(B) The chemical equation is: $CO + Cl_2 \rightleftharpoons COCl_2$First,calculate the molar concentrations at equilibrium:$[CO] = \frac{0.1 \ mol}{0.

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(B) The chemical equation is: $CO + Cl_2 \rightleftharpoons COCl_2$First,calculate the molar concentrations at equilibrium:$[CO] = \frac{0.1 \ mol}{0.5 \ L} = 0.2 \ M$$[Cl_2] = \frac{0.1 \ mol}{0.5 \ L} = 0.2 \ M$$[COCl_2] = \frac{0.2 \ mol}{0.5 \ L} = 0.4 \ M$The expression for the equilibrium constant $K_c$ is:$K_c = \frac{[COCl_2]}{[CO][Cl_2]}$Substituting the values:$K_c = \frac{0.4}{0.2 \times 0.2} = \frac{0.4}{0.04} = 10$

Class 11 Chemistry 1 Equilibrium: Chemical Equilibrium Kp and Kc Relationship

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In a $500 \ mL$ capacity vessel,$CO$ and $Cl_2$ are mixed to form $COCl_2$. At equilibrium,it contains $0.2 \ mol$ of $COCl_2$ and $0.1 \ mol$ of each of $CO$ and $Cl_2$. The equilibrium constant $K_c$ for the reaction $CO + Cl_2 \rightleftharpoons COCl_2$ is:

AIPMT 1998 All AIPMT

A
$5$
B
$10$
C
$15$
D
$20$

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1 Equilibrium: Chemical Equilibrium

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Kp and Kc Relationship

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For the reaction $CO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons CO_{2(g)}$,the value of $K_p/K_c$ is .....

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At $400 \ K$,in a $1.0 \ L$ vessel,$N_2O_4$ is allowed to attain equilibrium,$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At equilibrium,the total pressure is $600 \ mm \ Hg$,when $20 \%$ of $N_2O_4$ is dissociated. The value of $K_p$ for the reaction is

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$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$
The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).

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$A$ vessel at $1000 \ K$ contains $CO_2$ with a pressure of $0.5 \ atm$. Some of the $CO_2$ is converted into $CO$ on the addition of graphite. If the total pressure at equilibrium is $0.8 \ atm$,the value of $K_p$ is $..... \ atm$.

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In a $500 \ mL$ capacity vessel,$CO$ and $Cl_2$ are mixed to form $COCl_2$. At equilibrium,it contains $0.2 \ mol$ of $COCl_2$ and $0.1 \ mol$ of each of $CO$ and $Cl_2$. The equilibrium constant $K_c$ for the reaction $CO + Cl_2 \rightleftharpoons COCl_2$ is:

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For the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,$1 mol$ of $PCl_5$ is taken at $5 atm$ pressure. If $50\%$ of $PCl_5$ dissociates at equilibrium,calculate $K_p$.

For the reaction $CO_{(g)} + 1/2 O_{2(g)} \rightleftharpoons CO_{2(g)}$,the value of $K_p/K_c$ is .....

At $400 \ K$,in a $1.0 \ L$ vessel,$N_2O_4$ is allowed to attain equilibrium,$N_2O_{4(g)} \rightleftharpoons 2NO_{2(g)}$. At equilibrium,the total pressure is $600 \ mm \ Hg$,when $20 \%$ of $N_2O_4$ is dissociated. The value of $K_p$ for the reaction is

$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).

$A$ vessel at $1000 \ K$ contains $CO_2$ with a pressure of $0.5 \ atm$. Some of the $CO_2$ is converted into $CO$ on the addition of graphite. If the total pressure at equilibrium is $0.8 \ atm$,the value of $K_p$ is $..... \ atm$.

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$A_{(s)} \rightleftharpoons M_{(s)} + \frac{1}{2} O_{2(g)}$
The equilibrium constant for the reaction is $K_{p} = 4$. At equilibrium,the partial pressure of $O_{2}$ is $.... \ atm.$ (Round off to the nearest integer).