The value of K_P for the equilibrium reaction | vedclass

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(D) For the reaction: ${N_2}{O_4}_{(g)} \rightleftharpoons 2N{O_2}_{(g)}$Let the initial moles of ${N_2}{O_4}$ be $1$ and degree of dissociation be $\

Vedclass · Accepted Answer

$71$

Vedclass · Answer

(D) For the reaction: ${N_2}{O_4}_{(g)} ightleftharpoons 2N{O_2}_{(g)}$Let the initial moles of ${N_2}{O_4}$ be $1$ and degree of dissociation be $\alpha$.At equilibrium,moles of ${N_2}{O_4} = (1 - \alpha)$ and moles of $N{O_2} = 2\alpha$.Total moles at equilibrium $= (1 - \alpha) + 2\alpha = (1 + \alpha)$.Partial pressure of ${N_2}{O_4} = \frac{(1 - \alpha)}{(1 + \alpha)} 	imes P$.Partial pressure of $N{O_2} = \frac{2\alpha}{(1 + \alpha)} 	imes P$.$K_P = \frac{(P_{N{O_2}})^2}{P_{{N_2}{O_4}}} = \frac{[\frac{2\alpha}{1 + \alpha} 	imes P]^2}{\frac{1 - \alpha}{1 + \alpha} 	imes P} = \frac{4\alpha^2 P}{1 - \alpha^2}$.Given $K_P = 2$ and $P = 0.5 \ atm$:$2 = \frac{4 	imes \alpha^2 	imes 0.5}{1 - \alpha^2} = \frac{2\alpha^2}{1 - \alpha^2}$.$1 - \alpha^2 = \alpha^2 \implies 2\alpha^2 = 1 \implies \alpha^2 = 0.5$.$\alpha = \sqrt{0.5} \approx 0.707$.Percentage dissociation $= \alpha 	imes 100 = 70.7 \% \approx 71 \%$.

The value of $K_P$ for the equilibrium reaction ${N_2}{O_4}_{(g)} \rightleftharpoons 2N{O_2}_{(g)}$ is $2$. The percentage dissociation of ${N_2}{O_4}_{(g)}$ at a pressure of $0.5 \ atm$ is

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