(C) The given reaction is $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$.Since the volume of the vessel is $1 L$,the molar concentrations are

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(C) The given reaction is $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$.Since the volume of the vessel is $1 L$,the molar concentrations are equal to the number of moles:$[H_2S] = 0.5 mol L^{-1}$$[H_2] = 0.10 mol L^{-1}$$[S_2] = 0.4 mol L^{-1}$The expression for the equilibrium constant $K_c$ is:$K_c = \frac{[H_2]^2 [S_2]}{[H_2S]^2}$Substituting the values:$K_c = \frac{(0.10)^2 \times (0.4)}{(0.5)^2} = \frac{0.01 \times 0.4}{0.25} = \frac{0.004}{0.25} = 0.016 mol L^{-1}$.

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

Medium

An equilibrium mixture of the reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$ had $0.5 mol$ $H_2S$,$0.10 mol$ $H_2$,and $0.4 mol$ $S_2$ in a $1 L$ vessel. The value of the equilibrium constant $(K_c)$ in $mol L^{-1}$ is:

IIT 1992 All IIT

AIIMS 1998 All AIIMS

A
$0.004$
B
$0.008$
C
$0.016$
D
$0.16$

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6-1.Equilibrium (Chemical Equilibrium)

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An equilibrium mixture of the reaction $2H_2S_{(g)} \rightleftharpoons 2H_{2(g)} + S_{2(g)}$ had $0.5 mol$ $H_2S$,$0.10 mol$ $H_2$,and $0.4 mol$ $S_2$ in a $1 L$ vessel. The value of the equilibrium constant $(K_c)$ in $mol L^{-1}$ is:

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In the reaction,$A + B \rightleftharpoons 2C$,at equilibrium,the concentration of $A$ and $B$ is $0.20 \ mol \ L^{-1}$ each and that of $C$ was found to be $0.60 \ mol \ L^{-1}$. The equilibrium constant of the reaction is

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