An equilibrium mixture for the reaction- 2H2Sg - 2H2g - S2g had 1 mole of H2S- 0-2 mole of H2 and 0-8 mole of S2 in 2 L flask- The value of Kc in mol- lit-1 is-

The correct option is D 0.016The equilibrium reaction is 2H_2S(g) ⇔ #160; 2H_2(g) + S_2(g) .The expression for the equilibrium constant is K_c= [H_ ...

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Standard XII

Chemistry

Characteristics of Chemical Equilibrium

An equilibriu...

Question

An equilibrium mixture for the reaction: $2 H_{2} S (g) \Leftrightarrow 2 H_{2} (g) + S_{2} (g)$ had 1 mole of $H_{2} S$ , 0.2 mole of $H_{2}$ and 0.8 mole of $S_{2}$ in 2 L flask. The value of $K_{c}$ in $m o l . l i t^{- 1}$ is;

0.004

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0.08

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0.016

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0.16

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Solution

The correct option is D 0.016
The equilibrium reaction is $2 H_{2} S (g) \Leftrightarrow 2 H_{2} (g) + S_{2} (g)$ .

The expression for the equilibrium constant is $K_{c} = \frac{[H_{2}]^{2} [S_{2}]}{[H_{2} S]^{2}}$

The number of moles of $H_{2} S, H_{2} a n d S_{2}$ in the equilibrium mixture are $1, 0.2 a n d 0.8$ respectively.

The volume of the flask is 2 L.

Hence, the molar concentrations of $H_{2} S, H_{2}$ and $S_{2}$ in the equilibrium mixture are $0.5, 0.1 a n d 0.4$ respectively.

Substitute the values in the expression for the equilibrium constant.
$K_{c} = \frac{(0.1)^{2} \times 0.4}{(0.5)^{2}} = 0.016$

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An equilibrium mixture for the reaction: 2H2S(g)⇔2H2(g)+S2(g) had 1 mole of H2S, 0.2 mole of H2 and 0.8 mole of S2 in 2 L flask. The value of Kc in mol.lit−1 is;

An equilibrium mixture for the reaction: $2 H_{2} S (g) \Leftrightarrow 2 H_{2} (g) + S_{2} (g)$ had 1 mole of $H_{2} S$ , 0.2 mole of $H_{2}$ and 0.8 mole of $S_{2}$ in 2 L flask. The value of $K_{c}$ in $m o l . l i t^{- 1}$ is;