CHEM1101
2005-J-9
June 2005
 The value of the equilibrium constant, Kc, for the following reaction is 0.118 mol L–1.
2CO2(g) + N2(g)
Marks
2
2CO(g) + 2NO(g)
What is the equilibrium concentration of CO(g) if the equilibrium concentration of
[CO2(g)] = 0.392 M, [N2(g)] = 0.419 M and [NO(g)] = 0.246 M?
The equilibrium constant in terms of concentrations, Kc, is:
Kc =
[CO(g)]2 [NO(g)]2
[CO 2 (g)]2 [N 2 (g)]

[CO(g)2 ](0.246)2
(0.392)2 (0.419)
 0.118
Therefore, [CO(g)]2 = 0.126 M2 or [CO(g)] = 0.354 M
Answer: 0.354 M
 When hydrogen cyanide (HCN) is dissolved in water it dissociates into ions according
to the following equation.
HCN(aq)
H+(aq) + CN–(aq)
The equilibrium constant for this reaction is Kc = 6.2  10–10 mol L–1. If 1.00 mol of
HCN is dissolved to make 1.00 L of solution, calculate the percentage of HCN that
will be dissociated.
The initial concentration of HCN is 1.00 M. The reaction table is:
[initial]
change
[equilibrium]
HCN(aq)
1.00
-x
1.00-x
H+(aq)
0
+x
x
CN-(aq)
0
+x
x
[H  (aq)][CN  (aq)]
(x)(x)
x2


Kc =
= 6.2 × 10-10
[HCN(aq)]
(1.00  x) (1.00  x)
As Kc is very small, the amount of dissociation will be tiny and 1.00-x ~ 1.00.
Hence,
x2 ~ 6.2 × 10-10x, so x = 2.5 × 10-5 = [H+(aq)] = [CN-(aq)].
As [HCN(aq)] = 1.00 – x, [HCN(aq)] = 1.00 and the percentage dissociation is:
percentage dissociation =
2.5  105
 100  2.5 × 10-3 %
1.00
Answer: 2.5 × 10-3 %
3