(B) The equilibrium constant $K$ for a general reversible reaction $aA + bB \rightleftharpoons cC + dD$ is defined as the ratio of the product of the

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$K = \frac{[H_3O^{+}][CH_3COO^{-}]}{[H_2O][CH_3COOH]}$

(B) The equilibrium constant $K$ for a general reversible reaction $aA + bB \rightleftharpoons cC + dD$ is defined as the ratio of the product of the concentrations of the products to the product of the concentrations of the reactants,each raised to the power of their stoichiometric coefficients.For the reaction $CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}$,the expression is:$K = \frac{[H_3O^{+}][CH_3COO^{-}]}{[CH_3COOH][H_2O]}$Comparing this with the given options,option $B$ matches this expression.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Law of equilibrium and Equilibrium constant

Medium

For the reaction $CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}$,the equilibrium constant $K$ is given by:

A
$K = \frac{[H_3O^{+}][H_2O]}{[CH_3COO^{-}][CH_3COOH]}$
B
$K = \frac{[H_3O^{+}][CH_3COO^{-}]}{[H_2O][CH_3COOH]}$
C
$K = \frac{[H_3O^{+}][H_2O]}{[CH_3COOH][CH_3COO^{-}]}$
D
$K = \frac{[H_2O][CH_3COO^{-}]}{[H_2O][CH_3COOH]}$

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

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

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For the reaction $2 H_{2(g)} + 2 NO_{(g)} \rightarrow N_{2(g)} + 2 H_2O_{(g)}$,the observed rate expression is $rate = k_f [NO]^2 [H_2]$. The rate expression of the reverse reaction is:

MediumJEE MAIN 2020

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At $T \ K$,the $K_C$ value for the reaction $\frac{1}{3} N_{2(g)} + H_{2(g)} \rightleftharpoons \frac{2}{3} NH_{3(g)}$ is $50$. The $K_C$ value for the reaction $2 NH_{3(g)} \rightleftharpoons N_{2(g)} + 3 H_{2(g)}$ at the same temperature is:

MediumTS EAMCET 2023

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What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of $ICl$ was $0.78 \, M$?
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}; \, K_c = 0.14$

Medium

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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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For the reaction $CH_3COOH + H_2O \rightleftharpoons H_3O^{+} + CH_3COO^{-}$,the equilibrium constant $K$ is given by:

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

For the reaction $2 H_{2(g)} + 2 NO_{(g)} \rightarrow N_{2(g)} + 2 H_2O_{(g)}$,the observed rate expression is $rate = k_f [NO]^2 [H_2]$. The rate expression of the reverse reaction is:

At $T \ K$,the $K_C$ value for the reaction $\frac{1}{3} N_{2(g)} + H_{2(g)} \rightleftharpoons \frac{2}{3} NH_{3(g)}$ is $50$. The $K_C$ value for the reaction $2 NH_{3(g)} \rightleftharpoons N_{2(g)} + 3 H_{2(g)}$ at the same temperature is:

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of $ICl$ was $0.78 \, M$?$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}; \, K_c = 0.14$

$3 O_{2(g)} \rightleftharpoons 2 O_{3(g)}$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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What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of $ICl$ was $0.78 \, M$?
$2ICl_{(g)} \leftrightarrow I_{2(g)} + Cl_{2(g)}; \, K_c = 0.14$

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