At T(K), the equilibrium constant of A_{2}(g) | vedclass

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(D) If a chemical equation is multiplied by a factor '$n$',the new equilibrium constant $K'$ is given by $K' = K^{n}$.Here, the original reaction is $

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$3 	imes 10^{-2}$

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(D) If a chemical equation is multiplied by a factor '$n$',the new equilibrium constant $K'$ is given by $K' = K^{n}$.Here, the original reaction is $A_{2}(g) + B_{2}(g) ightleftharpoons C(g)$ with equilibrium constant $K = 2.7 	imes 10^{-5}$.The new reaction is $\frac{1}{3}A_{2}(g) + \frac{1}{3}B_{2}(g) ightleftharpoons \frac{1}{3}C(g)$, which is obtained by multiplying the original reaction by $n = 1/3$.Therefore, the new equilibrium constant $K'$ is $K^{1/3}$.$K' = (2.7 	imes 10^{-5})^{1/3} = (27 	imes 10^{-6})^{1/3}$.$K' = (27)^{1/3} 	imes (10^{-6})^{1/3} = 3 	imes 10^{-2}$.

At $T(K)$, the equilibrium constant of $A_{2}(g) + B_{2}(g) \rightleftharpoons C(g)$ is $2.7 \times 10^{-5}$. What is the equilibrium constant for $\frac{1}{3}A_{2}(g) + \frac{1}{3}B_{2}(g) \rightleftharpoons \frac{1}{3}C(g)$ at the same temperature?

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