For the gaseous reactions (I) and (II),the eq | vedclass

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(D) Given reactions:$I. \frac{1}{2} N_{2(g)} + O_{2(g)} \rightleftharpoons NO_{2(g)}$ with equilibrium constant $X$.$II. 2 NO_{2(g)} \rightleftharpoon

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$Z = \frac{1}{X^2Y}$

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(D) Given reactions:$I. \frac{1}{2} N_{2(g)} + O_{2(g)} ightleftharpoons NO_{2(g)}$ with equilibrium constant $X$.$II. 2 NO_{2(g)} ightleftharpoons N_2O_{4(g)}$ with equilibrium constant $Y$.Target reaction $(III): N_2O_{4(g)} ightleftharpoons N_{2(g)} + 2 O_{2(g)}$.To obtain reaction $(III)$,we reverse reaction $(II)$ and add it to twice the reverse of reaction $(I)$ (or simply reverse the sum of $2 	imes Eq(I) + Eq(II)$):$2 	imes Eq(I): N_{2(g)} + 2 O_{2(g)} ightleftharpoons 2 NO_{2(g)}$ with constant $X^2$.$Eq(II): 2 NO_{2(g)} ightleftharpoons N_2O_{4(g)}$ with constant $Y$.Adding these gives: $N_{2(g)} + 2 O_{2(g)} ightleftharpoons N_2O_{4(g)}$ with constant $K = X^2Y$.Since reaction $(III)$ is the reverse of this sum,its equilibrium constant $Z = \frac{1}{K} = \frac{1}{X^2Y}$.

Similar Questions

State the law of chemical equilibrium for an equilibrium process.

The equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $32$ at a given temperature. The equilibrium concentrations of $I_2$ and $HI$ are $0.5 \times 10^{-3} \ M$ and $8 \times 10^{-3} \ M$ respectively. The equilibrium concentration of $H_2$ is:

The equilibrium constant for the equilibrium $2HX_{(g)} \rightleftharpoons H_{2(g)} + X_{2(g)}$ is $1 \times 10^{-5}$. What will be the equilibrium concentration of $HX$ if the equilibrium concentrations for $H_2$ and $X_2$ are $1.2 \times 10^{-3} \ M$ and $1.2 \times 10^{-4} \ M$ respectively?

The equilibrium constant for the given reaction $H_2 + I_2 \rightleftharpoons 2HI$ is correctly given by the expression:

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