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1/13/2015
Ch. 15: Chemical Equilibrium
15.1 Chemical Equilibrium
• Not all chemical reactions go to completion. Every
reaction will achieve a state of equilibrium; most are
reversible and will have a combination of reactants and
products present when equilibrium is reached.
• Example: N2O4(g) ⇌ 2NO2(g) (note equilibrium arrows)
Chemistry: Atoms First
Julia Burdge & Jason Overby
• Equilibrium defined:
• Reactant (and product) concentrations no longer
change. It DOES NOT mean reactant and product
concentrations are equal!
• Rate of the forward reaction equals the rate of the
reverse reaction.
Chemical Equilibrium
• Equilibrium is a dynamic state—both the forward and
reverse reactions continue to occur, although there is no
net change in reactant and product concentration over
time.
• At equilibrium, the rates of the forward and reverse
reactions are equal.
• Equilibrium can be established starting with only
reactants, with only products, or with any mixture of
reactants and products.
• At a given temperature, the ratio of products to
reactants will always be the same!
Chemical Equilibrium
Chemical Equilibrium
ratef = kf[N2O4]
and
rater = kr[NO2]2
Starting
with N2O4
15.2 Equilibrium Constant
N2O4(g) ⇌ 2NO2(g)
rate forward = rate reverse
Starting
with NO2
kf[N2O4]eq = kr[NO2]2eq
The subscript “eq” denotes equilibrium concentration.
equilibrium expression
Rearranging
equilibrium constant
The ratio of two constants (kf/kr) is also a constant!
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Equilibrium Expression, Kc
 Write the equilibrium constant expressions (Kc) for the
following equations.
• CO(g) + H2(g)  CH4(g) + H2O(g)
• NH3(g)  N2(g) + H2(g)
• SO2(g) + O2(g)  SO3(g)
15.3 Gaseous Equilibrium
Homogeneous equilibrium: all substances in a reaction are in the
same phase (physical state).
Heterogenous equilibrium: the substances are in different phases.
Kc expressions: use concentrations. Only (g) and (aq).
Kp expressions: use pressures. Only (g).
Pure solids and pure liquids are never included (because they have
constant concentrations).
CO2(g) + C(s) ⇌ 2CO(g)
2Fe(s) + 3H2O(l) ⇌ Fe2O3(s) + 3H2(g)
Worked Examples 15.3 and 15.5
Using the Equilibrium Constant
Worked Example 15.3
Write equilibrium expressions (Kc and Kp) for each of
the following reactions:
(a) CaCO3(s) ⇌ CaO(s) + CO2(g)
(b) Hg(l) + Hg2+(aq) ⇌ Hg22+(aq)
(d) O2(g) + 2H2(g) ⇌ 2H2O(l)
Solution (a) Kc = [CO2]
(c) Kc = [H2]3
Calculate Kp for this reaction at 298 K when PH2 =
1.24 atm, PN2 = 2.17 atm, and PNH3 = 1.39 atm.
2+
(b) Kc = [Hg22+ ]
[Hg ]
(d) Kc =
We can calculate a value for Kc by plugging equilibrium
concentrations (or pressures) of each substance into the
equilibrium expression.
Calculate Kc for the reaction 3H2 (g) + N2 (g)  2NH3 (g)
if [H2]eq = 0.104 M, [N2]eq = 0.554 M, [NH3]eq = 0.418 M
at 298K.
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[O2][H2]2
Worked Examples 15.1, 15.2
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Equilibrium Constant
Equilibrium Constant
N2O4(g) ⇌ 2NO2(g)
N2O4(g) ⇌ 2NO2(g)
Calculate the equilibrium constant for each of the
following sets of equilibrium concentrations:
Trial
1
[N2O4], M
0.643
[NO2], M
0.0547
2
0.448
0.0457
3
0.491
0.0475
4
5
0.594
0.0898
0.0523
0.0204
Kc
Calculate the equilibrium constant for each of the
following sets of equilibrium concentrations:
Trial
[N2O4], M
[NO2], M
Kc
-3
1
0.643
0.0547
4.65x10
2
0.448
0.0457
4.66x10
3
0.491
0.0475
4.60x10
4
0.594
0.0523
4.60x10
5
0.0898
0.0204
4.63x10
-3
-3
-3
-3
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Equilibrium Constant
Equilibrium Constant
The equilibrium constant tells us the extent to which a reaction will
proceed at a particular temperature.
2) The reaction will not occur to any significant degree, and the
equilibrium mixture will consist predominantly of reactant (reactantfavored).
Three outcomes are possible:
1) The reaction will go essentially to completion and the equilibrium
mixture will consist predominately of products (product-favored).
Ag+(aq) + 2NH3(aq) ⇌ Ag(NH3)2+(g)
Kc = 1.5 x 107 (at 25oC)
Large Kc, product favored
N2(g) + O2(g) ⇌ 2NO(g)
Kc = 4.3 x 10–25 (at 25oC)
Small Kc, reactant favored
3) The reaction will proceed a significant degree but will not go to
completion, and the equilibrium mixture will contain comparable
amounts of both reactants and products. What will Kc be in this
case?
Equilibrium Constant
Equilibrium Constant
 A chemical reaction between X2 (red) and Y2 (blue)
 Which system below (all at equilibrium) has the largest
produces XY (red-blue). All compounds are in a gaseous
state. The picture shown here represents the equilibrium
mixture.
equilibrium constant? Why?
 What is the value of Kc?
Gaseous Equilibria
Worked Example 15.6
1) The equilibrium constant, Kc, for the reaction
For the equilibrium: N2O4(g) ⇌ 2NO2(g) we can write expressions
as
[NO ]2
Kc = [N O2 ]
2 4
or
(P
2NO2(g) ⇌ N2O4(g)
)2
KP = PNO2
N2O4
is
Because of PV = nRT, the relationship between Kc and KP can be
expressed as
2.16×102
at
25oC.
What is the value of KP?
Solution
KP = Kc 0.08206 L·atm ×T
K·mol
KP = Kc(RT)Δn)
where Δn = moles of gaseous products – moles of gaseous reactants
and R = 0.08206 L·atm/K·mol.
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= (2.16×102)(0.08206 × 298)-1
= 8.83
Practice at home: For the reaction 2H2S(g) + 3O2(g) ⇌
2H2O(g) + 2SO2(g), Kp = 1.2 x 104 at 25oC. Calculate Kc
for this reaction. Kc = 3.0 x 105
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Equilibrium Constant
Equilibrium Constant
The value of the reaction quotient, Q, changes as the reaction
progresses.
The reaction quotient (Qc ) is a ratio that is set up the same
way as the equilibrium constant (products over reactants;
coefficients are exponents) but the concentrations (or
pressures) are not necessarily at equilibrium.
2NO2(g) ⇌ N2O4(g)
At t = 0, Q = 0
aA + bB ⇌ cC + dD
What happens to Q
as the reaction proceeds?
(only at equilibrium)
Q increases until equilibrium is reached.
(Q = K)
15.4 Using Expressions
Q versus K
The equilibrium expression may be used to predict the direction of a
reaction and to calculate equilibrium concentrations.
Predictions are made based on comparisons between Qc and Kc.
There are three possibilities:
1) Q < K: The ratio of initial concentrations of products to reactants is too
small. To reach equilibrium, reactants must be converted to products.
The system proceeds in the forward direction.
2)
Q = K: The initial concentrations are equilibrium concentrations. The
system is at equilibrium.
3)
Q > K: The ratio of initial concentrations of products to reactants is too
large. To reach equilibrium products must be converted to reactants.
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The system proceeds in the reverse direction.
Equilibrium Constant
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Equilibrium Expressions
When a reversible chemical equation is manipulated, it is also
necessary to make appropriate changes in the equilibrium
expression and the equilibrium constant.
Calculate Q or K for the following examples.
1) N2O4 (g)  2 NO2 (g)
At 100oC, the following concentrations are measured at
equilibrium: [N2O4] = 0.0172 M, [NO2] = 0.00140 M. Calculate
K. Are you solving for Q or K?
K = 1.14 x 10-4
2) H2(g) + I2(g)  2 HI (g) Kc = 57.0 at 700 K
If [H2] = 0.10 M, [I2] = 0.20 M, and [HI] = 0.40 M, is this system
at equilibrium? Are you solving for Q or K?
Q = 8.0; Q < K; shifts right to reach equilibrium
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15.4 Calculating Equil. Conc.
Worked Example 15.4
The following reactions have the indicated equilibrium
constants at 100oC:
(1) 2NOBr(g) ⇌ 2NO(g) + Br2(g)
Kc1 = 0.014
(2) Br2(g) + Cl2(g) ⇌ 2BrCl(g)
Kc2 = 7.2
Determine the value of Kc for the following reactions at
2NO(g) + Br2(g) ⇌ 2NOBr(g)
Kc1' = 71
2BrCl(g) ⇌ Br2(g) + Cl2(g)
Kc2' = 0.14
 If you are given initial concentrations and K values, you
can set up a table to solve for the equilibrium
concentrations (or pressures) of all substances in a reaction.
 Use ICE table (Initial, Change, Equilibrium).
100oC:
 Example: If 1.00 M H2(g) is added to 1.00 M I2 (g), what
will all equilibrium concentrations be for this reaction?
Initial conc. (M)
H2(g) +
I2 (g)
 2 HI(g)
1.00 M
1.00 M
0M
Change in conc. (M)
Equilibrium conc. (M)
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Calculating Equil. Conc.
Perfect Squares
15.4 Calculating Equil. Conc.
 Calculate all equilibrium concentrations for this
 3 methods for solving equilibrium concentrations:
 Use perfect squares (H2 + I2  2HI)
 If initial concentrations are the same; both sides of the
system: H2 (g) + I2 (g)  HI (g), Kc = 50.5 starting
with 1.00 M H2 and 1.00 M I2.
expression become squares.
 Assume x is much smaller than initial concentration
 If K < 10-3, system is reactant-favored and change (x)
will be small
 Use quadratic if neither perfect squares or assuming
small x will work.
 Solution gives two values of x; one is usually negative
and incorrect.
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Calculating Equil. Conc.
Assuming Small x
N2(g) + O2(g)  NO(g), Kc = 1.0 x
0.80 M N2 and 0.20 M O2.
like this?
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Calculating Equil. Conc.
Quadratic
 Calculate all equilibrium concentrations for the system:
 Calculate all equilibrium concentrations for the system:
10-5
 How can you check your answers on questions
PCl5(g)  PCl3(g) + Cl2(g), Kc = 5.8 x 10-2 when 0.320
moles of PCl5 is placed in 2.00 L container.
starting with
 Not perfect squares and Kc isn’t very small. Have to use
quadratic. Also have to start with Molarity!
 For Fun: You can sing the formula to the tune of “Pop Goes the
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Weasel”. Check the interwebs….
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Calculating Equil. Conc.
Calculating Equil. Conc.
A 5.60 L reaction vessel is filled with 0.215 moles of H2,
0.215 moles of I2, and 6.13 moles of HI. Kc = 129 at 500 K.
In an experiment, 3.00 moles of CO and 5.00 moles of H2O
were placed in a 2.00 L flask.
CO (g) + H2O (g)  CO2 (g) + H2 (g)
At equilibrium, there were 2.00 mol of CO remaining.
What is the value of the equilibrium constant for this
reaction?
 2HI (g),
H2(g) + I2(g)
0.03839 M, 0.03839 M, 1.0946 M
Is it at equilibrium? In which direction will it proceed?
Initial
What are the equilibrium concentrations?
Change
Equilibrium
Worked Example 15.9
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15.5 Factors Affecting Equilibria
Le Châtelier’s principle states that when a stress is applied to a
system at equilibrium, the system, will respond by shifting in
the direction that minimizes the effect of the stress.
•Stress refers to any of the following:
• The addition of a reactant or product
• The removal of a reactant or product
• A change in temperature
• A change in volume of the system, resulting in a
change in concentration or partial pressure of the
reactants and products
CO (g) +
H2O (g) 
1.50 M
2.50 M
0
0
-x
-x
+x
+x
1.00 M
2.00 M
0.500 M
0.500 M
CO2 (g) +
H2 (g)
K = 0.125; Worked Example 15.9
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Le Chatelier: Concentrations
Consider the Haber process at 700 K:
N2(g) + 3H2(g) ⇌ 2NH3(g)
At equilibrium:
[N2] = 2.05 M
[H2] = 1.56 M
[NH3] = 1.52 M
Applying stress by the addition of N2 to give the following
concentrations:
[N2] = 3.51 M
[H2] = 1.56 M
[NH3] = 1.52 M
The reaction shifts to the right.
N2(g) + 3H2(g) ⇌ 2NH3(g)
Le Chatelier: Concentrations
Le Chatelier: Concentrations
N2(g) + 3H2(g) ⇌ 2NH3(g)
Addition of a reactant or removal of a product will cause an
equilibrium to shift to the right.
Addition of a product or removal of a reactant will cause an
equilibrium to shift to the left.
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Le Chatelier: Temperature
A change in temperature can alter the value of the
equilibrium constant. Changes in volume and concentration
do not change the value of the equilibrium constant.
Heat + N2O4(g) ⇌ 2NO2(g)
ΔH° = + 58.0 kJ/mol
Le Chatelier: Temperature
For any endothermic reaction, heat can be viewed as a
reactant:
heat + reactants ⇌ products
• Adding heat:  (Kc increases)
• Removing heat:  (Kc decreases)
ΔHo > 0 kJ/mol
For any exothermic reaction, heat can be viewed as a
product:
Because the processes is
endothermic, adding heat
shifts the equilibrium
toward products
Le Chatelier: Temperature
CoCl42- + 6H2O ⇌ Co(H2O)62+ + 4Clblue
pink
reactants ⇌ products + heat
•
•
ΔHo < 0 kJ/mol
Adding heat:  (Kc decreases)
Removing heat:  (Kc increases)
Le Chatelier: Volume/Pressure
When volume is decreased, the equilibrium is driven
toward the side with fewer moles of gas.
N2O4(g) ⇌ 2NO2(g)
The reaction shifts to the left.
N2O4(g) ⇌ 2NO2(g)
Volume decreases by half
Le Chatelier: Volume/Pressure
When pressure is decreased, volume increases.
Equilibrium is driven toward the side with more moles of
gas.
N2O4(g) ⇌ 2NO2(g)
The reaction shifts to the right.
N2O4(g) ⇌ 2NO2(g)
Volume is doubled
Le Chatelier: Volume/Pressure
In which direction will each of the following
equilibrium systems shift when the changes below are
made?
1) H2 (g) + Cl2 (g) ⇌ 2 HCl (g)
a) Volume is increased
b) Pressure is increased
2) C (s) + H2O (g) ⇌ CO (g) + H2 (g)
a) Volume is decreased
b) Pressure is decreased
Worked Examples 15.11, 15.12
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Factors Affecting Equilibria
•
•
Catalysts act by reducing the activation energy of a reaction
(Section 14.8) which occurs to the same extent for both the
forward and reverse reactions.
As a result the addition of a catalyst reduces the time
required to reach equilibrium but has no effect on
equilibrium constants of the position of equilibrium.
Factors Affecting Equilibria
 Determine how the equilibrium will shift if the following
changes are made:
 2 Cl2(g) + 2 H2O(g)  4 HCl(g) + O2(g)







Ho = +113 kJ
Temperature is increased
Volume is increased
Pressure is increased
HCl is added
Catalyst is added
Cl2 is added
H2O is removed
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