(A) For the reversible reaction $A \underset{k_2}{\overset{k_1}{\longleftrightarrow}} B$,at equilibrium,the rate of the forward reaction equals the ra

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$\frac{k_1 a - k_2 b}{k_1 + k_2}$

(A) For the reversible reaction $A \underset{k_2}{\overset{k_1}{\longleftrightarrow}} B$,at equilibrium,the rate of the forward reaction equals the rate of the backward reaction.$k_1 [A]_{eq} = k_2 [B]_{eq}$Given $[A]_{eq} = (a - x)$ and $[B]_{eq} = (b + x)$,we have:$k_1 (a - x) = k_2 (b + x)$$k_1 a - k_1 x = k_2 b + k_2 x$$k_1 a - k_2 b = k_1 x + k_2 x$$k_1 a - k_2 b = x(k_1 + k_2)$$x = \frac{k_1 a - k_2 b}{k_1 + k_2}$Thus,the correct option is $A$.

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

Medium

In a reversible reaction $A \underset{k_2}{\overset{k_1}{\longleftrightarrow}} B$,the initial concentrations of $A$ and $B$ are $a$ and $b$ in moles per litre,and the equilibrium concentrations are $(a - x)$ and $(b + x)$ respectively. Express $x$ in terms of $k_1, k_2, a,$ and $b$.

A
$\frac{k_1 a - k_2 b}{k_1 + k_2}$
B
$\frac{k_1 a - k_2 b}{k_1 - k_2}$
C
$\frac{k_1 a - k_2 b}{k_1 k_2}$
D
$\frac{k_1 a + k_2 b}{k_1 + k_2}$

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1 Equilibrium: Chemical Equilibrium

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Law of equilibrium and Equilibrium constant

For the reactions $SO_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}$ and $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$,if the equilibrium constants at $298 \ K$ are $K_1$ and $K_2$ respectively,then the correct relationship between them is .......

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The equilibrium constant $(K_c)$ for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$ at temperature $T$ is $4 \times 10^{-4}$. The value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)}$ at the same temperature is:

MediumAIEEE 2012

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$2$ moles of $PCl_5$ were heated in a closed vessel of $2 \ L$ capacity. At equilibrium,$40\%$ of $PCl_5$ is dissociated into $PCl_3$ and $Cl_2$. The value of equilibrium constant is

Medium

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Explain the use of the equilibrium constant to predict the extent of a reaction with an example.

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An equilibrium mixture of the reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$ had $0.5 mol$ $H_2S$,$0.10 mol$ $H_2$,and $0.4 mol$ $S_2$ in a $1 L$ vessel. The value of the equilibrium constant $(K_c)$ in $mol L^{-1}$ is:

MediumAIIMS 1998

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In a reversible reaction $A \underset{k_2}{\overset{k_1}{\longleftrightarrow}} B$,the initial concentrations of $A$ and $B$ are $a$ and $b$ in moles per litre,and the equilibrium concentrations are $(a - x)$ and $(b + x)$ respectively. Express $x$ in terms of $k_1, k_2, a,$ and $b$.

Similar Questions

For the reactions $SO_{2(g)} + \frac{1}{2}O_{2(g)} \rightleftharpoons SO_{3(g)}$ and $2SO_{3(g)} \rightleftharpoons 2SO_{2(g)} + O_{2(g)}$,if the equilibrium constants at $298 \ K$ are $K_1$ and $K_2$ respectively,then the correct relationship between them is .......

The equilibrium constant $(K_c)$ for the reaction $N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$ at temperature $T$ is $4 \times 10^{-4}$. The value of $K_c$ for the reaction $NO_{(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + \frac{1}{2} O_{2(g)}$ at the same temperature is:

$2$ moles of $PCl_5$ were heated in a closed vessel of $2 \ L$ capacity. At equilibrium,$40\%$ of $PCl_5$ is dissociated into $PCl_3$ and $Cl_2$. The value of equilibrium constant is

Explain the use of the equilibrium constant to predict the extent of a reaction with an example.

An equilibrium mixture of the reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$ had $0.5 mol$ $H_2S$,$0.10 mol$ $H_2$,and $0.4 mol$ $S_2$ in a $1 L$ vessel. The value of the equilibrium constant $(K_c)$ in $mol L^{-1}$ is:

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