The equilibrium constant for the reaction SO_ | vedclass

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(C) For the reaction $SO_{3(g)} \rightleftharpoons SO_{2(g)} + \frac{1}{2} O_{2(g)}$,the equilibrium constant is $K_c = \frac{[SO_2][O_2]^{1/2}}{[SO_3

Vedclass · Accepted Answer

$416$

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(C) For the reaction $SO_{3(g)} ightleftharpoons SO_{2(g)} + \frac{1}{2} O_{2(g)}$,the equilibrium constant is $K_c = \frac{[SO_2][O_2]^{1/2}}{[SO_3]} = 4.9 	imes 10^{-2}$.To obtain the reaction $2SO_{2(g)} + O_{2(g)} ightleftharpoons 2SO_{3(g)}$,we reverse the original reaction and multiply it by $2$.If a reaction is reversed,the new equilibrium constant becomes $1/K_c$. If it is multiplied by a factor $n$,the new equilibrium constant becomes $(K_c)^n$.Therefore,for the new reaction,$K_c' = \frac{1}{(K_c)^2} = \frac{1}{(4.9 	imes 10^{-2})^2}$.$K_c' = \frac{1}{24.01 	imes 10^{-4}} = \frac{10^4}{24.01} \approx 416.5$.

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 $2SO_{2(g)} + O_{2(g)} \rightleftharpoons 2SO_{3(g)}$ will be

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