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In the dissociation of HI, 20% of HI is ...

In the dissociation of HI, `20%` of HI is dissociated at equilibrium. Calculate `K_(p)` for
`HI(g) hArr 1//2 H_(2)(g)+1//2 I_(2)(g)`

Text Solution

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`HI(g) rarr 1/2H_(2)(g) + 1/2I_(2)(g)`
where `alpha` is degree of dissociation: let volume of container be V litre. Also, `alpha = 0.2 i.e., 20%`
`Deltan = 0`
`K_() = K_() = (alpha//2V)^(1//2)(alpha//2V)^(1//2)/[(1-alpha)//V] = alpha/2(1-alpha)`
`K_(p) = K_(c) = 0.2/(2xx(1-0.2)) = 0.125`
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