From the given data of the following reactions;

$C a O (s) + H_{2} (g) ⇌ C a (s) + H_{2} O (g)$ ; $K = 670$
$C a O (s) + C O (g) ⇌ C a (s) + C O_{2} (g)$ ; $K = 490$
The value of $K_{c}$ (nearest integer) for the given reaction is _____.

$C O_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)$

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Solution

$C a O (s) + H_{2} (g) ⇌ C a (s) + H_{2} O (g)$ ; $K = 670$ ......(1)
$C a O (s) + C O (g) ⇌ C a (s) + C O_{2} (g)$ ; $K = 490$ ......(2)

$C O_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)$ ......(3)

When the second reaction is subtracted from the first reaction, the third reaction is obtained.

Hence, the equilibrium constant for the third reaction is the ratio of the equilibrium constant for the first reaction and the equilibrium constant for the second reaction.

$K_{3} = \frac{K_{1}}{K_{2}}$ .

Substitute values in the above expression:

$K_{3} = \frac{670}{490} = 1.37$

Hence, the value of the equilibrium constant for the third reaction is 1.37.

The nearest integer is $1$ .

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Similar questions

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$C a O (s) + H_{2} (g) ⇌ C a (s) + H_{2} O (g)$ ; $K = 670$
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