Find the value of frac{P}{K_p} for the reacti | vedclass

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(B) For the reaction $2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2(g)}$,let the initial pressure of $NOBr$ be $P_0$ and at equilibrium,the pressur

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$81$

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(B) For the reaction $2NOBr_{(g)} ightleftharpoons 2NO_{(g)} + Br_{2(g)}$,let the initial pressure of $NOBr$ be $P_0$ and at equilibrium,the pressure of $Br_2$ be $x = \frac{P}{9}$.From the stoichiometry,the pressure of $NO$ is $2x = \frac{2P}{9}$ and the pressure of $NOBr$ is $P_0 - 2x$.The total pressure $P = P_{NOBr} + P_{NO} + P_{Br_2} = (P_0 - 2x) + 2x + x = P_0 + x$.Thus,$P_0 = P - x = P - \frac{P}{9} = \frac{8P}{9}$.Now,$P_{NOBr} = P_0 - 2x = \frac{8P}{9} - \frac{2P}{9} = \frac{6P}{9} = \frac{2P}{3}$.The equilibrium constant $K_p$ is given by $K_p = \frac{(P_{NO})^2 	imes P_{Br_2}}{(P_{NOBr})^2} = \frac{(\frac{2P}{9})^2 	imes \frac{P}{9}}{(\frac{2P}{3})^2} = \frac{\frac{4P^2}{81} 	imes \frac{P}{9}}{\frac{4P^2}{9}} = \frac{P}{81}$.Therefore,$\frac{P}{K_p} = \frac{P}{P/81} = 81$.

Find the value of $\frac{P}{K_p}$ for the reaction at a certain temperature: $2NOBr_{(g)} \rightleftharpoons 2NO_{(g)} + Br_{2(g)}$,where $P$ is the total pressure of gases at equilibrium and $P_{Br_2} = \frac{P}{9}$.

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