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(g)`
The value of `K_(c )` is `1.0xx10^(-6)`

Answer

B

Explanation

molar concentration of `H_(2)S=0.1/0.4 M=0.25 M`
Let `alpha` be the degree of ionisation of `H_(2)S`
`{:(,H_(2)S,hArr,2H^(o+),+,S_(2)^(-)),("Initial",0.25 M,,,,),("At equilibrium",0.25(1-alpha),,2xx0.25alpha,,0.25alpha),(,,,=0.5 alpha,,):}`
`K_(c)=([H^(o+)]^(2)[S^(2-)])/([H_(2)S])`
`10^(-6)=((0.5alpha)^(2)(0.25alpha))/(0.25alpha(1-alpha))=(0.25alpha^(2))/(1-alpha)`
Neglecting `alpha`,
`10^(-6)=0.25alpha^(2)`
`alpha^(2)=4xx10^(-6)`
`alpha=2xx10^(-3)=0.002=0.2%`