Download Chapter 15: Chemical Equilibrium

15
Chemical Equilibrium
Unit Outline
15.1
15.2
15.3
15.4
The Nature of the Equilibrium State
The Equilibrium Constant, K
Using Equilibrium Constants in Calculations
Disturbing a Chemical Equilibrium: Le Chatelier’s Principle
In This Unit…
Vladmir Fedorchuk/Fotolia.com
We now begin our coverage of chemical equilibria, and in other units we
will apply the concepts and tools learned in this unit to acid−base equilibria and the chemistry of insoluble ionic compounds. Chemical equilibrium
and the reversibility of chemical reactions plays an important role in one
of the most important chemical reactions in the human body: the binding
of oxygen molecules to heme groups in the protein hemoglobin. When
hemoglobin is exposed to oxygen in the lungs, oxygen molecules attach to
the heme groups. When the oxygenated hemoglobin reaches oxygendepleted cells, the oxygen is released. The process is reversible, and as you
will see in this unit, it is the reversible nature of chemical reactions that is
the basis of chemical equilibrium.
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15.1 The Nature of the Equilibrium State
15.1a Principle of Microscopic Reversibility
In Chemical Kinetics (Unit 14) we learned that chemical reactions proceed in a series of
steps called a reaction mechanism. The principle of microscopic reversibility tells us
that the elementary steps in a reaction mechanism are reversible; that is, the mechanism
of the reverse net reaction is reverse of the forward reaction mechanism (when the reaction conditions are the same). We define a reversible process as one in which it is possible to return to the starting conditions along the exact same path without altering the
surroundings. Consider the two-step mechanism for the reaction of nitrogen dioxide with
carbon monoxide.
Step 1
2 NO2(g) S NO(g) 1 NO3(g)
Step 2
NO3(g) 1 CO(g) S NO2(g) 1 CO2(g)
Net reaction: NO2(g) 1 CO(g) S NO(g) 1 CO2(g)
The reverse reaction has a two-step mechanism that is the reverse of the forward reaction mechanism.
Step 1
NO2(g) 1 CO2(g) S NO3(g) 1 CO(g) (Step 2 reversed)
Step 2
NO3(g) 1 NO(g) S 2 NO2(g)
(Step 1 reversed)
Net reaction: NO(g) 1 CO2(g) S NO2(g) 1 CO(g)
In chemical equations, an equilibrium arrow () is used to indicate that a reaction is
reversible.
NO2(g) 1 CO(g)  NO(g) 1 CO2(g)
The reversibility of chemical reactions can be very useful in chemical synthesis, such
as the conversion between alkenes and alcohols (Interactive Figure 15.1.1). Although all
chemical reactions are reversible, there are occasions when the use of an equilibrium arrow
is not appropriate. For example, when hydrogen and oxygen react to form water vapor,
essentially all reactants are converted to product and no noticeable amounts of reactants
are formed by the reverse reaction. The chemical equation representing this reaction therefore uses a single reaction arrow (S).
2 H2(g) 1 O2(g) S 2 H2O(g)
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Interactive Figure 15.1.1
Investigate the reversibility of chemical reactions.
H
H
H
H
H
C
C
H
C
C
+
O
C
+
H
H
H
H
H
O
O
H
H
H
H
H
C
C
C
H
H O H
H
H
H
H
H
H
H
C
H
O
H
O
H
H
H
H
H
H
H
H
H
H
C
C
C
+
H
H
O
H
H
H
H
C
C
C
H
H
C
C
C
H
O
H
H
H
H
+
O
H
H
H
H O H
H
H
H
H
H
H
O
H
O
H
H
H
Alkenes are hydrolyzed to form alcohols in a reversible reaction.
Also, if the reverse reaction is unlikely to occur, such as when one of the reaction products is physically separated from a reaction mixture as a gas (the reaction of a metal carbonate with acid, for example), a single reaction arrow is used.
CuCO3(s) 1 2 HCl(aq) S CuCl2(aq) 1 H2O(ℓ) 1 CO2(g)
15.1b The Equilibrium State
In Intermolecular Forces and the Liquid State (Unit 11) we described the nature of a
dynamic equilibrium between a liquid and its vapor in a sealed flask. ( Flashback to
Section 11.2 Vapor Pressure) When equilibrium was reached in the flask, the amount of
liquid did not change, but the vaporization and condensation processes continued. When
the rate of vaporization is equal to the rate of condensation, a state of dynamic equilibrium between the two phases is reached. We can apply this concept to reversible chemical
reactions and use it to describe the nature of a dynamic chemical equilibrium.
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Consider the reversible reaction between Fe31 and the thiocyanate ion.
31
21
Fe (aq) 1 SCN (aq)  FeSCN (aq)
2
Interactive Figure 15.1.2
Explore the equilibrium state.
Rate (forward) 5 kforward[Fe31][SCN2]
As the reaction proceeds, the product concentration increases and the reverse reaction
begins to take place.
FeSCN21(aq) S Fe31(aq) 1 SCN2(aq)
Rate (reverse) 5 kreverse[FeSCN21]
As the reaction continues, the rate of the forward reaction decreases (because [Fe31] and
[SCN2] decrease) and the rate of the reverse reaction increases (because [FeSCN21] increases). Eventually, a state of chemical equilibrium is reached where the rate of the
forward reaction is equal to the rate of the reverse reaction:
Rate (forward) 5 Rate (reverse)
Fe3+
SCN–
FeSCN2+
© 2013 Cengage Learning
Fe31(aq) 1 SCN2(aq) S FeSCN21(aq)
Concentration, mol/L
When the two reactants are mixed, they react in the forward direction to form FeSCN21 in
a second-order process.
Time
Graphical representation of progression toward chemical
equilibrium
kforward[Fe31][SCN2] 5 kreverse[FeSCN21]
When this equilibrium state is achieved, the concentrations of all the species in solution
are constant, even though the forward and reverse reactions continue to take place. Note
that when a system is at equilibrium, while the rates of the forward and reverse reactions
are equal, the rate constants for those reactions and the concentrations of reactant and
products are generally not equal.
The equilibrium state can be represented graphically by plotting the concentration
of reactants and products over time (Interactive Figure 15.1.2). Notice that during
the first stage of the reaction, the forward reaction is faster than the reverse reaction
and reactant concentrations decrease while the product concentration increases. When
the system reaches equilibrium, the forward and reverse rates are equal and concentrations of reactants and product do not change. Also notice that significant amounts of
both reactants and products (Fe31, SCN2, and FeSCN21) are present in the equilibrium
system.
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Section 15.1 Mastery
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15.2 The Equilibrium Constant, K
15.2a Equilibrium Constants
The relationship between forward and reverse rate constants for an equilibrium system is
shown in an equilibrium constant expression and quantified by an equilibrium constant (K). An equilibrium constant expression is written by rearranging the equation relating
forward and reverse reaction rates. When the forward and reverse reaction rates are equal,
the chemical system is at equilibrium and the ratio of forward and reverse rate constants is
equal to the equilibrium constant, K. For the Fe(SCN)21 equilibrium system, for example,
Rate (forward) 5 Rate (reverse)
kforward[Fe31][SCN2] 5 kreverse[FeSCN21]
3 FeSCN21 4
kforward
5K5
5 142 at 25 °C
3 Fe31 4 3 SCN2 4
kreverse
Note that because rate constants change with temperature, the equilibrium constant will
also vary with temperature.
The magnitude of the equilibrium constant provides information about the relative rate
constants of the forward and reverse reactions and the relative amounts of reactants and
products at equilibrium.
●
K  1
A large value of K (K  1) means that at equilibrium, the concentration of products
is much larger than the concentration of reactants. The rate constant for the forward
reaction is much larger than the reverse reaction rate constant (kforward  kreverse),
and at equilibrium the system contains mostly products. This is called a productfavored reaction. For example, consider the product-favored reaction between gaseous iodine and chlorine:
I2(g) 1 Cl2(g)  2 ICl(g)
K 5 2.1 3 105 at 25 ºC
This is a product-favored reaction, and at equilibrium there will be very little I2 or Cl2
in the reaction flask.
●
K  1
A small value of K (K  1) means that at equilibrium, the concentration of reactants
is much larger than the concentration of products. The rate constant for the reverse
reaction is much larger than the forward reaction rate constant (kreverse >> kforward),
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and at equilibrium the system contains mostly reactants. This is called a reactantfavored reaction. For example, consider the reaction of acetic acid with water:
K 5 1.8 3 10−5 at 25 ºC
CH3CO2H(aq) 1 H2O(ℓ)  CH3CO22(aq) 1 H3O1(aq)
This is a reactant-favored reaction, and at equilibrium there will be very little CH3CO22
or H3O1 in the reaction flask.
●
K≈1
A value of K very close to 1 (K ≈ 1) means that at equilibrium, significant amounts of
both reactants and product are found in the reaction vessel. The rate constants for the
forward and reverse reactions are comparable (kforward ≈ kreverse), and at equilibrium
the system contains a mixture of both reactants and products. For example, consider
the dimerization of nitrogen dioxide to form dinitrogen tetraoxide:
2 NO2(g)  N2O4(g)
K 5 1.4 at 50 ºC
The equilibrium constant has a value close to 1, so at equilibrium significant amounts
of both NO2 and N2O4 will be found in the reaction flask.
The difference between a product-favored reaction and a reactant-favored reaction can
be shown graphically, as in Interactive Figure 15.2.1.
Interactive Figure 15.2.1
Recognize the significance of the magnitude of the equilibrium constant.
Product-favored (K > 1)
Time
(a)
Reactants
Products
Time
© 2013 Cengage Learning
Reactants
Concentration
Concentration
Products
Reactant-favored (K < 1)
(b)
Graphical representations of (a) product-favored and (b) reactant-favored chemical reactions
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15.2b Writing Equilibrium Constant Expressions
For the general equilibrium system
aA1bBcC1dD
the equilibrium constant expression is written
K5
3 C 4c 3 D 4d
3 A 4a 3 B 4b
(15.1)
In the equilibrium constant expressions,
●
●
●
product concentrations are multiplied in the numerator, and each is raised to the
power of the stoichiometric coefficient from the balanced equation;
reactant concentrations are multiplied in the denominator, and each is raised to the
power of the stoichiometric coefficient from the balanced equation; and
the concentrations of A, B, C, and D are those measured when the reaction is at
equilibrium.
implied “1”
2 NO2(g)
N2O4(g) K 5
[N2O4]
[NO2]2
© 2013 Cengage Learning
For example, the equilibrium constant expression for the dimerization of nitrogen dioxide has the concentration of the product, N2O4, in the numerator and the concentration
of the reactant, NO2, in the denominator. Each concentration is raised to the power of the
corresponding stoichiometric coefficient from the balanced equation.
There are two cases when a reactant or product does not appear in the equilibrium
expression.
1. Pure solids do not appear in equilibrium constant expressions. When an equilibrium system includes a pure solid as a separate phase, it is observed that the equilibrium concentrations of the species in solution or in the gas phase are independent of
the amount of solid present. For example, consider the equilibrium between solid
ammonium chloride and gaseous ammonia and hydrogen chloride:
NH4Cl(s)  NH3(g) 1 HCl(g)
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K 5 [NH3][HCl]
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The equilibrium concentrations of ammonia and hydrogen chloride do not depend on the
amount of solid present as long as there is some solid present for the system to reach
equilibrium. The concentration of a pure solid depends only on the density of the substance, a constant that can be incorporated into the equilibrium constant.
2. Pure liquids and solvents do not appear in equilibrium constant expressions.
When an equilibrium system involves a pure liquid, often a solvent, its concentration
is typically very large compared with the concentrations of the other species present.
For example, at 25 ºC, the concentration of H2O in a dilute aqueous solution can be
as high as 56 mol/L! As the system reaches equilibrium, the solvent concentration remains essentially constant, equal to that of the pure liquid. For example, consider
the reaction that takes place in a solution of aqueous ammonia:
NH3(aq) 1 H2O(ℓ)  NH41(aq) 1 OH−(aq)
K5
2
3 NH1
4 4 3 OH 4
3 NH3 4
Water is the solvent for this reaction and it also participates in the chemical equilibrium.
Its concentration remains essentially constant during the equilibrium process, so it is
incorporated into the equilibrium constant and does not appear in the equilibrium
constant expression.
Example Problem 15.2.1 Write equilibrium constant expressions.
Write equilibrium constant expressions for the following reactions:
a. H2(g) 1 Cl2(g)  2 HCl(g)
b. C(s) 1 H2O(g)  H2(g) 1 CO(g)
c. CH3CO2H(aq) 1 H2O(ℓ)  H3O1(aq) 1 CH3CO2−(aq)
Solution:
You are asked to write an equilibrium expression for a given reaction.
You are given a chemical equation for a reaction.
a. K 5
3 HCl 4 2
3 H2 4 3 Cl2 4
Note that the HCl concentration is squared because of its stoichiometric coefficient (2) in
the balanced equation.
c
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
Example Problem 15.2.1 (continued)
b. K 5
3 H2 4 3 CO 4
3 H2O 4
Note that C(s) does not appear in the equilibrium constant expression. Also, notice that in
this example, H2O appears in the equilibrium constant expression because it is not a solvent.
c. K 5
3 H3O1 4 3 CH3CO22 4
3 CH3CO2H 4
Video Solution
Note that water, the solvent in this aqueous equilibrium system, is treated as a pure liquid.
Its concentration does not change significantly during the reaction, so it does not appear in
the equilibrium constant expression.
Tutored Practice
Problem 15.2.1
All of the equilibrium constant expression examples we have seen to this point express
reactant and product concentrations in molarity units (mol/L). For gas-phase equilibria, it is also
possible to express concentrations in terms of partial pressures of reactants and products. To
differentiate between these two expressions, we use the symbol Kc to indicate a constant calculated using molarity units and Kp to indicate a constant calculated using partial pressure units.
In this text, all equilibrium constants labeled K are Kc values.
For equilibria involving gases, Kc and Kp values are related by the equation
Kp 5 Kc(RT)∆n
(15.2)
where R is the ideal gas constant, T is temperature (in kelvin units), and ∆n is the change
in number of moles of gas in the reaction (∆n 5 moles gaseous products − moles gaseous
reactants).
Example Problem 15.2.2 Interconvert Kp and Kc values.
Calculate Kp for the following reactions at the indicated temperature.
a. 2 NOBr(g)  2 NO(g) 1 Br2(g)
b. NH4I(s)  NH3(g) 1 HI(g)
Kc 5 6.50 × 10−3 at 298 K
Kc 5 7.00 × 10−5 at 673 K
Solution:
You are asked to calculate Kp for an equilibrium reaction at a given temperature.
You are given a Kc value for an equilibrium reaction at a given temperature.
First determine the change in number of moles of gas in the reaction, then use Equation 15.2
to calculate Kp.
c
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
Example Problem 15.2.2 (continued)
a. ∆n 5 (3 mol gaseous products) 2 (2 mol gaseous reactants) 5 1
Kp 5 Kc(RT)∆n 5 (6.50 × 10−3)[(0.082057 L  atm/K  mol)(298 K)]1 5 0.159
b. ∆n 5 (2 mol gaseous products − 0 mol gaseous reactants) 5 2
Kp 5 Kc(RT)∆n 5 (7.00 × 10−5)[(0.082057 L  atm/K  mol)(673 K)]2 5 0.213
Video Solution
Tutored Practice
Problem 15.2.2
15.2c Manipulating Equilibrium Constant Expressions
The value of the equilibrium constant for a chemical system is a function of how the equilibrium expression is written. Consider, for example, the reaction between Cu21(aq) and
NH3(aq). The chemical equation that describes this reaction can be written as:
Cu21(aq) 1 4 NH3(aq)  Cu(NH3)421(aq)
K5
3 Cu 1NH32 21
4 4
5 6.8 3 1012 at 25 °C
3 Cu21 4 3 NH3 4 4
This equation can be manipulated by (1) multiplying the equation by a constant, (2)
writing the reaction in the reverse direction, and (3) combining the equation with another
chemical equation to describe a different equilibrium system. Each manipulation results in
a new equilibrium constant, each of which is a correct value for the reaction as described
by the chemical equation. The equation describing a chemical system can be written in
many ways, and each equation has its own K value. This means that it is important to be
specific when reporting an equilibrium constant by including the chemical equation along
with the value for K.
Charles D. Winters
1. Multiply the equation by a constant.
If the chemical equation for the copper-ammonia equilibrium is manipulated by multiplying the stoichiometric coefficients by ¼, the equilibrium constant expression is written as:
¼ Cu21(aq) 1 NH3(aq)  ¼ Cu(NH3)421(aq)
Knew 5
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1 /4
3 Cu 1NH32 21
4 4
3 Cu21 4 1/4 3 NH3 4
Aqueous Cu21 reacts with ammonia to form
deep blue Cu(NH3)421.
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The new equilibrium expression is equal to the original expression raised to the power
of the multiplication factor (¼):
Knew 5
1 /4
1 /4
3 Cu 1NH32 21
3 Cu 1NH32 21
4 4
4 4
5a
5 1Kold 2 1/4 5 16.8 3 10122 1/4 5 1.6 3 103
21 1/4
21
4b
3 Cu 4 3 NH3 4
3 Cu 4 3 NH3 4
In general, when an equilibrium equation is multiplied by a constant, n,
(15.3)
Knew 5 (Kold)n
2. Reverse the reaction direction.
If the chemical equation for the copper-ammonia equilibrium is manipulated by writing it
in the reverse direction, the equilibrium constant expression is written as:
Cu(NH3)421(aq)  Cu21(aq) 1 4 NH3(aq)
Knew 5
3 Cu21 4 3 NH3 4 4
3 Cu 1NH32 21
4 4
The new equilibrium expression is equal to the inverse of the original expression:
Knew 5
21
3 Cu 1NH32 21
3 Cu21 4 3 NH3 4 4
1
1
4 4
5
5
5 1.5 3 10213
21
4b
21 5 a
3 Cu 1NH32 4 4
3 Cu 4 3 NH3 4
Kold
6.8 3 1012
In general, when an equilibrium equation is written in the reverse direction,
1
Knew 5
Kold
(15.4)
3. Combine reactions.
If the chemical equation for the copper-ammonia equilibrium is manipulated by combining it with the copper(II) hydroxide dissolution equilibrium (K 5 1.6 × 10−19), the
new equilibrium is written as:
3 Cu 1NH32 21
4 4
3 Cu21 4 3 NH3 4 4
(1) Cu21(aq) 1 4 NH3(aq)  Cu(NH3)421(aq)
K1 5
(2) Cu(OH)2(s)  Cu21(aq) 1 2 OH−(aq)
K2 5 [Cu21][OH−]2
Net reaction:
Cu(OH)2(s) 1 4 NH3(aq)  Cu(NH3)421(aq) 1 2 OH−(aq)
Knet 5
2 2
3 Cu 1NH32 21
4 4 3 OH 4
4
3 NH3 4
The new equilibrium expression is equal to the product of the two original equilibrium
expressions:
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Knet 5
2 2
3 Cu 1NH32 21
3 Cu 1NH32 21
4 4 3 OH 4
4 4
5
3 3 Cu21 4 3 OH2 4 2
4
21
3 NH3 4
3 Cu 4 3 NH3 4 4
= K1 × K2 5 (6.8 × 1012)(1.6 × 10−19)
5 1.1 × 10−6
In general, when two equations representing equilibrium systems are added together,
Knew 5 K1 × K2
(15.5)
Example Problem 15.2.3 Manipulate equilibrium constant expressions.
The equilibrium constant for the reaction of hydrogen with oxygen to form water is 5.7 × 1040
at 25 ºC.
H2(g) 1 ½ O2(g)  H2O(g)
K 5 5.7 × 1040 at 25 ºC
Calculate the equilibrium constant for the following reactions.
a. 2 H2(g) 1 O2(g)  2 H2O(g)
b. H2O(g)  H2(g) 1 ½ O2(g)
Solution:
You are asked to calculate an equilibrium constant for a reaction.
You are given a chemical equation, with an equilibrium constant, that is related to the equation for the given reaction.
a. Multiplying the original equation by a factor of 2 results in the new equation, so the new
equilibrium constant is equal to the original constant squared (raised to the power of 2).
Koriginal 5
3 H2O 4
3 H2 4 3 O2 4 1/2
Knew 5
3 H2O 4 2
5 K 2original
3 H2 4 2 3 O2 4
Knew 5 K2original 5 15.7 3 10402 2 5 3.2 3 1081
b. Reversing the original equation results in the new equation, so the new equilibrium constant is equal to the inverse of the original constant.
Koriginal 5
3 H2O 4
3 H2 4 3 O2 4 1/2
Knew 5
Unit 15 Chemical Equilibrium
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Knew 5
3 H2 4 3 O2 4 1/2
1
5
3 H2O 4
Koriginal
1
1
5
5 1.8 3 10 241
Koriginal
5.7 3 1040
Video Solution
Tutored Practice
Problem 15.2.3
Section 15.2 Mastery
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15.3 Using Equilibrium Constants in Calculations
15.3a Determining an Equilibrium Constant
Using Experimental Data
The value of an equilibrium constant can be determined using the equilibrium concentrations of the chemical species involved in the equilibrium process. The concentrations are
substituted into the equilibrium expression and the constant is calculated. For any equilibrium system there are an infinite number of combinations of equilibrium concentrations, all
of which result in the same value of K.
Example Problem 15.3.1 Use equilibrium concentrations to calculate K.
Some sulfur trioxide is placed in a flask and heated to 1400 K. When equilibrium is reached,
the flask is found to contain SO3 (0.152 M), SO2 (0.0247 M), and O2 (0.0330 M). What is the
value of the equilibrium constant for the following at reaction at 1400 K?
2 SO3(g)  2 SO2(g) 1 O2(g)
Solution:
You are asked to calculate the equilibrium constant for a reaction at a given temperature.
You are given equilibrium concentrations of reactants and products at a given temperature.
Step 1. Write the equilibrium constant expression.
K5
3 SO2 4 2 3 O2 4
3 SO3 4 2
Step 2. Substitute the equilibrium concentrations into the equilibrium expression and calculate K.
3 SO2 4 2 3 O2 4
10.02472 2 10.03302
K5
5
5 8.71 3 1024
2
3 SO3 4
10.1522 2
Notice that we do not assign units to the equilibrium constant, even though the concentrations
have units of mol/L.
Video Solution
Tutored Practice
Problem 15.3.1
It is often difficult to accurately measure the concentration of all species in solution.
When this is the case, the equilibrium concentration of only one species is measured and
stoichiometric relationships are used to determine the concentrations of the other species
at equilibrium. We will use the ICE method to determine the concentration of species at
equilibrium. The ICE method involves the use of a table showing the Initial concentrations,
the Change in concentrations as the reaction approaches equilibrium, and the Equilibrium
concentrations. The ICE method is demonstrated in the following example problem.
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Example Problem 15.3.2 Use an ICE table to calculate K.
A mixture of nitrogen and hydrogen is allowed to react in the presence of a catalyst.
N2(g) 1 3 H2(g)  2 NH3(g)
The initial concentrations of the reactants are [N2] 5 0.1000 M and [H2] 5 0.2200 M. After the
system reaches equilibrium, it is found that the nitrogen concentration has decreased to
0.0271 M. Determine the value of the equilibrium constant, K, for this reaction.
Solution:
You are asked to calculate the equilibrium constant for a reaction.
You are given the initial concentrations of reactants and the equilibrium concentration of one
of the reactants.
Step 1. Write the equilibrium constant expression.
K5
3 NH3 4 2
3 N2 4 3 H2 4 3
Step 2. Set up an ICE table (initial concentrations, the change in concentrations as the reaction progresses, and the equilibrium concentrations) and fill in the initial concentrations.
N2(g)
Initial (M)
Change (M)
Equilibrium (M)
1
0.1000
3 H2(g)

0.2200
2 NH3(g)
0
Step 3. Use the stoichiometry of the balanced chemical equation to define the change in
concentration for each species as the reaction proceeds toward equilibrium in terms of the
unknown quantity, x. Thus, the concentration of N2 decreases by x, the concentration of H2
decreases by 3x, and the concentration of NH3 increases by 2x.
N2(g)
Initial (M)
Change (M)
Equilibrium (M)
0.1000
2x
1
3 H2(g)
0.2200
23x

2 NH3(g)
0
12x
Notice that the reactant concentrations decrease and the product concentration increases as
the reaction proceeds toward equilibrium.
c
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
Example Problem 15.3.2 (continued)
Step 4. Add the initial and change in concentrations to represent the equilibrium concentrations in terms of x.
N2(g)
Initial (M)
Change (M)
Equilibrium (M)
0.1000
2x
0.1000 2 x
1
3 H2(g)

2 NH3(g)
0.2200
23x
0.2200 2 3x
0
12x
2x
Step 5. Determine the value of x using the actual equilibrium concentration of N2 (given in
the problem, 0.0271 M) and the N2 equilibrium concentration described in terms of x.
0.1000 − x 5 0.0271
x 5 0.0729 M
Step 6. Use the value of x to determine the equilibrium concentrations of H2 and NH3.
[H2]equilibrium 5 0.2200 − 3x 5 0.2200 − 3(0.0729) 5 0.0013 M
[NH3]equilibrium 5 2x 5 2(0.0729) 5 0.146 M
Step 7. Substitute the equilibrium concentrations into the equilibrium constant expression
and calculate K.
3 NH3 4 2
10.1462 2
5
5 3.6 3 108
K5
3 N2 4 3 H2 4 3
10.02712 10.00132 3
Video Solution
Tutored Practice
Problem 15.3.2
15.3b Determining Whether a System Is at Equilibrium
Many chemical systems, particularly those found in nature, are not at equilibrium. We can
determine whether a system is at equilibrium by comparing the ratio of concentrations at a
specific point in the reaction process (the reaction quotient, Q) to the ratio found at
equilibrium (the equilibrium constant). A ratio of reactant and product concentrations,
called the reaction quotient expression, can be written for any reaction that may or may
not be at equilibrium. For the reaction
aA1bBcC1dD
the reaction quotient expression is written
Q5
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3 C 4c 3 D 4d
3 A 4a 3 B 4b
(15.6)
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Notice that the reaction quotient expression has the same form as the equilibrium
constant expression, with the exception that the concentrations of A, B, C, and D may
or may not be equilibrium concentrations.
As was true for equilibrium constant expressions, pure liquids and pure solids do not
appear in reaction quotient expressions.
Comparing Q to K for a specific reaction allows us to determine whether a system is at
equilibrium (Interactive Figure 15.3.1). There are three possible relationships between the
two values:
●
QK
The system is not at equilibrium. Reactants will be consumed and product concentration will increase until Q 5 K. The product concentration (in the numerator of the
Q expression) is too small, so the reaction will proceed in the forward direction (to the
right) to reach equilibrium.
●
QK
The system is not at equilibrium. Products will be consumed and reactant concentration will increase until Q 5 K. The product concentration (in the numerator of the
Interactive Figure 15.3.1
Use Q and K to determine if a system is at equilibrium.
System has too
much product
Q
Q
K
Q
System is at
equilibrium
K
Shift reaction
to the right
Q<K
K
Shift reaction
to the left
Q5K
Q>K
© 2013 Cengage Learning
System has too
much reactant
The relationship between Q, K, and the direction
a reaction proceeds to reach equilibrium
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Q expression) is too large, so the reaction will proceed in the reverse direction (to the
left) to reach equilibrium.
●
Q5K
The system is at equilibrium and no further change in reactant or product concentration will occur.
Example Problem 15.3.3 Use the reaction quotient, Q, to determine whether a
system is at equilibrium.
Consider the following system (K 5 22.3 at a given temperature):
Cl2(g) 1 F2(g)  2 ClF(g)
The concentrations of reactants and products are measured at a particular time and found to be:
[Cl2] 5 0.300 M
[F2] 5 0.620 M
[ClF] 5 0.120 M
Is the system at equilibrium? If not, in which direction will it proceed to reach equilibrium?
Solution:
You are asked to determine if a system is at equilibrium and, if it is not, the direction it will
proceed to reach equilibrium.
You are given the chemical equation and equilibrium constant for a reaction, and concentrations of reactants and products.
Substitute the concentrations into the reaction quotient expression and compare the calculated value of Q to the equilibrium constant.
Q5
3 ClF 4 2
10.1202 2
5
5 0.0774
3 Cl2 4 3 F2 4
10.3002 10.6202
Q  K, so the system is not at equilibrium. The reaction will proceed in the forward direction
(to the right), consuming chlorine and fluorine and forming additional ClF until equilibrium is
reached.
Video Solution
Tutored Practice
Problem 15.3.3
15.3c Calculating Equilibrium Concentrations
If a chemical system is not at equilibrium, the equilibrium constant can be used to determine the equilibrium concentrations of reactants and products. The ICE method is used to
determine equilibrium concentrations.
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Example Problem 15.3.4 Use K and initial concentrations to calculate equilibrium
concentrations.
Consider the equilibrium system involving two isomers of butane.
CH
CH33
HH33CC
H3CH3CCHCH
CH
2 2CHCH
2 2 CH
3 3
Butane
Butane
C
H
CH33
CH
[isobutane]
[isobutane]
5 2.5
at°C
20 °C
KK55 [butane]
at 20
[butane]5 2.5
Isobutane
Isobutane
A flask originally contains 0.200 M butane. Calculate the equilibrium concentrations of butane
and isobutane.
Solution:
You are asked to calculate equilibrium concentrations of reactants and products in a
reaction.
You are given a chemical equation and equilibrium constant and the initial concentration of a
reactant.
Step 1. Use the ICE method to determine equilibrium concentrations. Use the unknown
quantity x to define the equilibrium concentrations of butane and isobutane.
[butane]
Initial (M)
Change (M)
Equilibrium (M)

0.200
2x
0.200 2 x
[isobutane]
0
1x
x
Step 2. Substitute the equilibrium concentrations into the equilibrium constant expression
and determine the value of x, the change in butane and isobutane concentrations as the system approaches equilibrium.
K 5 2.5 5
x
0.200 2 x
x 5 0.143
Step 3. Use the value of x to calculate the equilibrium concentrations of butane and
isobutane.
[butane]equilibrium 5 0.200 − x 5 0.057 M
[isobutane]equilibrium 5 x 5 0.143 M
Is your answer reasonable? As a final check of your answer, confirm that these are equilibrium concentrations by substituting them into the equilibrium constant expression and calculating K.
K5
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3 isobutane 4
0.143
5
5 2.5
3 butane 4
0.057
Video Solution
Tutored Practice
Problem 15.3.4
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15.4 Disturbing a Chemical Equilibrium:
Le Chatelier’s Principle
15.4a Addition or Removal of a Reactant or Product
According to Le Chatelier’s principle, if a chemical system at equilibrium is disturbed
so that it is no longer at equilibrium, the system will respond by reacting in either the
forward or reverse direction so as to counteract the disturbance, resulting in a new equilibrium composition. We often refer to this as a reaction “shifting” to the right (forming
additional product) or to the left (forming additional reactant) as a result of an equilibrium disturbance. Typical disturbances include (1) the addition or removal of a reactant
or product, (2) a change in the volume of a system involving gases, and (3) a change in
temperature.
Consider the reversible reaction between Fe31 and SCN−.
Fe31(aq) 1 SCN−(aq)  FeSCN21(aq)
K5
3 FeSCN21 4
3 Fe31 4 3 SCN2 4
If the concentration of Fe31 is increased by the addition of some solid Fe(NO3)3, the system
is no longer at equilibrium. The increased Fe31 concentration results in a situation where
Q < K.
[Fe31]new > [Fe31]equilibrium
Q5
3 FeSCN21 4
, K
3 Fe31 4 new 3 SCN2 4
The reaction will shift to the right, consuming SCN− and some of the additional Fe31 and
forming additional FeSCN21. In terms of Le Chatelier’s principle, the system shifts to the
right, away from the disturbance of added reactant (Interactive Figure 15.4.1).
A common practice in the chemical industry involves the use of Le Chatelier’s principle
to drive a chemical reaction to completion. If one of the products is continually removed as
it is formed, a chemical reaction can be shifted to the right, forming more and more products until all reactants are consumed. The result of adding or removing reactants or products on the equilibrium composition is summarized in Table 15.4.1.
After a change in reactant or product concentration, new equilibrium concentrations
can be calculated using the ICE method, as shown in the following example.
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Interactive Figure 15.4.1
Predict effect of concentration change on an equilibrium system.
C1D
Concentration, mol/L
A1B
A
B
More of B added
Time
© 2013 Cengage Learning
C
D
Graphical representation of the addition of a reactant to an equilibrium system
Table 15.4.1 Effect of Concentration Change on an Equilibrium System
Change
System Response
Effect on K
Add reactant
Shifts to the right
Reactants S Products
No change
Add product
Shifts to the left
Reactants d Products
No change
Remove reactant
Shifts to the left
Reactants d Products
No change
Remove product
Shifts to the right
Reactants S Products
No change
Example Problem 15.4.1 Predict and calculate the effect of concentration changes
on an equilibrium system.
Some FeSCN21 is allowed to dissociate into Fe31 and SCN− at 25 ºC. At equilibrium, [FeSCN21]
5 0.0768 M, and [Fe31] 5 [SCN−] 5 0.0232 M. Additional Fe31 is added so that [Fe31]new 5
0.0300 M and the system is allowed to once again reach equilibrium.
Fe31(aq) 1 SCN−(aq)  FeSCN21(aq)
K 5 142 at 25 ºC
a. In which direction will the reaction proceed to reach equilibrium?
b. What are the new concentrations of reactants and products after the system reaches
equilibrium?
c
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
Example Problem 15.4.1 (continued)
Solution:
You are asked to predict the direction a reaction will proceed when additional reactant is
added and the new equilibrium concentrations of reactants and products.
You are given a chemical equation and equilibrium constant for a reaction, equilibrium concentrations of reactants and products, and a new (non-equilibrium) concentration of reactant.
a. Additional reactant has been added to the reaction flask, so the system will respond by
consuming the excess reactant. The reaction proceeds to the right, consuming reactants
(Fe31 and SCN−) and forming additional product (FeSCN21).
b. Step 1. Predict the direction the reaction will proceed to reach equilibrium. In this case,
additional reactant has been added to the system, so the reaction will proceed to the right,
forming additional product and consuming reactants.
Step 2. Use an ICE table to define equilibrium concentrations in terms of x. The initial
concentration of Fe31 is equal to the new iron(III) concentration, and the initial concentrations of the other species are equal to the first equilibrium concentrations.
Initial (M)
Change (M)
Equilibrium (M)
Fe31(aq)
0.0300
2x
0.0300 2 x
1
SCN2(aq)
0.0232
2x
0.0232 2 x

FeSCN21(aq)
0.0768
1x
0.0768 1 x
Step 3. Substitute the equilibrium concentrations into the equilibrium constant expression
and solve for x. (Note that you must use the quadratic equation to solve this problem.)
K5
3 FeSCN21 4
0.0768 1 x
5
5 142
10.0300 2 x2 10.0232 2 x2
3 Fe31 4 3 SCN2 4
0 5 142x2 − 8.55x 1 0.0220
x 5 0.00267 or 0.0575 (not a physically possible answer
because it results in negative equilibrium concentrations for Fe31 and SCN−)
Step 4. Use the value of x to calculate the new equilibrium concentrations.
[Fe31] 5 0.0300 − x 5 0.0273 M
[SCN−] 5 0.0232 − x 5 0.0205 M
[FeSCN21] 5 0.0768 1 x 5 0.0795 M
Is your answer reasonable? As a final check of your answer, substitute the new equilibrium
concentrations into the equilibrium constant expression and calculate K.
3 FeSCN21 4
0.0795
K5
5
5 142
10.02732 10.02052
3 Fe31 4 3 SCN2 4
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Video Solution
Tutored Practice
Problem 15.4.1
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15.4b Change in the Volume of the System
A change in system volume is particularly important for gas-phase equilibria. A change in
system volume does not affect systems involving only liquids or aqueous solutions because
liquids are not easily compressed or expanded.
Consider the equilibrium system involving solid carbon, carbon dioxide, and carbon
monoxide.
C(s) 1 CO2(g)  2 CO(g)
If this system is at equilibrium and the volume of the reaction container is decreased at
constant temperature, the pressure inside the container increases (recall that pressure and
volume are inversely proportional). When this occurs, the system shifts in the direction that
decreases the pressure inside the container—the direction that decreases the amount of
gas present in the container. In this case, the reaction will shift to the left, forming 1 mol of
CO2 for every 2 mol of CO consumed. The direction the reaction shifts is related to the
stoichiometry of the gaseous reactants and products, as demonstrated by the effect of volume change on the NO2/N2O4 equilibrium system shown in Interactive Figure 15.4.2.
Interactive Figure 15.4.2
© 2013 Cengage Learning
Explore the effect of volume change on an
equilibrium system.
Volume reduced
2 NO2(g)
N2O4(g)
K 5 171
[N2O4] increases when the reaction flask volume is reduced.
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The result of changing the volume on an equilibrium system involving gases is summarized in Table 15.4.2.
Changing the volume of a system involving gases effectively changes the concentrations of species in the reaction vessel. The ICE method can be used to calculate new equilibrium concentrations, as shown in the following example.
Table 15.4.2 Effect of Volume Change
on an Equilibrium System
Example Problem 15.4.2 Predict and calculate the effect of volume change on an
equilibrium system.
Change
System Response
Effect on K
Decrease
volume
Shifts to form fewer
moles of gas
No change
Increase
volume
Shifts to form more
moles of gas
No change
Consider the equilibrium between NO2 and N2O4.
N2O4(g)  2 NO2(g)
K 5 0.690 at 50 ºC
The reaction is allowed to reach equilibrium in a 2.00-L flask. At equilibrium, [NO2] 5 0.314 M
and [N2O4] 5 0.143 M.
a. Predict the change in NO2 concentration when the equilibrium mixture is transferred to a
1.00-L flask.
b. Calculate the new equilibrium concentrations that result when the equilibrium mixture is
transferred to a 1.00-L flask.
Solution:
You are asked to predict the change in product concentration when the reaction flask
volume is changed and to calculate the new equilibrium concentrations in the new reaction
flask.
You are given a chemical equation and equilibrium constant for a reaction, equilibrium concentrations and reaction flask volume, and the volume of the new reaction flask.
a. The reaction flask volume decreases from 2.00 L to 1.00 L. The equilibrium will shift to
form fewer moles of gas. In this reaction, there is 1 mol of gaseous reactants and 2 mol of
gaseous product, so the reaction shifts to the left, decreasing the NO2 concentration and
increasing the N2O4 concentration.
b. Step 1. Determine the new initial concentrations of NO2 and N2O4 after the volume is decreased but prior to reequilibration. Because the volume of the reaction vessel is halved,
the gas concentrations double.
New initial [NO2] 5 2(0.314 M) 5 0.628 M
New initial [N2O4] 5 2(0.143 M) 5 0.286 M
Step 2. Predict the direction the reaction will proceed to reach equilibrium. In this case,
the volume of the reaction flask decreased, so the reaction will proceed to the left, forming
fewer moles of gas.
c
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
Example Problem 15.4.2 (continued)
Step 3. Use the ICE method to define equilibrium concentrations in terms of x.
N2O4(g)
Initial (M)
Change (M)
Equilibrium (M)
2 NO2(g)

0.286
1x
0.286 1 x
0.628
22x
0.628 2 2x
Step 4. Substitute the equilibrium concentrations into the equilibrium constant expression
and solve for x. (Note that you must use the quadratic equation to solve this problem.)
K5
3 NO2 4 2
10.628 2 2x2 2
5
5 0.690
3 N2O4 4
0.286 1 x
0 5 4x2 − 3.20x 1 0.197
x 5 0.0672 and 0.733 (not a possible answer because it results
in a negative NO2 equilibrium concentration)
Step 5. Use the value of x to calculate the new equilibrium concentrations.
[NO2] 5 0.628 − 2x 5 0.494 M
[N2O4] 5 0.286 1 x 5 0.353 M
Is your answer reasonable? As a final check of your answer, substitute the new equilibrium
concentrations into the equilibrium constant expression and calculate K.
3 NO2 4
10.4942
5
5 0.691
3 N2O4 4
0.353
2
K5
2
Video Solution
Tutored Practice
Problem 15.4.2
15.4c Change in Temperature
Only in the case of a temperature change does the value of the equilibrium constant
for a reaction change. To understand the effect of a temperature change on an equilibrium system, consider heat as a reactant or product of a reaction. An exothermic reaction releases energy in the form of heat, so heat can be considered a product of an
exothermic reaction. In the same way, heat can be considered a reactant in an endothermic reaction.
Exothermic reaction: reactants  products 1 heat
Endothermic reaction: reactants 1 heat  products
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When heat is added or removed from an equilibrium system to change the temperature,
the system responds by shifting to offset the addition or loss of heat. When the temperature
increases, the equilibrium system shifts to consume the added energy.
Exothermic reaction shifts to the left:
reactants d products 1 heat
Endothermic reaction shifts to the right:
reactants 1 heat S products
When the temperature decreases, the equilibrium system shifts to produce additional energy.
Exothermic reaction shifts to the right:
reactants S products 1 heat
Endothermic reaction shifts to the left:
reactants 1 heat d products
Consider the equilibrium shown in Interactive Figure 15.4.3.
2 NO2(g)  N2O4(g)
K 5 170 at 298K
∆Hº 5 −57.1 kJ/mol
Because this reaction is exothermic, decreasing the temperature of this equilibrium system
results in a decrease in [NO2], as shown by the loss of color in the reaction vessel (NO2 is
dark red gas and N2O4 is colorless).
The result of temperature changes on an equilibrium system is summarized in Table
15.4.3.
Interactive Figure 15.4.3
Photos: Charles D. Winters; art: © 2013 Cengage Learning
Explore the effect of temperature changes on an equilibrium system.
[NO2] decreases when the reaction flask temperature is reduced.
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Table 15.4.3 Effect of Temperature Change on an Equilibrium System
Change
Reaction Type
System Response
Effect on K
Increase temperature
Exothermic
Shifts to the left (d)
Decreases
Increase temperature
Endothermic
Shifts to the right (S)
Increases
Decrease temperature
Exothermic
Shifts to the right (S)
Increases
Decrease temperature
Endothermic
Shifts to the left (d)
Decreases
It is possible to estimate the new value for the equilibrium constant when there is a
change in temperature by using the van’t Hoff equation,
lna
K2
DH° 1
1
2
b52
a
b
K1
R T2
T1
(15.7)
where K1 and K2 are the equilibrium constants at temperatures T1 and T2, respectively, ∆Hº
is the enthalpy change for the reaction, and R is the ideal gas constant in thermodynamic
units (8.3145 × 10−3 kJ/K  mol). This equation is similar to the Arrhenius equation and the
Clausius–Clapeyron equation.
Example Problem 15.4.3 Predict and calculate the effect of temperature change on
an equilibrium system.
Sulfur dioxide reacts with oxygen to form sulfur trioxide. The equilibrium constant, Kp, for
this reaction is 0.365 at 1150 K.
2 SO2(g) 1 O2(g)  2 SO3(g)
The standard enthalpy change for this reaction (∆Hº) is 2198 kJ/mol.
a. Predict the effect on the O2 concentration when the temperature of the equilibrium system
is increased.
b. Use the van’t Hoff equation to estimate the equilibrium constant for this reaction at 1260 K.
Solution:
You are asked to predict the effect on reactant concentration when the temperature of an
equilibrium system changes and to calculate the equilibrium constant for a reaction at a different temperature.
c
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
Example Problem 15.4.3 (continued)
You are given the chemical equation and equilibrium constant at a given temperature for a
reaction, the standard enthalpy change for the reaction, and a new temperature.
a. This is an exothermic reaction (∆Hº 5 2198 kJ/mol). Increasing the temperature of the
system will cause the equilibrium to shift to the left, consuming SO3 and forming additional
SO2 and O2. The O2 concentration will increase.
b. In this problem, T1 5 1150 K, T2 5 1260 K, and K1 5 0.365. Substitute these values into the
van’t Hoff equation and calculate the equilibrium constant at the new temperature.
lna
lna
K2
DH° 1
1
b52
a 2 b
K1
R T2 T1
K2
2198 kJ /mol
1
1
2
b52
a
b
23
#
0.365
8.3145 3 10 kJ /K mol 1260 K 1150 K
K2 5 0.0599
Is your answer reasonable? Notice that the equilibrium constant decreases as predicted for
an exothermic reaction at a higher temperature.
Video Solution
Tutored Practice
Problem 15.4.3
Section 15.4 Mastery
Unit Recap
Key Concepts
15.1 The Nature of the Equilibrium State
●
●
●
The principle of microscopic reversibility tells us that the elementary steps in a reaction mechanism are reversible (15.1a).
An equilibrium arrow () is used to indicate a reversible chemical reaction (15.1a).
When a chemical reaction reaches a state of chemical equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction (15.1b).
15.2 The Equilibrium Constant, K
●
●
The relationship between forward and reverse rate constants for an equilibrium system
is shown in an equilibrium constant expression and quantified by an equilibrium constant (K) (15.2a).
The magnitude of the equilibrium constant provides information about the relative
amounts of reactants and products at equilibrium (15.2a).
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●
●
●
●
A reactant-favored reaction has a small equilibrium constant (K << 1) and a productfavored reaction has a large equilibrium constant (K >> 1) (15.2a).
The equilibrium constant expression is related to the stoichiometry of the equilibrium
reaction and does not include concentrations of pure solids, pure liquids, or solvents
(15.2b).
There are two forms of equilibrium constant expressions: Kc, where reactant and product concentrations are expressed in mol/L, and Kp, where reactant and product concentrations are expressed in partial pressure units (15.2b).
Equilibrium constant expressions and the corresponding equilibrium constants can be
manipulated by multiplying by a constant, reversing the reaction direction, or combining with other equilibrium expressions (15.2c).
15.3 Using Equilibrium Constants in Calculations
●
●
Equilibrium constants are determined from experimental data, and ICE tables are often used to keep track of changes to reactant and product concentrations as the system
approaches equilibrium (15.3a).
The reaction quotient, Q, is used to determine whether a system is at equilibrium and
in which direction the reaction proceeds to reach equilibrium (15.3b).
15.4 Disturbing a Chemical Equilibrium: Le Chatelier’s Principle
●
●
●
Le Chatelier’s principle states that if a chemical system at equilibrium is disturbed so
that it is no longer at equilibrium, the system will respond by reacting in either the
forward or reverse direction so as to counteract the disturbance, resulting in a new
equilibrium composition (15.4).
Typical disturbances include the addition or removal of a reactant or product (15.4a),
a change in the volume of a system involving gases (15.4b), and a change in temperature (15.4c).
The van’t Hoff equation is used to quantify the relationship between the equilibrium
constant, reaction temperature, and the enthalpy change for the reaction (15.4c).
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Key Equations
(15.2)
When an equilibrium equation is written
in the reverse direction,
1
(15.4)
Knew 5
Kold
When an equilibrium equation is multiplied by a constant, n,
Knew 5 (Kold)n
(15.3)
When two equations representing equilibrium systems are added together,
(15.5)
Knew 5 K1 × K2
K5
3 C 4c 3 D 4d
3 A 4a 3 B 4b
Kp 5 Kc(RT)∆n
(15.1)
Q5
lna
3 C 4c 3 D 4d
3 A 4a 3 B 4b
K2
DH° 1
1
2
b52
a
b
K1
R T2
T1
(15.6)
(15.7)
Key Terms
15.1 The Nature of the Equilibrium State
principle of microscopic reversibility
reversible process
chemical equilibrium
15.3 Using Equilibrium Constants
in Calculations
reaction quotient (Q)
reaction quotient expression
15.4 Disturbing a Chemical Equilibrium:
Le Chatelier’s Principle
Le Chatelier’s principle
van’t Hoff equation
15.2 The Equilibrium Constant, K
equilibrium constant expression
equilibrium constant (K)
product-favored reaction
reactant-favored reaction
Unit 15
Review and Challenge Problems
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