$3 O_{2(g)} \rightleftharpoons 2 O_{3(g)}$
For the above reaction at $298 \ K$,$K_c$ is found to be $3.0 \times 10^{-59}$. If the concentration of $O_2$ at equilibrium is $0.040 \ M$,then the concentration of $O_3$ in $M$ is ...... .

  • A
    $1.9 \times 10^{-63}$
  • B
    $2.4 \times 10^{31}$
  • C
    $1.2 \times 10^{21}$
  • D
    $4.38 \times 10^{-32}$

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Similar Questions

In the figure shown below,reactant $A$ (represented by square) is in equilibrium with product $B$ (represented by circle). The equilibrium constant is:

At $T(K)$,the equilibrium constant of $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ is $49$. If $[H_2]$ and $[I_2]$ at equilibrium at the same temperature are $2.0 \times 10^{-2} \ M$ and $8.0 \times 10^{-2} \ M$ respectively,the $[HI]$ at equilibrium in $mol \ L^{-1}$ is:

The equilibrium constant expression for a gas reaction is,
$K_{C} = \frac{[NH_{3}]^{4}[O_{2}]^{5}}{[NO]^{4}[H_{2}O]^{6}}$
Write the balanced chemical equation corresponding to this expression.

The equilibrium concentrations of $HI$,$I_2$,and $H_2$ are $0.7 \ M$,$0.1 \ M$,and $0.1 \ M$ respectively. The equilibrium constant for the reaction $I_2 + H_2 \rightleftharpoons 2HI$ is:

For the reaction,$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the equilibrium constant is $K_1$. The equilibrium constant is $K_2$ for the reaction,$2NO_{(g)} + O_{2(g)} \rightleftharpoons 2NO_{2(g)}$. What is $K$ for the reaction,$NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$?

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