2NO_2 rightleftharpoons 2NO + O_2; K = 1.6 ti | vedclass

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Question

(C) Given reaction $(i)$: $2NO_2 ightleftharpoons 2NO + O_2$ with equilibrium constant $K = \frac{[NO]^2 [O_2]}{[NO_2]^2} = 1.6 	imes 10^{-12}$.Tar

Vedclass · Accepted Answer

$K' = \frac{1}{\sqrt{K}}$

Vedclass · Answer

(C) Given reaction $(i)$: $2NO_2 ightleftharpoons 2NO + O_2$ with equilibrium constant $K = \frac{[NO]^2 [O_2]}{[NO_2]^2} = 1.6 	imes 10^{-12}$.Target reaction $(ii)$: $NO + \frac{1}{2}O_2 ightleftharpoons NO_2$ with equilibrium constant $K' = \frac{[NO_2]}{[NO] [O_2]^{1/2}}$.Comparing the two reactions,we observe that reaction $(ii)$ is the reverse of reaction $(i)$ multiplied by a factor of $\frac{1}{2}$.Mathematically,$K' = (\frac{1}{K})^{1/2} = \frac{1}{\sqrt{K}}$.Therefore,the correct option is $C$.

$2NO_2 \rightleftharpoons 2NO + O_2$; $K = 1.6 \times 10^{-12}$. For $NO + \frac{1}{2}O_2 \rightleftharpoons NO_2$,$K' = $

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