Chemistry 100.12
Problem Session 4a
G. Marangoni
Date: November 8, 2011.
1. Obtain the equilibrium constant expressions for the following reactions.
a) WO3 (s) + 3H2 (g) ⇌W (s) + 3 H2O (g)
b) NH3 (aq) + H2O(l) ⇌ NH4+ (aq) + OH (aq)
c) 2 SO3 (g) ⇌ 2 SO2 (g) + O2 (g)
d) 2 FeCl3 (s) + 3H2O (g) ⇄ Fe2O3 (s) + 6 HCl (g)
2. For the equilibrium reaction,
2NO2  s 
N 2O4  g 
a) 1.500 atm of N2O4 (g) and 1.00 atm of NO2 (g) is placed into a flask and allowed
to come to equilibrium at 298 K. The Keq for this reaction is 0.150.
b) Predict the direction the reaction will take to get to equilibrium.
c) Calculate the composition of the system at equilibrium.
3. For the reaction
NH 4 HS  s 
NH 3  g   H 2 S  g 
the equilibrium constant Keq is 0.0712 at 22C. Calculate the partial pressure and the
total pressure of the gases in the mixture at equilibrium. What is the minimum mass
of ammonium hydrosulfide required in a 2.00 L vessel to establish the gas phase
equilibrium at 295K?
4. At 1400 K, the equilibrium constant for the reaction
CH 4  g   2 H 2 S  g 
4 H 2  g   CS 2  g 
is 33.0. A 10.0 L reaction vessel contains 4.0 mol of CH4 (g), 3.0 mol of H2S (g), 2.0
mol of H2 (g), and 2.0 mol of CS2 (g). Is the reaction mixture at equilibrium? If not, in
which direction does the reaction proceed to reach equilibrium?
5. For the reaction
2 NO  g   Cl2  g 
2 NOCl  g 
4.60 atm of NO and 1.92 atm of Cl2 were mixed in a container, yielding 3.47 atm of
NOCl (g) at equilibrium. Calculate the amount of material reacted, the composition
of the system at equilibrium, and the equilibrium constant Keq.
6. When 9.25 g of ClF3 was introduced into an empty 2.00 L container at 700.0 K,
19.8% of the ClF3 (g) decomposed to give an equilibrium mixture of ClF3 (g), ClF (g),
and F2 (g).
a) What is the value of the equilibrium constant at 700.0 K?
b) In a separate experiment, 39.4 g of ClF3 (g) was introduced into an empty 2.00 L
container at 700.0 K. Calculate the composition of the system at equilibrium.
ANSWERS
1. Note pure solid and pure liquids don’t appear in Keq
a)
P 

P 
3
H 2O
K eq
3
H2
b)
 NH 4   OH  
K 
 NH 3 
c)
P  P

P 
/
2
SO3
K eq
O2
2
SO3
d)
K
/
P 
 HCl
6
P 
3
H 2O
2. The equilibrium reaction is written
2NO2  s 
N 2O4  g 
a) Calculate the reaction quotient Q
PN2O4
1.500
Q

 1.500  K eq
2
2
1.00 
PNO2


Hence, the reaction has to shift to the left to reach equilibrium.
b) To calculate the system composition, we need to use the equilibrium data table.
Let x = the amount of N2O4 that reacts as we proceed towards equilibrium
Substance
Start
Change
m
K eq 
NO2
1.00
+2x
1.00+2x
PN2O4
P 
NO2
2

N2O4
1.50
-x
1.50-x
1.500  x
1.00  2 x 
2
 0.150
0.150  1.00  2 x   1.500  x
2


0.150  1.00  4 x  4 x 2  1.500  x
0.150  0.60 x  0.60 x 2  1.500  x
1.350  1.60 x  0.60 x 2  0
Note – using the quadratic formula with a = 0.60, b = 1.60, and c = 1.350
x
1.60  1.602  4  0.60  1.350
2  0.60
x 1  0.674; x  2   3.34
Hence, PN2O4  0.827 atm; PNO2  2.35 atm
3. The equilibrium reaction
NH 4 HS  s 
NH 3  g   H 2 S  g 
a) The equilibrium constant expression
K  PNH3 PH 2 S
Data Table. Let x = PH 2 S at equilibrium
Substance
Start
Change
m
NH4HS
NH3
---0
---+x
---x
K eq  PNH3 PH 2 S   x  x   x 2
0.0712  x 2
x   0.0712 
1
2
 0.267 atm
 P(H2S) = 0.267 atm; P(NH3) = 0.267 atm; PT = 0.534 atm
To calculate the amount of solid req’d to reach equilibrium.
From the reaction stoichiometry,
nNH3  nNH 4 HS
PNH3V  nNH3 RT
H2S
0
+x
x
0.267 atm  2.00 L
 0.0221 mol  nNH 4 HS
RT
0.08206 L atm
 295 K
K mol
 mass of solid  nNH 4 HS  molar mass
PNH 3V
 nNH3 

 0.0221 mol  65.14 g
4. For this reaction
CH 4  g   2 H 2 S  g 
mol
 1.44 g
4 H 2  g   CS 2  g 
Calculate the partial pressures of the gases
nCH4 RT
PCH4 
V
PCS2 

nCS2 RT
V
PH 2 S 
4.00 mol  0.08206 L atm

nH 2 S RT
V
K mol
10.00 L
2.00 mol  0.08206 L atm
1400 K
K mol
1400 K
10.00L

3.00 mol  0.08206 L atm
The reaction quotient Q 
 45.95 atm
K mol
 22.98 atm  PH2
1400 K
10.00 L
 
P 
PCS2 PH 2
PCH 4
H2S
4
2

22.98  22.98 
4
45.95  34.47 
2
 34.47 atm
 117.4  K eq
Hence, the reaction has to shift to the left to reach equilibrium.
5. The equilibrium reaction
2 NO  g   Cl2  g 
2 NOCl  g 
To calculate the system composition, we need to use the equilibrium data table.
Data Table
Note – let x = amount of Cl2 (in atm) that reacts
Substance
NO
Cl2
NOCl
Start
4.60
1.92
0
Change
-2x
-x
+2x
2x
m
4.60  2x
1.92  x
P 
Cl2 eq
At equilibrium,  PNOCl eq  3.47 atm  2 x  x  1.735 atm
 1.92 atm  1.74 atm  0.19 atm;  PNO eq  4.60 atm  2 1.74 atm  1.13 atm
 PNOCl 
Keq 
2
 PNO  PCl
2
Hence,
2
6. The equilibrium reaction
ClF3  g 
 3.47   51.0

2
1.13 0.19
2
ClF  g   F2  g 
a) Calculate the number of mole of ClF3 (g)  nClF3 
PClF3 
nClF3 RT
0.100 mol  0.08206 L atm

V
K eq 
Data Table
Note – let x = PF2
 
K mol
9.25 g
 0.100 mol
92.45 g mol 1
 700 K
2.00 L
PF2 PClF
 2.87 atm
PClF3
eq
Substance
Start
Change
m
ClF3
2.87 atm
-x
2.87  x
ClF
0
+x
x
F2
0
+x
x
Note we are told that 19.8 % of the ClF3 has decomposed. Hence, that means that the
pressure of ClF3 at equilibrium is 80.2 % of its original value.
 P   0.802   P   0.806  2.87 atm  2.30 atm
Hence, since  P    P   x  x  2.87 atm  2.30 atm  0.57 atm
 P    P   0.57 atm.
ClF3
ClF3
eq
ClF3
eq
ClF3
o
F2 eq
K eq 
ClF eq
PF2 PClF
PClF3
b) Obtain the mol of ClF3 nClF3 
o

 0.57  0.57   0.140
2.30
39.4 g
 0.426 mol
92.45 g mol 1
PClF3 
nClF3 RT
V
0.426 mol  0.08206 L atm

 700 K
2.00 L
 
Note – let x = PF2
K mol
 12.24 atm
Data Table
eq
Substance
Start
Change
m
ClF3
12.24 atm
-x
12.24  x
K eq 
PF2 PClF

ClF
0
+x
x
 x  x 
 0.140
12.24  x 
0.140  12.24  x   x 2
PClF3
1.714  0.140 x  x 2
x 2  0.140 x  1.714  0
Note – using the quadratic formula with a =1, b =0.140, and c = 1.80
x
0.140  0.1402  4 1 1.714
2 1
x 1  1.241; x  2   1.381
P 
ClF3
eq
 12.24 atm  1.24 atm  11.0 atm
P 
F2 eq
  PClF eq  1.24 atm.
F2
0
+x
x