = 2.000 * 10-3 M - 0.935 * 10-3 M = 1.065 * 10-3 M
H21g2
+
I21g2
∆
2 HI1g2
Initial concentration (M)
1.000 * 10-3
2.000 * 10-3
0
Change in concentration (M)
-0.935 * 10-3
-0.935 * 10-3
+ 1.87 * 10-3
Equilibrium concentration (M)
0.065 * 10-3
1.065 * 10-3
1.87 * 10-3
646
chapter 15 Chemical Equilibrium
Notice that the entries for the changes are negative when a reactant is consumed and positive when
a product is formed. Finally, we use the equilibrium-constant expression to calculate the equilibrium
constant:
Kc =
3HI42
3H243I24
=
11.87 * 10-322
10.065 * 10-3211.065 * 10-32
= 51
Comment The same method can be applied to gaseous equilibrium problems to calculate Kp, in which
case partial pressures are used as table entries in place of molar concentrations. Your instructor may refer
to this kind of table as an ICE chart, where ICE stands for Initial – Change – Equilibrium.
Practice Exercise 1
In Section 15.1, we discussed the equilibrium between N2O41g2 and NO21g2. Let’s return to that
equation in a quantitative example. When 9.2 g of frozen N2O4 is added to a 0.50 L reaction vessel and the vessel is heated to 400 K and allowed to come to equilbirum, the concentration of
N2O4 is determined to be 0.057 M. Given this information, what is the value of Kc for the reaction N2O41g2 ¡ 2 NO21g2 at 400 K? (a) 0.23, (b) 0.36, (c) 0.13, (d) 1.4, (e) 2.5.
Practice Exercise 2
The gaseous compound BrCl decomposes at high temperature in a sealed container:
2 BrCl1g2 ∆ Br21g2 + Cl21g2. Initially, the vessel is charged at 500 K with BrCl(g) at a partial
pressure of 0.500 atm. At equilibrium, the BrCl(g) partial pressure is 0.040 atm. Calculate the
value of Kp at 500 K.
15.6 | Applications of Equilibrium
Constants
We have seen that the magnitude of K indicates the extent to which a reaction proceeds. If K is very large, the equilibrium mixture contains mostly substances on the
product side of the equation for the reaction. That is, the reaction proceeds far to the
right. If K is very small (that is, much less than 1), the equilibrium mixture contains
mainly substances on the reactant side of the equation. The equilibrium constant also
allows us to (1) predict the direction in which a reaction mixture achieves equilibrium
and (2) calculate equilibrium concentrations of reactants and products.
Predicting the Direction of Reaction
For the formation of NH3 from N2 and H2 (Equation 15.6), Kc = 0.105 at 472 °C. Suppose we place 2.00 mol of H2, 1.00 mol of N2, and 2.00 mol of NH3 in a 1.00-L container
at 472 °C. How will the mixture react to reach equilibrium? Will N2 and H2 react to
form more NH3, or will NH3 decompose to N2 and H2?
To answer this question, we substitute the starting concentrations of
N2, H2, and NH3 into the equilibrium-constant expression and compare its value to the
equilibrium constant:
3NH342
3N243H24
3
=
12.0022
11.00212.0023
= 0.500 whereas Kc = 0.105[15.22]
To reach equilibrium, the quotient 3NH342 >3N243H243 must decrease from the starting value of 0.500 to the equilibrium value of 0.105. Because the system is closed, this
change can happen only if 3NH34 decreases and 3N24 and 3H24 increase. Thus, the reaction proceeds toward equilibrium by forming N2 and H2 from NH3; that is, the reaction
as written in Equation 15.6 proceeds from right to left.
This approach can be formalized by defining a quantity called the reaction quotient. The reaction quotient, Q, is a number obtained by substituting reactant and
product concentrations or partial pressures at any point during a reaction into an equilibrium-constant expression. Therefore, for the general reaction
aA + bB ∆ dD + eE
section 15.6 Applications of Equilibrium Constants
647
the reaction quotient in terms of molar concentrations is
Qc =
3D4d3E4e
3A4a3B4b
[15.23]
(A related quantity Qp can be written for any reaction that involves gases by using partial pressures instead of concentrations.)
Although we use what looks like the equilibrium-constant expression to calculate
the reaction quotient, the concentrations we use may or may not be the equilibrium
concentrations. For example, when we substituted the starting concentrations into the
equilibrium-constant expression of Equation 15.22, we obtained Qc = 0.500 whereas
Kc = 0.105. The equilibrium constant has only one value at each temperature. The
reaction quotient, however, varies as the reaction proceeds.
Of what use is Q? One practical thing we can do with Q is tell whether our reaction
really is at equilibrium, which is an especially valuable option when a reaction is very
slow. We can take samples of our reaction mixture as the reaction proceeds, separate
the components, and measure their concentrations. Then we insert these numbers into
Equation 15.23 for our reaction. To determine whether we are at equilibrium, or in
which direction the reaction proceeds to achieve equilibrium, we compare the values of
Qc and Kc or Qp and Kp. Three possible situations arise:
• Q 6 K: The concentration of products is too small and that of reactants too large.
The reaction achieves equilibrium by forming more products; it proceeds from left
to right.
• Q = K: The reaction quotient equals the equilibrium constant only if the system is
at equilibrium.
• Q 7 K: The concentration of products is too large and that of reactants too small.
The reaction achieves equilibrium by forming more reactants; it proceeds from
right to left.
These relationships are summarized in ▶ Figure 15.8.
At equilibrium
Q< K
Q
K
Reaction proceeds
to form more
products
Q
Q=K
K
Equilibrium
Q
Q >K
K
Reaction proceeds
to form more
reactants
▲ Figure 15.8 Predicting the direction of
a reaction by comparing Q and K at a given
temperature.
Sample
Exercise 15.9 Predicting the Direction of Approach to Equilibrium
At 448 °C, the equilibrium constant Kc for the reaction
H21g2 + I21g2 ∆ 2 HI1g2
is 50.5. Predict in which direction the reaction proceeds to reach equilibrium if we start with
2.0 * 10-2 mol of HI, 1.0 * 10-2 mol of H2, and 3.0 * 10-2 mol of I2 in a 2.00-L container.
Solution
Analyze We are given a volume and initial molar amounts of the species in a reaction and asked to determine in which direction the reaction must proceed to achieve equilibrium.
Plan We can determine the starting concentration of each species
in the reaction mixture. We can then substitute the starting concentrations into the equilibrium-constant expression to calculate the
reaction quotient, Qc. Comparing the magnitudes of the equilibrium
constant, which is given, and the reaction quotient will tell us in
which direction the reaction will proceed.
Solve The initial concentrations are
3HI4 = 2.0 * 10-2 mol>2.00 L = 1.0 * 10-2 M
3H24 = 1.0 * 10-2 mol>2.00 L = 5.0 * 10-3 M
3I24 = 3.0 * 10-2 mol>2.00 L = 1.5 * 10-2 M
The reaction quotient is therefore
Qc =
3HI42
3H243I24
=
11.0 * 10-222
15.0 * 10-3211.5 * 10-22
= 1.3
Because Qc 6 Kc, the concentration of HI must increase and the concentrations of H2 and I2 must decrease to reach equilibrium; the reaction as written proceeds left to right to attain equilibrium.
Practice Exercise 1
Which of the following statements accurately describes what
would happen to the direction of the reaction described in the
sample exercise above, if the size of the container were different
from 2.00 L? (a) The reaction would proceed in the opposite direction (from right to left) if the container volume were reduced sufficiently. (b) The reaction would proceed in the opposite direction if
the container volume were expanded sufficiently. (c) The direction
of this reaction does not depend on the volume of the container.
Practice Exercise 2
At 1000 K, the value of Kp for the reaction 2 SO31g2 ∆
2 SO21g2 + O21g2 is 0.338. Calculate the value for Qp, and
predict the direction in which the reaction proceeds toward
equilibrium if the initial partial pressures are PSO3 = 0.16 atm;
PSO2 = 0.41 atm; PO2 = 2.5 atm.
648
chapter 15 Chemical Equilibrium
Calculating Equilibrium Concentrations
Chemists frequently need to calculate the amounts of reactants and products present at
equilibrium in a reaction for which they know the equilibrium constant. The approach
in solving problems of this type is similar to the one we used for evaluating equilibrium
constants: We tabulate initial concentrations or partial pressures, changes in those concentrations or pressures, and final equilibrium concentrations or partial pressures. Usually, we end up using the equilibrium-constant expression to derive an equation that
must be solved for an unknown quantity, as demonstrated in Sample Exercise 15.10.
Sample
Exercise 15.10 Calculating Equilibrium Concentrations
For the Haber process, N21g2 + 3 H21g2 ∆ 2 NH31g2, Kp = 1.45 * 10-5, at 500 °C. In an equilibrium mixture of the three gases at 500 °C, the partial pressure of H2 is 0.928 atm and that of N2 is
0.432 atm. What is the partial pressure of NH3 in this equilibrium mixture?
Solution
Analyze We are given an equilibrium constant, Kp, and the equilibrium partial pressures of two of the three substances in the equation
1N2 and H22, and we are asked to calculate the equilibrium partial
pressure for the third substance 1NH32.
Plan We can set Kp equal to the equilibrium-constant expression and
substitute in the partial pressures that we know. Then we can solve for
the only unknown in the equation.
Solve We tabulate the equilibrium pressures:
N21g2 + 3 H21g2 ∆ 2 NH31g2
Equilibrium pressure 1atm2 0.432
0.928
x
Because we do not know the equilibrium pressure of NH3, we represent
it with x. At equilibrium, the pressures must satisfy the equilibriumconstant expression:
Kp =
1PNH322
3
PN21PH22
=
x2
= 1.45 * 10-5
10.432210.92823
We now rearrange the equation to solve for x:
x2 = 11.45 * 10-5210.432210.92823 = 5.01 * 10-6
x = 25.01 * 10-6 = 2.24 * 10-3 atm = PNH3
Check We can always check our answer by using it to recalculate the
value of the equilibrium constant:
12.24 * 10-322
Kp =
= 1.45 * 10-5
10.432210.92823
Practice Exercise 1
At 500 K, the reaction 2 NO1g2 + Cl21g2 ∆ 2 NOCl1g2 has
Kp = 51. In an equilibrium mixture at 500 K, the partial pressure
of NO is 0.125 atm and Cl2 is 0.165 atm. What is the partial pressure of NOCl in the equilibrium mixture? (a) 0.13 atm, (b) 0.36
atm, (c) 1.0 atm, (d) 5.1 * 10-5 atm, (e) 0.125 atm.
Practice Exercise 2
At 500 K, the reaction PCl51g2 ∆ PCl31g2 + Cl21g2 has
Kp = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PCl5 is 0.860 atm and that of PCl3 is 0.350 atm. What is the
partial pressure of Cl2 in the equilibrium mixture?
In many situations, we know the value of the equilibrium constant and the initial amounts of all species. We must then solve for the equilibrium amounts. Solving
this type of problem usually entails treating the change in concentration as a variable.
The stoichiometry of the reaction gives us the relationship between the changes in the
amounts of all the reactants and products, as illustrated in Sample Exercise 15.11. The
calculations frequently involve the quadratic formula, as you will see in this exercise.
Sample
Exercise 15.11 Calculating Equilibrium Concentrations from Initial Concentrations
A 1.000-L flask is filled with 1.000 mol of H21g2 and 2.000 mol of I21g2 at 448 °C. The value of the
equilibrium constant Kc for the reaction
H21g2 + I21g2 ∆ 2 HI1g2
at 448 °C is 50.5. What are the equilibrium concentrations of H2, I2, and HI in moles per liter?
Solution
Analyze We are given the volume of a container, an equilibrium constant, and starting amounts of reactants in the container and are asked
to calculate the equilibrium concentrations of all species.
Plan In this case, we are not given any of the equilibrium concentra-
tions. We must develop some relationships that relate the initial concentrations to those at equilibrium. The procedure is similar in many
regards to that outlined in Sample Exercise 15.8, where we calculated
an equilibrium constant using initial concentrations.
649
section 15.6 Applications of Equilibrium Constants
Solve
(1) We note the initial concentrations of
H2 and I2:
(2) We construct a table in which we tabulate the
initial concentrations:
3H24 = 1.000 M and 3I24 = 2.000 M
Initial concentration (M)
H21g2
I21g2
∆
H21g2 + I21g2
∆
+
1.000
2.000
2 HI1g2
0
Change in concentration (M)
Equilibrium concentration (M)
(3) We use the stoichiometry of the reaction to
determine the changes in concentration that
occur as the reaction proceeds to equilibrium. The H2 and I2 concentrations will decrease as equilibrium is established and that
of HI will increase. Let’s represent the change
in concentration of H2 by x. The balanced
chemical equation tells us the relationship
between the changes in the concentrations of
the three gases. For each x mol of H2 that reacts, x mol of I2 are consumed and 2x mol of
HI are produced:
(4) We use initial concentrations and changes in
concentrations, as dictated by stoichiometry,
to express the equilibrium concentrations.
With all our entries, our table now looks like
this:
(5) We substitute the equilibrium concentrations
into the equilibrium-constant expression and
solve for x:
If you have an equation-solving calculator, you can
solve this equation directly for x. If not, expand this
expression to obtain a quadratic equation in x:
Initial concentration (M)
Change in concentration (M)
1.000
2.000
0
-x
-x
+ 2x
Equilibrium concentration (M)
H21g2 +
Initial concentration (M)
1.000
-x
Change in concentration (M)
1.000 - x
Equilibrium concentration (M)
Kc =
3HI42
3H243I24
=
12x22
11.000 - x212.000 - x2
I21g2
∆
0
-x
+ 2x
2.000 - x
2x
= 50.5
4x2 = 50.51x2 - 3.000x + 2.0002
When we substitute x = 2.323 into the expressions
for the equilibrium concentrations, we find negative concentrations of H2 and I2. Because a negative
concentration is not chemically meaningful, we reject
this solution. We then use x = 0.935 to find the
equilibrium concentrations:
Check We can check our solution by putting these
numbers into the equilibrium-constant expression
to assure that we correctly calculate the equilibrium
constant:
x =
- 1-151.52 { 21- 151.522 - 4146.521101.02
2146.52
3H24 = 1.000 - x = 0.065 M
3I24 = 2.000 - x = 1.065 M
3HI4 = 2x = 1.87 M
Kc =
3HI42
3H243I24
=
11.8722
10.065211.0652
= 51
Comment Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions to the equation will give you a value that leads to negative concentrations and thus is not
chemically meaningful. Reject this solution to the quadratic equation.
2 HI1g2
2.000
46.5x2 - 151.5x + 101.0 = 0
Solving the quadratic equation (Appendix A.3) leads
to two solutions for x:
2 HI1g2
= 2.323 or 0.935
650
chapter 15 Chemical Equilibrium
Practice Exercise 1
For the equilibrium Br21g2 + Cl21g2 ∆ 2 BrCl1g2, the equilibrium constant Kp is 7.0 at
400 K. If a cylinder is charged with BrCl(g) at an initial pressure of 1.00 atm and the system is
allowed to come to equilibrium what is the final (equilibrium) pressure of BrCl? (a) 0.57 atm,
(b) 0.22 atm, (c) 0.45 atm, (d) 0.15 atm, (e) 0.31 atm.
Practice Exercise 2
For the equilibrium PCl51g2 ∆ PCl31g2 + Cl21g2, the equilibrium constant Kp is 0.497 at
500 K. A gas cylinder at 500 K is charged with PCl51g2 at an initial pressure of 1.66 atm. What
are the equilibrium pressures of PCl5, PCl3, and Cl2 at this temperature?
15.7 | Le Châtelier’s Principle
Many of the products we use in everyday life are obtained from the chemical industry. Chemists and chemical engineers in industry spend a great deal of time and
effort to maximize the yield of valuable products and minimize waste. For example,
when Haber developed his process for making ammonia from N2 and H2, he examined how reaction conditions might be varied to increase yield. Using the values
of the equilibrium constant at various temperatures, he calculated the equilibrium
amounts of NH3 formed under a variety of conditions. Some of Haber’s results are
shown in ▼ Figure 15.9.
Notice that the percent of NH3 present at equilibrium decreases with increasing
temperature and increases with increasing pressure.
We can understand these effects in terms of a principle first put forward by
Henri-Louis Le Châtelier* (1850–1936), a French industrial chemist: If a system at
equilibrium is disturbed by a change in temperature, pressure, or a component concentration, the ­system will shift its equilibrium position so as to counteract the effect of the
disturbance.
Go Figure
At what combination of pressure and temperature should you run the reaction to maximize NH3 yield?
Percent NH3
produced
.7%
47
.8%
.9%
.0%
35
.9%
.9%
.2%
42
.9%
54
.9%
16
32
20
.8%
37
48
.8%
.8%
550
60
.6%
re
500
450
400
Te
m
200
To
tal 300
pr
ess 400
ur
e(
atm 500
)
)
.4%
12
26
(°C
27
38
%
.9%
atu
8.8
18
Percent of NH3
decreases with
increasing
temperature
pe
r
Percent of NH3
increases with
increasing
pressure
▲ Figure 15.9 Effect of temperature and pressure on NH3 yield in the Haber process. Each
mixture was produced by starting with a 3 : 1 molar mixture of H2 and N2.
*Pronounced “le-SHOT-lee-ay.”
section 15.7 Le Châtelier’s Principle
Le Châtelier’s Principle
If a system at equilibrium is disturbed by a change in concentration, pressure, or temperature,
the system will shift its equilibrium position so as to counter the effect of the disturbance.
Concentration: adding or removing a reactant or product
If a substance is added to a system at equilibrium, the system reacts to consume some of the substance. If a substance is removed
from a system, the system reacts to produce more of substance.
Initial equilibrium
+
Substance added
Equilibrium reestablished
+
+
Substances react
Pressure: changing the pressure by changing the volume
At constant temperature, reducing the volume of a gaseous equilibrium
mixture causes the system to shift in the direction that reduces the
number of moles of gas.
Pressure
Initial volume
System shifts to
direction of fewer
moles of gas
Temperature:
If the temperature of a system at equilibrium is increased, the system reacts as if we added a reactant to an endothermic reaction
or a product to an exothermic reaction. The equilibrium shifts in the direction that consumes the “excess reactant,” namely heat.
Endothermic
Exothermic
Increasing T
Reaction shifts right
Increasing T
Reaction shifts left
Decreasing T
Reaction shifts left
Decreasing T
Reaction shifts right
In this section, we use Le Châtelier’s principle to make qualitative predictions
about how a system at equilibrium responds to various changes in external conditions.
We consider three ways in which a chemical equilibrium can be disturbed: (1) adding
or removing a reactant or product, (2) changing the pressure by changing the volume,
and (3) changing the temperature.
Change in Reactant or Product Concentration
A system at dynamic equilibrium is in a state of balance. When the concentrations
of species in the reaction are altered, the equilibrium shifts until a new state of balance is attained. What does shift mean? It means that reactant and product concentrations change over time to accommodate the new situation. Shift does not mean
that the equilibrium constant itself is altered; the equilibrium constant remains the
same. Le Châtelier’s principle states that the shift is in the direction that minimizes
or reduces the effect of the change. Therefore, if a chemical system is already at equilibrium and the concentration of any substance in the mixture is increased (either
reactant or product), the system reacts to consume some of that substance. Conversely,
if the concentration of a substance is decreased, the system reacts to produce some of
that substance.
There is no change in the equilibrium constant when we change the concentrations of reactants or products. As an example, consider our familiar equilibrium mixture of N2, H2, and NH3:
N21g2 + 3 H21g2 ∆ 2 NH31g2
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chapter 15 Chemical Equilibrium
Go Figure
Why does the nitrogen concentration decrease after hydrogen is added?
N2(g) + 3 H2(g)
Initial
equilibrium
2 NH3(g)
Equilibrium
reestablished
H2 added
Partial pressure
H2
NH3
N2
Time
▲ Figure 15.10 Effect of adding H2 to an equilibrium mixture of N2, H2, and NH3. Adding H2
causes the reaction as written to shift to the right, consuming some N2 to produce more NH3.
Adding H2 causes the system to shift so as to reduce the increased concentration of H2
(▲ Figure 15.10). This change can occur only if the reaction consumes H2 and simultaneously consumes N2 to form more NH3. Adding N2 to the equilibrium mixture likewise causes the reaction to shift toward forming more NH3. Removing NH3 also causes
a shift toward producing more NH3, whereas adding NH3 to the system at equilibrium
causes the reaction to shift in the direction that reduces the increased NH3 concentration: Some of the added ammonia decomposes to form N2 and H2. All of these “shifts”
are entirely consistent with predictions that we would make by comparing the reaction
quotient Q with the equilibrium constant K.
In the Haber reaction, therefore, removing NH3 from an equilibrium mixture of
N2, H2, and NH3 causes the reaction to shift right to form more NH3. If the NH3 can
be removed continuously as it is produced, the yield can be increased dramatically. In
the industrial production of ammonia, the NH3 is continuously removed by selectively
liquefying it (▶ Figure 15.11). (The boiling point of NH3, -33 °C, is much higher than
those of N2, -196 °C, and H2, -253 °C.) The liquid NH3 is removed, and the N2 and H2
are recycled to form more NH3. As a result of the product being continuously removed,
the reaction is driven essentially to completion.
Give It Some Thought
Does the equilibrium 2 NO1g2 + O21g2 ∆ 2 NO21g2 shift to the right (more
products) or left (more reactants) if
(a) O2 is added to the system?
(b) NO is removed?
Effects of Volume and Pressure Changes
If a system containing one or more gases is at equilibrium and its volume is decreased,
thereby increasing its total pressure, Le Châtelier’s principle indicates that the system
responds by shifting its equilibrium position to reduce the pressure. A system can reduce its pressure by reducing the total number of gas molecules (fewer molecules of gas
section 15.7 Le Châtelier’s Principle
Hot gases
N2
H2
1
Incoming N2 and
H2 gases
Heat
exchanger
Heat
exchanger
5
Unreacted N2 and
H2 recycled
3
4
As gas mixture cools,
NH3(g) liquifies
2
Heated gases pass over
catalyst, NH3 forms
N2 and H2 gases
heated to
approximately
500 °C
Liquid NH3
outlet
▲ Figure 15.11 Diagram of the industrial production of ammonia. Incoming N21g2 and H21g2 are
heated to approximately 500 °C and passed over a catalyst. When the resultant N2, H2, and NH3
mixture is cooled, the NH3 liquefies and is removed from the mixture, shifting the reaction to
produce more NH3.
exert a lower pressure). Thus, at constant temperature, reducing the volume of a gaseous
equilibrium mixture causes the system to shift in the direction that reduces the number of
moles of gas. Increasing the volume causes a shift in the direction that produces more
gas molecules (▼ Figure 15.12).
2 A(g)
B(g)
Decrease volume,
increase pressure
New equilibrium favors
products to reduce total
moles of gas
Increase volume,
decrease pressure
Initial volume
New equilibrium favors
reactants to increase total
moles of gas
▲ Figure 15.12 Pressure and Le Châtelier’s principle.
653
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chapter 15 Chemical Equilibrium
Give It Some Thought
What happens to the equilibrium 2 SO21g2 + O21g2 ∆ 2 SO31g2, if the
volume of the system is increased?
In the reaction N21g2 + 3 H21g2 ∆ 2 NH31g2, four molecules of reactant are
consumed for every two molecules of product produced. Consequently, an increase in
pressure (caused by a decrease in volume) shifts the reaction in the direction that produces fewer gas molecules, which leads to the formation of more NH3, as indicated in
Figure 15.9. In the reaction H21g2 + I21g2 ∆ 2 HI1g2, the number of molecules of
gaseous products (two) equals the number of molecules of gaseous reactants; therefore,
changing the pressure does not influence the position of equilibrium.
Keep in mind that, as long as temperature remains constant, pressure–volume
changes do not change the value of K. Rather, these changes alter the partial pressures
of the gaseous substances. In Sample Exercise 15.7, we calculated Kp = 2.79 * 10-5 for
the Haber reaction, N21g2 + 3 H21g2 ∆ 2 NH31g2, in an equilibrium mixture at
472 °C containing 7.38 atm H2, 2.46 atm N2, and 0.166 atm NH3. Consider what happens when we suddenly reduce the volume of the system by one-half. If there were no
shift in equilibrium, this volume change would cause the partial pressures of all substances to double, giving PH2 = 14.76 atm, PN2 = 4.92 atm, and PNH3 = 0.332 atm.
The reaction quotient would then no longer equal the equilibrium constant:
Qp =
1PNH322
3
PN21PH22
=
10.33222
14.922114.7623
= 6.97 * 10-6 ≠ Kp
Because Qp 6 Kp, the system would no longer be at equilibrium. Equilibrium would be reestablished by increasing PNH3 and/or decreasing PN2 and PH2 until Qp = Kp = 2.79 * 10-5.
Therefore, the equilibrium shifts to the right in the reaction as written, as Le Châtelier’s principle predicts.
It is possible to change the pressure of a system in which a chemical reaction is
running without changing its volume. For example, pressure increases if additional
amounts of any reacting components are added to the system. We have already seen
how to deal with a change in concentration of a reactant or product. However, the
total pressure in the reaction vessel might also be increased by adding a gas that is not
involved in the equilibrium. For example, argon might be added to the ammonia equilibrium system. The argon would not alter the partial pressures of any of the reacting
components and therefore would not cause a shift in equilibrium.
Effect of Temperature Changes
Changes in concentrations or partial pressures shift equilibria without changing the
value of the equilibrium constant. In contrast, almost every equilibrium constant
changes as the temperature changes. For example, consider the equilibrium established
when cobalt(II) chloride 1CoCl22 is dissolved in hydrochloric acid, HCl(aq), in the
endothermic reaction
Co1H2O262+ 1aq2 + 4 Cl-1aq2 ∆ CoCl42- 1aq2 + 6 H2O1l2 ∆H 7 0[15.24]
Pale pink
Deep blue
Because Co1H2O262+ is pink and CoCl42- is blue, the position of this equilibrium is readily apparent from the color of the solution (▶ Figure 15.13). When the
solution is heated it turns blue, indicating that the equilibrium has shifted to form
more CoCl42- . Cooling the solution leads to a pink solution, indicating that the equilibrium has shifted to produce more Co1H2O262+ . We can monitor this reaction by
spectroscopic methods, measuring the concentration of all species at the different
(Section 14.2) We can then calculate the equilibrium constant at
temperatures.
each temperature. How do we explain why the equilibrium constants and therefore the
position of equilibrium both depend on temperature?
section 15.7 Le Châtelier’s Principle
∆H > 0, endothermic reaction
Heat + Co(H2O)62+(aq) + 4 Cl−(aq)
Pink
CoCl42−(aq) + 6 H2O(l)
Blue
CoCl42+
Co(H2O)62–
Cool
Solution appears pink because
lowering the temperature shifts the
equilibrium to favor formation of
the pink Co(H2O)62+ ion.
Heat
Solution appears violet because
appreciable amounts of both pink
Co(H2O)62+ and blue CoCl42– are
present.
▲ Figure 15.13 Temperature and Le Châtelier’s principle. In the molecular level views, only
the CoCl42 - and Co1H2O262+ ions are shown for clarity.
We can deduce the rules for the relationship between K and temperature from
Le Châtelier’s principle. We do this by treating heat as a chemical reagent. In an endothermic (heat-absorbing) reaction, we consider heat a reactant, and in an exothermic
(heat-releasing) reaction, we consider heat a product:
Endothermic:
Exothermic:
Reactants + heat ∆ products
Reactants ∆ products + heat
When the temperature of a system at equilibrium is increased, the system reacts as if we
added a reactant to an endothermic reaction or a product to an exothermic reaction.
The equilibrium shifts in the direction that consumes the excess reactant (or product),
namely heat.
Give It Some Thought
Use Le Châtelier’s principle to explain why the equilibrium vapor pressure of a
liquid increases with increasing temperature.
In an endothermic reaction, such as Equation 15.24, heat is absorbed as reactants
are converted to products. Thus, increasing the temperature causes the equilibrium to
shift to the right, in the direction of making more products, and K increases. In an exothermic reaction, the opposite occurs: Heat is produced as reactants are converted to
Solution appears blue because
raising the temperature shifts the
equilibrium to favor formation of
the blue CoCl42– ion.
655
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chapter 15 Chemical Equilibrium
products. Thus, increasing the temperature in this case causes the equilibrium to shift
to the left, in the direction of making more reactants, and K decreases.
Endothermic:
Increasing T results in higher K value
Exothermic:
Increasing T results in lower K value
Cooling a reaction has the opposite effect. As we lower the temperature, the equilibrium shifts in the direction that produces heat. Thus, cooling an endothermic reaction shifts the equilibrium to the left, decreasing K, as shown in Figure 15.13, and
cooling an exothermic reaction shifts the equilibrium to the right, increasing K.
Sample
Exercise 15.12 Using Le Châtelier’s Principle to Predict Shifts in Equilibrium
Consider the equilibrium
N2O41g2 ∆ 2 NO21g2
∆H ° = 58.0 kJ
In which direction will the equilibrium shift when (a) N2O4
is added, (b) NO2 is removed, (c) the pressure is increased by addition of N21g2, (d) the volume is increased, (e) the temperature is
decreased?
Solution
Analyze We are given a series of changes to be made to a system at
equilibrium and are asked to predict what effect each change will have
on the position of the equilibrium.
(e) The reaction is endothermic, so we can imagine heat as a reagent
on the reactant side of the equation. Decreasing the temperature
will shift the equilibrium in the direction that produces heat, so
the equilibrium shifts to the left, toward the formation of more
N2O4. Note that only this last change also affects the value of the
equilibrium constant, K.
Plan Le Châtelier’s principle can be used to determine the effects of
Practice Exercise 1
For the reaction
Solve
4 NH31g2 + 5 O21g2 ∆ 4 NO1g2 + 6 H2O1g2
each of these changes.
(a) The system will adjust to decrease the concentration of the added
N2O4, so the equilibrium shifts to the right, in the direction of
product.
(b) The system will adjust to the removal of NO2 by shifting to the
side that produces more NO2; thus, the equilibrium shifts to the
right.
(c) Adding N2 will increase the total pressure of the system, but N2
is not involved in the reaction. The partial pressures of NO2 and
N2O4 are therefore unchanged, and there is no shift in the position of the equilibrium.
(d) If the volume is increased, the system will shift in the direction
that occupies a larger volume (more gas molecules); thus, the
equilibrium shifts to the right.
∆H ° = - 904 kJ
which of the following changes will shift the equilibrium to the
right, toward the formation of more products? (a) Adding more
water vapor, (b) Increasing the temperature, (c) Increasing the volume of the reaction vessel, (d) Removing O2(g), (e) Adding 1 atm
of Ne(g) to the reaction vessel.
Practice Exercise 2
For the reaction
PCl51g2 ∆ PCl31g2 + Cl21g2
∆H° = 87.9 kJ
in which direction will the equilibrium shift when (a) Cl21g2 is
removed, (b) the temperature is decreased, (c) the volume of the
reaction system is increased, (d) PCl31g2 is added?
Sample
Exercise 15.13 Predicting the Effect of Temperature on K
(a) Using the standard heat of formation data in Appendix C, determine the standard enthalpy
change for the reaction
N21g2 + 3 H21g2 ∆ 2 NH31g2
(b) Determine how the equilibrium constant for this reaction should change with temperature.
Solution
Analyze We are asked to determine the standard enthalpy change of a
reaction and how the equilibrium constant for the reaction varies with
temperature.
Plan
(a) We can use standard enthalpies of formation to calculate ∆H° for
the reaction.
(b) We can then use Le Châtelier’s principle to determine what effect
temperature will have on the equilibrium constant.
Solve
(a) Recall that the standard enthalpy change for a reaction
is given by the sum of the standard molar enthalpies
of formation of the products, each multiplied by its
coefficient in the balanced chemical equation, minus the
same quantities for the reactants.
(Section 5.7) At
25 °C, ∆Hf° for NH31g2 is -46.19 kJ>mol. The ∆Hf° values
for H21g2 and N21g2 are zero by definition because the
enthalpies of formation of the elements in their normal
section 15.7 Le Châtelier’s Principle
657
with changes in temperature and that it is larger at lower
temperatures.
states at 25 °C are defined as zero.
(Section 5.7) Because
2 mol of NH3 is formed, the total enthalpy change is
Comment The fact that Kp for the formation of NH3 from N2 and H2
12 mol21-46.19 kJ>mol2 - 0 = - 92.38 kJ
(b) Because the reaction in the forward direction is exothermic,
we can consider heat a product of the reaction. An increase in
temperature causes the reaction to shift in the direction of less
NH3 and more N2 and H2. This effect is seen in the values for Kp
presented in ▼ Table 15.2. Notice that Kp changes markedly
decreases with increasing temperature is a matter of great practical
importance. To form NH3 at a reasonable rate requires higher temperatures. At higher temperatures, however, the equilibrium constant
is smaller, and so the percentage conversion to NH3 is smaller. To
compensate for this, higher pressures are needed because high pressure favors NH3 formation.
Table 15.2 Variation in K p with Temperature
for n2 + 3h 2 ∆ 2 nh 3
Temperature 1 °C 2
Kp
300
4.34 * 10-3
400
1.64 * 10-4
450
4.51 * 10
-5
500
1.45 * 10-5
550
5.38 * 10-6
600
2.25 * 10-6
Practice Exercise 1
The standard enthalpy of formation of HCl(g) is - 92.3 kJ>mol.
Given only this information, in which direction would you expect
the equilibrium for the reaction H21g2 + Cl21g2 ∆ 2 HCl1g2
to shift as the temperature increases: (a) to the left, (b) to the right,
(c) no shift in equilibrium?
Practice Exercise 2
Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction
2 POCl31g2 ∆ 2 PCl31g2 + O21g2
Use this result to determine how the equilibrium constant for the
reaction should change with temperature.
The Effect of Catalysts
What happens if we add a catalyst to a chemical system that is at equilibrium? As shown
in ▼ Figure 15.14, a catalyst lowers the activation barrier between reactants and products. The activation energies for both the forward and reverse reactions are lowered. The
catalyst thereby increases the rates of both forward and reverse reactions. Since K is the
ratio of the forward and reverse rate constants for a reaction, you can predict, correctly,
that the presence of a catalyst, even though it changes the reaction rate, does not affect
the numeric value of K (Figure 15.14). As a result, a catalyst increases the rate at which
equilibrium is achieved but does not change the composition of the equilibrium mixture.
The rate at which a reaction approaches equilibrium is an important practical consideration. As an example, let’s again consider the synthesis of ammonia from
Go Figure
What quantity dictates the speed of a reaction: (a) the energy difference between the initial state and the transition state or
(b) the energy difference between the initial state and the final state?
Catalyzed reaction
proceeds more rapidly
Transition states
Catalyzed reaction has
lower activation energy
[B]
Energy
[B]eq
A
The two reactions reach the
same equilibrium mixture,
but the catalyzed reaction
achieves equilibrium faster
B
Reaction pathway
Time
▲ Figure 15.14 An energy profile for the reaction A ∆ B (left), and the change in
concentration of B as a function of time (right), with and without a catalyst. Green curves show
the reaction with a catalyst; black curves show the reaction without a catalyst.
658
chapter 15 Chemical Equilibrium
N2 and H2. In designing his process, Haber had to deal with a rapid decrease in the
equilibrium constant with increasing temperature (Table 15.2). At temperatures sufficiently high to give a satisfactory reaction rate, the amount of ammonia formed was
too small. The solution to this dilemma was to develop a catalyst that would produce a
reasonably rapid approach to equilibrium at a sufficiently low temperature, so that the
equilibrium constant remained reasonably large. The development of a suitable catalyst
thus became the focus of Haber’s research efforts.
After trying different substances to see which would be most effective, Carl Bosch
settled on iron mixed with metal oxides, and variants of this catalyst formulation are still
used today.
(Section 15.2, “The Haber Process”) These catalysts make it possible to
obtain a reasonably rapid approach to equilibrium at around 400 to 500 °C and 200 to
600 atm. The high pressures are needed to obtain a satisfactory equilibrium amount of
NH3. If a catalyst could be found that leads to sufficiently rapid reaction at temperatures
lower than 400 °C, it would be possible to obtain the same extent of equilibrium conversion at pressures much lower than 200 to 600 atm. This would result in great savings
in both the cost of the high-pressure equipment and the energy consumed in the production of ammonia. It is estimated that the Haber process consumes approximately 1%
of the energy generated in the world each year. Not surprisingly chemists and chemical
engineers are actively searching for improved catalysts for the Haber process. A breakthrough in this field would not only increase the supply of ammonia for fertilizers, it
would also reduce the global consumption of fossil fuels in a significant way.
Give It Some Thought
Can a catalyst be used to increase the amount of product produced for a
reaction that reaches equilibrium quickly without a catalyst?
Sample
Integrative Exercise Putting Concepts Together
At temperatures near 800 °C, steam passed over hot coke (a form of carbon obtained from coal)
reacts to form CO and H2:
C1s2 + H2O1g2 ∆ CO1g2 + H21g2
The mixture of gases that results is an important industrial fuel called water gas. (a) At 800 °C
the equilibrium constant for this reaction is Kp = 14.1. What are the equilibrium partial pressures of H2O, CO, and H2 in the equilibrium mixture at this temperature if we start with solid
carbon and 0.100 mol of H2O in a 1.00-L vessel? (b) What is the minimum amount of carbon
required to achieve equilibrium under these conditions? (c) What is the total pressure in the
vessel at equilibrium? (d) At 25 °C the value of Kp for this reaction is 1.7 * 10-21. Is the
reaction exothermic or endothermic? (e) To produce the maximum amount of CO and H2 at
equilibrium, should the pressure of the system be increased or decreased?
Solution
(a) To determine the equilibrium partial pressures, we use the ideal-gas equation, first determining the starting partial pressure of water.
PH2O =
nH2ORT
V
=
10.100 mol210.08206 L@atm>mol@K211073 K2
1.00 L
= 8.81 atm
We then construct a table of initial partial pressures and their changes as equilibrium is achieved:
C1s2
+
H2O1g2
8.81
0
H21g2
-x
+x
+x
8.81 - x
x
x
Initial partial pressure (atm)
Change in partial pressure (atm)
Equilibrium partial pressure (atm)
∆
CO1g2
+
0
There are no entries in the table under C(s) because the reactant, being a solid, does not appear
in the equilibrium-constant expression. Substituting the equilibrium partial pressures of the
other species into the equilibrium-constant expression for the reaction gives
Kp =
PCOPH2
PH2O
=
1x21x2
18.81 - x2
= 14.1
section 15.7 Le Châtelier’s Principle
659
Multiplying through by the denominator gives a quadratic equation in x:
x2 = 114.1218.81 - x2
x2 + 14.1x - 124.22 = 0
Solving this equation for x using the quadratic formula yields x = 6.14 atm. Hence,
the equilibrium partial pressures are PCO = x = 6.14 atm, PH2 = x = 6.14 atm, and
PH2O = 18.81 - x2 = 2.67 atm.
(b) Part (a) shows that x = 6.14 atm of H2O must react for the system to achieve equilibrium.
We can use the ideal-gas equation to convert this partial pressure into a mole amount.
n =
16.14 atm211.00 L2
PV
= 0.0697 mol
=
RT
10.08206 L@atm>mol@K211073 K2
Thus, 0.0697 mol of H2O and the same amount of C must react to achieve equilibrium. As a
result, there must be at least 0.0697 mol of C (0.836 g C) present among the reactants at the start
of the reaction.
(c) The total pressure in the vessel at equilibrium is simply the sum of the equilibrium partial
pressures:
Ptotal = PH2O + PCO + PH2 = 2.67 atm + 6.14 atm + 6.14 atm = 14.95 atm
(d) In discussing Le Châtelier’s principle, we saw that endothermic reactions exhibit an increase
in Kp with increasing temperature. Because the equilibrium constant for this reaction increases
as temperature increases, the reaction must be endothermic. From the enthalpies of formation
given in Appendix C, we can verify our prediction by calculating the enthalpy change for the
reaction,
∆H° = ∆Hf°1CO1g22 + ∆Hf°1H21g22 - ∆Hf°1C(s, graphite2 - ∆Hf°1H2O1g22 = +131.3 kJ
The positive sign for ∆H° indicates that the reaction is endothermic.
(e) According to Le Châtelier’s principle, a decrease in the pressure causes a gaseous equilibrium to shift toward the side of the equation with the greater number of moles of gas. In this
case, there are 2 mol of gas on the product side and only one on the reactant side. Therefore, the
pressure should be decreased to maximize the yield of the CO and H2.
Chemistry Put to Work
Controlling Nitric
Oxide Emissions
The formation of NO from N2 and O2,
1
2
N21g2 +
1
2
O21g2 ∆ NO1g2
Go Figure
∆H ° = 90.4 kJ[15.25]
provides an interesting example of the practical importance of the fact
that equilibrium constants and reaction rates change with temperature. By applying Le Châtelier’s principle to this endothermic reaction
and treating heat as a reactant, we deduce that an increase in temperature shifts the equilibrium in the direction of more NO. The equilibrium constant Kp for formation of 1 mol of NO from its elements at
300 K is only about 1 * 10-15 (▶ Figure 15.15). At 2400 K, however, the equilibrium constant is about 0.05, which is 1013 times larger
than the 300 K value.
Figure 15.15 helps explain why NO is a pollution problem. In the
cylinder of a modern high-compression automobile engine, the temperature during the fuel-burning part of the cycle is approximately
2400 K. Also, there is a fairly large excess of air in the cylinder. These
conditions favor the formation of NO. After combustion, however,
the gases cool quickly. As the temperature drops, the equilibrium in
Equation 15.25 shifts to the left (because the reactant heat is being
removed). However, the lower temperature also means that the
reaction rate decreases, so the NO formed at 2400 K is essentially
“trapped” in that form as the gas cools.
The gases exhausting from the cylinder are still quite hot, perhaps
1200 K. At this temperature, as shown in Figure 15.15, the equilibrium
constant for formation of NO is about 5 * 10-4, much smaller than the
Estimate the value of Kp at 1200 K, the exhaust gas
temperature.
1
2
1
1
N2(g) + 2 O2(g)
1 × 10−5
Kp
Exhaust gas
temperature
1 × 10−10
NO(g)
Cylinder temperature
during combustion
1 × 10−15
0
1000
2000
Temperature (K)
▲ Figure 15.15 Equilibrium and temperature. The equilibrium
constant increases with increasing temperature because the reaction
is endothermic. It is necessary to use a log scale for Kp because the
values vary over such a large range.
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chapter 15 Chemical Equilibrium
value at 2400 K. However, the rate of conversion of NO to N2 and O2 is
too slow to permit much loss of NO before the gases are cooled further.
As discussed in the “Chemistry Put to Work” box in Section 14.7,
one of the goals of automotive catalytic converters is to achieve rapid
conversion of NO to N2 and O2 at the temperature of the exhaust gas.
Some catalysts developed for this reaction are reasonably effective
under the grueling conditions in automotive exhaust systems. Nevertheless, scientists and engineers are continuously searching for new
materials that provide even more effective catalysis of the decomposition of nitrogen oxides.
Chapter Summary and Key Terms
The Concept of Equilibrium (Section 15.1) A chemical reaction can achieve a state in which the forward and reverse processes are
occurring at the same rate. This condition is called chemical equilibrium,
and it results in the formation of an equilibrium mixture of the reactants
and products of the reaction. The composition of an equilibrium mixture
does not change with time if temperature is held constant.
The Equilibrium Constant (Section 15.2) An equilib-
rium that is used throughout this chapter is the reaction N21g2 +
3 H21g2 ∆ 2 NH31g2. This reaction is the basis of the Haber
process for the production of ammonia. The relationship between
the concentrations of the reactants and products of a system at
equilibrium is given by the law of mass action. For an equilibrium
equation of the form a A + b B ∆ d D + e E, the equilibriumconstant expression is written as
Kc =
3D4d3E4e
3A4a3B4b
where Kc is a dimensionless constant called the equilibrium constant.
When the equilibrium system of interest consists of gases, it is often
convenient to express the concentrations of reactants and products in
terms of gas pressures:
Kp =
1PD2d1PE2e
1PA2a1PB2b
Kc and Kp are related by the expression Kp = Kc1RT2∆n. To do this
conversion properly, use R = 0.08206 L@atm>mol@K and temperature
in kelvins.
Understanding and Working with Equilibrium Constants (Section 15.3) The value of the equilibrium constant
changes with temperature. A large value of Kc indicates that the equilibrium mixture contains more products than reactants and therefore lies
toward the product side of the equation. A small value for the equilibrium constant means that the equilibrium mixture contains less products
than reactants and therefore lies toward the reactant side. The equilibrium-constant expression and the equilibrium constant of the reverse of
a reaction are the reciprocals of those of the forward reaction. If a reaction is the sum of two or more reactions, its equilibrium constant will
be the product of the equilibrium constants for the individual reactions.
Heterogeneous Equilibria (Section 15.4) Equilibria for
which all substances are in the same phase are called homogeneous
equilibria; in heterogeneous equilibria, two or more phases are present.
Because their activities are exactly 1 the concentrations of pure solids
and liquids are left out of the equilibrium-constant expression for a
heterogeneous equilibrium.
Calculating Equilibrium Constants (Section 15.5) If
the concentrations of all species in an equilibrium are known, the
equilibrium-constant expression can be used to calculate the equilibrium constant. The changes in the concentrations of reactants and
products on the way to achieving equilibrium are governed by the
stoichiometry of the reaction.
Applications of Equilibrium Constants (Section
15.6) The reaction quotient, Q, is found by substituting reactant and
product concentrations or partial pressures at any point during a reaction into the equilibrium-constant expression. If the system is at equilibrium, Q = K. If Q ≠ K, however, the system is not at equilibrium.
When Q 6 K, the reaction will move toward equilibrium by forming more products (the reaction proceeds from left to right); when
Q 7 K, the reaction will move toward equilibrium by forming more
reactants (the reaction proceeds from right to left). Knowing the value
of K makes it possible to calculate the equilibrium amounts of reactants and products, often by the solution of an equation in which the
unknown is the change in a partial pressure or concentration.
Le Châtelier’s Principle (Section 15.7) Le Châtelier’s prin-
ciple states that if a system at equilibrium is disturbed, the equilibrium
will shift to minimize the disturbing influence. Therefore, if a reactant or product is added to a system at equilibrium, the equilibrium
will shift to consume the added substance. The effects of removing
reactants or products and of changing the pressure or volume of a reaction can be similarly deduced. For example, if the volume of the system is
reduced, the equilibrium will shift in the direction that decreases the number of gas molecules. While changes in concentration or pressure lead to
shifts in the equilibrium concentrations they do not change the value of
the equilibrium constant, K.
Changes in temperature affect both the equilibrium concentrations and the equilibrium constant. We can use the enthalpy change for
a reaction to determine how an increase in temperature affects the equilibrium: For an endothermic reaction, an increase in temperature shifts
the equilibrium to the right; for an exothermic reaction, a temperature
increase shifts the equilibrium to the left. Catalysts affect the speed at
which equilibrium is reached but do not affect the magnitude of K.
Learning Outcomes After studying this chapter, you should be able to:
• Explain what is meant by chemical equilibrium and how it relates
to reaction rates. (Section 15.1)
• Write the equilibrium-constant expression for any reaction.
(Section 15.2)
• Given the value of Kc convert to Kp and vice versa. (Section 15.2)
• Relate the magnitude of an equilibrium constant to the relative
amounts of reactants and products present in an equilibrium
mixture. (Section 15.3)
• Manipulate the equilibrium constant to reflect changes in the
chemical equation. (Section 15.3)
Exercises
661
• Write the equilibrium-constant expression for a heterogeneous re-
• Calculate equilibrium concentrations given the equilibrium con-
• Calculate an equilibrium constant from concentration measure-
• Calculate equilibrium concentrations, given the equilibrium con-
• Predict the direction of a reaction given the equilibrium constant
• Use Le Châtelier’s principle to predict how changing the concentra-
action. (Section 15.4)
ments. (Section 15.5)
and the concentrations of reactants and products. (Section 15.6)
stant and all but one equilibrium concentration. (Section 15.6)
stant and the starting concentrations. (Section 15.6)
tions, volume, or temperature of a system at equilibrium affects
the equilibrium position. (Section 15.7)
Key Equations
3D4d3E4e
• Kc =
• Kp =
• Kp = Kc1RT2∆n
[15.14]Relating the equilibrium constant based on pressures to the equilibrium constant based on concentration
• Qc =
[15.23]The reaction quotient. The concentrations are for any time during
a reaction. If the concentrations are equilibrium concentrations,
then Qc = Kc.
3A4a3B4b
1PD2d1PE2e
1PA2a1PB2b
3D4d3E4e
3A4a3B4b
[15.8]The equilibrium-constant expression for a general reaction of the
type a A + b B ∆ d D + e E, the concentrations are equilibrium concentrations only
[15.11]The equilibrium-constant expression in terms of equilibrium partial pressures
Exercises
Visualizing Concepts
Potential energy
15.1 (a) Based on the following energy profile, predict whether
kf 7 kr or kf 6 kr. (b) Using Equation 15.5, predict whether
the equilibrium constant for the process is greater than 1 or
less than 1. [Section 15.1]
Reactants
Products
Reaction progress
15.2The following diagrams represent a hypothetical reaction
A ¡ B, with A represented by red spheres and B represented by blue spheres. The sequence from left to right
represents the system as time passes. Does the system reach
equilibrium? If so, in which diagram is the system in equilibrium? [Sections 15.1 and 15.2]
1
2
greater or smaller than 1 if the volume is 1 L and each atom/
molecule in the diagram represents 1 mol? [Section 15.2]
3
4
15.4The following diagram represents a reaction shown going to
completion. Each molecule in the diagram represents 0.1 mol
and the volume of the box is 1.0 L. (a) Letting A = red spheres
and B = blue spheres, write a balanced equation for the reaction. (b) Write the equilibrium-constant expression for the reaction. (c) Calculate the value of Kc. (d) Assuming that all of
the molecules are in the gas phase, calculate ∆n, the change
in the number of gas molecules that accompanies the reaction.
(e) Calculate the value of Kp. [Section 15.2]
5
15.3The following diagram represents an equilibrium mixture
produced for a reaction of the type A + X ∆ AX. Is K
15.5Snapshots of two hypothetical reactions, A1g2 + B1g2 ∆
AB1g2 and X1g2 + Y1g2 ∆ XY1g2 at five different times
662
chapter 15 Chemical Equilibrium
are shown here. Which reaction has a larger equilibrium constant? [Sections 15.1 and 15.2]
t = 10 s
20 s
30 s
A(g) + B(g)
t = 10 s
20 s
30 s
X(g) + Y(g)
40 s
50 s
AB(g)
40 s
50 s
XY(g)
15.6Ethene 1C2H42 reacts with halogens 1X22 by the following
reaction:
C2H41g2 + X21g2 ∆ C2H4X21g2
The following figures represent the concentrations at equilibrium at the same temperature when X2 is Cl2 (green), Br2
(brown), and I2 (purple). List the equilibria from smallest to
largest equilibrium constant. [Section 15.3]
5.0 g PbO2(g)
in both vessels
Vessel A
V = 50 mL
15.8The reaction A2 + B2 ∆ 2 AB has an equilibrium constant Kc = 1.5. The following diagrams represent reaction
mixtures containing A2 molecules (red), B2 molecules (blue),
and AB molecules. (a) Which reaction mixture is at equilibrium? (b) For those mixtures that are not at equilibrium, how
will the reaction proceed to reach equilibrium? [Sections 15.5
and 15.6]
(i)
(a)
(b)
Vessel B
V = 100 mL
(ii)
(iii)
15.9The reaction A21g2 + B1g2 ∆ A1g2 + AB1g2 has an
equilibrium constant of Kp = 2. The accompanying diagram
shows a mixture containing A atoms (red), A2 molecules, and
AB molecules (red and blue). How many B atoms should be
added to the diagram to illustrate an equilibrium mixture?
[Section 15.6]
(c)
15.7When lead (IV) oxide is heated above 300 °C it decomposes according to the following reaction PbO21s2 ∆
PbO1s2 + O21g2. Consider the two sealed vessels of PbO2
shown here. If both vessels are heated to 400 °C and allowed to
come to equilibitum which of the following statements is true?
(a) There will be less PbO2 remaining in vessel A, (b) There
will be less PbO2 remaining in vessel B, (c) The amount of
PbO2 remaining in each vessel will be the same. [Section 15.4]
15.10The diagram shown here represents the equilibrium state for
the reaction A21g2 + 2 B1g2 ∆ 2AB1g2. (a) Assuming
the volume is 2 L, calculate the equilibrium constant Kc for the
reaction. (b) If the volume of the equilibrium mixture is
decreased, will the number of AB molecules increase or decrease? [Sections 15.5 and 15.7]
Exercises
663
15.15Write the expression for Kc for the following reactions. In
each case indicate whether the reaction is homogeneous or
heterogeneous.
(a) 3 NO1g2 ∆ N2O1g2 + NO21g2
(b) CH41g2 + 2 H2S1g2 ∆ CS21g2 + 4 H21g2
(c) Ni1CO241g2 ∆ Ni1s2 + 4 CO1g2
(d) HF1aq2 ∆ H+1aq2 + F-1aq2
(e) 2 Ag1s2 + Zn2+1aq2 ∆ 2 Ag +1aq2 + Zn1s2
(f) H2O1l2 ∆ H+1aq2 + OH-1aq2
(g) 2 H2O1l2 ∆ 2 H+1aq2 + 2 OH-1aq2
15.11The following diagrams represent equilibrium mixtures for
the reaction A2 + B ∆ A + AB at 300 K and 500 K. The
A atoms are red, and the B atoms are blue. Is the reaction exothermic or endothermic? [Section 15.7]
15.16Write the expressions for Kc for the following reactions. In
each case indicate whether the reaction is homogeneous or
heterogeneous.
(a) 2 O31g2 ∆ 3 O21g2
(b) Ti1s2 + 2 Cl21g2 ∆ TiCl41l2
(c) 2 C2H41g2 + 2 H2O1g2 ∆ 2 C2H61g2 + O21g2
(d) C1s2 + 2 H21g2 ∆ CH41g2
(e) 4 HCl1aq2 + O21g2 ∆ 2 H2O1l2 + 2 Cl21g2
(f) 2 C8H181l2 + 25 O21g2 ∆ 16 CO21g2 + 18 H2O1g2
(g) 2 C8H181l2 + 25 O21g2 ∆ 16 CO21g2 + 18 H2O1l2
15.17When the following reactions come to equilibrium, does
the equilibrium mixture contain mostly reactants or mostly
products?
300 K
500 K
[AB]
15.12The following graph represents the yield of the compound AB
at equilibrium in the reaction A1g2 + B1g2 ¡ AB1g2 at
two different pressures, x and y, as a function of temperature.
P=y
P=x
Temperature
(a) Is this reaction exothermic or endothermic? (b) Is P = x
greater or smaller than P = y? [Section 15.7]
Equilibrium; The Equilibrium Constant (Sections
15.1–15.4)
15.13Suppose that the gas-phase reactions A ¡ B and
B ¡ A are both elementary processes with rate constants
of 4.7 * 10-3 s-1 and 5.8 * 10-1 s-1, respectively. (a) What
is the value of the equilibrium constant for the equilibrium
A1g2 ∆ B1g2? (b) Which is greater at equilibrium, the
partial pressure of A or the partial pressure of B?
15.14Consider the reaction A + B ∆ C + D. Assume that
both the forward reaction and the reverse reaction are elementary processes and that the value of the equilibrium
constant is very large. (a) Which species predominate at equilibrium, reactants or products? (b) Which reaction has the
larger rate constant, the forward or the reverse?
(a) N21g2 + O21g2 ∆ 2 NO1g2; Kc = 1.5 * 10-10
(b) 2 SO21g2 + O21g2 ∆ 2 SO31g2; Kp = 2.5 * 109
15.18Which of the following reactions lies to the right, favoring the
formation of products, and which lies to the left, favoring formation of reactants?
(a) 2 NO1g2 + O21g2 ∆ 2 NO21g2; Kp = 5.0 * 1012
(b) 2 HBr1g2 ∆ H21g2 + Br21g2; Kc = 5.8 * 10-18
15.19Which of the following statements are true and which are false?
(a) The equilibrium constant can never be a negative number. (b) In reactions that we draw with a single-headed arrow,
the equilibrium constant has a value that is very close to zero.
(c) As the value of the equilibrium constant increases the
speed at which a reaction reaches equilibrium increases.
15.20Which of the following statements are true and which are
false? (a) For the reaction 2 A1g2 + B1g2 ∆ A2B1g2 Kc
and Kp are numerically the same. (b) It is possible to distinguish Kc from Kp by comparing the units used to express the
equilibrium constant. (c) For the equilibrium in (a), the value
of Kc increases with increasing pressure.
15.21If Kc = 0.042 for PCl31g2 + Cl21g2 ∆ PCl51g2 at 500 K,
what is the value of Kp for this reaction at this temperature?
15.22Calculate Kc at 303 K for SO21g2 + Cl21g2 ∆ SO2Cl21g2
if Kp = 34.5 at this temperature.
15.23The equilibrium constant for the reaction
2 NO1g2 + Br21g2 ∆ 2 NOBr1g2
is Kc = 1.3 * 10-2 at 1000 K. (a) At this temperature does
the equilibrium favor NO and Br2, or does it favor NOBr?
(b) Calculate Kc for 2 NOBr1g2 ∆ 2 NO1g2 + Br21g2.
(c) Calculate Kc for NOBr1g2 ¡ NO1g2 + 12 Br21g2.
15.24Consider the following equilibrium:
2 H21g2 + S21g2 ∆ 2 H2S1g2 Kc = 1.08 * 107at 700 °C
664
chapter 15 Chemical Equilibrium
(a) Calculate Kp. (b) Does the equilibrium mixture contain
mostly H2 and S2 or mostly H2S? (c) Calculate the value of Kc
if you rewrote the equation H21g2 + 12 S21g2 ∆ H2S1g2.
15.25At 1000 K, Kp = 1.85 for the reaction
SO21g2 +
1
2
O21g2 ∆ SO31g2
(a) What is the value of Kp for the reaction SO31g2 ∆
SO21g2 + 12 O21g2? (b) What is the value of Kp for the reaction
2 SO21g2 + O21g2 ∆ 2 SO31g2? (c) What is the value of
Kc for the reaction in part (b)?
15.26Consider the following equilibrium, for which Kp = 0.0752
at 480 °C:
2 Cl21g2 + 2 H2O1g2 ∆ 4 HCl1g2 + O21g2
(a) What is the value of Kp for the reaction
4 HCl1g2 + O21g2 ∆ 2 Cl21g2 + 2 H2O1g2?
(b) What is the value of Kp for the reaction
Cl21g2 + H2O1g2 ∆ 2 HCl1g2 + 12 O21g2?
(c) What is the value of Kc for the reaction in part (b)?
15.27The following equilibria were attained at 823 K:
CoO1s2 + H21g2 ∆ Co1s2 + H2O1g2 Kc = 67
CoO1s2 + CO1g2 ∆ Co1s2 + CO21g2 Kc = 490
Based on these equilibria, calculate the equilibrium constant
for H21g2 + CO21g2 ∆ CO1g2 + H2O1g2 at 823 K.
15.28Consider the equilibrium
N21g2 + O21g2 + Br21g2 ∆ 2 NOBr1g2
Calculate the equilibrium constant Kp for this reaction, given
the following information (at 298 K):
2 NO1g2 + Br21g2 ∆ 2 NOBr1g2
Kc = 2.0
2 NO1g2 ∆ N21g2 + O21g2 Kc = 2.1 * 1030
15.29Mercury(I) oxide decomposes into elemental mercury and
e l e m e n t a l o x y g e n : 2 Hg2O1s2 ∆ 4 Hg1l2 + O21g2.
(a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this
reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibrium-constant expression
in terms of molarities for the reaction, using (solv) to indicate
solvation.
15.30Consider the equilibrium Na2O1s2 + SO21g2 ∆ Na2SO31s2.
(a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) All the compounds in
this reaction are soluble in water. Rewrite the equilibriumconstant expression in terms of molarities for the aqueous
reaction.
Calculating Equilibrium Constants
(Section 15.5)
15.31Methanol 1CH3OH2 is produced commercially by the
catalyzed reaction of carbon monoxide and hydrogen:
CO1g2 + 2 H21g2 ∆ CH3OH1g2. A n e q u i l i b r i u m
mixture in a 2.00-L vessel is found to contain 0.0406 mol
CH3OH, 0.170 mol CO, and 0.302 mol H2 at 500 K. Calculate
Kc at this temperature.
15.32Gaseous hydrogen iodide is placed in a closed container at
425 °C, where it partially decomposes to hydrogen and iodine: 2 HI1g2 ∆ H21g2 + I21g2. At equilibrium it is
found that 3HI4 = 3.53 * 10-3 M, 3H24 = 4.79 * 10-4 M,
and 3I24 = 4.79 * 10-4 M. What is the value of Kc at this
temperature?
15.33The equilibrium 2 NO1g2 + Cl21g2 ∆ 2 NOCl1g2 is established at 500 K. An equilibrium mixture of the three gases
has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for
NO, Cl2, and NOCl, respectively. (a) Calculate Kp for this reaction at 500.0 K. (b) If the vessel has a volume of 5.00 L, calculate Kc at this temperature.
15.34Phosphorus trichloride gas and chlorine gas react to form
phosphorus pentachloride gas: PCl31g2 + Cl21g2 ∆
PCl51g2. A 7.5-L gas vessel is charged with a mixture of
PCl31g2 and Cl21g2, which is allowed to equilibrate at 450
K. At equilibrium the partial pressures of the three gases are
PPCl3 = 0.124 atm, PCl2 = 0.157 atm, a n d PPCl5 = 1.30 atm.
(a) What is the value of Kp at this temperature? (b) Does the
equilibrium favor reactants or products? (c) Calculate Kc for
this reaction at 450 K.
15.35A mixture of 0.10 mol of NO, 0.050 mol of H2, and 0.10 mol of
H2O is placed in a 1.0-L vessel at 300 K. The following equilibrium is established:
2 NO1g2 + 2 H21g2 ∆ N21g2 + 2 H2O1g2
At equilibrium 3NO4 = 0.062 M. (a) Calculate the equilibrium concentrations of H2, N2, and H2O. (b) Calculate Kc.
15.36A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a
2.00-L vessel at 700 K. These substances react according to
H21g2 + Br21g2 ∆ 2 HBr1g2
At equilibrium, the vessel is found to contain 0.566 g of H2.
(a) Calculate the equilibrium concentrations of H2, Br2, and
HBr. (b) Calculate Kc.
15.37A mixture of 0.2000 mol of CO2, 0.1000 mol of H2, and 0.1600
mol of H2O is placed in a 2.000-L vessel. The following equilibrium is established at 500 K:
CO21g2 + H21g2 ∆ CO1g2 + H2O1g2
(a) Calculate the initial partial pressures of CO2, H2, and H2O.
(b) At equilibrium PH2O = 3.51 atm. Calculate the equilibrium partial pressures of CO2, H2, and CO. (c) Calculate Kp
for the reaction. (d) Calculate Kc for the reaction.
15.38A flask is charged with 1.500 atm of N2O41g2 and 1.00 atm
NO21g2 at 25 °C, and the following equilibrium is achieved:
N2O41g2 ∆ 2 NO21g2
After equilibrium is reached, the partial pressure of NO2 is
0.512 atm. (a) What is the equilibrium partial pressure of
N2O4? (b) Calculate the value of Kp for the reaction. (c) Calculate Kc for the reaction.
15.39Two different proteins X and Y are dissolved in aqueous solution at 37 °C. The proteins bind in a 1:1 ratio to form XY. A
solution that is initially 1.00 mM in each protein is allowed
to reach equilibrium. At equilibrium, 0.20 mM of free X and
0.20 mM of free Y remain. What is Kc for the reaction?
15.40A chemist at a pharmaceutical company is measuring equilibrium constants for reactions in which drug candidate
molecules bind to a protein involved in cancer. The drug
Exercises
molecules bind the protein in a 1:1 ratio to form a drug–
protein complex. The protein concentration in aqueous solution at 25 °C is 1.50 * 10-6M. Drug A is introduced into the
protein solution at an initial concentration of 2.00 * 10-6M.
Drug B is introduced into a separate, identical protein solution at an initial concentration of 2.00 * 10-6M. At equilibrium, the drug A-protein solution has an A-protein complex
concentration of 1.00 * 10-6M, and the drug B solution has a
B-protein complex concentration of 1.40 * 10-6M. Calculate
the Kc value for the A-protein binding reaction and for the Bprotein binding reaction. Assuming that the drug that binds
more strongly will be more effective, which drug is the better
choice for further research?
Applications of Equilibrium Constants
(Section 15.6)
15.47At 1285 °C, the equilibrium constant for the reaction
Br21g2 ∆ 2 Br1g2 is Kc = 1.04 * 10-3. A 0.200-L vessel
containing an equilibrium mixture of the gases has 0.245 g
Br21g2 in it. What is the mass of Br1g2 in the vessel?
15.48For the reaction H21g2 + I21g2 ∆ 2 HI1g2, Kc = 55.3 at
700 K. In a 2.00-L flask containing an equilibrium mixture of
the three gases, there are 0.056 g H2 and 4.36 g I2. What is the
mass of HI in the flask?
15.49At 800 K, the equilibrium constant for I21g2 ∆ 2 I1g2 is
Kc = 3.1 * 10-5. If an equilibrium mixture in a 10.0-L vessel
contains 2.67 * 10-2 g of I(g), how many grams of I2 are in
the mixture?
15.50For 2 SO21g2 + O21g2 ∆ 2 SO31g2, Kp = 3.0 * 104 at
700 K. In a 2.00-L vessel, the equilibrium mixture contains
1.17 g of SO3 and 0.105 g of O2. How many grams of SO2 are
in the vessel?
15.51At 2000 °C, the equilibrium constant for the reaction
15.41(a) If Qc 6 Kc, in which direction will a reaction proceed in
order to reach equilibrium? (b) What condition must be satisfied so that Qc = Kc?
15.42(a) If Qc 7 Kc, how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants
are present; no products have been formed. What is the value
of Qc at this point in the reaction?
15.43At 100 °C, the equilibrium constant for the reaction
COCl21g2 ∆ CO1g2 + Cl21g2 h a s t h e v a l u e Kc =
2.19 * 10-10. Are the following mixtures of COCl2, CO, and
Cl2 at 100 °C at equilibrium? If not, indicate the direction that
the reaction must proceed to achieve equilibrium.
(a) 3COCl24 = 2.00 * 10-3 M, 3CO4 = 3.3 * 10-6 M,
3Cl24 = 6.62 * 10-6 M
(b) 3COCl24 = 4.50 * 10-2 M, 3CO4 = 1.1 * 10-7 M,
3Cl24 = 2.25 * 10-6 M
-6
(c) 3COCl24 = 0.0100 M, 3CO4 = 3Cl24 = 1.48 * 10 M
15.44As shown in Table 15.2, Kp for the equilibrium
N21g2 + 3 H21g2 ∆ 2 NH31g2
is 4.51 * 10-5 at 450 °C. For each of the mixtures listed here,
indicate whether the mixture is at equilibrium at 450 °C. If
it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mixture must shift to
achieve equilibrium.
(a) 98 atm NH3, 45 atm N2, 55 atm H2
(b) 57 atm NH3, 143 atm N2, no H2
(c) 13 atm NH3, 27 atm N2, 82 atm H2
15.45At 100 °C, Kc = 0.078 for the reaction
SO2Cl21g2 ∆ SO21g2 + Cl21g2
In an equilibrium mixture of the three gases, the concentrations of SO2Cl2 and SO2 are 0.108 M and 0.052 M, respectively. What is the partial pressure of Cl2 in the equilibrium
mixture?
15.46At 900 K, the following reaction has Kp = 0.345:
2 SO21g2 + O21g2 ∆ 2 SO31g2
665
In an equilibrium mixture the partial pressures of SO2 and O2
are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial pressure of SO3 in the mixture?
2 NO1g2 ∆ N21g2 + O21g2
is Kc = 2.4 * 103. If the initial concentration of NO is
0.175 M, what are the equilibrium concentrations of NO,
N2, and O2?
15.52For the equilibrium
Br21g2 + Cl21g2 ∆ 2 BrCl1g2
at 400 K, Kc = 7.0. If 0.25 mol of Br2 and 0.55 mol of Cl2 are
introduced into a 3.0-L container at 400 K, what will be the
equilibrium concentrations of Br2, Cl2, and BrCl?
15.53At 373 K, Kp = 0.416 for the equilibrium
2 NOBr1g2 ∆ 2 NO1g2 + Br21g2
If the pressures of NOBr(g) and NO(g) are equal, what is the
equilibrium pressure of Br21g2?
15.54At 218 °C, Kc = 1.2 * 10-4 for the equilibrium
NH4SH1s2 ∆ NH31g2 + H2S1g2
Calculate the equilibrium concentrations of NH3 and H2S if a
sample of solid NH4SH is placed in a closed vessel at 218 °C
and decomposes until equilibrium is reached.
15.55Consider the reaction
CaSO41s2 ∆ Ca2+1aq2 + SO42-1aq2
At 25 °C, the equilibrium constant is Kc = 2.4 * 10-5 for
this reaction. (a) If excess CaSO41s2 is mixed with water at
25 °C to produce a saturated solution of CaSO4, what are
the equilibrium concentrations of Ca2+ and SO42-? (b) If the
resulting solution has a volume of 1.4 L, what is the minimum mass of CaSO41s2 needed to achieve equilibrium?
15.56At 80 °C, Kc = 1.87 * 10-3 for the reaction
PH3BCl31s2 ∆ PH31g2 + BCl31g2
(a) Calculate the equilibrium concentrations of PH3 and BCl3 if
a solid sample of PH3BCl3 is placed in a closed vessel at 80 °C
and decomposes until equilibrium is reached. (b) If the flask has
a volume of 0.250 L, what is the minimum mass of PH3BCl31s2
that must be added to the flask to achieve equilibrium?
15.57For the reaction I2 + Br21g2 ∆ 2 IBr1g2, Kc = 280 at
150 °C. Suppose that 0.500 mol IBr in a 2.00-L flask is allowed
to reach equilibrium at 150 °C. What are the equilibrium concentrations of IBr, I2, and Br2?
666
chapter 15 Chemical Equilibrium
15.58At 25 °C, the reaction
CaCrO41s2 ∆ Ca2+1aq2 + CrO42-1aq2
has an equilibrium constant Kc = 7.1 * 10-4. What are the
equilibrium concentrations of Ca2 + and CrO42- in a saturated solution of CaCrO4?
15.59Methane, CH4, reacts with I2 according to the reaction
CH41g2 + l21g2 ∆ CH3l1g2 + HI1g2. At 630 K, Kp for
this reaction is 2.26 * 10-4. A reaction was set up at 630 K
with initial partial pressures of methane of 105.1 torr and of
7.96 torr for I2. Calculate the pressures, in torr, of all reactants
and products at equilibrium.
15.60The reaction of an organic acid with an alcohol, in organic solvent, to produce an ester and water is commonly done in the
pharmaceutical industry. This reaction is catalyzed by strong
acid (usually H2SO4). A simple example is the reaction of acetic
acid with ethyl alcohol to produce ethyl acetate and water:
CH3COOH1solv2 + CH3CH2OH1solv2 ∆
CH3COOCH2CH31solv2 + H2O1solv2
where “(solv)” indicates that all reactants and products are
in solution but not an aqueous solution. The equilibrium
constant for this reaction at 55 °C is 6.68. A pharmaceutical
chemist makes up 15.0 L of a solution that is initially 0.275
M in acetic acid and 3.85 M in ethanol. At equilibrium, how
many grams of ethyl acetate are formed?
Le Châtelier’s Principle (Section 15.7)
15.61Consider the following equilibrium for which ∆H 6 0
2 SO21g2 + O21g2 ∆ 2 SO31g2
How will each of the following changes affect an equilibrium
mixture of the three gases: (a) O21g2 is added to the system;
(b) the reaction mixture is heated; (c) the volume of the reaction vessel is doubled; (d) a catalyst is added to the mixture;
(e) the total pressure of the system is increased by adding a
noble gas; (f) SO31g2 is removed from the system?
15.62Consider the reaction
4 NH31g2 + 5 O21g2 ∆
4 NO1g2 + 6 H2O1g2, ∆H = - 904.4 kJ
Does each of the following increase, decrease, or leave
unchanged the yield of NO at equilibrium? (a) increase 3NH34; (b) increase 3H2O4; (c) decrease 3O24;
(d) decrease the volume of the container in which the reaction occurs; (e) add a catalyst; (f) increase temperature.
15.63How do the following changes affect the value of the equilibrium constant for a gas-phase exothermic reaction: (a) removal of a reactant, (b) removal of a product, (c) decrease in
the ­volume, (d) decrease in the temperature, (e) addition of a
catalyst?
15.64For a certain gas-phase reaction, the fraction of products in
an equilibrium mixture is increased by either increasing the
temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does
the balanced chemical equation have more molecules on the
reactant side or product side?
15.65Consider the following equilibrium between oxides of
nitrogen
3 NO1g2 ∆ NO21g2 + N2O1g2
(a) Use data in Appendix C to calculate ∆H° for this
­reaction. (b) Will the equilibrium constant for the reaction
increase or decrease with increasing temperature? (c) At
constant temperature, would a change in the volume of the
container affect the fraction of products in the equilibrium
mixture?
15.66Methanol 1CH3OH2 can be made by the reaction of CO
with H2:
CO1g2 + 2 H21g2 ∆ CH3OH1g2
(a) Use thermochemical data in Appendix C to calculate ∆H°
for this reaction. (b) To maximize the equilibrium yield of
methanol, would you use a high or low temperature? (c) To
maximize the equilibrium yield of methanol, would you use a
high or low pressure?
15.67Ozone, O3, decomposes to molecular oxygen in the stratosphere according to the reaction 2 O31g2 ¡ 3 O21g2.
Would an increase in pressure favor the formation of ozone or
of oxygen?
15.68The water–gas shift reaction CO1g2 + H2O1g2 ∆
CO21g2 + H21g2 is used industrially to produce hydrogen.
The reaction enthalpy is ∆H° = - 41 kJ. (a) To increase
the equilibrium yield of hydrogen would you use high or
low temperature? (b) Could you increase the equilibrium
yield of hydrogen by controlling the pressure of this reaction? If so would high or low pressure favor formation of
H21g2?
Additional Exercises
15.69 Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps:
CO1g2 + Cl21g2 ∆ COCl1g2 + Cl1g2
flask and is found to contain 8.62 g of CO, 2.60 g of H2,
43.0 g of CH4, and 48.4 g of H2O. Assuming that equilibrium has been reached, calculate Kc and Kp for the reaction
CH41g2 + H2O1g2 ∆ CO1g2 + 3 H21g2.
At 25 °C, the rate constants for the forward and reverse reactions are 1.4 * 10-28 M -1 s-1 and 9.3 * 1010 M -1 s-1, respectively. (a) What is the value for the equilibrium constant at 25 °C?
(b) Are reactants or products more plentiful at equilibrium?
15.72When 2.00 mol of SO2Cl2 is placed in a 2.00-L flask at 303 K,
56% of the SO2Cl2 decomposes to SO2 and Cl2 :
15.70If Kc = 1 for the equilibrium 2 A1g2 ∆ B1g2, what is the
relationship between [A] and [B] at equilibrium?
(a) Calculate Kc for this reaction at this temperature.
(b) Calculate Kp for this reaction at 303 K. (c) According to
Le Châtelier’s principle, would the percent of SO2Cl2 that decomposes increase, decrease or stay the same if the mixture
15.71A mixture of CH4 and H2O is passed over a nickel catalyst at 1000 K. The emerging gas is collected in a 5.00-L
SO2Cl21g2 ∆ SO21g2 + Cl21g2
Additional Exercises
were transferred to a 15.00-L vessel? (d) Use the equilibrium
constant you calculated above to determine the percentage of
SO2Cl2 that decomposes when 2.00 mol of SO2Cl2 is placed in
a 15.00-L vessel at 303 K.
15.73 A mixture of H2, S, and H2S is held in a 1.0-L vessel at 90 °C
and reacts according to the equation:
H21g2 + S1s2 ∆ H2S1g2
At equilibrium, the mixture contains 0.46 g of H2S and 0.40
g H2. (a) Write the equilibrium-constant expression for this
reaction. (b) What is the value of Kc for the reaction at this
temperature?
15.74 A sample of nitrosyl bromide (NOBr) decomposes according
to the equation
2 NOBr1g2 ∆ 2 NO1g2 + Br21g2
An equilibrium mixture in a 5.00-L vessel at 100 °C contains
3.22 g of NOBr, 3.08 g of NO, and 4.19 g of Br2. (a) Calculate
Kc. (b) What is the total pressure exerted by the mixture of gases?
(c) What was the mass of the original sample of NOBr?
667
At 700 K, the equilibrium constant Kp for this reaction is 0.26.
Predict the behavior of each of the following mixtures at this
temperature and indicate whether or not the mixtures are at
equilibrium. If not, state whether the mixture will need to
produce more products or reactants to reach equilibrium.
(a) PNO = 0.15 atm, PCl2 = 0.31 atm, PNOCl = 0.11 atm
(b) PNO = 0.12 atm, PCl2 = 0.10 atm,
PNOCl = 0.050 atm
(c) PNO = 0.15 atm, PCl2 = 0.20 atm,
PNOCl = 5.10 * 10-3 atm
15.82At 900 °C, Kc = 0.0108 for the reaction
CaCO31s2 ∆ CaO1s2 + CO21g2
A mixture of CaCO3, CaO, and CO2 is placed in a 10.0-L vessel at 900 °C. For the following mixtures, will the amount of
CaCO3 increase, decrease, or remain the same as the system
approaches equilibrium?
(a) 15.0 g CaCO3, 15.0 g CaO, and 4.25 g CO2
(b) 2.50 g CaCO3, 25.0 g CaO, and 5.66 g CO2
(c) 30.5 g CaCO3, 25.5 g CaO, and 6.48 g CO2
15.75 Consider the hypothetical reaction A1g2 ∆ 2 B1g2. A
flask is charged with 0.75 atm of pure A, after which it is allowed to reach equilibrium at 0 °C. At equilibrium, the partial
pressure of A is 0.36 atm. (a) What is the total pressure in the
flask at equilibrium? (b) What is the value of Kp? (c) What
could we do to maximize the yield of B?
15.83When 1.50 mol CO2 and 1.50 mol H2 are placed in a
3.00-L container at 395 °C, the following reaction occurs:
CO21g2 + H21g2 ∆ CO1g2 + H2O1g2. If Kc = 0.802,
what are the concentrations of each substance in the equilibrium mixture?
15.76As shown in Table 15.2, the equilibrium constant for the reaction
N21g2 + 3 H21g2 ∆ 2 NH31g2 is Kp = 4.34 * 10-3 at
300 °C. Pure NH3 is placed in a 1.00-L flask and allowed to reach
equilibrium at this temperature. There are 1.05 g NH3 in the equilibrium mixture. (a) What are the masses of N2 and H2 in the
equilibrium mixture? (b) What was the initial mass of ammonia
placed in the vessel? (c) What is the total pressure in the vessel?
15.84The equilibrium constant Kc for C1s2 + CO21g2 ∆
2 CO1g2 is 1.9 at 1000 K and 0.133 at 298 K. (a) If excess C is
allowed to react with 25.0 g of CO2 in a 3.00-L vessel at 1000
K, how many grams of CO are produced? (b) How many
grams of C are consumed? (c) If a smaller vessel is used for
the reaction, will the yield of CO be greater or smaller? (d) Is
the reaction endothermic or exothermic?
15.77For the equilibrium
15.85NiO is to be reduced to nickel metal in an industrial process
by use of the reaction
2 IBr1g2 ∆ I21g2 + Br21g2
Kp = 8.5 * 10-3 at 150 °C. If 0.025 atm of IBr is placed in a
2.0-L container, what is the partial pressure of all substances
after equilibrium is reached?
15.78For the equilibrium
PH3BCl31s2 ∆ PH31g2 + BCl31g2
Kp = 0.052 at 60 °C. (a) Calculate Kc. (b) After 3.00 g of solid
PH3BCl3 is added to a closed 1.500-L vessel at 60 °C, the vessel
is charged with 0.0500 g of BCl31g2. What is the equilibrium
concentration of PH3?
[15.79] Solid NH4SH is introduced into an evacuated flask at 24 °C.
The following reaction takes place:
NH4SH1s2 ∆ NH31g2 + H2S1g2
At equilibrium, the total pressure (for NH3 and H2S taken together) is 0.614 atm. What is Kp for this equilibrium at 24 °C?
[15.80] A 0.831-g sample of SO3 is placed in a 1.00-L container and
heated to 1100 K. The SO3 decomposes to SO2 and O2:
2 SO31g2 ∆ 2 SO21g2 + O21g2
At equilibrium, the total pressure in the container is 1.300 atm.
Find the values of Kp and Kc for this reaction at 1100 K.
15.81 Nitric oxide (NO) reacts readily with chlorine gas as follows:
2 NO1g2 + Cl21g2 ∆ 2 NOCl1g2
NiO1s2 + CO1g2 ∆ Ni1s2 + CO21g2
At 1600 K, the equilibrium constant for the reaction is
Kp = 6.0 * 102. If a CO pressure of 150 torr is to be
­employed in the furnace and total pressure never exceeds
760 torr, will reduction occur?
15.86Le Châtelier noted that many industrial processes of his time
could be improved by an understanding of chemical equilibria. For example, the reaction of iron oxide with carbon monoxide was used to produce elemental iron and CO2 according
to the reaction
Fe2O31s2 + 3 CO1g2 ∆ 2 Fe1s2 + 3 CO21g2
Even in Le Châtelier’s time, it was noted that a great deal of
CO was wasted, expelled through the chimneys over the furnaces. Le Châtelier wrote, “Because this incomplete reaction
was thought to be due to an insufficiently prolonged contact
between carbon monoxide and the iron ore [oxide], the dimensions of the furnaces have been increased. In England, they
have been made as high as 30 m. But the proportion of carbon
monoxide escaping has not diminished, thus demonstrating, by
an experiment costing several hundred thousand francs, that
the reduction of iron oxide by carbon monoxide is a limited
reaction. Acquaintance with the laws of chemical equilibrium
would have permitted the same conclusion to be reached more
rapidly and far more economically.” What does this anecdote
tell us about the equilibrium constant for this reaction?
668
chapter 15 Chemical Equilibrium
[15.87] At 700 K, the equilibrium constant for the reaction
CCl41g2 ∆ C1s2 + 2 Cl21g2
is Kp = 0.76. A flask is charged with 2.00 atm of CCl4, which
then reaches equilibrium at 700 K. (a) What fraction of the
CCl4 is converted into C and Cl2? (b) What are the partial
pressures of CCl4 and Cl2 at equilibrium?
[15.88] The reaction PCl31g2 + Cl21g2 ∆ PCl51g2 has Kp =
0.0870 at 300 °C. A flask is charged with 0.50 atm PCl3, 0.50
atm Cl2, and 0.20 atm PCl5 at this temperature. (a) Use the reaction quotient to determine the direction the reaction must
proceed to reach equilibrium. (b) Calculate the equilibrium
partial pressures of the gases. (c) What effect will increasing
the volume of the system have on the mole fraction of Cl2
in the equilibrium mixture? (d) The reaction is exothermic.
What effect will increasing the temperature of the system
have on the mole fraction of Cl2 in the equilibrium mixture?
[15.89] An equilibrium mixture of H2, I2, and HI at 458 °C contains
0.112 mol H2, 0.112 mol I2, and 0.775 mol HI in a 5.00-L vessel.
What are the equilibrium partial pressures when equilibrium is
reestablished following the addition of 0.200 mol of HI?
[15.90] Consider the hypothetical reaction A1g2 + 2 B1g2 ∆
2 C1g2, for which Kc = 0.25 at a certain temperature. A 1.00-L
reaction vessel is loaded with 1.00 mol of compound C, which
is allowed to reach equilibrium. Let the variable x represent
the number of mol>L of compound A present at equilibrium.
(a) In terms of x, what are the equilibrium concentrations of
compounds B and C? (b) What limits must be placed on the value
of x so that all concentrations are positive? (c) By putting the
equilibrium concentrations (in terms of x) into the equilibriumconstant expression, derive an equation that can be solved for x.
(d) The equation from part (c) is a cubic equation (one that
has the form ax3 + bx2 + cx + d = 0). In general, cubic
equations cannot be solved in closed form. However, you can
estimate the solution by plotting the cubic equation in the allowed range of x that you specified in part (b). The point at
which the cubic equation crosses the x-axis is the solution.
(e) From the plot in part (d), estimate the equilibrium concentrations of A, B, and C. (Hint: You can check the accuracy of
your answer by substituting these concentrations into the equilibrium expression.)
15.91At 1200 K, the approximate temperature of automobile exhaust gases (Figure 15.15), Kp for the reaction
2 CO21g2 ∆ 2 CO1g2 + O21g2
is about 1 * 10-13. Assuming that the exhaust gas (total
pressure 1 atm) contains 0.2% CO, 12% CO2, and 3% O2 by
volume, is the system at equilibrium with respect to the CO2
reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst
that speeds up the CO2 reaction? Recall that at a fixed pressure and temperature, volume % = mol %.
15.92Suppose that you worked at the U.S. Patent Office and a patent
application came across your desk claiming that a newly developed catalyst was much superior to the Haber catalyst for ammonia synthesis because the catalyst led to much greater equilibrium
conversion of N2 and H2 into NH3 than the Haber catalyst under
the same conditions. What would be your response?
Integrative Exercises
15.93Consider the reaction IO4-1aq2 + 2 H2O1l2 ∆ H4IO6-1aq2;
Kc = 3.5 * 10-2. If you start with 25.0 mL of a 0.905 M solution of NaIO4, and then dilute it with water to 500.0 mL, what
is the concentration of H4IO6- at equilibrium?
[15.94] Silver chloride, AgCl(s), is an “insoluble” strong electrolyte. (a) Write the equation for the dissolution of AgCl(s) in
H2O1l2. (b) Write the expression for Kc for the reaction in
part (a). (c) Based on the thermochemical data in Appendix C
and Le Châtelier’s principle, predict whether the solubility of
AgCl in H2O increases or decreases with increasing temperature. (d) The equilibrium constant for the dissolution of AgCl
in water is 1.6 * 10-10 at 25 °C. In addition, Ag +1aq2 can react with Cl-1aq2 according to the reaction
[15.96] The phase diagram for SO2 is shown here. (a) What does this
diagram tell you about the enthalpy change in the reaction
SO21l2 ¡ SO21g2? (b) Calculate the equilibrium constant
for this reaction at 100 °C and at 0 °C. (c) Why is it not possible to calculate an equilibrium constant between the gas
and liquid phases in the supercritical region? (d) At which
of the three points marked in red does SO21g2 most closely
approach ideal-gas behavior? (e) At which of the three red
points does SO21g2 behave least ideally?
10
Ag +1aq2 + 2 Cl-1aq2 ∆ AgCl2-1aq2
Supercritical
region
2
Liquid
Critical point
where Kc = 1.8 * 10 at 25 °C. Although AgCl is “not soluble” in water, the complex AgCl2- is soluble. At 25 °C, is the
solubility of AgCl in a 0.100 M NaCl solution greater than the
solubility of AgCl in pure water, due to the formation of soluble AgCl2- ions? Or is the AgCl solubility in 0.100 M NaCl
less than in pure water because of a Le Châtelier-type argument? Justify your answer with calculations. (Hint: Any form
in which silver is in solution counts as “solubility.”)
[15.95] Consider the equilibrium A ∆ B in which both the forward and reverse reactions are elementary (single-step) reactions. Assume that the only effect of a catalyst on the reaction
is to lower the activation energies of the forward and reverse
reactions, as shown in Figure 15.14. Using the Arrhenius
equation (Section 14.5), prove that the equilibrium constant is
the same for the catalyzed reaction as for the uncatalyzed one.
Pressure (atm)
5
10
1
Gas
10−1
0
100
200
Temperature (°C)
300
Design an Experiment
[15.97] Write the equilibrium-constant expression for the equilibrium
C1s2 + CO21g2 ∆ 2 CO1g2
The table that follows shows the relative mole percentages
of CO21g2 and CO(g) at a total pressure of 1 atm for several
temperatures. Calculate the value of Kp at each temperature.
Is the reaction exothermic or endothermic?
850
Co2 1mol % 2
6.23
Co 1mol % 2
950
1.32
98.68
1050
0.37
99.63
1200
0.06
99.94
Temperature 1 °C 2
93.77
15.98In Section 11.5, we defined the vapor pressure of a liquid in
terms of an equilibrium. (a) Write the equation representing the
equilibrium between liquid water and water vapor and the corresponding expression for Kp. (b) By using data in Appendix B,
give the value of Kp for this reaction at 30 °C. (c) What is the
value of Kp for any liquid in equilibrium with its vapor at the
normal boiling point of the liquid?
669
15.99 Water molecules in the atmosphere can form hydrogenbonded dimers, 1H2O22. The presence of these dimers is
thought to be important in the nucleation of ice crystals in
the atmosphere and in the formation of acid rain. (a) Using
VSEPR theory, draw the structure of a water dimer, using
dashed lines to indicate intermolecular interactions. (b) What
kind of intermolecular forces are involved in water dimer formation? (c) The Kp for water dimer formation in the gas phase
is 0.050 at 300 K and 0.020 at 350 K. Is water dimer formation
endothermic or exothermic?
15.100 The protein hemoglobin (Hb) transports O2 in mammalian blood. Each Hb can bind 4 O2 molecules. The equilibrium constant for the O2 binding reaction is higher in fetal
hemoglobin than in adult hemoglobin. In discussing protein
oxygen-binding capacity, biochemists use a measure called
the P50 value, defined as the partial pressure of oxygen at
which 50% of the protein is saturated. ­Fetal hemoglobin has a
P50 value of 19 torr, and adult hemoglobin has a P50 value
of 26.8 torr. Use these data to ­e stimate how much larger
Kc is for the aqueous reaction 4 O21g2 + Hb1aq2 ¡
3Hb1O2241aq24.
Design an Experiment
The reaction between hydrogen and iodine to form hydrogen iodide was used to illustrate Beer’s law in Chapter 14 (Figure 14.5). The reaction can be monitored using visible-light spectroscopy because I2 has a violet color while H2 and HI are colorless. At 300 K,
the equilibrium constant for the reaction H21g2 + I21g2 ∆ 2 HI1g2 is Kc = 794.
To answer the following questions assume you have access to hydrogen, iodine, hydrogen iodide, a transparent reaction vessel, a visible-light spectrometer, and a means for
changing the temperature. (a) Which gas or gases concentration could you readily monitor with the spectrometer? (b) To use Beer’s law (Equation 14.5) you need to determine
the extinction coefficient, e, for the substance in question. How would you determine e?
(c) Describe an experiment for determining the equilibrium constant at 600 K. (d) Use the bond
enthalpies in Table 8.4 to estimate the enthalpy of this reaction. (e) Based on your answer to part
(d), would you expect Kc at 600 K to be larger or smaller than at 300 K?
16
Acid–Base Equilibria
The acids and bases that you have used so far in the laboratory are
probably solutions of relatively simple inorganic substances, such as
hydrochloric acid, sulfuric acid, sodium hydroxide, and the like.
But acids and bases are important even when we are not in the lab.
They are ubiquitous, including in the foods we eat. The characteristic flavor of the
grapes shown in the opening photograph is largely due to tartaric acid 1H2C4H4O62
and malic acid 1H2C4H4O52 (Figure 16.1), two closely related (they differ by only one
O atom) organic acids that are found in biological systems. Fermentation of the sugars
in the grapes ultimately forms vinegar, the tangy, sour flavor of which is due to acetic
acid 1CH3COOH2, a substance we discussed in Section 4.3. The sour taste of oranges,
lemons, and other citrus fruits is due to citric acid 1H3C6H5O72, and, to a lesser extent,
ascorbic acid 1H2C6H6O62, better known as Vitamin C.
Acids and bases are among the most important substances in chemistry, and
they affect our daily lives in innumerable ways. Not only are they present in our
foods, but acids and bases are also crucial components of living systems, such as
the amino acids that are used to synthesize proteins and the nucleic acids that code
genetic information. Both citric and malic acids are among several acids involved
in the Krebs cycle (also called the citric acid cycle) that is used to generate energy
in aerobic organisms. The application of acid–base chemistry has also had critical
roles in shaping modern society, including such human-driven activities as industrial manufacturing, the creation of advanced pharmaceuticals, and many aspects
of the environment.
The impact of acids and bases depends not only on the type of acid or base, but
also on how much is present. The time required for a metal object immersed in water
to corrode, the ability of an aquatic environment to support fish and plant life, the fate
of pollutants washed out of the air by rain, and even the rates of reactions that maintain
What’s
Ahead
16.1 Acids and Bases: A Brief Review We begin by
reviewing the Arrhenius definition of acids and bases.
16.2 Brønsted–Lowry Acids and Bases We learn that
a Brønsted–Lowry acid is a proton donor and a Brønsted–Lowry
base is a proton acceptor. Two species that differ by the presence
or absence of a proton are known as a conjugate acid–base pair.
16.3 The Autoionization of Water We see that the
autoionization of water produces small quantities of H3O+
and OH− ions. The equilibrium constant for autoionization,
▶ Clusters of grapes and balsamic
vinegar. Grapes contain several acids that
contribute to their characteristic flavor. The
distinctive flavor of all vinegars is due to
acetic acid. Balsamic vinegar is obtained by
fermenting grapes.
Kw = [H3O+][OH−] defines the relationship between H3O+ and
OH− concentrations in aqueous solutions.
16.4 The pH Scale We use the pH scale to describe the
acidity or basicity of an aqueous solution. Neutral solutions have a
pH = 7, acidic solutions have pH below 7, and basic solutions have
pH above 7.
16.5 Strong Acids and Bases We categorize acids
and bases as being either strong or weak electrolytes. Strong
acids and bases are strong electrolytes, ionizing or dissociating
completely in aqueous solution. Weak acids and bases are weak
electrolytes and ionize only partially.