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Calculate the precentage dissociation of...

Calculate the precentage dissociation of `H_(2)S(g)` if `0.1` mol of `H_(2)S` is kept in a `0.5 L` vessel at `1000 K`. The value of `K_(c )` for the reaction
`2H_(2)ShArr2H_(2)(g)+S_(2)(g)`
is `1.0xx10^(-7).`

Text Solution

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`2H_(2)S(g) hArr 2H_(2)(g)+S_(2)(g)`, Volume of vessel = V = 0.5 L
Let `alpha` be the degree of dissociation

`K_(C)=([H_(2)]^(2)[S_(2)])/([H_(2)S]^(2))=(((0.1alpha)/(V))^(2)((0.1alpha)/(2V)))/(((0.1-0.1alpha)/(V))^(2))=10^(-7) rArr (alpha^(3))/(2V)=10^(-6) rArr alpha=0.01 rArr 1%` dissociationof `H_(2)S`.
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