At 400 K,in a 1.0 L vessel,N_2O_4 is allowe | vedclass

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(B) The reaction is $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$.Let the initial moles of $N_2O_4$ be $1$.At equilibrium,$20 \%$ of $N_2O_4$ is dissoci

Vedclass · Accepted Answer

$100$

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(B) The reaction is $N_2O_{4(g)} ightleftharpoons 2NO_{2(g)}$.Let the initial moles of $N_2O_4$ be $1$.At equilibrium,$20 \%$ of $N_2O_4$ is dissociated,so the moles of $N_2O_4$ remaining $= 1 - 0.2 = 0.8$.The moles of $NO_2$ formed $= 2 	imes 0.2 = 0.4$.Total moles at equilibrium $= 0.8 + 0.4 = 1.2$.Partial pressure of $N_2O_4$ $(P_{N_2O_4})$ $= \frac{0.8}{1.2} 	imes 600 \ mm \ Hg = 400 \ mm \ Hg$.Partial pressure of $NO_2$ $(P_{NO_2})$ $= \frac{0.4}{1.2} 	imes 600 \ mm \ Hg = 200 \ mm \ Hg$.$K_p = \frac{(P_{NO_2})^2}{P_{N_2O_4}} = \frac{(200)^2}{400} = \frac{40000}{400} = 100$.

At $400 \ K$,in a $1.0 \ L$ vessel,$N_2O_4$ is allowed to attain equilibrium,$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At equilibrium,the total pressure is $600 \ mm \ Hg$,when $20 \%$ of $N_2O_4$ is dissociated. The value of $K_p$ for the reaction is

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