Name: ____ KEY _________________
CHM 152 – Equilbrium (Ch. 13)
c
p
1.
Once a system has reached equilibrium, are the following true or false? a.
The reaction is finished, no more products are forming. __ false __ b.
The concentrations of the reactants and the products are equal. false __ c.
The concentrations are no longer changing. _ true _ d.
The reaction is not over, but will continue forever if isolated. _ true _ e.
The speed at which products are made equals the speed at which reactants form. true
2.
What is equal at equilibrium? ____ forward and reverse rates _____
3. What general information can be gathered by observing the magnitude of the equilibrium constant?
Whether a reaction is reactant- or product-favored.
4.
Answer the following for the reaction of NO gas with chlorine gas to produce NOCl gas. Write out the balanced reaction, and the K c
and K p
expressions.
2 NO(g) + Cl
2
(g)  2 NOCl(g) K c
=
[ NOCl]
[NO]
2
[ Cl
2
2
]
K p
=
P
NOCl
2
(P
NO
2
)(P
Cl 2
)
5.
Write the equilibrium expression K c
for the reaction between potassium phosphate and calcium nitrate in water. Show the balanced reaction.
2 K
3
PO
4
(aq) + 3 Ca(NO
3
)
2
(aq)  6 KNO
3
(aq) + Ca
3
(PO
4
)
2
(s)
K c
=
[K
3
PO
4
[KNO
3
]
6
]
2
[Ca(NO
3
)
2
]
3
K p
= NA
6.
Write the equilibrium expression K c
for the reaction between sodium carbonate and calcium hydroxide in water.
Na
2
CO
3
(aq) + Ca(OH)
2
(aq)

2 NaOH(aq) + CaCO
3
(s)
K c
= [NaOH] 2 / [Na
2
CO
3
][ Ca(OH)
2
]
7 .Write the expression for K c
for the reaction: Ag
+
(aq) + 2 NH
3
(aq)

Ag(NH
3
)
2
+
(aq)
K c
= [Ag(NH
3
)
2
+ ] / [Ag + ][(NH
3
) 2 ]
8. Write the expression for K p
for the reaction : H
2
(g) + Br
2
(l)

2 HBr(g)
K p
= P
HBr
2 / (P
H2
)
9. Write the expression for K p
for the reaction: CO
2
(g) + CaO(s)

CaCO
3
(s)
K p
= 1 / P
CO2
CHM 152 Chemical Equilibrium Page 1 of 8

Name: ____ KEY _________________
10. Write K c
expressions for the following reactions: a) 3 O
2
(g)

2 O
3
(g) b) N
2
(g) + 3 H
2
(g)

2 NH
3
(g)
K c
= [(O
3
) 2 ] / [(O
2
) 3 ]
K c
= [(NH
3
) 2 ] / [N
2
][(H
2
) 3 ] c) H
2
(g) + I
2
(g)

2 HI (g) d) PCl
5
(g)

PCl
3
(g) + Cl
2
(g)
K c
= [(HI) 2 ] / [H
2
][I
2
]
K c
= [PCl
3
][Cl
2
] / [PCl
5
] e) SO
2
(g) + ½ O
2
(g)

SO
3
(g) K c
= [SO
3
] / [SO
2
][(O
2
) 1/2 ]
11. Write K p
expressions for each of the following reactions: a) Ni(s) + 4CO(g)

Ni(CO)
4
(g) K p
= P
Ni(CO)4
/ (P
CO
) 4 b) 5CO(g) + I
2
O
5
(s)

I
2
(g) + 5CO
2
(g) K p
= (P
I2
)(P
CO2
) 5 / (P
CO
) 5 c) Ca(HCO
3
)
2
(aq)

CaCO
3
(s) + H
2
O(l) + CO
2
(g) K p
= P
CO2 d) AgCl(s)

Ag + (aq) + Cl

(aq) K p
= 1
12. Arrange the reactions in order of their increasing tendency to proceed toward completion: _ B _ _ C _ _ D _ _ A _
(a) 4NH
3
(g) + 3O
2
(g)

2N
2
(g) + 6H
2
O(g) K p
= 1 x 10 228 atm
(b) N
2
(g) + O
2
(g)

2NO(g)
(c) 2HF(g)

H
2
(g) + F
2
(g)
K
K p p
= 5 x 10
= 1 x 10
-31
-13
(d) 2NOCl(g)

2NO(g) + Cl
2
(g) K p
= 4.7 x 10 -4 atm
13. Nitrogen dioxide dimerizes to form dinitrogen tetraoxide: 2 NO
2
(g)

N
2
O
4
(g)
Calculate the value of K c
, given that the gas phase equilibrium constant, K p
, for the reaction is 1.3
× 10 3
at 273 K. ( R = 0.08206 L·atm/mol·K)
 n = -1; K p
= K c
(RT)
 n = 2.9 x 10 4
CHM 152 Chemical Equilibrium Page 2 of 8

Name: ____ KEY _________________
14.
Answer the following for the reaction of NO gas with chlorine gas to produce NOCl gas.
Calculate K p
and K c
at 25.0
o
C given the pressure of NO gas is 1.56 atm, chlorine gas is 0.887 atm, and NOCl gas is 3.45 atm.
K p
= = 5.51398 = 5.51
K c
= = 135
15.
Calculate the equilibrium constant for this reaction: 2 PO
2
Br (aq)

2 PO
2
(aq) + Br
2
(aq)
Given: [PO
2
Br] = 0.0255M, [PO
2
] = 0.155M, and [Br
2
] = 0.00351M at equilibrium.
K = [PO
2
] 2 [Br
2
] / [PO
2
Br] 2 = (0.155) 2 (0.00351) / (0.0255) 2 = 0.130
16.
Calculate the equilibrium constant K c
for this unbalanced reaction: 2 SO
3
(g)

2 SO
2
(g) + O
2
(g)
Given: [SO
3
] = 0.0255M, [SO
2
] = 1.08M, and [O
2
] = 1.45M at equilibrium.
K c
= = = 2.60 x 10 3
17. Use the equilibrium reaction: H
2
(g) + I
2
(g)

2 HI (g) to calculate K c
for each of the sets of equilibrium concentrations below. a. [H
2
] = 0.0505 M
[I
2
] = 0.0498 M
[HI] = 0.349 M b. [H
2
] = 0.00560 M
[I
2
] = 0.000590 M
[HI] = 0.0127 M
K
K c c
= 48.4
= 48.8
c. [H
2
] = 0.00460 M
[I
2
] = 0.000970 M
[HI] = 0.0147 M K c
= 48.4
What do you notice about the 3 K c
values?
They’re all about the same value (or should be close).
18. For the combination reaction: H
2
(g) + I
2
(g)

2 HI (g), calculate all three equilibrium concentrations when [H
2
] o
= [I
2
] o
= 0.200 M and K c
= 64.0.
ICE table: x = 0.160 M
[H
2
] = 0.200 - x = 0.040 M
[I
2
] = 0.200 - x = 0.040 M
[HI] = 2 (x) = 0.320 M
19. For the combination reaction, PCl
3
(g) + Cl
2
(g)

PCl
5
(g), calculate all three equilibrium concentrations when K c
= 16.0 and [PCl
5
] o
= 1.00 M.
CHM 152 Chemical Equilibrium Page 3 of 8

Name: ____ KEY _________________
Bad problem – no solution. You can set up the ICE table and equilibrium expressions, but the solution doesn’t work out with these numbers. I’ll fix it later….
ICE table: K c
= (1.00 - x) / x 2 = 16.0
20. For the decomposition reaction, COCl
2
(g)

CO (g) + Cl
2
(g), calculate all three equilibrium concentrations when K c
= 0.680 with [CO] o
= 0.500 and [Cl
2
] o
= 1.00 M.
ICE table: Kc = (0.500 – x)(1.00 – x) / x = 0.680 x = (2.18 ± 1.66) / 2 = 0.26048
[COCl
2
]
eq
= 0.260 M , [CO]
eq
= 0.500 – 0.260 = 0.240 M , [Cl
2
]
eq
= 1.00 – 0.260 = 0.740 M
21. We place 0.064 mol N
2
O
4
(g) in a 4.00 L flask at 200 K. After reaching equilibrium, the concentration of NO
2
(g) is 0.0030 M. What is K c
for the reaction N
2
O
4
(g)

2 NO
2
(g)? x = 0.0015
[N
2
O
4
] eq
= 0.0145, [NO
2
] eq
= 0.0030
K c
= 6.2 x 10 -4
22. Carbonyl bromide decomposes to carbon monoxide and bromine: COBr
2
(g)

CO(g) + Br
2
(g)
K c
is 0.190 at 73 o
C. If an initial concentration of 0.330 M COBr
2
is allowed to equilibrate, what are the equilibrium concentrations of COBr
2
, CO, and Br
2
?
K c
= x 2 / (0.330 – x) = 0.190 x = 0.1728
[COBr
2
] eq
= 0.330 – 0.1728 = 0.157 M ; [CO] eq
= [Br
2
] eq
= 0.173 M
23. H
2(g)
+ CO
2(g)

H
2
O
(g)
+ CO
(g) a) It is found at 986 o
C that there are 11.2 atm each of CO and water vapor and 8.8atm each of H
2
and
CO
2
at equilibrium. Calculate the equilibrium constant.
K p
= (11.2)(11.2) / (8.8)(8.8) = 1.6198 = 1.62
b) If there were 8.8 moles of H
2
and CO
2
in a 500.0mL container at equilibrium, how many moles of
CO
(g)
and H
2
O
(g)
would be present?
K c
= 1.62 = x
2
/ y
2
y = 8.8 moles / 0.5000 L = 17.6 1.62 = x
2
/ 309.76 x = 11.2 moles
24. Consider the equilibrium: 2N
2
O(g) + O
2
(g) 4NO(g)
3.00 moles of NO(g) are introduced into a 1.00-Liter evacuated flask. When the system comes to equilibrium, 1.00 mole of N
2
O(g) has formed. Determine the equilibrium concentrations of each substance. Calculate the K c
for the reaction based on these data.
CHM 152 Chemical Equilibrium Page 4 of 8

Name: ____ KEY _________________
2N
2
O(g) + O
0
+2x
2
(g)
0
+x
4NO(g)
3.00 M
-4x
2x x 3.00 – 4x K c
= (3.00 – 4x)
4
/ (2x)
2
(x)
1.00 mole of N
2
O in 1.00 L = 1.00 M; Therefore, 2x = 1.00 M and x = 0.50 M
K c
= (3.00 – 4x) 4 / (2x) 2 (x) = (3.00 – 4*0.5) 4 / (2*0.5) 2 (0.5) = 2
[NO] eq
= [N2O] eq
= 1.00 M; [O
2
] eq
= 0.50 M
25. For the reaction: SiH
4(g)
+ O
2(g)

SiO
2(g)
+ H
2
O
(g) a. Balance the equation. SiH
4(g)
+ 2 O
2(g)

SiO
2(g)
+ 2 H
2
O
(g) b. Write the equilibrium expression for the forward reaction:
K c
= [SiO
2
][H
2
O] 2 / [SiH
4
][O
2
] 2 c. Write the equilibrium expression for the reverse reaction:
K c
= [SiH
4
][O
2
] 2 / [SiO
2
][H
2
O] 2 d. What is the equilibrium constant in the forward direction if [SiH
4
] = 0.45M; [O
2
] = 0.25M; [SiO
2
] =
0.15M; and [H
2
O] = 0.10M at equilibrium?
K f
= (0.15)(0.10) 2 / (0.45)(0.25) 2 = 0.053 e. What is the equilibrium constant in the reverse reaction? K f
= (0.45)(0.25) 2 / (0.15)(0.10) 2 = 19 f. If [SiH
4
] = 0.34M; [O
2
] = 0.22M; [SiO
2
] = 0.35M; and [H
2
O] = 0.20M, what would be the reaction quotient (Q) in the forward direction?
Q = (0.35)(0.20) 2 / (0.34)(0.22) 2 = 0.85 g. Which direction will the reaction in part f go? (Toward products or reactants?)
Toward reactants
26.
Phosphorus pentachloride decomposes into phosphorus trichloride and chlorine gas. What is the initial concentration of phosphorus pentachloride if at equilibrium the concentration of chlorine gas is
0.500M? Given: K c
= 10.00 (Hint: ICE table)
PCl
5

PCl
3
+ Cl
2
I x 0 0 10.00 = (0.500) 2 / (x-0.500)
C -0.500 +0.500 +0.500
E (x-0.500) (0.500) (0.500)
10.00 x – 5 = 0.25 x = 0.525 M = [PCl
5
]
27.
A 1.000 L flask is initially filled with 1.000 mole of hydrogen gas and 2.000 moles of iodine gas at
448 o C. At this temperature K c
is 50.5. Calculate the equilibrium concentrations for all the chemical species in the reaction, which is hydrogen gas and iodine gas produce HI gas.
H
2
(g) + I
2
(g)  2 HI(g)
CHM 152 Chemical Equilibrium Page 5 of 8

Name: ____ KEY _________________
I 1.000M 2.000M 0
C -x -x +2x
E (1.000-x) (2.000-x) (2x)
K c
= = = 50.5
50.0(1.000-x)(2.000-x) = 50.5(2.000-3.000+x 2 ) = (2x) 2
101-151.5x + 50.5x
2 = 4x 2
46.5x
2 – 151.5x + 101.0 = 0
Plug into quadratic x = {+151.5 +/- } / 2(46.5)
X = 2.3231 or 0.93498 (the first gives negative concentrations so is wrong)
[H
2
] = 1.000-x = 0.065M
[I
2
] = 2.000-x = 1.065M
[HI] = 2x = 1.870M
28.
Answer the following three questions for this reaction: the formation of hydrogen chloride gas. a. Write the K c
expression.
H
2
(g) + Cl
2
(g)

2 HCl(g)
K c
= [HCl] 2 / [H
2
][Cl
2
] b. Given K c
= 3.56 x 10 5 calculate K p
at 25.0
o C.
K p
= (3.56 x 10 5 )(RT) 0 = 3.56 x 10 5 c. If the original concentrations are [H
2
] = 1.55M and [Cl
2
] = 1.55M, calculate all the equilibrium concentrations for each species in the reaction in moles/L.
H
2
(g) + Cl
2
(g)

2 HCl(g)
I 1.55 1.55 0
C -x -x +2x
E (1.55-x) (1.55-x) (2x)
3.56 x 10 5 = (2x) 2 / (1.55-x) 2
Take square root both sides, solve for x = 1.54
[H
2
] = [Cl
2
] = 0.01M and [HCl] = 3.08M
29. A 0.0240 mol sample of N
2
O
4
(g) is allowed to reach equilibrium with NO
2
(g) in a 0.372 L flask at
25.0
o
C. Calculate the concentration of N
2
O
4
(g) at equilibrium.
N
2
O
4
(g)

2 NO
2
(g) K c
= 4.61 x 10
-3
I 0.06452M 0
C -x +2x
E (0.06452-x) (2x)
4.61 x 10 -3 = 4x 2 / (0.06452 – x)
4x 2 = -4.61 x 10 -3 x + 2.974 x 10 -4
4x 2 + 4.61 x 10 -3 x - 2.974 x 10 -4 = 0 quadratic formula with a = 4, b = +4.61 x 10 -3 and c = - 2.974 x 10 -4 x = 8.065 x 10 -3 (the other solution is negative and can’t have negative concentration) thus [N
2
O
4
] = 0.0565M
CHM 152 Chemical Equilibrium Page 6 of 8

Name: ____ KEY _________________
30.
For the reaction 2 HI(g)

H
2
(g) + I
2
(g) K c
= 12.3. If [H
2
] = [I
2
] = [HI] = 3.21 x 10
-3
M, which one of the following statements is true? a.
The concentrations of H
2
and HI will decrease as the system approaches equilibrium b.
The concentrations of H
2
and I
2
will increase as the system approaches equilibrium c.
The system is at equilibrium, so the concentrations will not change d.
The concentration of HI will rise as the system approaches equilibrium e.
We need pressures to solve this problem
Q = (3.21 x 10 -3 ) (3.21 x 10 -3 ) / (3.21 x 10 -3 ) 2 = 1
Q < K so reaction goes forward so there will be more products
31.
Consider this endothermic reaction: 3 O
2
(g)

2 O
3
(g). To shift this reaction to the reactants: a.
You could ___ decrease ___ the pressure. ( increase or decrease) b.
You could ____ increase ___ the volume. ( increase or decrease) c.
You could ____ remove ____ oxygen gas. ( add or remove) d.
You could ___ decrease __ the temperature. ( increase or decrease)
32.
Which of the following, if increasing , will change the value of the equilibrium constant? ________ a.
Pressure b. Volume c. [Product] d.
Temperature e. [Reactant]
33.
Consider this reaction: 2 SO
2
(g) + O
2
(g)

2 SO
3
(g). To shift this reaction towards the products: a.
You could ___ increase ___ the pressure. ( increase or decrease) b.
You could ___ decrease ____ the volume. (increase or decrease ) c.
You could ___ add ____ oxygen gas. ( add or remove)
36. For each system described below, indicate in which direction the equilibrium will shift when each stress is added or removed. Also explain how the system will react to alleviate the stress. a) N
2
(g) + 3 H
2
(g)

2 NH
3
(g): more H
2
is added to this reaction at equilibrium.
Reaction will shift toward products to offset the additional H
2 b) Using the same reaction, some NH
3
is removed from the reaction when it is at equilibrium.
Reaction will shift toward products to produce more NH
3 c) 2 SO
2
(g) + O
2
(g)

2 SO
3
(g) + heat: the system temperature goes up (heat is added).
Toward reactants; heat is a product, adding heat shifts reaction left. d) Using the same reaction, heat is removed (that is, the temperature goes down).
Toward products: heat is a product; removing some shifts reaction right. e) PCl
3
(g) + Cl
2
(g)

PCl
5
(g): volume is reduced by half.
Toward products: reducing volume shifts rxn to side with fewer moles. f) Using the same system as above, a catalyst is added to the system.
No change; catalysts do not affect equilibrium. g) H
2
(g) + Cl
2
(g)

2 HCl (g): volume is doubled.
CHM 152 Chemical Equilibrium Page 7 of 8

Name: ____ KEY _________________
No change; changing volume or pressure will not affect this system; same # moles on both sides. h) Using the same system as above, some neon is added to the system.
No change; neon is an inert gas; it won’t react with or affect the system.
37. Explain how the following changes in reaction conditions will affect the position of the equilibrium below, and explain your reasoning.
A
(g)
+ B
(aq)

C
(s)
ΔH rxn
= -453 kJ/mol
1) The pressure of A in the reaction chamber is increased.
The reaction is pushed toward products.
2) The temperature of the reaction is increased by 20
0
C.
Because heat can be thought of as being a product, the reaction will be pushed toward reactants.
3) A catalyst is added to the system.
No change. A catalyst doesn’t change the equilibrium position, it only changes how quickly equilibrium is reached.
4) As the reaction progresses, more of compound B is steadily added to the reaction chamber.
The reaction is pushed toward products.
5) An inhibitor is added to the reaction chamber.
No change, though the reaction will move more slowly.
6) Argon gas is added to the reaction chamber, doubling the pressure.
No change. If the partial pressure of gaseous comopunds is changed, the equilibrium will shift position. However, adding argon gas doesn’t change the partial pressures of A, so the equilibrium position is unaffected.
CHM 152 Chemical Equilibrium Page 8 of 8