For the reactions N_{2(g)} + O_{2(g)} rightle | vedclass

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(A) The first reaction is: $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ with equilibrium constant $K_1 = \frac{[NO]^2}{[N_2][O_2]}$.The second r

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$K_1 = (1/K_2)^2$

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(A) The first reaction is: $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ with equilibrium constant $K_1 = \frac{[NO]^2}{[N_2][O_2]}$.The second reaction is: $NO_{(g)} \rightleftharpoons 1/2N_{2(g)} + 1/2O_{2(g)}$ with equilibrium constant $K_2 = \frac{[N_2]^{1/2}[O_2]^{1/2}}{[NO]}$.Observe that the first reaction is the reverse of the second 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 $K^n$.Therefore,$K_1 = (1/K_2)^2$.

For the reactions $N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$ and $NO_{(g)} \rightleftharpoons 1/2N_{2(g)} + 1/2O_{2(g)}$,the equilibrium constants are $K_1$ and $K_2$ respectively. What is the relationship between $K_1$ and $K_2$?

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