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Question

# Calculate the percent dissociation of H2S(g) if 0.1 mole of H2S is kept in 0.4 L vessel at 1000 K. For the reaction, the value of Kc is 1.0×10−6. The reaction is : 2H2S(s)⇌2H2(g)+S2(g)

A
2.0
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B
4.0
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C
6.0
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D
None of the above
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Solution

## The correct option is A 2.0 At equilibrium 2H2S(s)⇌2H2(g)+S2(g) At equi. (0.1−x) x x2 conc. (0.1−x)0.4 x0.4 x0.8 Equilibrium constant, Kc=[H2]2[S2][H2S]2 putting the values, Kc=[x0.4]2[x0.8][(0.1−x)0.4]2 or x30.8(0.1−x)2=1.0×10−6 Now, as x is very small ⇒0.1−x→0.1 ⇒x=2×10−3 So, percent dissociation =2×10−30.1×100=2.0

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