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_2S(g) hArr , 2H_2(g) + , S_2(g)),("Initial conc.", 0.1 , 0,0),("Equi. conc.",0.1(1-x) , 0.1x , 0.1x//2):}`
Molar concentrations are :
`[H_2S]=(0.1(1-x))/0.4 , [H_2]=(0.1x)/0.4, [S_2]=(0.1x)/(2xx0.4)=(0.1x)/0.8`
`K_c=([H_2]^2[S_2])/[H_2S]`
`=((0.1x//0.4)^2(0.1x//0.8))/([0.1(1-x)//0.4]^2)=(0.1x)^3/(0.8(1-x))^2`
If x is small , 1-x `approx` 1
`K_c=1.0xx10^(-6)=(0.1x^3)/0.8`
`therefore x^3=8xx1xx10^(-6)`
or `x=2xx10^(-2)`
`therefore` % Dissociation = `2xx10^(-2) xx100` =2%