Calculate the percent dissociation of H2S(g) if 0.1 mol of H2S is kept in a 0.4 litre vessel at 1000 K. For the reaction, 2H2S(g)⇌2H2(g)+S2(g) the value of Kc is 1×10−6.
A
2%
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B
20%
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C
200%
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D
50%
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Solution
The correct option is A2% 2H2S(g)⇌2H2(g)+S2(g)(Initially)0.100(At equilibrium)0.1−xxx2 Kc=[H2]2[S2][H2S]2=(x0.4)2(x2×0.4)(0.1−x0.4)2 Assuming 0.1−x≈0.1 Since, ′x′ is small because Kc=10−6 ∴x32×0.4×(0.1)2=10−6 or, x=2×10−3 This is degree of dissociation for 0.1 mol ∴ Degree of dissociation for 1 mol =10×(2×10−3)=2×10−2 Hence, % dissociation =2×10−2×100=2