For the reaction $SO_{3(g)} \rightleftharpoons SO_{2(g)} + \frac{1}{2}O_{2(g)}$,the equilibrium constant $K_C = 4.9 \times 10^{-2}$. Calculate the $K_C$ for the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.
(416) Given reaction $I$: $SO_{3(g)} \rightleftharpoons SO_{2(g)} + \frac{1}{2}O_{2(g)}$ with $K_{C1} = 4.9 \times 10^{-2}$.
Target reaction $II$: $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.
Reaction $II$ is obtained by reversing reaction $I$ and multiplying by $2$.
Therefore,$K_{C2} = (1 / K_{C1})^2$.
$K_{C2} = (1 / (4.9 \times 10^{-2}))^2 = (100 / 4.9)^2 \approx (20.408)^2 \approx 416.49$.
Target reaction $II$: $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.
Reaction $II$ is obtained by reversing reaction $I$ and multiplying by $2$.
Therefore,$K_{C2} = (1 / K_{C1})^2$.
$K_{C2} = (1 / (4.9 \times 10^{-2}))^2 = (100 / 4.9)^2 \approx (20.408)^2 \approx 416.49$.
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