2NOBr_{(g)} rightleftharpoons 2NO_{(g)} + Br_ | vedclass

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Question

(A) For the reaction $2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2_{(g)}}$,let the initial moles of $NOBr$ be $1$. At equilibrium,the moles are $(

Vedclass · Accepted Answer

$45$

Vedclass · Answer

(A) For the reaction $2NOBr_{(g)} ightleftharpoons 2NO_{(g)} + Br_{2_{(g)}}$,let the initial moles of $NOBr$ be $1$. At equilibrium,the moles are $(1-\alpha)$,$\alpha$,and $\alpha/2$. Total moles at equilibrium = $1 - \alpha + \alpha + \alpha/2 = 1 + \alpha/2$. Given $\alpha = 0.4$,total moles = $1 + 0.2 = 1.2$. Partial pressures at total pressure $P = 0.30 	ext{ atm}$: $P_{NOBr} = \frac{1-0.4}{1.2} 	imes 0.3 = \frac{0.6}{1.2} 	imes 0.3 = 0.15 	ext{ atm}$. $P_{NO} = \frac{0.4}{1.2} 	imes 0.3 = 0.1 	ext{ atm}$. $P_{Br_2} = \frac{0.2}{1.2} 	imes 0.3 = 0.05 	ext{ atm}$. For the reverse reaction $2NO_{(g)} + Br_{2_{(g)}} ightleftharpoons 2NOBr_{(g)}$,the equilibrium constant $K_p'$ is the reciprocal of the forward reaction $K_p$: $K_p' = \frac{(P_{NOBr})^2}{(P_{NO})^2 \cdot P_{Br_2}} = \frac{(0.15)^2}{(0.1)^2 \cdot 0.05} = \frac{0.0225}{0.01 \cdot 0.05} = \frac{0.0225}{0.0005} = 45$.

$2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2_{(g)}}$. If $NOBr$ is $40\%$ dissociated at a certain temperature and a total pressure of $0.30 \text{ atm}$,the $K_p$ for the reaction $2NO_{(g)} + Br_{2_{(g)}} \rightleftharpoons 2NOBr_{(g)}$ is:

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