Calculate the percent dissociation of `H_(2)S(g)` if `0.1 mol` of `H_(2)S` is kept in `0.4 L` vessel at `1000 K`. For the reaction:
`2H_(2)S(g) hArr 2H_(2)(g)+S_2(g)`
The value of `K_(c )` is `1.0xx10^(-6)`

Explanation

`{:(,2H_(2)S_((g)),hArr,2H_(2),+,S_(2(g))),("Mole before dissociation",0.1,,0,,0),("Mole after dissociation ",(0.1-a),,a,,(a)/(2)):}`
`:' K_(c)=1.0xx10^(-6)=([H_(2)]^(2)[S_(2)])/([H_(2)S])=([(a)/(V)]^(2)xx(a)/(2V))/([(0.1-a)/(V)]^(2))`
or `(a^(3))/(2V(0.1-a)^(2))=1.0xx10^(-6)`
or `=(a^(3))/(2xx0.4(0.1-a)^(2))=1.0xx10^(-6)`
or `(a^(3))/((0.1-a)^(2))=0.8xx10^(-6)`
`:. (a^(3))/((0.1)^(2))=0.8xx10^(-6)`
(`:' a` is small `:. 0.1-a~~0.1`)
`:. a^(3)=8xx10^(-9)`
`:. a=2xx10^(-3)`
`:. alpha =(a)/(0.1)xx100`
`=(2xx10^(-3)xx100)/(0.1)=2%`