Calculate the equation constant for the reaction: $H_{2} (g) + {C O}_{2} (g) ⇌ H_{2} O (g) + C O$ at 1395 K, if the equilibrium constants at 1395 K for the following are
$2 H_{2} O (g) ⇌ 2 H_{2} (g) + O_{2} (g) K_{1} = 2.1 \times 10^{- 13}$
$2 {C O}_{2} (g) ⇌ 2 C O (g) + O_{2} (g) K_{2} = 1.4 \times 10^{- 12}$

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Solution

$2 C O_{2} (g) ⇌ 2 C O (g) + O_{2} (g); K_{2} = 1.4 \times 10^{- 12}$
$i i ⟵ C O_{2} (g) ⇌ C O (g) + \frac{1}{2} O_{2} (g); K_{2}^{1} = {(1.4 \times 10^{- 12})}^{\frac{1}{2}}$
$2 H_{2} O (g) ⇌ 2 H_{2} (g) + O_{2} (g); K_{1} = 2.1 \times 10^{- 13}$
$i ⟵ H_{2} (g) + \frac{1}{2} O_{2} (g) ⇌ H_{2} O (g); K_{1}^{1} = + {(2.1 \times 10^{- 13})}^{\frac{1}{2}}$
$H_{2} (g) ⇌ H_{2} (g); K$

$K = K_{1}^{1} \times K_{1}^{2} = + {(2.1 \times 10^{- 13})}^{\frac{1}{2}} \times {(1.4 \times 10^{- 12})}^{\frac{1}{2}}$
$K = 1.71 \times 10^{- 12}$
Therefore, the equilibrium canstant for the given reaction is $1.71 \times 10^{- 12}$ .

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