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If the equilibrium constant for the reaction, $2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3$ is 64 at $500 \mathrm{~K}$, then the equilibrium constant for the reaction $\mathrm{SO}_3 \rightleftharpoons \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2$ at the same temperature is
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Verified Answer
The correct answer is:
$\frac{1}{8}$
Given, $2 \mathrm{SO}_2+\mathrm{O}_2 \rightleftharpoons 2 \mathrm{SO}_3, K=64$
$\mathrm{SO}_3 \rightleftharpoons \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2, K^{\prime}=?$
$\because$ New $(K)=[K]^{1 / n}$ i.e. $K^{\prime}$
where, $n=$ factor to the new equilibrium constant, which is $n$th root of the previous value and
$K^{\prime}=\left[\frac{1}{K}\right]^n,\left(n=\frac{1}{2}\right)$
$\therefore$ Hence, new $K$ i.e. $\left(K^{\prime}\right)=\left[\frac{1}{64}\right]^{1 / 2}=\frac{1}{8} \quad\left(\because n=\frac{1}{2}\right)$
$\therefore$ Option (b) is the correct answer.
$\mathrm{SO}_3 \rightleftharpoons \mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2, K^{\prime}=?$
$\because$ New $(K)=[K]^{1 / n}$ i.e. $K^{\prime}$
where, $n=$ factor to the new equilibrium constant, which is $n$th root of the previous value and
$K^{\prime}=\left[\frac{1}{K}\right]^n,\left(n=\frac{1}{2}\right)$
$\therefore$ Hence, new $K$ i.e. $\left(K^{\prime}\right)=\left[\frac{1}{64}\right]^{1 / 2}=\frac{1}{8} \quad\left(\because n=\frac{1}{2}\right)$
$\therefore$ Option (b) is the correct answer.
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