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Standard XII

Chemistry

Derivation of Kp and Kc

An equilibriu...

Question

An equilibrium mixture of the reaction $2 H_{2} S (g) ⇌ 2 H_{2} (g) + S_{2} (g)$ had 0.5 mole $H_{2} S, 0.10 m o l e H_{2}$ and 0.4 mole $S_{2}$ in one litre vessel. The value of equilibrium constant (K) in mole $l i t r e^{- 1}$ is

0.004

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0.008

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0.016

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0.16

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Solution

The correct option is C 0.016
$K = \frac{[H_{2}]^{2} [S_{2}]}{[H_{2} S]^{2}} = \frac{[0.10]^{2} [0.4]}{[0.5]^{2}} = 0.016$

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Similar questions

Q. An equilibrium mixture of the reation

2 H_{2} S (g) ⇌ 2 H_{2} (g) + S_{2} (g)

had

0.5 m o l H_{2} S, 0.10 m o l H_{2}

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Q. An equilibrium mixture for the reaction:

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had 1 mole of

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Q. An equilibrium mixture for the reaction

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has

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An equilibrium mixture of the reaction 2H2S(g)⇌2H2(g)+S2(g) had 0.5 mole H2S,0.10 mole H2 and 0.4 mole S2 in one litre vessel. The value of equilibrium constant (K) in mole litre−1 is

The correct option is C 0.016K=[H2]2[S2][H2S]2=[0.10]2[0.4][0.5]2=0.016

An equilibrium mixture of the reaction $2 H_{2} S (g) ⇌ 2 H_{2} (g) + S_{2} (g)$ had 0.5 mole $H_{2} S, 0.10 m o l e H_{2}$ and 0.4 mole $S_{2}$ in one litre vessel. The value of equilibrium constant (K) in mole $l i t r e^{- 1}$ is

The correct option is C 0.016
$K = \frac{[H_{2}]^{2} [S_{2}]}{[H_{2} S]^{2}} = \frac{[0.10]^{2} [0.4]}{[0.5]^{2}} = 0.016$