(D) The equilibrium constant expression for the reaction $3 O_{2(g)} \rightleftharpoons 2 O_{3(g)}$ is given by $K_c = \frac{[O_3]^2}{[O_2]^3}$.Given

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(D) The equilibrium constant expression for the reaction $3 O_{2(g)} \rightleftharpoons 2 O_{3(g)}$ is given by $K_c = \frac{[O_3]^2}{[O_2]^3}$.Given $K_c = 3.0 \times 10^{-59}$ and $[O_2] = 0.040 \ M = 4 \times 10^{-2} \ M$.Substituting the values into the expression:$3.0 \times 10^{-59} = \frac{[O_3]^2}{(4 \times 10^{-2})^3}$$[O_3]^2 = 3.0 \times 10^{-59} \times (64 \times 10^{-6})$$[O_3]^2 = 192 \times 10^{-65} = 1.92 \times 10^{-63}$$[O_3] = \sqrt{1.92 \times 10^{-63}} = \sqrt{19.2 \times 10^{-64}} \approx 4.38 \times 10^{-32} \ M$.

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Law of equilibrium and Equilibrium constant

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$3 O_{2(g)} \rightleftharpoons 2 O_{3(g)}$
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 ...... .

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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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1 Equilibrium: Chemical Equilibrium

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Law of equilibrium and Equilibrium constant

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Write the uses of equilibrium constant.

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The compounds $A$ and $B$ are mixed in equimolar proportion to form the products,$A + B \rightleftharpoons C + D$. At equilibrium,one-third of $A$ and $B$ are consumed. The equilibrium constant for the reaction is:

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For the following equilibrium,$K_{C} = 6.3 \times 10^{14}$ at $1000 \ K$:
$NO_{(g)} + O_{3(g)} \longleftrightarrow NO_{2(g)} + O_{2(g)}$
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $K_{C}$ for the reverse reaction?

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

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At a given temperature,the equilibrium constant for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ is $2.4 \times 10^{-3}$. At the same temperature,the equilibrium constant for the reaction $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$ is:

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$3 O_{2(g)} \rightleftharpoons 2 O_{3(g)}$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 ...... .

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Write the uses of equilibrium constant.

The compounds $A$ and $B$ are mixed in equimolar proportion to form the products,$A + B \rightleftharpoons C + D$. At equilibrium,one-third of $A$ and $B$ are consumed. The equilibrium constant for the reaction is:

For the following equilibrium,$K_{C} = 6.3 \times 10^{14}$ at $1000 \ K$:$NO_{(g)} + O_{3(g)} \longleftrightarrow NO_{2(g)} + O_{2(g)}$Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $K_{C}$ for the reverse reaction?

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

At a given temperature,the equilibrium constant for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$ is $2.4 \times 10^{-3}$. At the same temperature,the equilibrium constant for the reaction $PCl_{3(g)} + Cl_{2(g)} \rightleftharpoons PCl_{5(g)}$ is:

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$3 O_{2(g)} \rightleftharpoons 2 O_{3(g)}$
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 ...... .

For the following equilibrium,$K_{C} = 6.3 \times 10^{14}$ at $1000 \ K$:
$NO_{(g)} + O_{3(g)} \longleftrightarrow NO_{2(g)} + O_{2(g)}$
Both the forward and reverse reactions in the equilibrium are elementary bimolecular reactions. What is $K_{C}$ for the reverse reaction?