Keq and ICE Problems Worksheet
1. Calculate the equilibrium constant, Keq, for the following reaction at 25 °C, if [NO]eq = 0.106 M,
[O2]eq = 0.122 M and [NO2]eq = 0.129 M.
(Answer: Keq = 12.1)
2 NO (g)
+
E 0.106
K eq 
[NO2 ]2
2 NO2 (g) ; K eq 
[ NO]2 [O 2 ]

O2 (g)
0.122
0.129
[0.129] 2
 12.1
[0.106 ]2 [0.122 ]
2. Given the balanced equation and the value for Keq from #1, calculate new value of Keq for the
following:
(Answer: a. Keq = 1.87, b. Keq = 0.0826, Keq = 0.287)
a. ½ NO (g)
+

¼ O2 (g)
½ NO2 (g)
Multiply original reaction by ¼, then K'eq  K eq
1/ 4
b. 2 NO2 (g)

2 NO (g)
+
 4 12.1  1.87
O2 (g)
1
-1
Multiply original reaction by -1 then K'eq  K eq  (12.1)  0.0826
c. NO2 (g)

NO (g)
+
½ O2 (g)
Multiply original reaction by - ½ then K'eq  K eq
1 / 2

1
 0.287
12.1
3. Find the equilibrium constant, Keq, for the following equilibrium. The initial concentrations of AB
and A2D are 0.30 M before they are mixed and when equilibrium is reached, the equilibrium
concentration of A2D is 0.20 M. Be sure to show an ICE table for your calculation.
(Answer: Keq = 0.80)
I
C
E
2 AB (g) +
0.30
+2x
0.30 + 2x
C2D (s)
n/a

[A 2 D]  0.30 - x  0.20  x  0.10
[AB]  0.30  2(0.10)  0.50
[A D] 0.20
K eq  2 2 
 0.80
[AB]
0.50 2
A2D (g) + 2 CB (s)
0.30
n/a
-x
0.30 –x = 0.20
4. If 0.50 mol of NO2 is placed in a 2.0L flask to create NO and O2, calculate [ ]eq if Keq = 1.2 x 10-5.
2 NO 2
[NO]2 [O 2 ]
; K eq =
[NO 2 ]2
 2 NO  O 2
I 0.25
C -2 x
0
+2x
0
+x
E 0.25 - 2 x + 2 x
+x
= 1.2 x 10 -5
Q  0  K eq
100 Rule  applies
0.25
= 20 833  100
1.2 x 10 -5
 0.25 - 2 x  0.25
(2 x) 2 (x )
K eq =
= 1.2 x 10 -5 then
2
(0.25)
4 x3
 1.2 x 10 -5
0.0625
x  5.7 x 10 -3  0.0057
 [O 2 ]eq  0.0057 M, [NO]eq  2(0.0057)  0.0114 M
and [NO2 ]eq = (0.25 - 2(0.0057))  0.24 M
(0.0114) 2 (0.0057 )
Check : Q 
 1.3 x 10 -5  K eq
2
(0.24 )
5. For the system, if we start with 0.100mol/L of CO2 and H2, what are the concentrations of the
reactants and products at equilibrium given that Keq = 0.64 at 900K?
CO 2

I 0.100
C -x
E 0.100 - x
H2

0.100
-x
0.100 - x
[CO][H 2 O]
 0.64
[H 2 ][CO 2 ]
CO 
H 2 O
K eq 
0
x
x
0
x
x
Q  0  Keq  
100 Rule - doesn' t apply
(x)(x)
K eq 
 0.64
(0.100 - x)(0.100 - x)
then
x2
 0.64
(0.100 - x) 2
x
 0.80
0.100 - x
x  0.080 - 0.80x
1.80x  0.080 and x  0.044
 [CO]eq  [H 2 O]eq  0.044 M and [CO 2 ]eq  [H 2 ]eq  0.100 - 0.044  0.056 M
(0.044) 2
Check : Q 
 0.62
(0.056 ) 2
6. For the system, if we start with 0.010 mol/L of H2 and I2 and 0.096 mol/L of HI, what are their
concentrations at equilibrium given that Keq = 0.016?
2 HI
I

H2 
0.096
C - 2x
E 0.096 - 2x
[H 2 ][I 2 ]
 0.016
[HI]2
I2
K eq 
0.010
0.010
0.010 2
Q
 0.010  Keq  
0.096 2
x
0.010  x
x
0.010  x
(0.010  x)(0.010  x)
K eq 
 0.016
(0.096 - 2x) 2
then
(0.010  x) 2
 0.016
(0.096 - 2x) 2
0.010  x
 0.126
0.096 - 2x
0.010  x  0.012 - 0.252x
1.252x  0.002
x  0.0016
 [HI]eq  0.096 - 2(0.0016)  0.093 M and [H 2 ]eq  [I 2 ]eq  0.010  0.0016  0.012 M
(0.012) 2
Check : Q 
 0.017
(0.093 ) 2
7. At 650°C, the reaction below has a Keq value of 0.771. If 2.00 mol of both hydrogen and carbon
dioxide are placed in a 4.00 L container and allowed to react, what will be the equilibrium
concentrations of all four gases?
H2

I 0.500
C -x
E 0.500 - x
CO 2
 CO 
0.500
-x
0.500 - x
0
x
x
[CO][H 2 O]
 0.771
[CO 2 ][H 2 ]
H 2 O
K eq 
0
x
x
Q  0  Keq  
(x)(x)
K eq 
 0.771 then
(0.500 - x)(0.500 - x)
x2
 0.771
(0.500 - x) 2
x
 0.878
0.500 - x
x  0.439 - 0.878x t hen
x  0.234
 [CO]eq  0.260 M , [H 2O]eq  0.234 M and [H 2 ]eq  [CO 2 ]eq  0.500 - 0.234  0.266 M
(0.234) 2
Check : Q 
 0.774
(0.266 ) 2
8. Carbonyl bromide, COBr2, can be formed by reacting CO with Br2. A mixture of 0.400 mol CO,
0.300 mol Br2, and 0.0200 mol COBr2 is sealed in a 5.00L flask. Calculate equilibrium
concentrations for all gases.

CO
Br2

COBr2 
K eq 
[COBr2 ]
 0.680
[CO][Br2 ]
0.0600
0.00400
Q  0.83  Keq  
x
-x
100 Rule - doesn' t apply
0.0600 - x
0.00400 - x
0.00400 - x
K eq 
 0.680
(0.0800  x)(0.0600  x)
I 0.0800
C x
E 0.0800  x
0.00400 - x  0.680 (0.0048  0.14x  x 2 )
0.00400 - x  0.0033  0.095x  0.680x 2
0.680x 2  1.095x - 0.0007  0
a  0.680 b  1.095 c  - 0.0007
- 1.095  1.20
1.36
- 1.095  1.096
x
1.36
x  0.0007 or x  - 1.611 (no negatives)
 [CO]eq  0.0807 M, [Br2 ]eq  0.0607 M and [COBr2 ]eq  0.00400  0.0007  0.0033M
x
Check : Q 
(0.0033)
 0.674
(0.0807 )(0.0607 )