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(B) Given reaction: $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$,$K = 64$. Target reaction: $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$,$K' = ?$. To

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$\frac{1}{8}$

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(B) Given reaction: $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$,$K = 64$. Target reaction: $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$,$K' = ?$. To obtain the target reaction,we first reverse the given reaction: $2 SO_3 \rightleftharpoons 2 SO_2 + O_2$. The equilibrium constant for this reversed reaction is $K_{rev} = \frac{1}{K} = \frac{1}{64}$. Next,we multiply the coefficients of the reversed reaction by $\frac{1}{2}$: $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$. When a reaction is multiplied by a factor $n$,the new equilibrium constant is $(K_{original})^n$. Here,$n = \frac{1}{2}$. Therefore,$K' = (K_{rev})^{1/2} = (\frac{1}{64})^{1/2} = \frac{1}{8} = 0.125$. Thus,option $(b)$ is the correct answer.

If the equilibrium constant for the reaction,$2 SO_2 + O_2 \rightleftharpoons 2 SO_3$ is $64$ at $500 \ K$,then the equilibrium constant for the reaction $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$ at the same temperature is

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