When `3.06 g` of solid `NH_(4)HS` is introduced into a two-litre evacuated flask at `27^(@)C, 30%` of the solid decomposes into gaseous ammonia and hydrogen sulphide. (i) Calculate `K_(c )` and `K_(p)` for the reaction at `27^(@)C`. (ii) What would happen to the equilibrium when more solid `NH_(4)HS` is introduced into the flask?

Explanation

Moles of `NH_4HS` introduced into the flask =`3.06/51`=0.06
`{:(,NH_4HS(s)hArr, NH_3(g)+,H_2S(g)),("Initial conc.",0.06,0,0),("Equi. conc.",0.06 (1-x), 0.06x,0.06x):}`
Since x is 30% i.e., =0.3
`{:("At equi.",0.06(1-0.3),0.06 xx 0.3,0.06 xx 0.3),(,0.042,0.018,0.018):}`
Vol. of flask =2 L
`[NH_3]=0.018/2`=0.009
`[H_2S]=0.018/2`=0.009
`K_c=[NH_3][H_2S]`
`=0.009xx0.009=8.1xx10^(-5)`
`K_p=K_c(RT)^(Deltan_g)`
`Deltan_g` =1+1-0 =2
`K_p=8.1xx10^(-5)xx(0.082xx300)^2`
=0.049
(ii) Addition of `NH_4HS` does not change the position of equilibrium because `K_c` does not depend upon the amount of `NH_4HS(s)`