The reaction 2 NO_{2(g)} rightleftharpoons N_ | vedclass

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(B) The reaction is $2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$.Let the initial moles of $NO_2$ be $n_0$. At equilibrium,let $x$ moles of $N_2O_4$ be

Vedclass · Accepted Answer

$1.34$

Vedclass · Answer

(B) The reaction is $2 NO_{2(g)} ightleftharpoons N_2O_{4(g)}$.Let the initial moles of $NO_2$ be $n_0$. At equilibrium,let $x$ moles of $N_2O_4$ be formed.Moles at equilibrium: $n_{NO_2} = n_0 - 2x$,$n_{N_2O_4} = x$.Total mass: $46(n_0 - 2x) + 92x = 64.4 \implies 46n_0 = 64.4 \implies n_0 = 1.4 \ mol$.Total moles $n_T = (1.4 - 2x) + x = 1.4 - x$.$P_{NO_2} = \frac{1.4-2x}{1.4-x} P_T$ and $P_{N_2O_4} = \frac{x}{1.4-x} P_T$.$K_p = \frac{P_{N_2O_4}}{(P_{NO_2})^2} = \frac{x(1.4-x)}{(1.4-2x)^2 P_T} = 6.67$.Using $P_T V = n_T RT \implies P_T = \frac{(1.4-x) 	imes 0.082 	imes 300}{15} = 1.64(1.4-x)$.Substituting $P_T$ into $K_p$ expression and solving for $x$ gives $x \approx 0.58 \ mol$.Then $n_T = 1.4 - 0.58 = 0.82 \ mol$.$P_T = \frac{0.82 	imes 0.082 	imes 300}{15} = 1.34 \ atm$.

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