(C) For the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$,the equilibrium constant is $K_1 = 5 \times 10^{-2}$.For the reac

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(C) For the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$,the equilibrium constant is $K_1 = 5 \times 10^{-2}$.For the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$,we observe that the reaction is the reverse of the first reaction multiplied by a factor of $2$.Therefore,the new equilibrium constant $K_2$ is given by $K_2 = \frac{1}{K_1^2}$.Substituting the value of $K_1$:$K_2 = \frac{1}{(5 \times 10^{-2})^2} = \frac{1}{25 \times 10^{-4}} = \frac{10^4}{25} = 400 = 4 \times 10^2$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Law of equilibrium and Equilibrium constant

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The equilibrium constant for the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$ is $5 \times 10^{-2} \ atm^{-1/2}$. The equilibrium constant of the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$ would be

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$100 \ atm$
B
$200 \ atm$
C
$4 \times 10^2 \ atm$
D
$6.25 \times 10^4 \ atm$

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

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The equilibrium constant for the reaction $SO_{2(g)} + \frac{1}{2} O_{2(g)} \rightleftharpoons SO_{3(g)}$ is $5 \times 10^{-2} \ atm^{-1/2}$. The equilibrium constant of the reaction $2 SO_{3(g)} \rightleftharpoons 2 SO_{2(g)} + O_{2(g)}$ would be

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