The equilibrium constants for the following reaction at 1000 K are given :
$C o O (s) + H_{2 (g)} ⇌ C o (s) + H_{2} O_{(g)} K_{1} = 62.5$
$C o O (s) + C O_{(g)} ⇌ C o (s) + C O_{2 (g)} K_{2} = 500$
Then the equilibrium constant [K] for the below given reaction is :
$C O_{(g)} + H_{2} O_{(g)} ⇌ C O_{2 (g)} + H_{2 (g)}$

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Solution

The correct option is A 8
First equilibrium reaction is $C o O (s) + H_{2 (g)} ⇌ C o (s) + H_{2} O_{(g)}, K_{1} = 62.5$ .
The expression for the equilibrium constant is $K_{1} = \frac{[H_{2} O]}{[H_{2}]} = 62.5$ .
Second equilibrium reaction is $C o O (s) + C O_{(g)} ⇌ C o (s) + C O_{2 (g)}, K_{2} = 500$
The expression for the equilibrium constant is $K_{2} = \frac{[C O_{2}]}{[C O]} = 500$ .
When first equilibrium reaction is reversed and added to second equilibrium reaction, third equilibrium reaction is obtained.
$C O_{(g)} + H_{2} O_{(g)} ⇌ C O_{2 (g)} + H_{2 (g)}$
The expression for the equilibrium constant is $K = \frac{[C O_{2}] [H_{2}]}{[C O] [H_{2} O]} = \frac{K_{2}}{K_{1}} = \frac{500}{62.5} = 8$ .

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