For the reactions:2NO + O_2 rightleftharpoons | vedclass

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Question

(D) Given reactions:$(i) \ 2NO + O_2 \rightleftharpoons 2NO_2 ; K_1 = \frac{[NO_2]^2}{[NO]^2 [O_2]}$$(ii) \ 4NO + 2Cl_2 \rightleftharpoons 4NOCl ; K_2

Vedclass · Accepted Answer

$\sqrt{K_2} / K_1$

Vedclass · Answer

(D) Given reactions:$(i) \ 2NO + O_2 ightleftharpoons 2NO_2 ; K_1 = \frac{[NO_2]^2}{[NO]^2 [O_2]}$$(ii) \ 4NO + 2Cl_2 ightleftharpoons 4NOCl ; K_2 = \frac{[NOCl]^4}{[NO]^4 [Cl_2]^2}$$(iii) \ NO_2 + 1/2 Cl_2 ightleftharpoons NOCl + 1/2 O_2 ; K_3 = \frac{[NOCl][O_2]^{1/2}}{[NO_2][Cl_2]^{1/2}}$To find the relationship,consider $K_3^2$:$K_3^2 = \left( \frac{[NOCl][O_2]^{1/2}}{[NO_2][Cl_2]^{1/2}} ight)^2 = \frac{[NOCl]^2 [O_2]}{[NO_2]^2 [Cl_2]}$Now,observe the expression for $K_2 / K_1$:$\frac{K_2}{K_1} = \frac{[NOCl]^4}{[NO]^4 [Cl_2]^2} 	imes \frac{[NO]^2 [O_2]}{[NO_2]^2} = \frac{[NOCl]^4 [O_2]}{[NO]^2 [Cl_2]^2 [NO_2]^2}$Actually,let us manipulate the equations:$2 	imes (iii) \Rightarrow 2NO_2 + Cl_2 ightleftharpoons 2NOCl + O_2 ; K_{eq} = K_3^2$From $(i)$,$2NO_2 ightleftharpoons 2NO + O_2 ; K = 1/K_1$Adding these: $2NO + Cl_2 ightleftharpoons 2NOCl ; K = K_3^2 / K_1$Comparing with $(ii)$,$4NO + 2Cl_2 ightleftharpoons 4NOCl$ is $2 	imes (2NO + Cl_2 ightleftharpoons 2NOCl)$,so $K_2 = (K_3^2 / K_1)^2$$K_2 = K_3^4 / K_1^2$ $\Rightarrow K_3^4 = K_2 K_1^2$ $\Rightarrow K_3^2 = \sqrt{K_2} K_1$ (Wait,re-evaluating).Correct derivation: $K_3 = \frac{K_2^{1/4}}{K_1^{1/2}} \Rightarrow K_3^2 = \frac{K_2^{1/2}}{K_1} = \frac{\sqrt{K_2}}{K_1}$.

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