(B) For the reaction:$CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$The equilibrium constant expression is:$K_c = \frac{[COCl_2]}{[CO][Cl_2]} \d

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(B) For the reaction:$CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$The equilibrium constant expression is:$K_c = \frac{[COCl_2]}{[CO][Cl_2]} \dots (I)$Given volume $V = 500 \ mL = 0.5 \ L$.Concentrations at equilibrium are:$[COCl_2] = \frac{0.3 \ mol}{0.5 \ L} = 0.6 \ mol \ L^{-1}$$[CO] = \frac{0.1 \ mol}{0.5 \ L} = 0.2 \ mol \ L^{-1}$$[Cl_2] = \frac{0.1 \ mol}{0.5 \ L} = 0.2 \ mol \ L^{-1}$Substituting these values into expression $(I)$:$K_c = \frac{0.6}{(0.2)(0.2)} = \frac{0.6}{0.04} = 15$

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

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The equilibrium of formation of phosgene is represented as:
$CO_{(g)} + Cl_{2(g)} \rightleftharpoons COCl_{2(g)}$
The reaction is carried out in a $500 \ mL$ flask. At equilibrium,$0.3 \ mol$ of phosgene,$0.1 \ mol$ of $CO,$ and $0.1 \ mol$ of $Cl_2$ are present. The equilibrium constant $(K_c)$ of the reaction is:

A
$30$
B
$15$
C
$5$
D
$25$

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

At $T(K)$,the equilibrium constant of $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ is $49$. If $[H_2]$ and $[I_2]$ at equilibrium at the same temperature are $2.0 \times 10^{-2} \ M$ and $8.0 \times 10^{-2} \ M$ respectively,the $[HI]$ at equilibrium in $mol \ L^{-1}$ is:

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For the reaction $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$,the units of $K_c$ are

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Given the following equilibria:
$K_1 : N_2 + 3H_2 \rightleftharpoons 2NH_3$
$K_2 : N_2 + O_2 \rightleftharpoons 2NO$
$K_3 : H_2 + 1/2 O_2 \rightleftharpoons H_2O$
The equilibrium constant for the reaction $2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $K_1, K_2$,and $K_3$ is:

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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

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For the reaction $2HI \rightleftharpoons H_2 + I_2$,the equilibrium constant is $K_1 = 0.25$. What is the value of $K_2$ for the reaction $H_2 + I_2 \rightleftharpoons 2HI$?

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At $T(K)$,the equilibrium constant of $H_{2(g)} + I_{2(g)} \rightleftharpoons 2 HI_{(g)}$ is $49$. If $[H_2]$ and $[I_2]$ at equilibrium at the same temperature are $2.0 \times 10^{-2} \ M$ and $8.0 \times 10^{-2} \ M$ respectively,the $[HI]$ at equilibrium in $mol \ L^{-1}$ is:

For the reaction $2 SO_2 + O_2 \rightleftharpoons 2 SO_3$,the units of $K_c$ are

Given the following equilibria: $K_1 : N_2 + 3H_2 \rightleftharpoons 2NH_3$ $K_2 : N_2 + O_2 \rightleftharpoons 2NO$ $K_3 : H_2 + 1/2 O_2 \rightleftharpoons H_2O$ The equilibrium constant for the reaction $2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $K_1, K_2$,and $K_3$ 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

For the reaction $2HI \rightleftharpoons H_2 + I_2$,the equilibrium constant is $K_1 = 0.25$. What is the value of $K_2$ for the reaction $H_2 + I_2 \rightleftharpoons 2HI$?

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Given the following equilibria:
$K_1 : N_2 + 3H_2 \rightleftharpoons 2NH_3$
$K_2 : N_2 + O_2 \rightleftharpoons 2NO$
$K_3 : H_2 + 1/2 O_2 \rightleftharpoons H_2O$
The equilibrium constant for the reaction $2NH_3 + 5/2 O_2 \rightleftharpoons 2NO + 3H_2O$ in terms of $K_1, K_2$,and $K_3$ is: