K_1, K_2 and K_3 are the equilibrium constant | vedclass

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(D) For reaction $(I): N_2 + 2O_2 \rightleftharpoons 2NO_2$,the equilibrium constant is $K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2}$.For reaction $(II): 2NO_

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$K_1 = \frac{1}{K_2} = \frac{1}{K_3^2}$

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(D) For reaction $(I): N_2 + 2O_2 \rightleftharpoons 2NO_2$,the equilibrium constant is $K_1 = \frac{[NO_2]^2}{[N_2][O_2]^2}$.For reaction $(II): 2NO_2 \rightleftharpoons N_2 + 2O_2$,the equilibrium constant is $K_2 = \frac{[N_2][O_2]^2}{[NO_2]^2}$.Clearly,$K_2 = \frac{1}{K_1}$,which implies $K_1 = \frac{1}{K_2}$.For reaction $(III): NO_2 \rightleftharpoons \frac{1}{2}N_2 + O_2$,the equilibrium constant is $K_3 = \frac{[N_2]^{1/2}[O_2]}{[NO_2]}$.Comparing $K_3$ with $K_1$,we see that $K_3 = \sqrt{\frac{[N_2][O_2]^2}{[NO_2]^2}} = \sqrt{\frac{1}{K_1}} = \frac{1}{\sqrt{K_1}}$.Squaring both sides,$K_3^2 = \frac{1}{K_1}$,so $K_1 = \frac{1}{K_3^2}$.Thus,the correct relation is $K_1 = \frac{1}{K_2} = \frac{1}{K_3^2}$.

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$K_1, K_2$ and $K_3$ are the equilibrium constants of the following reactions $(I), (II)$ and $(III)$,respectively :$(I) \ N_2 + 2O_2 \rightleftharpoons 2NO_2$$(II) \ 2NO_2 \rightleftharpoons N_2 + 2O_2$$(III) \ NO_2 \rightleftharpoons \frac{1}{2}N_2 + O_2$The correct relation from the following is