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At $400 \mathrm{~K}$, in a $1.0 \mathrm{~L}$ vessel, $\mathrm{N}_2 \mathrm{O}_4$ is allowed to attain equilibrium, $\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})$ At equilibrium, the total pressure is $600 \mathrm{~mm} \mathrm{Hg}$, when $20 \%$ of $\mathrm{N}_2 \mathrm{O}_4$ is dissociated. The value of $K_n$ for the reaction is
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100
$\begin{array}{lcc}\text { } & \mathrm{N}_2 \mathrm{O}_4 \rightleftharpoons 2 \mathrm{NO}_2 \\ \text { Initial moles } & 1 & 0 \\ \text { At equilibrium } & (1-0.2) & 2 \times 0.2=0.8 \\ \text { Total moles }=0.4+0.8=12\end{array}$
$K_P=\frac{\left(P_{\mathrm{NO}_2}\right)^2}{\left(P_{\mathrm{N}_2 \mathrm{O}_4}\right)} \Rightarrow K_P=\frac{\left(\frac{0.4}{1.2} \times 600\right)^2}{\left(\frac{0.8}{1.2} \times 600\right)}=100$
$K_P=\frac{\left(P_{\mathrm{NO}_2}\right)^2}{\left(P_{\mathrm{N}_2 \mathrm{O}_4}\right)} \Rightarrow K_P=\frac{\left(\frac{0.4}{1.2} \times 600\right)^2}{\left(\frac{0.8}{1.2} \times 600\right)}=100$
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