For the reaction equilibrium 2NOBr g -2NO g - Br2 g - if PBr2- P9 at equilibrium and P is the initial total pressure- then the ratio KPP is equal to -

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Derivation of Kp and Kc

For the react...

Question

For the reaction equilibrium $2 N O B r_{(g)} ⇋ 2 N O_{(g)} + B r_{2 (g)}$ , if $P_{B r_{2}} = \frac{P}{9}$ at equilibrium and P is the initial total pressure, then the ratio $\frac{K_{P}}{P}$ is equal to :

\frac{1}{9}

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\frac{4}{441}

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\frac{1}{27}

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\frac{1}{3}

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Solution

The correct option is B $\frac{4}{441}$
The equilibrium reaction is $2 N O B r_{(g)} ⇋ 2 N O_{(g)} + B r_{2 (g)} .$
The partial pressure of bromine is $\frac{P}{9}$ . The partial pressures of $N O B r$ and $N O$ will be $\frac{7 P}{9}$ and $\frac{2 P}{9}$ respectively.
The expression for the equilibrium constant will be $K_{p} = \frac{P_{N O}^{2} P_{B r_{2}}}{P_{N O B r}^{2}} = \frac{(\frac{2 P}{9})^{2} (\frac{P}{9})}{(\frac{7 P}{9})^{2}}$ .
Thus, $\frac{K_{p}}{P} = \frac{4}{441}$ .

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For the reaction equilibrium; $2 N O B r_{(g)} ⇌ 2 N O_{g} + B r_{2 (g)}$ , if $P_{B r_{2}} = \frac{P}{9}$ at equilibrium and 'P' is total pressure, the ratio is equal to:

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For the reaction equilibrium 2NOBr(g)⇋2NO(g)+Br2(g), if PBr2=P9 at equilibrium and P is the initial total pressure, then the ratio KPP is equal to :

The correct option is B 4441The equilibrium reaction is 2NOBr(g)⇋2NO(g)+Br2(g).The partial pressure of bromine is P9. The partial pressures of NOBr and NO will be 7P9 and 2P9 respectively.The expression for the equilibrium constant will be Kp=P2NOPBr2P2NOBr=(2P9)2(P9)(7P9)2.Thus, KpP=4441.

For the reaction equilibrium $2 N O B r_{(g)} ⇋ 2 N O_{(g)} + B r_{2 (g)}$ , if $P_{B r_{2}} = \frac{P}{9}$ at equilibrium and P is the initial total pressure, then the ratio $\frac{K_{P}}{P}$ is equal to :