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Question

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 1×106.

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Solution

2H2S(g)2H2(g)+S2(g)(Initially)0.100(At equilibrium)0.1xxx2
Kc=[H2]2[S2][H2S]2=(x0.4)2(x2×0.4)(0.1x0.4)2
Assuming
0.1x0.1
Since, x is small because Kc=106
x32×0.4×(0.1)2=106 or,
x=2×103
This is degree of dissociation for 0.1 mol
Degree of dissociation for 1 mol =10×(2×103)=2×102
Hence, % dissociation =2×102×100=2%

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