For the reactions:2NO + O_2 rightleftharpoons | vedclass

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(C) Given reactions:$(i) 2NO + O_2 \rightleftharpoons 2NO_2$ with equilibrium constant $K_1$$(ii) 4NO + 2Cl_2 \rightleftharpoons 4NOCl$ with equilibri

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$K_2 / K_1^2$

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(C) Given reactions:$(i) 2NO + O_2 ightleftharpoons 2NO_2$ with equilibrium constant $K_1$$(ii) 4NO + 2Cl_2 ightleftharpoons 4NOCl$ with equilibrium constant $K_2$$(iii) NO_2 + \frac{1}{2}Cl_2 ightleftharpoons NOCl + \frac{1}{2}O_2$ with equilibrium constant $K_3$To obtain reaction $(iii)$,we manipulate reactions $(i)$ and $(ii)$:Reverse reaction $(i)$ and divide by $2$:$NO_2 ightleftharpoons NO + \frac{1}{2}O_2$; $K' = \frac{1}{\sqrt{K_1}}$Divide reaction $(ii)$ by $4$:$NO + \frac{1}{2}Cl_2 ightleftharpoons NOCl$; $K'' = (K_2)^{1/4}$Adding these two reactions gives reaction $(iii)$:$NO_2 + \frac{1}{2}Cl_2 ightleftharpoons NOCl + \frac{1}{2}O_2$Therefore,$K_3 = K' 	imes K'' = \frac{(K_2)^{1/4}}{\sqrt{K_1}}$Squaring both sides:$K_3^2 = \frac{(K_2)^{1/2}}{K_1} = \sqrt{\frac{K_2}{K_1^2}}$Wait,checking the options provided,the standard relation is $K_3 = \frac{K_2^{1/4}}{K_1^{1/2}}$. Thus $K_3^2 = \frac{K_2^{1/2}}{K_1}$. Given the options,there is a discrepancy in the provided choices. Based on the derivation,$K_3^2 = \sqrt{K_2}/K_1$.

For the reactions:
$2NO + O_2 \rightleftharpoons 2NO_2$; $K_1$
$4NO + 2Cl_2 \rightleftharpoons 4NOCl$; $K_2$
$NO_2 + \frac{1}{2}Cl_2 \rightleftharpoons NOCl + \frac{1}{2}O_2$; $K_3$
Where $K_1, K_2, K_3$ are equilibrium constants,then $K_3^2$ is equal to:

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For the reactions:$2NO + O_2 \rightleftharpoons 2NO_2$; $K_1$$4NO + 2Cl_2 \rightleftharpoons 4NOCl$; $K_2$$NO_2 + \frac{1}{2}Cl_2 \rightleftharpoons NOCl + \frac{1}{2}O_2$; $K_3$Where $K_1, K_2, K_3$ are equilibrium constants,then $K_3^2$ is equal to: