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Question

Calculate the percentage dissociation of H2S(g) if 0.1 mole of H2S is kept in 0.4 litre vessel at 1000 K for the reaction,
2H2S(g)2H2(g)+S2(g)
The value of Kc is 1.0×106.

A
2.0
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B
2.00
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C
2
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Solution

Given reaction is:
2H2S(g)2H2(g)+ S2(g)
At equilibrium
(0.1x) x x2
Molar conc.
(0.1x)0.4 x0.4 x0.8

Kc=[H2]2[S2][H2S]2=(x0.4)2(x0.8)(0.1x0.4)21.0×106=x30.8(0.1x)2
as x is very small: 0.1x0.1x30.8×(0.1)2=1.0×106x3=8×109x=2×103α=x0.1=2×1030.1
So, percentage dissociation =2×1030.1×100=2.0%

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