The equilibrium constant for the reaction SO_ | vedclass

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(D) The given reaction is $SO_{3(g)} ightleftharpoons SO_{2(g)} + \frac{1}{2} O_{2(g)}$ with $K_{C} = 4.9 	imes 10^{-2}$.The target reaction is $2

Vedclass · Accepted Answer

$416$

Vedclass · Answer

(D) The given reaction is $SO_{3(g)} ightleftharpoons SO_{2(g)} + \frac{1}{2} O_{2(g)}$ with $K_{C} = 4.9 	imes 10^{-2}$.The target reaction is $2 SO_{2(g)} + O_{2(g)} ightleftharpoons 2 SO_{3(g)}$.First,reverse the original reaction: $SO_{2(g)} + \frac{1}{2} O_{2(g)} ightleftharpoons SO_{3(g)}$. The new equilibrium constant is $K_{C1} = \frac{1}{K_{C}} = \frac{1}{4.9 	imes 10^{-2}}$.Next,multiply the reversed reaction by $2$: $2 SO_{2(g)} + O_{2(g)} ightleftharpoons 2 SO_{3(g)}$. The new equilibrium constant is $K_{C}^{\prime} = (K_{C1})^2 = \left( \frac{1}{4.9 	imes 10^{-2}} ight)^2$.$K_{C}^{\prime} = \left( \frac{100}{4.9} ight)^2 \approx (20.408)^2 \approx 416.49 \approx 416$.

The equilibrium constant for the reaction $SO_{3(g)} \rightleftharpoons SO_{2(g)} + \frac{1}{2} O_{2(g)}$ is $K_{C} = 4.9 \times 10^{-2}$. The value of $K_{C}$ for the reaction $2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$ is:

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