When 3.06g of solid `NH_4HS` is introduced into a 21 evacuated flask at `27^(@)`C, 30% of the solid decomposes into gaseous ammonia and hydrogen sulphide.
Calculate `K_c` and `K_p` for the reaction at `27^(@)`C

Explanation

The following equilibrium establishes when `NH_4HS` undergoes decomposition:
`NH_4HS(s) 3.06g of `NH_4HS-=(3.06)/(51)-=0.16mol` of `NH_4HS`
Given that 30% `NH_4HS` gets dissociated.
So at equilibrium no. of moles of
`NH_3=0.06xx(30)/(100)=0.018mol`
`H_2S=0.06xx(30)/(100)=0.018mol`
[`:'` 1 mol of `NH_4HS-=` 1mol of `NH_3-=1` mol of `H_2S`]
`:.` In equilibrium mixture,
`[NH_3]=(0.018)/(2)=9xx10^(-3)xx9xx10^(-3)mol^2*L^(-2)`
`=8.1xx10^(-5)mol^2*L^(-2)`
For the given reaction, `/_\n=(1+1)-0=2`
`K_p=K_c(RT)^(/_\n)`
`=81xx10^(-5)(0.0821xx300)^2atm^2=0.049atm^2`