(C) $1$. Initial moles of $N_2O_4 = \frac{9.2 \ g}{92 \ g \ mol^{-1}} = 0.1 \ mol$. $2$. Since the volume of the vessel is $1 \ L$,the initial concent

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(C) $1$. Initial moles of $N_2O_4 = \frac{9.2 \ g}{92 \ g \ mol^{-1}} = 0.1 \ mol$. $2$. Since the volume of the vessel is $1 \ L$,the initial concentration $[N_2O_4]_0 = 0.1 \ mol \ L^{-1}$. $3$. At equilibrium,$50\%$ of $N_2O_4$ is dissociated,so the amount dissociated is $0.1 \times 0.5 = 0.05 \ mol$. $4$. The equilibrium concentrations are: $[N_2O_4]_{eq} = 0.1 - 0.05 = 0.05 \ mol \ L^{-1}$ and $[NO_2]_{eq} = 2 \times 0.05 = 0.1 \ mol \ L^{-1}$. $5$. The equilibrium constant $K_c = \frac{[NO_2]^2}{[N_2O_4]} = \frac{(0.1)^2}{0.05} = \frac{0.01}{0.05} = 0.2 \ mol \ L^{-1}$.

Class 11 Chemistry 6-1.Equilibrium (Chemical Equilibrium)Kp and Kc Relationship

Medium

$9.2 \ g$ of $N_2O_{4(g)}$ is taken in a closed $1 \ L$ vessel and heated until the following equilibrium is reached: $N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At equilibrium,$50\%$ of $N_2O_{4(g)}$ is dissociated. What is the equilibrium constant $K_c$ (in $mol \ L^{-1}$)? (Molecular weight of $N_2O_4 = 92$)

A
$0.1$
B
$0.4$
C
$0.2$
D
$2$

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6-1.Equilibrium (Chemical Equilibrium)

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Kp and Kc Relationship

The equilibrium concentrations of $X$,$Y$,and $YX_2$ are $4 \ mol/L$,$2 \ mol/L$,and $2 \ mol/L$ respectively for the equilibrium $2X + Y \rightleftharpoons YX_2$. The value of $K_c$ is:

Medium

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In the reaction,$A_{(s)} + B_{(g)} \rightleftharpoons 2C_{(s)} + 2D_{(g)}$,equilibrium is established. If the pressure of $B$ is doubled,then to reestablish the equilibrium,how many times does the pressure of $D$ become as compared to its initial pressure?

Medium

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For the reaction $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$,$K_C$ at $427\ ^oC$ is $3 \times 10^{-6}\ mol\ L^{-1}$. The value of $K_P$ is nearly $....... \times 10^{-4}$.

MediumAIIMS 2005

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The values of $K_p/K_c$ for the following reactions at $300 \ K$ are respectively (At $300 \ K, RT = 24.62 \ dm^3 \ atm \ mol^{-1}$):
$(i) \ N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$
$(ii) \ N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$
$(iii) \ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

DifficultJEE MAIN 2019

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The equilibrium constant $(K_p)$ for the formation of ammonia from its constituent elements at $27^{\circ} C$ is $1.2 \times 10^{-4}$ and at $127^{\circ} C$ is $0.60 \times 10^{-4}$. Calculate the mean heat of formation of ammonia per mole in this temperature range. (in $cal$)

DifficultTS EAMCET 2021

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The equilibrium concentrations of $X$,$Y$,and $YX_2$ are $4 \ mol/L$,$2 \ mol/L$,and $2 \ mol/L$ respectively for the equilibrium $2X + Y \rightleftharpoons YX_2$. The value of $K_c$ is:

In the reaction,$A_{(s)} + B_{(g)} \rightleftharpoons 2C_{(s)} + 2D_{(g)}$,equilibrium is established. If the pressure of $B$ is doubled,then to reestablish the equilibrium,how many times does the pressure of $D$ become as compared to its initial pressure?

For the reaction $2NOCl_{(g)} \rightleftharpoons 2NO_{(g)} + Cl_{2(g)}$,$K_C$ at $427\ ^oC$ is $3 \times 10^{-6}\ mol\ L^{-1}$. The value of $K_P$ is nearly $....... \times 10^{-4}$.

The values of $K_p/K_c$ for the following reactions at $300 \ K$ are respectively (At $300 \ K, RT = 24.62 \ dm^3 \ atm \ mol^{-1}$):$(i) \ N_{2(g)} + O_{2(g)} \rightleftharpoons 2NO_{(g)}$$(ii) \ N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$$(iii) \ N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}$

The equilibrium constant $(K_p)$ for the formation of ammonia from its constituent elements at $27^{\circ} C$ is $1.2 \times 10^{-4}$ and at $127^{\circ} C$ is $0.60 \times 10^{-4}$. Calculate the mean heat of formation of ammonia per mole in this temperature range. (in $cal$)

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