For the reaction $N_2O_2 \rightleftharpoons 2NO$,the value of the concentration equilibrium constant is .......

  • A
    $\frac{x^2}{(a - x)V}$
  • B
    $\frac{x^2P}{a^2 - x^2}$
  • C
    $\frac{4x^2V^2}{(a - x)(b - x)}$
  • D
    $\frac{4x^2}{(a - x)V}$

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Similar Questions

If the equilibrium constant for the reaction ${N_2}_{(g)} + {O_2}_{(g)} \rightleftharpoons 2NO_{(g)}$ is $K$,then what is the equilibrium constant for the reaction $\frac{1}{2}{N_2}_{(g)} + \frac{1}{2}{O_2}_{(g)} \rightleftharpoons NO_{(g)}$?

What is the equilibrium expression for the reaction $P_{4(s)} + 5O_{2(g)} \rightleftharpoons P_4O_{10(s)}$?

The value of $K_{C}$ is $64$ at $800 \ K$ for the reaction $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$.
The value of $K_{C}$ for the following reaction is:
$NH_{3(g)} \rightleftharpoons \frac{1}{2}N_{2(g)} + \frac{3}{2}H_{2(g)}$

For the gaseous reaction,equilibrium constant is given:
$XeF_6 + H_2O \rightleftharpoons XeOF_4 + 2HF, K_1$
$XeO_4 + XeF_6 \rightleftharpoons XeOF_4 + XeO_3F_2, K_2$
The equilibrium constant for the reaction:
$XeO_4 + 2HF \rightleftharpoons XeO_3F_2 + H_2O$ will be

The equilibrium constant $K_C$ of the reaction,$2 A \rightleftharpoons B + C$ is $0.5$ at $25^{\circ} C$. The reaction will proceed in the backward direction,when concentrations $[A], [B]$ and $[C]$ are,respectively:

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