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Chapter 13 Review Key
Fill in the blanks:
Chemical equilibrium is the state where the concentrations of both reactants and products
is __1__ over time. At equilibrium the __2__ of the forward and reverse reactions are equal.
The law of __3__ states that for a chemical reaction j A + k B  l C + m D, the equilibrium
expression is given by __4__. K is called the __5__.
An equilibrium position never depends on the amount of a pure __6__ or __7__, so these
are omitted from the equilibrium expression. A small value of K means that the equilibrium lies
far to the __8__. When calculating equilibrium amounts, a(n) __9__ is used. This shows the
given __10__ concentrations or pressures and defines the __11__ needed to reach equilibrium.
The amounts can be expressed in __12__ or __13__.
For a gas phase reaction, either Kc or Kp can be used; the relationship between them is
defined by the equation __14__, where Δn represents __15__ minus __16__. __17__ allows us
to qualitatively predict the effects of changes in __18__, __19__ or __20__ on a system at
equilibrium.
1. constant
11. change
2. rates
12. mol/L
3. mass action
13. pressure units (atm)
4. [C]l [D]m / [A]j [B]k
14. Kp = Kc (RT)Δn
5. equilibrium constant
15. moles of gaseous products
6. solid
16. moles of gaseous reactants
7. liquid
17. LeChâtelier’s Principle
8. left
18. pressure/volume/temp/concentration
9. ICE chart
19. pressure/volume/temp/concentration
10. initial
20. pressure/volume/temp/concentration
1
1. The reaction H2(g) + I2(g)  2 HI(g) has Kp = 45.9 at 763 K. A particular equilibrium
mixture at that temperature contains gaseous HI at a partial pressure of 4.00 atm and hydrogen
gas at a partial pressure of 0.200 atm. What is the partial pressure of I2?
Kp =
[HI]2
[H2][I2]
45.9 = (4.00)2
(0.200)x
x = PI2 = 1.74 atm
2. Calculate Kp for H2O(g) + ½ O2(g)  H2O2(g) at 600 K, using the following data:
H2(g) + O2(g)  H2O2(g)
Kp1 = 2.3 x 106 at 600 K
2 H2(g) + O2(g)  2 H2O(g)
Kp2 = 1.8 x 1037 at 600 K
need Kp = [H2O2]
[H2O][O2]½
Kp1 = [H2O2]
[H2][O2]
Kp2 =
[H2O]2
[H2]2 [O2]
flip Kp2 and take square root: 1 / √ Kp2 = [H2][O2]½
[H2O]
now multiply Kp1 x 1 / √ Kp2 =
[H2O2] x
[H2][O2]
[H2][O2]½ = [H2O2]
[H2O]
[H2O][O2]½
= Kp
so Kp = Kp1 / √ Kp2 = 2.3 x 106 / √ (1.8 x 1037) = 5.4 x 10–13
3. Given the equation 2 NOCl(g)  2 NO(g) + Cl2(g) the equilibrium constant Kc is 0.0150 at
115ºC. Calculate Kp.
Kp = Kc (RT)Δn
Kp = (0.0150) [(0.0821)(388)]1 = 0.478
4. For the gaseous equilibrium PCl5(g) + heat  PCl3(g) + Cl2(g) give the effect of each of the
following:
a. increased temperature
shifts right
b. increased pressure
shifts left
c. higher concentration of Cl2
shifts left
d. higher concentration of PCl5
shifts right
2
5. In a 10.0 L evacuated chamber, 0.500 mol H2 and 0.500 mol I2 are reacted at 448ºC.
H2(g) + I2(g)  2 HI(g)
At the given temperature, Kc = 50.0 for the reaction.
a. What is the value of Kp?
Kp = Kc (RT)Δn
Δn = zero, so Kp = Kc = 50.0
b. What is the total pressure in the chamber?
initial PH2 = initial PI2 = nRT/V = (0.500 mol)(0.0821 L atm/mol K)(721 K) = 2.96 atm
10.0 L
I
C
E
H2(g)
+
2.96
–x
2.96 – x
(2x)2 = 50.0
(2.96-x)2
I2(g)

2.96
–x
2.96 – x
2 HI(g)
0
+ 2x
2x
2x = 7.07 →→ x = 2.31 atm
2.96-x
equilibrium pressures:
PH2 = PI2 = 0.65 atm
PHI = 4.62 atm
Ptotal = 5.92 atm
3
6. A 3:1 starting mixture of H2 and N2 comes to equilibrium at 450.ºC. The mixture at
equilibrium is 9.6% NH3, 22.6% N2, and 67.8% H2 at 60.0 atm.
a. What are the values of Kp and Kc for this reaction?
equilibrium pressures:
PNH3 = 0.096 (60.0 atm) = 5.8 atm
PN2 = 0.226 (60.0 atm) = 13. 6 atm
PH2 = 0.678 (60.0 atm) = 40.7 atm
N2(g) + 3 H2(g)  2 NH3(g)
Kp =
(PNH3)2
(PN2)(PH2)3
Kc = Kp / (RT)Δn =
=
(5.8)2
(13.6)(40.7)3
=
3.6 x 10–5
3.6 x 10–5
= 0.13
[(0.0821)(723)]–2
b. Determine the initial pressure of each component.
make ICE chart and work backwards from E:
N2(g)
I
C
E
16.4
– 2.88
13.56
+
3 H2(g) 
2 NH3(g)
49.3
– 8.64
40.68
0
+ 5.76
5.76
initial pressures:
PN2 = 16.4 atm
PH2 = 49.3 atm
PNH3 = 0 atm
4
7. At a particular temperature, assume that Kc = 1.00 x 102 for the reaction
H2(g) + F2(g)  2 HF(g)
a. In an experiment, 2.00 mol H2 and 2.00 mol F2 are introduced into a 1.00-L flask. Calculate
the concentrations of all species when equilibrium is reached.
I
C
E
H2(g)
+
2.00
–x
2.00 – x
(2x)2 = 100.
(2.00-x)2
F2(g)

2.00
–x
2.00 – x
2 HF(g)
0
+ 2x
2x
2x = 20.0 –10.0x
12.0 x = 20.0
x = 1.67 M
2x = 10.0
(2.00-x)
[H2] = [F2] = 0.33 M
[HF] = 3.34 M
b. To the equilibrium mixture in part a, an additional 0.50 mol H2 is added. Calculate the new
equilibrium concentrations of H2, F2, and HF.
I
C
E
H2(g)
+
0.83
–x
0.83 – x
F2(g)

0.33
–x
0.33 – x
2 HF(g)
3.34
+ 2x
2x
(3.34 +2x)2 = 100.
(0.83-x)(0.33-x)
11.16 + 13.36 x + 4 x2 = 27.39 – 116 x + 100 x2
96 x2 – 129.4 x + 16.23 = 0
quadratic: x = 0.14 or 1.2
[H2] = 0.69 M
[F2] = 0.19 M
[HF] = 3.62 M
5
8. The molar solubility of silver carbonate is 0.032 M. Calculate Ksp.
Ag2CO3(s)  2 Ag+(aq) + CO32–(aq)
0
0
+ 2x
+x
0.064
0.032
x = 0.032 M
Ksp = (0.064)2 (0.032) = 1.3 x 10–4
9. Calculate the molar solubility of Ag2SO4 (Ksp = 1.5 x 10–5) in the following solutions:
(a) pure water
(b) 0.15 M Al2(SO4)3
(a) Ag2SO4(s) 
(b) Ag2SO4(s) 
2 Ag+(aq) + SO42–(aq)
0
0
+2x
+x
2x
x
(2x)2 x = 1.5 x 10–5
x = mol sol = 0.016 M
2 Ag+(aq) + SO42–(aq)
0
0.45
+ 2x
+x
2x
0.45 + x
(2x)2 (0.45) = 1.5 x 10–5
x = 2.9 x 10–3 M
10. At 35°C, Kc = 1.6 x 10–5 for the reaction 2 NOCl(g)  2 NO(g) + Cl2(g)
If 2.0 mol NO and 1.0 mol Cl2 are placed into a 1.0-L flask, calculate the equilibrium
concentrations of all species. (Remember to think about what a small K value means, and
THINK about whether your answers make sense!)
At equilibrium there will be very few products since K is so small.
To solve this question, assume that the reaction proceeds ALL the way to reactants,
then do an ICE chart:
I
C
E
2 NOCl(g) 
0
+2.0
2.0
–2x
2.0–2x
2 NO(g) +
2.0
–2.0
0
+2x
2x
(2x)2 x = 1.6 x 10–5
(2.0)2
[NOCl] = 2.0 M
Cl2(g)
1.0
–1.0
0
+x
x
x = 0.025 M (1.3%)
[NO] = 0.050 M
[Cl2] = 0.025 M
6