1
NAME:
CHM 2046
Quiz 1
Spring, 0000
Give your final answer with the correct units, if any, and to the correct sig. figs.
Show your work. Useful Information: 0 ºC ≈ 273 K, R = 0.0820 L. atm/mol. K
1.
a) (2 points each) Balance the following reactions and write down their
mass-action expression, Qc.
N2 H2 3
2 NH3 (g) ⇌ N2 (g) + 3 H2 (g)
NH3 2
Qc =
2 H2O2 (g) + 2 NO (g) ⇌ 2 H2O (g) + N2O4 (g)
H2O2 N2O4 
H2O2 2 NO2
Qc =
b) (5 points) At a particular temperature, Kp = 279 for the reaction
2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g)
In a particular reaction carried out, the pressures of the gases at a
particular time are 0.12 atm (atmospheres) of SO2, 1.2 atm of O2, and 2.4
atm of SO3. Has the reaction reached equilibrium yet? Explain.
Qp 
p(SO3 ) 2
(2.4)2

 333
p(SO2 ) 2 p(O2 ) (0.12)2 (1.2)
 Qp  Kp not at equilibrium
If not, in which direction is it proceeding? Explain.
Qp > Kp  reaction proceeding to the left.
c) (4 points) N2O4 (g) was introduced into a 2.00 L flask and allowed to
reach equilibrium at 100 ºC. At equilibrium, the flask contained 0.38 mol
N2O4 (g) ⇌ 2 NO2 (g)
of N2O4 and 0.40 mol of NO2. What is Kc for this reaction at 100 ºC?
Kc 
2.
(6 points total)
[NO2 ]2 (0.20) 2

 0.21
[N2O4 ]
0.19
a) At 350 K, the reaction below has Kc = 1.05. What is Kp at 350 K?
2 CH2Cl2 (g) ⇌ CH4 (g) + CCl4 (g)
Kp = Kc
Kp = 1.05
b) At 300 K, the reaction below has Kc = 279. What is Kp at 300 K?
2 SO2 (g) + O2 (g) ⇌ 2 SO3 (g)
Δn = -1 therefore
Kc
279
Kp 

 11.3
RT (0.082)(30 0)
3.
Kp = 11.3
(3 points each) The reaction below has Kc = 12. What are the Kc values for
the equilibria in a) and b)? Explain how you got your answer.
CO(g) + H2O(g) ⇌ CO2(g) + H2(g)
a)
2CO(g) + 2H2O(g) ⇌ 2CO2(g) + 2H2(g)
Kc’ = Kc2
b)
Kc = 144
CO2(g) + H2(g) ⇌ CO(g) + H2O(g)
Kc’ = 1/Kc
Kc = 0.083
2
4.
a) (6 points)
2HI (g) ⇌ H2 (g) + I2 (g)
A 2.00 L flask is filled with 0.300 mol of HI and allowed to reach equilibrium
at a particular temperature. At equilibrium, [HI] = 0.104 M. Calculate Kc at
this temperature.
I
C
E
2 HI
⇌ H2 + I 2
0.150 M
0
0
-2x
+x +x
(0.150-2x)
x
x
(0.150-2x) = 0.104
 x = 0.023
[H2 ][I2 ]
x2
(0.023)2
Kc 


[HI]2
(0.15 - 2x)2 (0.104)2
Kc  4.89 x 10 -2
b) (7 points)
CO2 (g) + H2 (g) ⇌ CO (g) + H2O (g)
(Kc = 2.4)
0.500 mol each of CO2 and H2 are placed in a 10.0 L flask and allowed to
reach equilibrium at some temperature. Calculate the equilibrium
concentrations of all species.
CO2
+
I
0.0500
C
-x
H2
⇌ CO + H2O
0.0500
-x
0
0
+x
+x
E (0.0500-x) (0.0500-x)
Kc  2.4 
x2
(0.0500 - x) 2
x
x
x
0.0500 - x
 1.549x  0.077  x
  1.549 
 2.549x  0.077
5.
 x  0.030
[CO2 ]  [H2 ]  0.020
[CO]  [H2O]  0.030
(12 points) Consider the reaction in Question 4b again.
CO2 (g) + H2 (g) ⇌ CO (g) + H2O (g)
(Kc = 2.4)
This time 0.500 mol of CO2 and 1.00 mol of H2 are placed in a 10.0 L flask
and allowed to reach equilibrium at some temperature. Calculate the
equilibrium concentrations of all species.
CO2
+
H2
⇌ CO + H2O
I 0.0500
0.100
0
0
C
-x
-x
+x
+x
E (0.0500-x) (0.100-x)
x
x
2
x
x2
K c  2.4 

 2.4
(0.0500 - x)(0.100 - x) (0.005 - 0.15x  x 2 )
 2.4x2 – 0.36 x + 0.012 = x2
 1.4 x2 = 0.36 x + 0.012 = 0
x
x
0.36  0.1296 - 4(1.4)(0.012)
2.8
0.36  0.250

0.36  0.0624
2.8
 0.039 or 0.218
2.8
We choose x = 0.039
 [CO2] = 0.011M
[H2] = 0.061M
[CO] = [H2O] = 0.039M