For the reaction: 2(g) + (g) ⇌ 2NOCl(g), and are initially taken in mole ratio of 2:1. The total pressure at equilibrium is found to be 1 atm. If the moles of NOCl are one-fourth of that of at equilib

Question

For the reaction: 2$NO$(g) + $Cl_2$(g) ⇌ 2NOCl(g), $NO$ and $Cl_2$ are initially taken in mole ratio of 2:1. The total pressure at equilibrium is found to be 1 atm. If the moles of NOCl are one-fourth of that of $Cl_2$ at equilibrium, $K_p$ for the reaction is:

Prashant Kotian · Accepted Answer

Correct answer: B. Reaction: Initial ratio NO : Cl = 2 : 1. Let initial moles be 2 and 1, with total pressure . ICE table (let mol of Cl react at equilibrium): \begin{center} \begin{tabular}{lccc} & 2NO & Cl & 2NOCl \ Initial & 2 & 1 & 0 \ Change & & & \ Equil. & & & \ \end{tabular} \end{center} Total moles at equilibrium: From the hint: . Total moles . Mole fractions: $$ x_{\text{NOCl}} = \frac{2/9}{26/9} = \frac{1}{13}, \quad x_{\text{NO}} = \frac{16/9}{26/9} = \frac{8}{13}, \quad x_{\text{Cl}_2} = \frac{8/9}{26/9} = \frac{4}{13} $$ $$ K_p = \frac{p_{\text{NOCl}}^2}{p_{\text{NO}}^2 \cdot p_{\text{Cl}_2}} = \frac{(P/13)^2}{(8P/13)^2 \cdot (4P/13)} = \frac{P^2/169}{64P^2/169 \cdot 4P/13} = \frac{P^2/169}{256P^3/2197} $$ $$ K_p = \frac{P^2}{169} \cdot \frac{2197}{256P^3} = \frac{13}{256P} $$ \textbf{Answer: B} —

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