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At 298 K equilibrium constant $K_{1}$ and $K_{2}$ of the following reaction
${SO}_{2} (g) + 1 / 2 O_{2 (g)} ⇌ {SO}_{3} (g)$ ……(1)
$2 {SO}_{3} (g) ⇌ 2 {SO}_{2 (g)} + O_{2} (g)$ …… (2)
The relation between $K_{1}$ and $K_{2}$ is

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$K_{1}$ = $K_{2}$

$K_{2} = K_{1}^{2}$

$K_{2} = 1 / K_{1}^{2}$

$K_{2}$ = 1/ $K_{1}$

answer is C.

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Detailed Solution

$K_{2} = \frac{1}{{(K_{1})}^{2}}$

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