(C) The equilibrium reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$.Given volume $V = 10 \ L$.Equilibrium concentrations are:$[PCl

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(C) The equilibrium reaction is: $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$.Given volume $V = 10 \ L$.Equilibrium concentrations are:$[PCl_5] = \frac{0.1 \ mol}{10 \ L} = 0.01 \ M$$[PCl_3] = \frac{0.2 \ mol}{10 \ L} = 0.02 \ M$$[Cl_2] = \frac{0.2 \ mol}{10 \ L} = 0.02 \ M$The equilibrium constant $K_c$ is given by:$K_c = \frac{[PCl_3][Cl_2]}{[PCl_5]}$$K_c = \frac{0.02 \times 0.02}{0.01} = \frac{0.0004}{0.01} = 0.04$.

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

Medium

$A$ quantity of $PCl_5$ was heated in a $10 \ L$ vessel at $250 \ ^oC$; $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. At equilibrium,the vessel contains $0.1 \ mol$ of $PCl_5$,$0.20 \ mol$ of $PCl_3$,and $0.2 \ mol$ of $Cl_2$. The equilibrium constant $(K_c)$ of the reaction is:

A
$0.02$
B
$0.05$
C
$0.04$
D
$0.025$

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6-1.Equilibrium (Chemical Equilibrium)

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Law of equilibrium and Equilibrium constant

If the equilibrium constant for the reaction ${N_2}_{(g)} + {O_2}_{(g)} \rightleftharpoons 2NO_{(g)}$ is $K$,then what is the equilibrium constant for the reaction $\frac{1}{2}{N_2}_{(g)} + \frac{1}{2}{O_2}_{(g)} \rightleftharpoons NO_{(g)}$?

Medium

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The decomposition of $N_2O_4$ to $NO_2$ is carried out at $280 \ K$ in chloroform. When equilibrium has been established,$0.2 \ mol$ of $N_2O_4$ and $2 \times 10^{-3} \ mol$ of $NO_2$ are present in $2 \ L$ solution. The equilibrium constant for the reaction $N_2O_4 \rightleftharpoons 2NO_2$ is:

MediumAIIMS 1984

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The following concentrations were observed at $500 \ K$ for the formation of $NH_3$ from $N_2$ and $H_2$. At equilibrium: $[N_2] = 2 \times 10^{-2} \ M$,$[H_2] = 3 \times 10^{-2} \ M$ and $[NH_3] = 1.5 \times 10^{-2} \ M$. The equilibrium constant for the reaction is:

DifficultJEE MAIN 2024

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If the equilibrium constant for the reaction $2AB \rightleftharpoons A_2 + B_2$ is $49$,then the equilibrium constant for the reaction $AB \rightleftharpoons \frac{1}{2}A_2 + \frac{1}{2}B_2$ will be:

Medium

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For the reactions:
$A \rightleftharpoons B$ $K_C = 2$
$B \rightleftharpoons C$ $K_C = 3$
$C \rightleftharpoons D + E$ $K_C = 5$
$K_C$ for the reaction $A \rightleftharpoons D + E$ is:

Medium

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$A$ quantity of $PCl_5$ was heated in a $10 \ L$ vessel at $250 \ ^oC$; $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. At equilibrium,the vessel contains $0.1 \ mol$ of $PCl_5$,$0.20 \ mol$ of $PCl_3$,and $0.2 \ mol$ of $Cl_2$. The equilibrium constant $(K_c)$ of the reaction is:

Similar Questions

If the equilibrium constant for the reaction ${N_2}_{(g)} + {O_2}_{(g)} \rightleftharpoons 2NO_{(g)}$ is $K$,then what is the equilibrium constant for the reaction $\frac{1}{2}{N_2}_{(g)} + \frac{1}{2}{O_2}_{(g)} \rightleftharpoons NO_{(g)}$?

The decomposition of $N_2O_4$ to $NO_2$ is carried out at $280 \ K$ in chloroform. When equilibrium has been established,$0.2 \ mol$ of $N_2O_4$ and $2 \times 10^{-3} \ mol$ of $NO_2$ are present in $2 \ L$ solution. The equilibrium constant for the reaction $N_2O_4 \rightleftharpoons 2NO_2$ is:

The following concentrations were observed at $500 \ K$ for the formation of $NH_3$ from $N_2$ and $H_2$. At equilibrium: $[N_2] = 2 \times 10^{-2} \ M$,$[H_2] = 3 \times 10^{-2} \ M$ and $[NH_3] = 1.5 \times 10^{-2} \ M$. The equilibrium constant for the reaction is:

If the equilibrium constant for the reaction $2AB \rightleftharpoons A_2 + B_2$ is $49$,then the equilibrium constant for the reaction $AB \rightleftharpoons \frac{1}{2}A_2 + \frac{1}{2}B_2$ will be:

For the reactions:$A \rightleftharpoons B$ $K_C = 2$$B \rightleftharpoons C$ $K_C = 3$$C \rightleftharpoons D + E$ $K_C = 5$$K_C$ for the reaction $A \rightleftharpoons D + E$ is:

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For the reactions:
$A \rightleftharpoons B$ $K_C = 2$
$B \rightleftharpoons C$ $K_C = 3$
$C \rightleftharpoons D + E$ $K_C = 5$
$K_C$ for the reaction $A \rightleftharpoons D + E$ is: