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\tCalculate the % dissociation of H2S(g) ,if 0.1 mole of H2S is kept in 0.4 litre vessel at 1000K.For the reaction 2H2S = 2H2+S2 , the value of kc is 1×10−6

zyan roy , 6 Years ago
Grade 11
anser 1 Answers
Nilabh

Last Activity: 6 Years ago

(0.1-x )    (  x )    (x/2)
2H2S  >
 
At equilibrium:-  (0.1-x)/V
Therefore ::::   Kc= {( x/V)^2(x/2V)} / (0.1-x/v)^2
    =>  1.0×10^-6 =( x^3) / 2V(0.1)^2
    =>   x^3 = 2×0.4×0.1×10^-6
    =>   x = 2×10 ^-3
   So % dissociation = (2×10^-3×100) / 0.1
                                   = 2%

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