Solid `NH_(4)HS(s)` (ammonium hydrogen sulphate) dissociates to give `NH_(3)(g)` and `H_(2)S(g)` and is allowed to attain equilibrium at `100^(@)C`. If the value of `K_(p)` for its dissociation is found to be `0.34`, find the total pressure at equilibrium at `100^(@)C`. If the value of `K_(p)` for its dissociation is found to be `0.34`, find the total pressure at equilibrium and partial pressure of each component.

Explanation

`NH_(4)SH(s) hArr NH_(3)(g)+H_(2)S(g)`, since `NH_(4)SH` is a solid, hence `a_(NH_(4)SH)=1` and its un-dissociated amount will not effect the total pressure (pressure will be due to gaseous `NH_(3)` and `H_(2)S` only). Let 'x' be its moles decomposed at equilibrium and P be the equilibrium pressure.

Total moles at equilibrium = moles of `NH_(3)+H_(2)S=2x`
`P=?" " K_(p)=0.34`
`P_(H_(2)S)=(x)/(2x)P=(P)/(2) and P_(NH_(3))=(x)/(2x)P=(P)/(2)`
(for equimolar proportions, partial pressures are equal)
`K_(P)=P_(H_(2)S). P_(NH_(3))" " (a_(NH_(4)HS)=1)`
`rArr 0.34=(P)/(2)xx(P)/(2)`
`rArr (P^(2))/(4)=0.34 rArr P= sqrt(4 xx 0.34)=1.17`
`rArr P_(NH_(3))=(P)/(2)=(1.17)/(2)=0.585` atm
and `P_(H_(2)S)=(P)/(2)=(1.17)/(2)=0.585` atm