For the reactions NO_{(g)} + 1/2 O_{2(g)} rig | vedclass

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(C) The first reaction is: $NO_{(g)} + 1/2 O_{2(g)} ightleftharpoons NO_{2(g)}$ with equilibrium constant $K_1 = 4 	imes 10^{-3}$.The second reacti

Vedclass · Accepted Answer

$6.25 	imes 10^{4}$

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(C) The first reaction is: $NO_{(g)} + 1/2 O_{2(g)} ightleftharpoons NO_{2(g)}$ with equilibrium constant $K_1 = 4 	imes 10^{-3}$.The second reaction is: $2NO_{2(g)} ightleftharpoons 2NO_{(g)} + O_{2(g)}$ with equilibrium constant $K_2$.Notice that the second reaction is the reverse of the first reaction multiplied by $2$.Therefore,the relationship between the equilibrium constants is $K_2 = (1/K_1)^2$.Substituting the value of $K_1$: $K_2 = 1 / (4 	imes 10^{-3})^2 = 1 / (16 	imes 10^{-6}) = 10^6 / 16 = 0.0625 	imes 10^6 = 6.25 	imes 10^4$.

For the reactions $NO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons NO_{2(g)}$ and $2NO_{2(g)} \rightleftharpoons 2NO_{(g)} + O_{2(g)}$,the equilibrium constants at a given temperature are $K_1$ and $K_2$ respectively. If the value of $K_1$ is $4 \times 10^{-3}$,then the value of $K_2$ will be ....

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