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

`H_(2) S = (0*1 )/(0*4) "mol" L^(-1) = 0*25 "mol" L^(-1) `
Suppose degree of dissociation of `H_(2) S = alpha ` . Then
` {: (,2 H_(2)S ,hArr,2H_(2),+,S_(2)),(" Intial conc.",0*25 M,,,,),("Conc.at eqm.",0*25 (1-alpha),,0*25 alpha = 0* 125 alpha ,,(0*25)/2alpha=0*125alpha):}`
`K_(c) = ([H_(2)]^(2) [S_(2)])/([H_(2)S]^(2)):. 10^(-6) = ((0*25 alpha)^(2) (0*125 alpha))/[ 0*25 (1-alpha)]^(2)`
Neglecting `alpha` in comparison to 1 , we get ` 10^(-6) = ((0*25 alpha)^(2) ( 0*125 alpha))/(0*25 )^(2) `
` 10^(-6)= 0*125 alpha^(3) or alpha^(3) = 8 xx 10^(-6) or alpha = 2 xx 10^(2) = 0*02 `
` :. % " age dissociation "= 0*02 xx 100 = 2 % `