The equilibrium constant for the folowing reactions have been measured at $823$ .
$C o O (s) + H_{2} (g) ⇌ C o (s) + H_{2} O (g); K = 67$
$C o O (s) + C O (g) ⇌ C o (s) + {C O}_{2} (g); K = 490$
The value of equilibrium constant of the reaction following reaction is
$2 {C O}_{2} (g) + 2 H_{2} (g) ⇌ 2 C O (g) + 2 H_{2} O (g)$

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Solution

2nd reaction is reversed so $K_{2}^{1} = \frac{1}{K_{2}}$
$C o O (s) + H_{2} (g) ⇌ C o (s) + H_{2} O (g) K_{1} = 67$
${C O}_{2} (g) + C o (s) ⇌ C o O (s) + C O (g) K_{2}^{1} = \frac{1}{490}$
${C O}_{2} + H_{2} ⇌ C O + H_{2} O K = K_{1} \times K_{2}^{1}$
$= \frac{67}{490} = 0.136$
as reaction is doubled, $K$ is squared $K^{1} = K^{2} = 0.0186$

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