(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 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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Class 11

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1 Equilibrium: Chemical Equilibrium

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Law of equilibrium and Equilibrium constant

For the reaction,$N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$,the equilibrium constant is $K_1$. The equilibrium constant is $K_2$ for the reaction,$2NO_{(g)} + O_{2(g)} \rightleftharpoons 2NO_{2(g)}$. What is $K$ for the reaction,$NO_{2(g)} \rightleftharpoons \frac{1}{2} N_{2(g)} + O_{2(g)}$?

DifficultAIPMT 2011

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

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

Easy

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For the reaction equilibrium $N_2O_4(g) \rightleftharpoons 2NO_2(g)$,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. The value of $K_c$ for the reaction is:

MediumAIEEE 2003

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Two moles of $NH_3$ when put into a previously evacuated vessel $(1 \ L)$,partially dissociate into $N_2$ and $H_2$. If at equilibrium one mole of $NH_3$ is present,the equilibrium constant is

Medium

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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

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

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

For the reaction equilibrium $N_2O_4(g) \rightleftharpoons 2NO_2(g)$,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. The value of $K_c$ for the reaction is:

Two moles of $NH_3$ when put into a previously evacuated vessel $(1 \ L)$,partially dissociate into $N_2$ and $H_2$. If at equilibrium one mole of $NH_3$ is present,the equilibrium constant is

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