For the reaction equilibrium $N_2O_4(g) \rightleftharpoons 2NO_2(g)$,the concentrations of $N_2O_4$ and $NO_2$ at equilibrium are $4.8 \times 10^{-2} \ mol \ L^{-1}$ and $1.2 \times 10^{-2} \ mol \ L^{-1}$ respectively. The value of $K_c$ for the reaction is:

  • A
    $3.3 \times 10^2 \ mol \ L^{-1}$
  • B
    $3 \times 10^{-1} \ mol \ L^{-1}$
  • C
    $3 \times 10^{-3} \ mol \ L^{-1}$
  • D
    $3 \times 10^3 \ mol \ L^{-1}$

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Similar Questions

Before any reaction takes place,the value of the reaction quotient $(Q_c)$ is ......

$N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}; K_1$
$NH_{3(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)}; K_2$
$\frac{1}{2} N_{2(g)} + \frac{3}{2} H_{2(g)} \rightleftharpoons NH_{3(g)}; K_3$
$2NH_{3(g)} \rightleftharpoons N_{2(g)} + 3H_{2(g)}; K_4$
If $K_1 = K_2^x = K_3^y = K_4^z$,then the correct values of $x, y,$ and $z$ are respectively:

At $400 \ K$ for the reaction $2NO_{2(g)} \rightleftharpoons N_{2}O_{4(g)}$,the concentration of $NO_{2}$ is $0.710 \ M$ and $N_{2}O_{4}$ is $0.145 \ M$. Find the equilibrium constant $K_{c}$. (in $M^{-1}$)

At $400 \ K$ in a closed vessel,the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ takes place. At equilibrium,the concentration of $H_2$ is $0.6 \ mol \ L^{-1}$,the concentration of $I_2$ is $0.8 \ mol \ L^{-1}$,and the concentration of $HI$ is $0.14 \ mol \ L^{-1}$. Calculate the equilibrium constant $(K_c)$.

$5 \ moles$ each of $H_2$ and $I_2$ were heated in a sealed $10 \ L$ vessel. At equilibrium,$2 \ moles$ of $HI$ were found. The equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is:

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