Equilibrium constants K_1 and K_2 for the fol | vedclass

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(D) For the first equilibrium: $NO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons NO_{2(g)}$,the equilibrium constant is $K_1 = \frac{[NO_2]}{[NO][O_2]^{1/2}

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$K_2 = 1 / K_1^2$

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(D) For the first equilibrium: $NO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons NO_{2(g)}$,the equilibrium constant is $K_1 = \frac{[NO_2]}{[NO][O_2]^{1/2}}$.For the second equilibrium: $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,the equilibrium constant is $K_2 = \frac{[NO]^2[O_2]}{[NO_2]^2}$.Observe that the second reaction is the reverse of the first reaction multiplied by $2$.If a reaction is reversed,the equilibrium constant becomes $1/K$. If a reaction is multiplied by a factor $n$,the equilibrium constant becomes $K^n$.Here,the first reaction is reversed and multiplied by $2$,so $K_2 = (1/K_1)^2 = 1/K_1^2$.

Equilibrium constants $K_1$ and $K_2$ for the following equilibria are given:
$NO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons NO_{2(g)}$
$2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$
What is the relationship between $K_1$ and $K_2$?

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Equilibrium constants $K_1$ and $K_2$ for the following equilibria are given:$NO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons NO_{2(g)}$$2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$What is the relationship between $K_1$ and $K_2$?