(A) The reaction quotient $Q_c$ is given by the expression: $Q_c = \frac{[C]^2}{[A][B]^2}$.Given concentrations in a $10 \, L$ vessel:$[A] = \frac{2 \

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(A) The reaction quotient $Q_c$ is given by the expression: $Q_c = \frac{[C]^2}{[A][B]^2}$.Given concentrations in a $10 \, L$ vessel:$[A] = \frac{2 \, \text{mol}}{10 \, L} = 0.2 \, M$$[B] = \frac{3 \, \text{mol}}{10 \, L} = 0.3 \, M$$[C] = \frac{1 \, \text{mol}}{10 \, L} = 0.1 \, M$Substituting these values into the expression for $Q_c$:$Q_c = \frac{(0.1)^2}{(0.2)(0.3)^2} = \frac{0.01}{0.2 \times 0.09} = \frac{0.01}{0.018} = \frac{10}{18} \approx 0.556$.Since $Q_c (0.556) < K_c (3.6)$,the reaction will proceed in the forward direction.

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

Medium

For the reaction $A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)} + D_{(s)}$,$2 \, \text{moles}$ of $A$,$3 \, \text{moles}$ of $B$ and $1 \, \text{mole}$ of $C$ are present in a $10 \, L$ vessel. If $K_c$ for the reaction is $3.6$,the reaction will proceed in:

A
Forward direction
B
Backward direction
C
Neither direction
D
None of these

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

$K_1, K_2$ and $K_3$ are the equilibrium constants of the following reactions $(I), (II)$ and $(III)$ respectively.
$(I) \, N_2 + 2O_2 \rightleftharpoons 2NO_2$
$(II) \, 2NO_2 \rightleftharpoons N_2 + 2O_2$
$(III) \, NO_2 \rightleftharpoons \frac{1}{2} N_2 + O_2$
The correct relation from the following is

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In the reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $B$ was $1.5$ times that of $[A]$. At equilibrium,the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is:

Medium

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For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the concentrations of $N_2O_4$ and $NO_2$ at equilibrium are $4.8 \times 10^{-2} \, mol \, L^{-1}$ and $1.2 \times 10^{-2} \, mol \, L^{-1}$ respectively. Calculate the value of $K_c$.

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If the equilibrium constant for $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$ is $K$,then the equilibrium constant for $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$ will be :

Medium

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For the reaction ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$,the equilibrium constant is $K$. For the reaction $2{N_2} + 6{H_2} \rightleftharpoons 4N{H_3}$,the equilibrium constant is $K'$. Then $K'$ is equal to:

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For the reaction $A_{(g)} + 2B_{(g)} \rightleftharpoons 2C_{(g)} + D_{(s)}$,$2 \, \text{moles}$ of $A$,$3 \, \text{moles}$ of $B$ and $1 \, \text{mole}$ of $C$ are present in a $10 \, L$ vessel. If $K_c$ for the reaction is $3.6$,the reaction will proceed in:

Similar Questions

In the reaction $A + 2B \rightleftharpoons 2C + D$,the initial concentration of $B$ was $1.5$ times that of $[A]$. At equilibrium,the concentrations of $A$ and $B$ became equal. The equilibrium constant for the reaction is:

For the reaction $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$,the concentrations of $N_2O_4$ and $NO_2$ at equilibrium are $4.8 \times 10^{-2} \, mol \, L^{-1}$ and $1.2 \times 10^{-2} \, mol \, L^{-1}$ respectively. Calculate the value of $K_c$.

If the equilibrium constant for $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$ is $K$,then the equilibrium constant for $SO_3 \rightleftharpoons SO_2 + \frac{1}{2} O_2$ will be :

For the reaction ${N_2} + 3{H_2} \rightleftharpoons 2N{H_3}$,the equilibrium constant is $K$. For the reaction $2{N_2} + 6{H_2} \rightleftharpoons 4N{H_3}$,the equilibrium constant is $K'$. Then $K'$ is equal to:

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