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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

Verified by Experts

`2H_(2)S(g)hArr2H_(2)(g)+S_(2)(g),`
Volume of vessel `=V=0.05 L`
Let `x` be the of dissociation

`K_(c )=([H_(2)]^(2)[S_(2)])/([H_(2)S]^(2))`
`(((0.1x)/V)^(2)((0.1x)/(2V)))/(((0.1-0.1x)/V)^(2))=10^(-7)`
Assuming `0.1-0.1x~~0.1`, we get
`rArr x^(3)/(2V)=10^(-6)rArr x=0.01`
Degree of dissociation `(alpha)`
`=("Amount dissociated")/("Initial amount")=(0.1x)/0.1=x`
`rArr 1%` dissociation of `H_(2)S`
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