The equilibrium constant for the reaction SO_ | vedclass

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(A) Given the reaction: $SO_{3(g)} ightleftharpoons SO_{2(g)} + 1/2 O_{2(g)}$ with $K_1 = 4.9 	imes 10^{-2}$.The target reaction is: $2SO_{2(g)} +

Vedclass · Accepted Answer

$416$

Vedclass · Answer

(A) Given the reaction: $SO_{3(g)} ightleftharpoons SO_{2(g)} + 1/2 O_{2(g)}$ with $K_1 = 4.9 	imes 10^{-2}$.The target reaction is: $2SO_{2(g)} + O_{2(g)} ightleftharpoons 2SO_{3(g)}$.First,reverse the given reaction: $SO_{2(g)} + 1/2 O_{2(g)} ightleftharpoons SO_{3(g)}$. The equilibrium constant for this reversed reaction is $K_2 = 1 / K_1 = 1 / (4.9 	imes 10^{-2})$.Next,multiply the stoichiometric coefficients by $2$: $2SO_{2(g)} + O_{2(g)} ightleftharpoons 2SO_{3(g)}$. The equilibrium constant for this reaction is $K_{eq} = (K_2)^2 = (1 / K_1)^2 = 1 / (K_1)^2$.$K_{eq} = 1 / (4.9 	imes 10^{-2})^2 = 1 / (24.01 	imes 10^{-4}) = 1 / 0.002401 \approx 416.49$.Thus,the value is approximately $416$.

The equilibrium constant for the reaction $SO_{3(g)} \rightleftharpoons SO_{2(g)} + 1/2 O_{2(g)}$ is $4.9 \times 10^{-2}$. Find the equilibrium constant for the reaction $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$.

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