Question

The equilibrium constants for the following reaction at 1000 K are given :

$CoO\left(s\right)+H_{2\left(g\right)}\rightleftharpoons Co\left(s\right)+H_2{O}_{\left(g\right)}{K}_1=62.5$

$CoO\left(s\right)+C{O}_{\left(g\right)}\rightleftharpoons Co\left(s\right)+C{O}_{2\left(g\right)}{K}_2=500$

Then the equilibrium constant [K] for the below given reaction is :

$C{O}_{\left(g\right)}+H_2{O}_{\left(g\right)}\rightleftharpoons C{O}_{2\left(g\right)}+H_{2\left(g\right)}$

• A. 10
• B. 8
• C. 12
• D. 9

Answer 1

Answer: (B)

8

First equilibrium reaction is $CoO\left(s\right)+H_{2\left(g\right)}\rightleftharpoons Co\left(s\right)+H_2{O}_{\left(g\right)},K_1=62.5$.

The expression for the equilibrium constant is $K_1=\frac{\left[H_{2}O\right]}{\left[H_2\right]}=62.5$.

Second equilibrium reaction is $CoO\left(s\right)+C{O}_{\left(g\right)}\rightleftharpoons Co\left(s\right)+C{O}_{2\left(g\right)},K_2=500$

The expression for the equilibrium constant is $K_2=\frac{\left[C{O}_2\right]}{\left[CO\right]}=500$.

When first equilibrium reaction is reversed and added to second equilibrium reaction, third equilibrium reaction is obtained.

$C{O}_{\left(g\right)}+H_2{O}_{\left(g\right)}\rightleftharpoons C{O}_{2\left(g\right)}+H_{2\left(g\right)}$
The expression for the equilibrium constant is $K=\frac{\left[C{O}_2\right]\left[H_2\right]}{\left[CO\right]\left[H_{2}O\right]}=\frac{K_2}{K_1}=\frac{500}{62.5}=8$.