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(D) For the reaction $AB_{(g)} \rightleftharpoons A_{(g)} + B_{(g)}$,let the initial moles of $AB$ be $1$. $33.3\%$ dissociation means $\alpha = \frac

Vedclass · Accepted Answer

$P = 8K_p$

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(D) For the reaction $AB_{(g)} ightleftharpoons A_{(g)} + B_{(g)}$,let the initial moles of $AB$ be $1$. $33.3\%$ dissociation means $\alpha = \frac{1}{3}$. At equilibrium,the moles are: $n_{AB} = 1 - \alpha = 1 - \frac{1}{3} = \frac{2}{3}$ $n_{A} = \alpha = \frac{1}{3}$ $n_{B} = \alpha = \frac{1}{3}$ Total moles at equilibrium $= \frac{2}{3} + \frac{1}{3} + \frac{1}{3} = \frac{4}{3}$. Partial pressures are given by $p_i = \frac{n_i}{n_{total}} 	imes P$: $p_{AB} = \frac{2/3}{4/3} 	imes P = \frac{P}{2}$ $p_{A} = \frac{1/3}{4/3} 	imes P = \frac{P}{4}$ $p_{B} = \frac{1/3}{4/3} 	imes P = \frac{P}{4}$ $K_p = \frac{p_{A} 	imes p_{B}}{p_{AB}} = \frac{(P/4) 	imes (P/4)}{P/2} = \frac{P^2/16}{P/2} = \frac{P}{8}$. Thus,$P = 8K_p$.

For the reaction $AB_{(g)} \rightleftharpoons A_{(g)} + B_{(g)}$,$AB$ is $33.3\%$ dissociated at a total equilibrium pressure of $P$. Therefore,$P$ is correctly related to $K_p$ by which of the following options?

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