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%