Download Unit 8 Student Notes

General Chemistry
Unit 8 Note Packet – Part 1: Equilibrium
Name:
Date:
Per.:
CHEMICAL EQUILIBRIUM
As you know, equilibrium involves a kind of balance in a system. Here are some examples:
 if an elevator is full and going up to the top floor and four people get off on the seventh floor, then four
people may get on; if two additional people want to get on, then two more must get off
 in a team sport, if three players enter the field or court, three who have been playing must leave because
the rules allow for only “X” players on the field or court at a time
 in a closed bottle of water, some liquid molecules at the surface escape from the intermolecular forces
and go into the gas phase; an equal number of gas molecules colliding with each other, the water surface,
and the container walls condense into the liquid phase
 in a saturated solution, some solute continues to dissolve while other solute particles already in solution
regenerate the solid
In all of these cases,
Reversible Reactions
Most chemical reactions can occur in both directions. Chemical reactions that are reversible can operate in both
directions
.
Example:
Smog is given its characteristic brown color by nitrogen dioxide. Nitrogen dioxide reacts to form dinitrogen
tetroxide:
2NO2 (g)  N2O4 (g)
The reverse reaction also occurs:
N2O4 (g)  2NO2 (g)
To show that both reactions occur at the same time, use two half arrows:
or
Chemical equilibrium does
necessarily mean that there are equal numbers of molecules/moles of
reactants and products. Rather, it means that the
of the forward reaction is equal to the
of the reverse reaction.
Consider the generic reaction while studying the diagram: A  B.
You’re beginning with only reactants (A) and you have zero products (B). The
concentration of the products [B] begins to increase as the concentration of the
reactants [A] decreases. As soon as product particles exist, the reverse reaction may
begin. Before equilibrium is established, both reactions are occurring, but the rate of
the forward reaction (AB) is decreasing while the rate of the reverse reaction
(BA) is increasing. Chemical equilibrium is reached when the rates are equal. Notice
that at equilibrium, [A] does not equal [B].
873996897
-1-
Chemical Equilibrium:
The Law of Chemical Equilibrium
 developed from careful empirical or experimental evidence
 the Law of Mass Action - expresses the relative concentrations of reactants and products at equilibrium
in terms of a quantity called the equilibrium constant, K eq
Consider the generic reaction:
aA + bB  cC + dD
where the a, b, c, and d represent coefficients and A, B, C, and D are products and reactants
Keq =
Note that because reaction varies with temperature, Keq is also affected by temperature. Any given value of K eq is
therefore specific to a particular temperature.
The Law of Chemical Equilibrium states that
Problems:
Write the equilibrium expression for each reaction:
1.
2NO2(g)  N2O4(g)
2. 2CO(g) + O2(g)  2CO2 (g)
3. 2SO2(g) + O2(g)  2SO3(g)
4. CO(g) + 3H2 (g)  CH4(g) + H2O(g)
5. H2O(g) + CO(g)  H2(g) + CO2(g)
6. 2H2(g) + O2 (g)  2H2O(g)
7. 2NO(g) + Br2(g)  2NOBr(g)
Study the reactions in the problems above. What do they all have in common?
873996897
-2-
HOMOGENEOUS AND HETEROGENEOUS EQUILIBRIA
– an equilibrium system in which all products and reactants are in the same state
– an equilibrium system in which more than one state of matter is present
How is the concentration of a solid or liquid (NOT a solution!) expressed?


the density of a pure liquid or solid is essentially constant, so the concentration of a liquid or solid
component is effectively incorporated in the equilibrium constant, Keq, so they are not in the equilibrium
expression
consider the following reaction:
C(s) + 2S(g)  CS2(g)
Keq =
Problems:
Write the equilibrium expression for each of the following reactions:
1.
C(s) + H2O(g)  CO(g) + H2 (g)
2.
NH4NO3(s)  N2O(g) + 2H2O(g)
3.
ZnCO3(s)  ZnO(s) + CO2 (g)
4.
SnO2 (s) + 2CO(g)  Sn(s) + 2CO2 (g)
5.
C(S) + CO2 (g)  2CO(g)
CALCULATING Keq
The equilibrium constant, Keq, for a given temperature is calculated using experimental data. Consider the
following experiment: Nitrogen dioxide is converted to dinitrogen tetroxide in the following reversible reaction:
2NO2(g)  N2O4(g)
Three experiments are conducted with different starting concentrations of NO 2 and N2O4 as shown in Table 1
below. Each system was allowed to reach equilibrium in a sealed test tube and the equilibrium concentrations were
measured as shown in Table 2. Calculate the equilibrium quotient for each experiment using the equilibrium
expression for the reaction. Note that there are no units reported for the equilibrium constant.
Table 1. Initial Concentrations
Experiment Initial [NO2]
(M)
1
0.0200
2
0.0300
3
0.0
Initial [N2O4]
(M)
0.0
0.0
0.0200



Table 2. Equilibrium Concentrations
Equilibrium [NO2]
Equilibrium [N2O4]
(M)
(M)
0.0172
0.00140
0.0243
0.00280
0.0310
0.00452
How do the three Keq values compare?
Does the initial concentration have any impact on the equilibrium constant?
Note that each set of equilibrium concentrations is called an equilibrium position.
873996897
-3-
Keq



Keq << 1 at equilibrium the system contains mostly reactants; “equilibrium lies to the left”
Keq  1 considerable amounts of both reactants and products are present at equilibrium
Keq >> 1 at equilibrium the system contains mostly products; “equilibrium lies to the right”
THE REACTION QUOTIENT
Thus far in this unit, we’ve been considering reversible chemical reactions in equilibrium. What if the system is
not in equilibrium? How does one determine when equilibrium has been reached? How can one predict the
predominant direction of the reaction given existing concentrations of reactants and products?
Answers to all these questions can be found by using the reaction quotient (Q) and comparing the reaction
quotient to the equilibrium constant (Keq).
The reaction quotient, Q, is calculated like Keq is calculated, but instead of inserting equilibrium concentrations,
we insert
.
Consider the following reaction for the synthesis of ammonia:
N2(g) + 3H2 (g)  2NH3(g)
If you’re running an experiment at 472ºC and you want to find out if the system is at equilibrium, you will have to
take samples to find out the concentration of each reactant and product. You collect the following data:
[H2] = 0.10 M
[N2] = 0.0020 M
Keq =
[NH3] = 0.15 M
Substitute the measured concentrations into the equilibrium expression above:
Q=
=
Compare this value to the known value of Keq at 472ºC, which is 0.105
 if Q = Keq, then the system is at equilibrium
 if Q < Keq, then the system is not at equilibrium; there are too many reactants compared to products so
reactants will continue to form more products until the system reaches equilibrium (the reaction proceeds
to the right)
 if Q > Keq, then the system is not at equilibrium; there are too many products compared to reactants so
products will continue to form more reactants until the system reaches equilibrium (the reaction proceeds
to the left)
Problems:
1.
At elevated temperatures, phosphorous pentachloride decomposes partially to phosphorous trichloride and
chlorine. All of these substances are gases. The products and reactants are sampled. The concentration
of PCl5 is found to be 0.80 mol/L, the concentration of PCl3 is 0.20 mol/L, and the concentration of Cl2 is
2.5 mol/L. Given that the equilibrium constant for the reaction is 0.555, is the reaction at equilibrium? If
not, in which direction will it proceed?
Step 1: Write the complete, balanced equation for the reaction.
Step 2: Write the equilibrium expression for the reaction.
Step 3: Substitute the measured concentrations in the equilibrium expression to find Q.
873996897
-4-
2.
Step 4: Compare the value of Q to the value of Keq given in the problem.
At 740ºC, Keq = 0.0060 for the decomposition of calcium carbonate (CaCO3), which is described by the
equation:
CaCO3 (s)  CaO(s) + CO2 (g)
Find Q and predict how the reaction will proceed if [CO2] = 0.0004M.
3.
For the reaction:
CO(g) + H2O(g)  H2 (g) + CO2 (g)
Keq = 5.10 at 527ºC.
If [CO] = 0.15M, [H2O] = 0.25M, [H2] = 0.42M and [CO2] = 0.37M, calculate Q and determine how the
reaction will proceed.
4.
A vial contains 0.150M NO2 and 0.300M N2O4. Calculate Q for the reaction 2NO2 (g)  N2O4(g)
5.
At 448ºC, Keq = 50.5 for the reaction:
H2 (g) + I2 (g) > 2HI(g).
Find Q and predict how the reaction proceeds if [H2] = 0.150M, [I2] = 0.175M, and [HI] = 1.20M
6.
For the reaction A + C  B, Keq = 0.025. What is the equilibrium constant for the reverse reaction?
Explain your answer by showing the equilibrium expression for both the forward and reverse reactions.
873996897
-5-
LE CHATELIER’S PRINCIPLE
Le Chatelier’s Principle:



a change in conditions for a system at equilibrium is often called a
another way to state Le Chatelier’s Principle is to say that a system will respond to a stress in a way that
relieves that stress
stresses include:
1.
change in concentration of a reactant or product
2.
change in pressure of the entire system
3.
change in temperature of the entire system
Changes in Concentration:
 there are two ways to change concentration
1. add product or reactant
a. if a reactant is added so its concentration is increased, the system will return to
equilibrium by consuming some of the reactant(s) and producing more product
b. the same is true if product is added
2. remove product or reactant
a. if a reactant is removed so its concentration is decreased, the system will return to
equilibrium by consuming some product(s) and producing more reactant (s)
b. the same is true if product is added

consider the smog example again where
2NO2 (g)  N2O4 (g)
and
Keq =
If additional NO2 is injected into the system and the reaction quotient, Q, is calculated, the denominator
will be larger and Q < Keq. To return to equilibrium, the reaction will
. If
additional N2O4 is added to the system and Q is calculated, the numerator will be larger and Q > Keq. To
return to equilibrium, the reaction will
. Removing either substance
will have the opposite effect.

Summary:


adding a substance drives the system to consume that substance
removing a substance drives the system to produce more of the substance
note that the value of Keq does not change when concentrations are changed;
only the equilibrium position shifts.
Changes in Pressure:
 Normally a change in pressure is accomplished by increasing or decreasing volume
 Equilibrium will shift to relieve the stress so that…
if pressure is increased, the system will shift in the direction that produces fewer molecules
873996897
-6-

if pressure is decreased, the system will shift in the direction that produces more molecules
Consider the smog example where
2NO2 (g)  N2O4 (g) (2 moles of reactant and 1 mole of product)
if the volume is decreased, which way will the equilibrium shift?
if the volume is increased, which way will the equilibrium shift?

What if the equilibrium system is heterogeneous and there are one or more solid or liquid substances?

Consider the following reaction:
NH4NO3 (s)  N2O (g) + 2H2O (g)
-moles of gas on reactant side =
-moles of gas on product side =
If the volume is decreased, which way will the equilibrium shift?
If the pressure is decreased, which way will the equilibrium shift?

If there are equal numbers of gas molecules/moles on both sides of the equation, changes in pressure have
no impact on the equilibrium position
note that the value of Keq does not change when pressure is changed;
only the equilibrium position shifts
Changes in Temperature:
 Temperature is extremely important in many chemical reactions
 Keq is temperature-dependent
 You can predict how temperature will affect the equilibrium position based on the reaction being
endothermic or exothermic
 Recall that heat can be considered a reactant or product as the following reaction:
H2(g) + I2 (g)  2 HI (g) + heat
Is this reaction endothermic or exothermic?
For this reaction at 400 ºC, Keq = 54.5
For this reaction at 490 ºC, Keq = 45.9
This means that the reaction is less complete (smaller amounts of product) at higher temperature.
WHY?


if heat is produced (exothermic reaction), adding heat by raising the temperature causes the system to
shift to the left
if heat is consumed (endothermic reaction), adding heat by raising the temperature causes the system to
shift to the right
note that the value of Keq DOES change when temperature is changed;
a new equilibrium must be established for the new temperature
873996897
-7-
SOLUBILITY EQUILIBRIA
The Dissolving Process:
Ionic substances are composed of ions with positive and negative charges. Cations are attracted to the negative
ends of nearby water molecules; anions are attracted to the positive ends of nearby water molecules. If the
attraction is strong enough, the anion will be pulled away from the surface of the crystal. This process is called
dissolution (the general term for all solvents and solutes is dissociation). The ion in solution will be surrounded by
a shell of water molecules. An equation can be written to describe this process:
CaCl2(s)  Ca2+(aq) + 2Cl-(aq)
The Precipitation Process:
Ions in solution are in constant motion and therefore are constantly colliding with other particles (other ions or
solvent particles). If an ion of calcium (Ca2+) collides with the surface of a salt crystal, it may lose its surrounding
shell of water molecules and join the crystal structure. If this new ion on the crystal surface creates a charge
imbalance, then a negative ion (Cl-) will be attracted to an adjacent spot in the crystal lattice. The process of ions
joining to make an ionic solid is called precipitation. The solid itself is called a precipitate.
Ca2+ (aq) + 2Cl- (aq)  CaCl2 (s)
Solubility Equilibria:
When the process of dissolution and precipitation are occurring at the same time and at the same rate,
equilibrium is established. The equilibrium can be shown by an equation with two arrows or a double arrow:
CaCl2 (s)  Ca2+ (aq) + 2Cl- (aq)
Note that the equation must be balanced just like other equations so equal numbers of ions (and charges) appear
on both sides of the equation.
Write the solubility equilibrium equation for each of the following ionic compounds:
K2S(s) 
NH4NO3(s) 
As you know, not all ionic substances dissolve in water to the same extent. Those substances that dissolve easily
are said to be
. Those that dissolve only partially are said to be
.
Substances that do not dissolve well are said to be
. Recall that you can use
in your Reference Tables packet to look up qualitative information on the solubility of
many ionic compounds.
Solubility Product:
The qualitative descriptions of solubility above are helpful, but it’s sometimes necessary to know precisely how
much of an ionic compound will dissolve. We use the solubility product constant, Ksp, to describe how much solid
dissolves and therefore how concentrated the resulting solution will be.
Consider the dissolution of zinc sulfide:
ZnS(s)  Zn2+ (aq) + S2- (aq)
Ksp =
As with the equilibrium constant for chemical reactions, the concentrations of ions (like products) are multiplied
in the numerator with coefficients as exponents. The concentration of the solid (like reactants) is in the
denominator. Recall that if a substance is in the solid state, then its concentration is constant and it effectively
gets incorporated into the constant K.
873996897
-8-
Write the equilibrium equation and then the expression for the solubility product constant for the following
compounds when they dissolve in water:
Cu3(PO4)2
SrSO4
Finding solubility products:
As is the case for Keq, determining the value of Ksp must be done
. When the
concentrations of ions in a solution have come to equilibrium, the solution is sampled and analyzed. Consider the
following example for the partial dissolution of silver chloride:
AgCl(s)  Ag+ (aq) + Cl- (aq)
The equilibrium solution is found to contain 1.3 x 10 -5 mol/L of silver ion and the same concentration of chloride
ions. These equilibrium concentrations are inserted into the solubility product expression:
Ksp =
As with Keq, there are no units for Ksp.
- a small value of Ksp indicates low solubility
- a large value of Ksp indicates high solubility
Note the difference between solubility and the solubility product constant. Solubility, you will recall, is the
amount of solute that will dissolve in a given amount of solvent. Solubility is usually expressed as the number of
grams of solute per 100 grams of solvent.
Assuming that there are no complicating factors (such as other ions in solution), the concentration of only one ion
in the solution must be measured to determine the value of K sp. This is because the ratio ions in solution is
constant and can therefore be easily determined from the one measured value. Consider the following example:
A sample of GaF3(s) is added to pure water and allowed to come to equilibrium at 25ºC. The concentration of Ga 3+
is 0.049 M at equilibrium. What is the value of K sp for GaF3?
Step 1. Write the solubility equation.
Step 2. Write the solubility product expression.
Step 3. Determine the relationship between the ion measured and the ion that was not measured to determine the
concentration of this second, unmeasured, ion.
Step 4. Insert the concentrations into the solubility product expression to determine the value for Ksp.
Problems:
A sample of SrCO3(s) is added to pure water and allowed to come to equilibrium at 25ºC. The concentration of
Sr2+ is 4.0 x 10-5 M at equilibrium. What is the value of Ksp for SrCO3?
873996897
-9-
At 18ºC, the concentration of Pb2+ ions in a saturated solution of lead oxalate (PbC2O4) is 5.23 x 10-6 M. What is
the value of Ksp for PbC2O4?
Using solubility products: (TABLE M in your reference tables)
You have seen how to calculate the solubility product constant, K sp, given an equilibrium concentration of an ion in
solution. The reverse process can be used to predict the equilibrium concentrations of ions in a saturated solution
if you know the value of Ksp. Consider the following problem:
Calcium sulfate has a Ksp value of 9.1 x 10-6 at 25ºC. Determine the equilibrium concentrations of the calcium and
sulfate ions in a saturated solution at this temperature.
Step 1. Write the solubility equation.
Step 2. Write the solubility product expression.
Step 3. Determine the relationship between the ions in solution (in other words, how many sulfate ions are there
for every calcium ion?) and simplify the equation by substituting a single variable, x, for the unknown
concentrations.
Step 4. Solve for x.
Problems:
What are the equilibrium concentrations of dissolved ions in a saturated solution of Ag 2SO4 at 25ºC?
(Ksp = 1.2 x 10-5) (Answer: [Ag+] = 2.9 x 10-2 M and [SO42-] = 1.4 x 10-2 M)
What are the equilibrium concentrations of the dissolved ions in a saturated solution of Fe(OH)2 at 25ºC
considering that Ksp = 1.8 x 10-15? (Answer: [Fe2+] = 7.7 x 10-6M and [OH-] = 1.5 x 10-5M)
873996897
- 10 -
What will be the equilibrium concentrations of the dissolved ions in a saturated solution of In(IO 3)3 at 25º C?
(Ksp = 1.05 x 10-4) (Answer: [In3+] = 0.0444 M and [IO3-] = 0.133 M)
PRECIPITATION – USING Ksp AND Q TO PREDICT THE FORMATION OF A PRECIPITATE
Review of Precipitates and Precipitation



precipitate – a solid formed from an aqueous solution
precipitation reactions are usually double replacement reactions
example:
lead (II) nitrate + potassium iodide  lead (II) iodide + potassium nitrate
The reactants are both clear solutions; when mixed, a bright yellow solid forms in a clear solution.
According to Table E, the bright yellow solid is
while the clear solution must be
according to Table E, it is
.



because it is
because,
unsaturated solution – a solution in which additional solute can be dissolved
saturated solution – a solution in which the maximum amount of solute is dissolved for the existing
conditions
supersaturated solution – a solution in which more than the normal amount of solute is dissolved; this is an
unstable, nonequilibrium state
*a precipitate will form from a supersaturated solution
Predicting the Formation of a Precipitate


Ksp (
solution for a saturated, equilibrium state
Q(
actually in solution
) describes the solute ion concentrations in
) describes the solute ion concentrations
Q (the ion product) can be compared to Ksp to determine if an aqueous solution of ions is supersaturated and will
form a precipitate (just as Q could be compared to K eq to determine if a system is at equilibrium)
Q = Ksp…
Q > Ksp…
873996897
- 11 -
Q < Ksp…
Example problem:
Determine if a precipitate will form when 0.010 mol of PbCl2 is dissolved in 150 mL of hot water and then cooled
slowly to 25ºC where the Ksp value is 1.6 x 10-5.
Step 1.
Write the balanced equation for the dissolution of lead (II) chloride.
Step 2.
Write the solubility product expression.
Step 3.
Determine the concentration (molarity) of each ion in solution.
Step 4.
Insert the concentrations from step 3 above into the solubility product expression to find the
value of the ion product (Q).
Step 5.
Compare Q to Ksp to determine if precipitation will occur.
Practice Problems:
1.
Suppose 0.0085 mol of CaSO4 is dissolved in 100.0 mL of distilled water at 60ºC. Will a precipitate form
when it is cooled to 25ºC if Ksp = 2.4 x 10-5 at that temperature?
873996897
- 12 -
2.
0.015 mol of barium hydroxide is dissolved in 500.0 mL of water at 75ºC. Will precipitation occur when
the solution cools to 25ºC? (Ksp = 5.0 x 10-3)
Precipitation Reactions - Net Ionic Equations – Common Ion Effect
Precipitation Reactions:
 most precipitation reactions are double replacement reactions
 a driving force must be present for a reaction to occur (products must be more stable than reactants)
 a reaction will normally occur if one of the following three products is formed
1.
2.
3.
 use the solubility rules (and Table E) to predict if a precipitate will form
 Will a double replacement reaction occur between solutions of copper (II) nitrate and sodium hydroxide?
Even if a precipitation reaction can proceed, actual precipitate formation depends on the concentrations of the
dissolved ions after the solutions are mixed. If the final solution is too dilute, no precipitate can form and the
double replacement reaction doesn’t occur. Only if the ion product (Q) exceeds the solubility product (Ksp) will a
precipitate form. Precipitation will continue until the ion concentrations decrease to the equilibrium levels.
Problem: Will AgBr(s) precipitate if 20.0 mL of 0.010 M AgNO3 and 20.0 mL of 3.0 x 10-4 M KBr are mixed. (Ksp
for AgBr = 5.0 x 10-13) (Hint: you must determine the concentrations of the pertinent ions, Ag+ and Br-, after
the solutions have been mixed.)
AgNO3 (aq) + KBr(aq)  AgBr(s) + KNO3(aq)
Ionic Equations:
The reaction above can be written to show the individual ions instead of showing the overall compounds in solution.
Complete ionic equation:
note that:
 the total charge on each side of the equation is zero
 only the silver ions and bromide ions undergo a chemical change in the reaction
873996897
- 13 -


the potassium ions and nitrate ions are unchanged
spectator ions – ions that do not take part in a chemical reaction
Net ionic equation:
The Common Ion Effect:
Le Chatelier’s Principle can be applied to dissolution and precipitation. In the problem above, suppose sodium
bromide is added to the system. This increases the concentration of the bromide ion and the equilibrium shifts to
the right (more precipitate forms). This is called the “common ion effect.” Likewise, if bromide ions or silver ions
are removed from the solution, the equilibrium will shift to the left and silver bromide will dissolve.
873996897
- 14 -