AP Chemistry 10: Equilibrium—General
A.
The Equilibrium State (15.1 to 15.6)
1. N2O4(g)  2 NO2(g)
a.
reversible because reactants and products are
B.
confined in same state and Ea  Ea’
b. reaction rates change as [R] and [P] change until
ratef = kf[N2O4] = rater = kr[NO2]2
EQUAL REACTION RATES NOT CONCENTRATIONS
c. law of mass action:
1. K = kf/kr = [NO2]2/[N2O4]
2. K = [Products]/[Reactants]
3. liquid and solid are not included in expression
because [ ] can’t change (part of K)
4. same as “Q” in Nernst equation
2. equilibrium constant, K
a. unit-less quantity
b. independent of original concentrations
c. depends on temperature
d. Kc vs. Kp for gaseous systems
1. concentration in mol/L, then Kc
2. concentration in atm, then Kp
3. Kp = Kc x (RT)n(gas) R = 0.0821
e. size of K and the reactant/product balance
1. large K, means [products] > [reactants]
2. small K, means [products] < [reactants]
f. if reaction is written backwards, then: K' = K-1
g. if coefficient x factor "n", then K’' = Kn
h. combining reactions to form a new equilibrium
2 NOBr(g)  2 NO(g) + Br2(g)
K1
Br2(g) + Cl2(g)  2 BrCl(g)
K2
2 NOBr(g) + Cl2(g)  2 NO(g) + 2 BrCl(g) K1 x K2
i. K, Go and Eo
1. equilibrium: G = E = 0, but Go and Eo  0
2. "o" [ ] = 1 M and P = 1 atm (not always 298 K)
3. Go = -RTlnK and Eo = (RT/nF)lnK
(R = 8.31, T in K, n = moles e-, F = 96,500)
4. when K > 1, then Go < 0, and Eo > 0
3. equilibrium problems
a. determine direction ( or  from [ ]o

substitute [ ]o into equilibrium expression = Q

if Q > K, then , if Q < K, then 
b. determine K, given [ ]E

write expression from equation

substitute [ ]E, solve for K
c. determine K, given [ ]o and one [ ]E

set up an "ICE Box" (shaded boxes are given)
o I is initial concentration
o C is change in concentration
o E is equilibrium concentration
[]
A
+
2B

C
+
3D
[A]o
[B]o
[C]o
[D]o
I
C
-x
-2x
[C]E – [C]o = x
3x
E
[A]o - x
[B]o – 2x
[C]E
[D]o + 3x

write expression from balanced equation

substitute [ ]E , solve for K
d. determine one [ ]E, given other [ ]E and K

write expression from equation

substitute given [ ]E and K, solve for missing [ ]E
e.
determine [ ]E, given [ ]o and K

set up an "ICE Box" (shaded boxes are given)
[]
A
+
2B

C
+
3D
[A]o
[B]o
[C]o
[D]o
I
-x
-2x
+x
+3x
C
[A]o – x
[B]o – 2x
[C]o + x
[D]o + 3x
E

write expression from balanced equation

substitute [ ]E and K, solve for x

substitute x back into formulas solve for [ ]E
Le Chatelier's Principle (15.7)
1. when a system at equilibrium is disturbed, the system
shifts ( or ) to minimize the disturbance
2. determine response using Le Chatelier’s principle
a. adding reactant or product: system shifts away from
added species (K is unchanged)
1. increase [R]: Q < K  shifts to the right
a. increase [ ]E of products + added reactant
b. decrease [ ]E of other reactants 
c. example: 3 H2(g) + N2(g)  2 NH3(g)
b.
c.
d.
2. increase [P]: Q > K  shifts to the left
removing reactant or product: system, shifts toward
removed species (K is unchanged)
1. decrease [R]: Q > K  shifts to the left
2. decrease [P]: Q < K  shifts to the right
change temperature: system shifts to restore
equilibrium temperature (K changes)
1. increase T: initial rateendo > initial rateexo
(shifts toward +H or away from –H)
2. decrease T: initial rateendo < initial rateexo
(shifts toward –H or away from +H)
3. predict change in K
a. shifts to the right  K increased
b. shift to the left  K decreased
change volume of the container: system shifts to
restore equilibrium [ ]gases (K is unchanged)
1. ngas reactant > ngas product
a. increase V decreases [ ]gases: Q > K
shifts to the left (toward more gas)
b. decrease V increases [ ]gases: Q < K
shifts to the right (toward less gas)
2. ngas reactant < ngas product
a. increase V decreases [ ]gases: Q < K
shifts to the right (toward more gas)
b. decrease V increases [ ]gases: Q > K
shifts to the left (toward less gas)
3. ngas reactants = ngas product: Q = K
 system is unresponsive
e.
adding species that are neither reactants or
products has no effect on equilibrium
1. catalyst (affect both rates equally)
2. inert substances ([ ] and T are unaffected)
AA
Practice Problems
A. The Equilibrium State
Write equilibrium expressions for the following gas systems.
__[SO3]2__
2 SO2(g) + O2(g)  2 SO3(g)
[SO2]2[O2]
[Ni(CO)4]
Ni(s) + 4 CO(g)  Ni(CO)4(g)
[CO]4
[NO2]4[H2O]6
4 NH3(g) + 7 O2(g)  4 NO2(g) + 6 H2O(g)
[NH3]4[O2]7
2. The following represents the equilibrium: A  B. Initially
the container (on the left) holds pure A. Each frame
represents 1 minute of time.
0 min
1 min
2 min
3 min
4 min
A
A
A
A
A
B
A
B
A
B
A
A
A
B
B
A
A
A
B
A
B
A
B
B
A
A
A
A
A
A
A
A
B
A
B
A
B
B
B
A
A
A
A
A
B
A
A
B
B
B
a. At what time does the system reach equilibrium?
3 minutes.
b. Calculate the equilibrium constant, K.
AA
B
AA
a.
initial state
Calculate Kc.
b.
6.
7.
For the reaction: N2(g) + 3 F2(g)  2 NF3(g) at equilibrium,
PN2 = 0.021 atm, PF2 = 0.063 atm and PNF3 = 0.48 atm.
a. Calculate Kp.
Kp = (PNF3)2/(PN2)(PF2)3
Kp = (0.48 atm)2/(0.021 atm)(0.063 atm)3 = 4.4 x 104
b. Determine the equilibrium concentrations of N2, F2 and
NF3 in mol/L at 25oC.
M = P/RT
MN2 = 0.021/(0.0821)(298) = 8.6 x 10-4 M
MF2 = 0.063/(0.0821)(298) = 2.6 x 10-3 M
MNF3 = 0.48/(0.0821)(298) = 2.0 x 10-2 M
c. Calculate Kc.
Kc = [NF3]/[N2][F2]3
Kc = (2.0 x 10-2 M)2/(8.6 x 10-4 M)(2.6 x 10-3 M)3
Kc = 2.6 x 107
d. Calculate Kp using the equation, KP = Kc x (RT)ng.
Kp = Kc x (RT)ng
Kp = (2.6 x 107)[(0.0821)(298)]-2 = 4.3 x 104
The following represents the system reaching equilibrium.
A2(g) + B(g)  A2B(g)

equilibrium state
2 SO3(g)  2 SO2(g) + O2(g)
Kc = 1/0.120 = 8.33
8.
2 SO2(g) + O2(g)  2 SO3(g)
Calculate Kc for
a. SO3(g)  SO2(g) + ½ O2(g)
Kc = 56
Kc = 56-½ = 0.13
b.
Calculate K for experiment 1.
K = [SO3]2/[SO2]2[O2] = (0.260)2/(0.590)2(0.0450) = 4.32
AAB
AAB
Given the following Kc values, calculate the corresponding
values of Kp at 298 K.
a. CO2(g) + H2(g)  CO(g) + H2(g)
Kc = 0.0431
Kp = Kc(RT)ng
KP = 0.0431[(0.0821)(298)]0 = 0.0431
b. 2 SO3(g)  2 SO2(g) + O2(g)
Kc = 3.2 x 10-4
Kp = Kc(RT)ng
Kp = 3.2 x 10-4[(0.0821)(298)]1 = 0.0078
2 SO2(g) + O2(g)  2 SO3(g)
Kc = 0.120
What are the equilibrium constants for the following?
a. SO2(g) + ½ O2(g)  SO3(g)
b.
Calculate K for experiment 2.
B
Kp = Kc x (RT)ng = (4) x [(0.0821)(298)]-1 = 0.16
The following results were collected for the equilibrium:
2 SO2(g) + O2(g)  2 SO3(g) for different initial [ ]o.
Experiment 1
Experiment 2
[ ]o
[]
[ ]o
[]
2.00 M
1.50 M
0.500 M
0.590 M
SO2
1.50 M
1.25 M
0M
0.0450 M
O2
3.00 M
3.50 M
0.350 M
0.260 M
SO3
a. Write the equilibrium expression for the reaction
c.
AAB
Calculate Kp at 25oC.
Kc = (0.120)½ = 0.346
K = [SO3]2/[SO2]2[O2] = (3.50)2/(1.50)2(1.25) = 4.36

B
K = [B]/[A] = 6/4 = 1.5
b.
5.
B
AA
Kc = [A2B]/[A2][B] = (4)/(1)(1) = 4
K = [SO3]2/[SO2]2[O2]
4.
AAB
AA
B
1.
3.
B
AA
4 SO2(g) + 2 O2(g)  4 SO3(g)
Kc = 562 = 3100
Given the following data at 25oC
2 NO(g)  N2(g) + O2(g)
Kc = 1 x 1030
2 NO(g) + Br2(g)  2 NOBr(g)
Kc = 2 x 103
Calculate Kc for the reaction:
N2(g) + O2(g) + Br2(g)  2 NOBr(g)
N2(g) + O2(g)  2 NO(g)
K = (1 x 1030)-1
2 NO(g) + Br2(g)  2 NOBr(g)
K = 2 x 103
N2(g) + O2(g) + Br2(g)  2 NOBr(g)
K = 2 x 10-27
10. Given:
H2(g) + S(s)  H2S(g)
Kc = 1.0 x 10-3
S(s) + O2(g)  SO2(g)
Kc = 5.0 x 106
calculate Kc for the reaction:
H2(g) + SO2(g)  H2S(g) + O2(g).
SO2(g)  S(s) + O2(g)
Kc = (5.0 x 106)-1
H2(g) + S(s)  H2S(g)
Kc = 1.0 x 10-3
H2(g) + SO2(g)  H2S(g) + O2(g)
Kc = 2.0 x 10-10
11.
Cl2(g)  2 Cl(g)
Kpo is 1.0 x 10-37
Calculate Go for the above reaction.
G = -RTlnKp
G = -(8.31)(298)(ln1.0 x 10-37) = 2.1 x 105 J
12.
2 SO2(g) + O2(g)  2 SO3(g)
Kpo = 2.3
a. What is Go?
Go = -RTlnKp
Go = -(8.31)(298)ln(2.3) = -2100 J
b. What is Eo?
Eo = (RT/nF)lnK
Eo = (8.31)(298)/(4)(96,500)ln2.3 = 5.3 x 10-3 V
9.
13. Given the following data at 25oC
2 NO(g) + Br2(g)  2 NOBr(g)
Kc = 100
2 NO(g)  N2(g) + O2(g)
Kc = 0.50
a. What is Kc for the following equilibrium?
N2(g) + O2(g) + Br2(g)  2 NOBr(g)
2 NO(g) + Br2(g)  2 NOBr(g)
Kc = 100
N2(g) + O2(g)  2 NO(g)
Kc = 2
N2(g) + O2(g) + Br2(g)  2 NOBr(g)
Kc = 200
b. What is Kp at 25oC?
Kp = Kc x (RT)ng = (200) x [(0.0821)(298)]-1 = 8.17
c.
What is Go for the equilibrium at standard conditions?
Go = -RTlnK = -(8.31)(298)ln(8.17) = -5.2 kJ
COCl2(g)  CO(g) + Cl2(g)
Kc = 1.0 x 10-5
What direction will the reaction proceed if 2.0 x 10-3 M
COCl2, 3.3 x 10-6 M CO, and 6.6 x 10-6 M Cl2 are mixed?
Q = [CO](Cl2]/[COCl2]
Q = (3.3 x 10-6)(6.6 x 10-6)/(2.0 x 10-3)
Q = 1.1 x 10-8 < KC 
15. The following represents three initial states for the system:
A2 + B2  2 AB (Kc = 1.5). Indicate the direction that the
system proceeds from its initial state to reach equilibrium.
=


AB
AA
AB
AA
BB
AB
BB
AB
BB
AB
BB
AB
AB
AB
AA
AB
AA AB
BB
AA
AA
BB
AA
AB
14.
N2(g) + O2(g)  2 NO(g)
Kc = 6.2 x 10-4
What is the initial direction of reaction when 0.052 mol N2,
0.0124 mol O2, and 0.0020 mol NO are added to a 5 L flask?
Q = [NO]2/[N2][O2]
Q = (0.0020 mol/5)2/(0.052 mol/5 L)( 0.0124 mol/5 L)
Q = 0.0064 > Kc 
17.
H2(g) + CO2(g)  H2O(g) + CO(g)
[H2] = 0.61 M, [CO2] = 1.6 M, [H2O] = 1.1 M and [CO] = 1.4 M.
Calculate Kc.
16.
Kc = [H2O][CO]/[H2][CO2] = (1.1)(1.4)/(0.61)(1.6) = 1.6
18.
Br2(g)  2 Br(g)
Given the equilibrium concentration: [Br2] = 0.97 M and
[Br] = 0.034 M, calculate Kc.
Kc = [Br]2/[Br2] = (0.034)2/(0.97) = 1.2 x 10-3
19.
A(g)  B(g) + 2 C(g).
Complete the chart, then calculate Kc for the equilibrium.
A

B
+
2C
[]
0.75
0.0
0.0
I
-0.15
+0.15
+0.30
C
0.60
0.15
E
0.30
Kc = [C]2[B]/[A] = (0.30)2(0.15)/(0.60) = 0.023
20.
SO2(g) + ½ O2(g)  SO3(g)
0.012 moles of O2 and 0.027 moles of SO2 are placed in a
sealed one-liter container. At equilibrium, the
concentration of SO3 is 0.020 M. Calculate Kc.
[]
SO2
+
½ O2
SO3
.027
.012
.00
I
-.02
-.01
.02
C
.007
.002
.02
E
Kc = [SO3]/[SO2][O2]½ = (.020)/(.007)(.002)½ = 64
21.
2 NOCl(g)  2 NO(g) + Cl2(g)
At 220oC in a sealed flask, initial concentrations are 0.520 M
NOCl, 0.010 M NO, and 0.053 M Cl2. At equilibrium is
4.23% of the NOCl is decomposed. Calculate Kc.
2 NOCl

2 NO
+
Cl2
[]
. 520
.010
.053
I
-.022
.022
.011
C
.498
.032
.064
E
Kc = [NO]2[Cl2]/[NOCl]2
Kc = (0.032)2(0.064)/(0.498)2 = 2.6 x 10-4
22.
2 NO(g) + Br2(g)  2 NOBr(g)
0.100 mol NO and 0.050 mol Br2 are placed in a 1.0-L flask.
At equilibrium [NOBr] = 0.060 M. Calculate Kc.
2 NO(g)
+
Br2(g)

2 NOBr(g)
[]
0.100
0.050
0
I
-0.060
-0.030
+0.060
C
0.040
0.020
0.060
E
Kc = [NOBr]2/[NO]2[Br2] = (0.060)2/(0.040)2(0.020) = 110
23.
H2(g) + I2(g)  2 HI(g)
0.200 mol H2 and 0.100 mol I2 are placed in a 2.0-L flask. At
equilibrium, 48.0 % of the H2 is consumed. Calculate Kc.
[]
H2
+
I2

2 HI
.100
.050
0
I
-.048
-.048
+.096
C
.052
.002
.096
E
Kc = [HI]2/[H2][I2] = (.096)2/(.052)(.002) = 89
24.
2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g)
The initial concentrations are 0.100 M NO, 0.050 M H2,
and 0.100 M H2O. At equilibrium [NO] = 0.070 M.
25.
26.
27.
28.
29.
30.
31.
a. Calculate the [H2]E, [N2]E, and [H2O]E.
2 NO(g) + 2 H2(g)  N2(g) + 2 H2O(g)
[]
0.100
0.050
0.000
0.100
I
-0.030
-0.030
+0.015
+0.030
C
0.070
0.020
0.015
0.130
E
b. Calculate Kc.
Kc = [N2][H2O]2/[NO]2[H2]2
Kc = (0.015)(0.130)2/(0.070)2(0.020)2 = 130
The following represents the system: A2 + B  A + AB.
AA
A
AA
AB
AB
AB
A
AB
Given Kp = 2, how many Bs should be added to the
diagram in order to represent equilibrium?
Kp = (PA)(PAB)/(PA2)(PB)
2 = (2)(4)/(2)PB  PB = 2  2 B atoms
2 BrCl(g)  Br2(g) + Cl2(g)
Kc = 0.040
What is [BrCl] when [Cl2] = [Br2] = 0.0325 M?
Kc = [Br2][Cl2]/[BrCl]2
0.040 = [0.0325]2/[BrCl]2  [BrCl] = 0.16 M
2 N2O(g) + O2(g)  4 NO(g) Kc = 2.9 x 10-26
What is [NO] when [N2O] = 0.0035 M and [O2] = 0.0027 M?
Kc = [NO]4/[N2O]2[O2]
2.9 x 10-26 = [NO]4/(0.0035)2(0.0027)  [NO] = 5.6 x 10-9 M
N2(g) + 3 H2(g)  2 NH3(g)
Kp = 1.5 x 10-5
At equilibrium PN2 = 0.40 atm, PH2 = 0.10 atm, what is PNH3?
Kp = [NH3]2/[N2][H2]3
1.5 x 10-5 = [NH3]2/(0.40)(0.10)3  [NH3] = 7.7 x 10-5 atm
H2(g) + I2(g)  2 HI(g)
Kc = 81
What are the equilibrium concentrations of the three gases?
[]
H2
+
I2

2 HI
0.50
0.50
0
I
-x
-x
+2x
C
0.50 – x
0.50 – x
2x
E
K = [HI]2/[H2][I2]
81 = (2x)2/(.50 – x)2  x = 0.41
[H2] = [I2] = 0.50 – 0.41 = 0.09 M
[HI] = 2x = 2(0.41) = 0.82 M
2 HI(g)  H2(g) + I2(g)
Kc = 0.012
[HI]o = 0.050 M, [H2]o = [I2]o = 0.100 M. What are the
equilibrium concentrations of all species?
[]
2 HI(g)

H2(g)
+
I2(g)
0.050
0.100
0.100
I
+2x
-x
-x
C
0.05 + 2x
0.1 – x
0.1 – x
E
Kc = [H2][I2]/[HI]2
0.012 = (0.1 – x)2/(0.050 + 2x)2  x = 0.077
[H2] = [I2] = 0.1 – 0.077 = .023 M
[HI] = 0.05 + 2(0.077) = .204 M
SO2(g) + NO2(g)  NO(g) + SO3(g)
Kc = 85
What will be the equilibrium concentrations of the four gases
if the initial concentrations of SO2 and NO2 are 0.075 M?
SO2(g) + NO2(g)  NO(g) + SO3(g)
[]
0.075
0.075
0
0
I
-x
-x
+x
+x
C
0.075-x
0.075-x
x
x
E
Kc = [NO][SO3]/[SO2][NO2]
85 = x2/(0.075 – x)2  x = 0.068 M
[NO] = [SO3] = 0.068 M
[SO2] = [NO2] = 0.075 – 0.068 = 0.007M
PCl5(g)  PCl3(g) + Cl2(g)
Kc = 0.050
A sealed 4.0-L vessel initially contains 0.025 mol PCl5.
a. What are the equilibrium concentrations of all species?
PCl5

PCl3
+
Cl2
[]
0.00625
0
0
I
-x
x
x
C
0.00625 - x
x
x
E
Kc = [PCl3][Cl2]/[PCl5]
0.050 = (x)2/(0.00625 – x)  x = 0.0056
[PCl3] = [Cl2] = 0.0056 M
[PCl5] = 0.00625 - 0.0056 = 0.00063 M
b. What is the equilibrium pressure at 250oC?
Mtot = 2(0.0056 M) + 0.00063 M = 0.0118 M
P = MRT = (0.0118)(0.0821)(523) = 0.51 atm
B. Le Chatelier's Principle
33. Determine the shift in an equilibrium system when it is
stressed using Le Chatelier's principle.
Condition
Stress
Shift (, ) Change K
32.
All
Exothermic
Endothermic
2 A(g)  B(g)
A(g)  2 B(g)
A(g)  B(g)
 Reactant

 Reactant

 Product

 Product

 Temperature


 Temperature


 Temperature


 Temperature


 Volume

 Volume

 Volume

 Volume

 Volume
0
 Volume
0
All
Add Inert Gas
0
34. Consider the reaction: 2 SO2(g) + O2(g)  2 SO3(g).
a. What is H? (HfoSO2 = -296.8 kJ, HfoSO3 = -395.7 kJ)
H = 2 HfoSO3 – 2 HfoSO2 – HfoO2
H = 2(-395.7) – 2(-296.8) – 0.0 = -197.8 kJ
b. How will each shift the equilibrium position and change K?
Change
Equilibrium shift
Change K
No change
Add O2

Remove SO3

No change
Increase V

No change
Add He(g)
no response
No change
K decreases
Increase T

35. The following represents the system: A2 + B  A + AB at
300 K (left side) and 500 K (right side). Is this an
exothermic or endothermic process? Explain your
reasoning.
300 K
500 K
B
A
B
AB
AB
AB
AB
A
AA
A
AB
AA
AA
A
AA
AB
AB
B
AB
B
Endothermic. Higher temperature favors the
endothermic reaction. When comparing the left with the
right, there is more product on the right  the reaction
proceeded in the forward direction  the forward
reaction is endothermic.
Questions 36-38 Indicate the direction ( or ) that the
equilibrium system will shift in response to each disruption.
36.
A(g) + 2 B(g) + Heat  C(g) + D(g)
Disruption
Response
Chemical A is added

37.
Chemical C is taken away

Temperature is increased

Volume is decreased

A(g) + B(g)  2 C(g) + 3 D(g) + Heat
Disruption
Response
Chemical A is removed

Chemical C is added

Temperature is decreased

Volume is decreased

38.
3A(g) + Heat  2 C(g) + D(g)
Disruption
Response
Chemical A is removed

Chemical C is added

Temperature is decreased

Volume is decreased
No response
Practice Multiple Choice
Briefly explain why the answer is correct in the space provided.
Questions 1-10. Consider the equilibrium: 2 A(g)  B(g) + C(g).
1. Which is the correct equilibrium expression?
(A) K = [B][C]/[A]2
(B) K = [A]2/[B][C]
(C) K = [B][C] – [A]
(D) K = [A]2 + [B][C]
K = [products]/reactants
A
K = [B][C]/[A]2
2. Calculate Kc at 25oC given the equilibrium concentrations:
[A] = 0.50 M, [B] = 1.0 M, and [C] = 4.0 M.
(A) 8.0
(B) 0.013 (C) 200
(D) 16
K = [B][C]/[A]2
D
K =(1.0)(4.0)/(0.50)2 = 16
3. Calculate Kp for the equilibrium at 25oC.
(A) 4.5
(B) 2000
(C) 8.0
(D) 16
D
4.
B
Kp = Kc x (R)(T)ngas = Kc = 16
Calculate Go for the equilibrium at 25oC when all the
species are at 1 atm partial pressure.
(A) -(0.0821)(298)lnK
(B) -(8.31)(298)lnK
(C) (8.31)(298)lnK
(D) -(8.31)(25)lnK
G = -RTlnK = (8.31 J/mol•K)(298 K)lnK
5.
In another experiment 1.0 mol of B and 1.0 mol of C are
placed in an evacuated 1.0-L vessel at 25oC. What is the
concentration of A after equilibrium is established?
(A) 0.050 M (B) 0.11 M (C) 0.22 M (D) 0.35 M
[A]E = 2x, [B]E = [C]E = (1 – x), K = [B][C]/[A]2
C
K = (1.0 – x)2/(2x)2 = 16  x = 0.11 and [A] = 0.22 M
6. 1.0 M A is placed in a vessel at 100oC. At equilibrium, [C]E
= 0.40 M. What is the equilibrium concentration of A?
(A) 0.20 M (B) 0.55 M (C) 0.40 M (D) 0.90 M
A
[A]E = 1.0 – 2(.40) = 0.20 M
What is Kc at 100oC?
(A) 0.25
(B) 4.0
(C) 16
K = [B][C]/[A]2
B
K = (0.40)(0.40)/(0.20)2 = 4
7.
(D) 2.0
8.
Which of the following is correct concerning this equilibrium?
(A) The forward reaction is exothermic because raising
temperature favors the exothermic reaction.
(B) The forward reaction is endothermic because raising
temperature favors the exothermic reaction.
(C) The forward reaction is exothermic because raising
temperature favors the endothermic reaction.
(D) The forward reaction is endothermic because raising
temperature favors the endothermic reaction.
K  when T  higher T favored reverse reaction and
C
higher T favors endothermic reaction  exothermic.
9. Which is correct concerning this equilibrium?
(A) Increase [A] shifts the equilibrium to the left.
(B) Increase [B] shifts the equilibrium to the right.
(C) Decrease [C] shifts the equilibrium to the left.
(D) Increasing the volume has no effect on the equilibrium.
 [A] shifts right, [B] shifts left,  [C] shifts right,
D
increase volume has no effect gas = 0
10. What is Kc at 25oC for the equilibrium A(g)  ½ B(g) + ½ C(g)?
(A) 4.0
(B) 0.063 (C) 8.0
(D) 260
A
The coefficients are halved  Kc' = (Kc)½ = (16)½ = 4.0
11. Which must be true for an equilibrium initially at standard
state that proceeds spontaneously in the forward direction?
(A) Go > 0 and Keq > 1 (B) Go > 0 and Keq < 1
(C) Go < 0 and Keq > 1 (D) Go < 0 and Keq < 1
Spontaneous from standard state Go < 0. [ ] products
C
> [ ] reactants  Keq = [P]/[R] > 1
12. The equilibrium constant for a reaction is 200. What is K
for the reverse reaction at the same temperature?
(A) -200
(B) -0.005 (C) 0.002 (D) 0.005
The equilibrium constant for the reverse reaction = 1/K.
D
1/200 = 0.005
13.
4 HCI(g) + O2(g)  2 Cl2(g) + 2 H2O(g)
Equal moles of HCI and O2 are added to an evacuated
vessel. Which must be true for the system at equilibrium?
(A) [HCI] < [Cl2]
(B) [HCI] > [Cl2]
(C) [HCl] > [O2].
(D) [Cl2] = [H2O]
Same coefficients  [Cl2] = [H2O]. [HCl] decreases
D
faster than [O2]  [O2] > [HCl]. [HCl] ? [Cl2] w/o K
14.
Cu + 2 Ag+  Cu2+ + 2 Ag
If the equilibrium constant is 3.7 x 1015, which of the
following correctly describes the standard voltage, Eo, and
the standard free energy change, Go, for this reaction?
(A) Eo > 0 and Go < 0
(B) Eo < 0 and Go > 0
(C) Eo > 0 and Go > 0
(D) Eo < 0 and Go < 0
When K > 1  spontaneously from standard condition.
A
Eo > 0 and Go < 0 because spontaneous process.
15. For the reaction A(g)  B(g) + C(g), Kp, = 2 x 10-4 at 25oC.
A mixture of the three gases at 25oC is placed in a flask
and the initial pressures are PA = 2 atm, PB = 0.5 atm, and
PC = 1 atm. Which is true at the instant of mixing?
(A) G < 0 (B) G > 0 (C) S = 0 (D) Go = 0
Q = [B]o[C]o/[A]o = (0.5)(1)/(2) = 0.25 > K  reaction
B
proceeds to the left and G > 0.
16.
CuO + H2(g)  Cu + H2O
H = -2.0 kJ
The equilibrium can be shifted to favor the products by
(A) increasing the volume at constant temperature.
(B) increasing the pressure by adding an inert gas.
(C) decreasing the temperature.
(D) releasing H2(g) at constant pressure and temperature.
Increase PH2, decreasing temperature or adding H2(g)
C
will shift the equilibrium to the right.
17.
D
18.
B
19.
A
2 SO3(g)  2 SO2(g) + O2(g)
After equilibrium is established, some pure O2(g) is injected
into the reaction vessel at constant temperature. After
equilibrium is reestablished, which of the following has a
lower value compared to its original equilibrium value?
(A) Keq for the reaction
(B) The amount of SO3(g) in the reaction vessel
(C) The amount of O2(g) in the reaction vessel
(D) The amount of SO2(g) in the reaction vessel
Adding O2 shifts left. SO3 goes up and SO2 goes down.
O2 goes up then down but ends up higher. K is same
2 NO(g) + O2(g)  2 NO2(g)
H < 0
Which of the following changes alone would cause a
decrease in the value of Keq for the reaction above?
(A) Decreasing the temperature
(B) Increasing the temperature
(C) Decreasing the volume of the reaction vessel
(D) Increasing the volume of the reaction vessel
Only changing temperature changes K. Exothermic
 T shift system to the left, which lowers K.
2 SO2(g) + O2(g)  2 SO3(g)
When 0.40 mole of SO2 and 0.60 mole of O2 are placed in
an evacuated 1.00 L flask. After reaching equilibrium the
flask contains 0.30 mole of SO3. Kc, is
(A) (0.30)2/(0.45)(0.10)2 (B) (0.30)2/(0.60)(0.40)2
(C) (0.60)/(0.45)(0.20)
(D) (0.30)/(0.45)(0.10)
.30 mol SO2 + .15 mol O2  .30 mol SO3 
[SO2]E = .10 M, [O2]E = .45 M and K = (0.30)2/(0.10)2(0.45)
Practice Free Response
1.
Given:
2 NO(g) + Br2(g)  2 NOBr(g)
Kc = 2.0
2 NO(g)  N2(g) + O2(g)
Kc = 2.1 x 1030
What is Kc for the following equilibrium?
N2(g) + O2(g) + Br2(g)  2 NOBr(g)
The second reaction is reversed and added to the first
 K = (2.0)(2.1 x 1030)-1 = 9.5 x 10-31
2. The equilibrium constant for the reaction is Kc = 2.19 x 10-10.
COCl2(g)  CO(g) + Cl2(g)
Is a mixture containing 2.0 x 10-3 M COCl2, 3.3 x 10-6 M CO,
and 6.6 x 10-6 M Cl2 at equilibrium? If not, what direction will
the reaction proceed to achieve equilibrium?
Q = [CO](Cl2]/[COCl2]
Q = (3.3 x 10-6)(6.6 x 10-6)/(2.0 x 10-3) = 1.1 x 10-8 > KC 
3. Consider the equilibrium at 25oC: A(g) + B(g)  2 C(g).
a. Write the equilibrium expression for this equilibrium.
Kc = [C]2/[A][B]
Calculate Kc at 25oC given [A]E = 0.025 M,
[B]E = 0.100 M, and [C]E = 0.500 M.
Kc = [C]2/[A][B]
Kc = (0.500)2/(0.025)(0.10) = 100
c. Calculate Kp for the equilibrium at 25oC.
Kp = Kc x (RT)ng
Kp = (100) x [(0.0821)(298)]0 = 100
d. State which direction the equilibrium will shift (, )
to relieve the following stresses on the system.
Increase V
Increase P
Increase [A]
No shift
No shift

e. What is Kc at 25oC for C(g)  ½ A(g) + ½ B(g)?
b.
Kc = (100)-½ = 0.10
Calculate Go for the equilibrium at 25oC when all the
species are at 1 atm partial pressure.
Go = -RTlnK
Go = -(8.31)(298)ln(100) = -11 kJ
f.
g.
In another experiment 1.00 mol of A and 1.00 mol of
B are placed in an evacuated 1.00-L vessel at 25oC.
What are the concentrations of the three gases after
equilibrium is established?
A
+
B

2C
[]
1.00
1.00
0
I
-x
-x
+2x
C
1.00 – x
1.00 – x
2x
E
K = [C]2/[A][B]  100 = (2x)2/(1.00 – x)2  x = 0.83
[A] = [B] = 1.00 – 0.83 = 0.17 M
[C] = 2x = 2(0.83) = 1.66 M
h. A mixture of 1.00 M A and 0.50 M B are placed in a
vessel at 100oC. At equilibrium, [C] = 0.90 M. What
are the equilibrium concentrations of A and B?
A
+
B

2C
[]
1.00
0.50
0
I
-0.45
-0.45
+0.90
C
0.55
0.05
0.90
E
i. What is Kc at 100oC?
Kc = [C]2/[A][B]2
Kc = (0.90)2/(0.55)(0.05) = 29
j. Is the forward reaction endothermic or exothermic.
Explain your reasoning.
K is reduced when T increases  higher T favored
reverse reaction and higher T favors endothermic
reaction  forward reaction is exothermic.
4.
N2(g) + 3 H2(g)  2 NH3(g) Kp = 1.45 x 10-5
What is the partial pressure of NH3 if the partial pressures
of N2 and H2 are 0.432 atm and 0.928 atm respectively?
Kp = (PNH3)2/(PN2)(PH2)3
1.45 x 10-5 = (PNH3)2/(.432)(.928)3 PNH3 = 2.24 x 10-3 atm
5.
H2(g) + CO2(g)  H2O(g) + CO(g)
When H2 is mixed with CO2 at 2,000 K, equilibrium is
achieved according to the equation. In one experiment,
the following equilibrium concentrations were measured.
[H2] = 0.20 M, [CO2] = 0.30 M, [H2O] = [CO] = 0.55 M
a. Calculate the value of Kc.
Kc = [H2O][CO]/[H2][CO2]
Kc = (0.55)(0.55)/(0.20)(0.30) = 5.0
b. Determine Kp in terms of Kc for this system.
Kp = Kc x (RT)ng = 5.0
c.
When the system is cooled to a lower temperature,
30.0 percent of the CO is converted back to CO2.
(1) Calculate the change in [CO]
30% of [CO] = 0.55 x 0.30 = 0.17
(2) Calculate Kc at this lower temperature.
H2(g) + CO2(g)  H2O(g) + CO(g)
[]
0.20
0.30
0.55
0.55
I
+0.17
+0.17
-0.17
-0.17
C
0.37
0.47
0.38
0.38
E
Kc = [H2O][CO]/ [H2][CO2]
Kc = (.38)(.38)/(.37)(.47) = .83
d. In a different experiment, 0.500 mole of H2 is mixed with
0.500 mole of CO2 in a 3.00-liter vessel at 2,000 K.
(1) Calculate the initial concentration of H2 and CO2.
0.500 mol H2/3.00 L = 0.167 M
0.500 mol CO2/3.00 L = 0.167 M
(2) Calculate the equilibrium concentrations.
H2(g) +
CO2(g)  H2O(g) +
CO(g)
[]
0.167
0.167
0
0
I
-x
-x
+x
+x
C
0.167 – x
0.167 – x
x
x
E
Kc = [H2O][CO]/[H2][CO2]
Kc = (x)(x)/(0.167 – x)( 0.167 – x) = 5.0  x = 0.12
[H2] = [CO2] = 0.167 – 0.12 = 0.047 M, [H2O] = [CO] = 0.12 M
6.
Answer the following questions regarding the
decomposition of arsenic pentafluoride, AsF5(g).
a. A 55.8 g sample of AsF5(g) is introduced into an
evacuated 10.5 L container at 105oC.
(1) What is the initial molar concentration of AsF5(g)?
55.8 g AsF5 x 1 mol/169.9 g AsF5 = 0.328 mol AsF5
0.328 mol AsF5/10.5 L = 0.0313 M AsF5
(2) What is the initial pressure, in atmospheres, of the
AsF5(g) in the container?
PV = nRT  P = MRT
P = (0.0313 mol/L)(.0821 L•atm/mol•K)(378 K) = 0.969 atm
At 105oC, AsF5(g) decomposes into AsF3(g) and F2(g)
according to the following chemical equation.
AsF5(g)  AsF3(g) + F2(g)
b. In terms of molar concentrations, write the equilibriumconstant expression for the decomposition of AsF5(g).
K = [AsF3][F2]/[AsF5]
c.
When equilibrium is established, 27.7 percent of the
original number of moles of AsF5(g) has decomposed.
(1) Calculate the molar concentration of AsF5(g) at
equilibrium.
100 % - 27.7 % = 72.3 % 
0.0313 M x 0.723 = 0.0226 M AsF5
(2) Using molar concentrations, calculate the value
of the equilibrium constant, Keq, at 105oC.
M AsF3 = M F2 = (0.0313 – 0.0226) = 8.7 x 10-3 M
K = [AsF3][F2]/[AsF5] = (8.7 x 10-3)2/(0.0226) = 0.00335
d. Calculate the mole fraction of F2(g) in the container at
equilibrium.
8.7 x 10-3/2(8.7 x 10-3)+ (0.0226) = 0.217
7.
Consider bromine chloride. It is formed by the reaction
between red-orange bromine vapor and yellow chlorine
gas; BrCl is itself a gas. The reaction is endothermic.
a. Write the chemical equation for the formation of BrCl,
using simplest whole-number coefficients.
Br2(g) + Cl2(g)  2 BrCl(g)
b.
Write the equilibrium expression for Kc.
Kc = [BrCl]2/[Br2][Cl2]
c.
At 400oC, after the reaction reached equilibrium, the
mixture contained 0.82 M BrCl, 0.20 M Br2 and 0.48 M
Cl2. Calculate Kc for the reaction.
Kc = [BrCl]2/[Br2][Cl2] = (0.82)2/(0.20)(0.48) = 7.0
d.
What is Kp at 400oC?
Kp = Kc x (RT)n(gas) = 7.0 x [(0.08210)(673)]0 = 7.0
e.
Initially, a 2.00-L flask contains Cl2 with partial
pressure 0.51 atm and Br2 with partial pressure 0.34
atm. After equilibrium is established, the partial
pressure of BrCl is 0.46 atm. Calculate the equilibrium
pressures of Cl2 and Br2.
PCl2 = 0.51 atm – ½(0.46 atm) = 0.28 atm
PBr2 = 0.34 atm – ½(0.46 atm) = 0.11 atm
f. Initially, a 1.00-L flask contains 0.15 mol of each gas.
What are the equilibrium concentrations of each gas at
400oC?
Kc = [BrCl]2/[Br2][Cl2] = (0.15 + 2x)2/(0.15 - x)2 = 7.0
x = 0.054  [BrCl] = 0.26 M, [Br2] = [Cl2] = 0.10 M
g. In what direction will the system shift if at equilibrium?
no change
the volume is increased?
helium gas is added?
no change
the temperature is increased?
