(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 6-1.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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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

$3.1 \ mol$ of $FeCl_3$ and $3.2 \ mol$ of $NH_4SCN$ are added to $1 \ L$ of water. At equilibrium,$3.0 \ mol$ of $FeSCN^{2+}$ is formed. The equilibrium constant $K_c$ for the reaction is:
$Fe^{3+} + SCN^{-} \rightleftharpoons FeSCN^{2+}$

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Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

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The equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $32$ at a given temperature. The equilibrium concentrations of $I_2$ and $HI$ are $0.5 \times 10^{-3} \ M$ and $8 \times 10^{-3} \ M$ respectively. The equilibrium concentration of $H_2$ is:

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At $200^{\circ} C$,nitric oxide reacts with oxygen to form nitrogen dioxide as follows: $2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g)$,$K_C = 3 \times 10^6$. In a mixture of the three species at equilibrium,we can accurately predict that:

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At $T \ K$,the equilibrium constant for the reaction $a A_{(g)} \rightleftharpoons b B_{(g)}$ is $K_c$. If the reaction takes place in the following form $2a A_{(g)} \rightleftharpoons 2b B_{(g)}$,its equilibrium constant is $K_c^{\prime}$. The correct relationship between $K_c$ and $K_c^{\prime}$ is

MediumAP EAMCET 2023

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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

$3.1 \ mol$ of $FeCl_3$ and $3.2 \ mol$ of $NH_4SCN$ are added to $1 \ L$ of water. At equilibrium,$3.0 \ mol$ of $FeSCN^{2+}$ is formed. The equilibrium constant $K_c$ for the reaction is:$Fe^{3+} + SCN^{-} \rightleftharpoons FeSCN^{2+}$

Explain why pure liquids and solids can be ignored while writing the equilibrium constant expression?

The equilibrium constant for the reaction $H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$ is $32$ at a given temperature. The equilibrium concentrations of $I_2$ and $HI$ are $0.5 \times 10^{-3} \ M$ and $8 \times 10^{-3} \ M$ respectively. The equilibrium concentration of $H_2$ is:

At $200^{\circ} C$,nitric oxide reacts with oxygen to form nitrogen dioxide as follows: $2 NO(g) + O_2(g) \rightleftharpoons 2 NO_2(g)$,$K_C = 3 \times 10^6$. In a mixture of the three species at equilibrium,we can accurately predict that:

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$3.1 \ mol$ of $FeCl_3$ and $3.2 \ mol$ of $NH_4SCN$ are added to $1 \ L$ of water. At equilibrium,$3.0 \ mol$ of $FeSCN^{2+}$ is formed. The equilibrium constant $K_c$ for the reaction is:
$Fe^{3+} + SCN^{-} \rightleftharpoons FeSCN^{2+}$