Calculate the proton affinity of `NH_(3)(g)` from the following data (in `kJ mol^(-1))`:
`DeltaH^(Theta)` dissociation: `H_(2)(g) = 218`
`DeltaH^(Theta)` formation: `NH_(3)(g) =- 46`
Lattice energy of `NH_(4)CI(s) = 683`
Ionisation energy of `H = 1310`
Electron affinity of `CI =- 348`
Bond dissociation energy `CI_(2)(g) = 124`
`Delta_(f)H^(Theta) (NH_(4)CI) =- 314`

Explanation

We have to calculate `DeltaH` for the following equation
`NH_(3)(g)+H^(o+)(g)rarr 2H(g)`
Given
`H_(2)(g) +2H(g) rarr 2H(g) DeltaH_(1) = 218 kJ mol^(-1)`
`(1)/(2)N_(2)(g) +(3)/(2)H_(2)(g) rarr NH_(3)(g) DeltaH_(2) =- 46 kJ mol^(-1)`
`NH_(4)CI(s)rarr NH_(4)^(o+)(g)+CI^(Theta)(g) DeltaH_(3) = 683kJ mol^(-1)`
`H(g)rarr H(g)^(**) DeltaH_(4) = 1310 kJ mol^(-1)`
`CI(g) rarr CI^(Theta) DeltaH_(5) =- 348 kJ mol^(-1)`
`CI_(2)(g) rarr 2CI(g) DeltaH_(6) = 124 kJ mol^(-1)`
`(1)/(2)N_(2)(g) +2H_(2)(g)+ (1)/(2)CI_(2)(g) rarr NH_(4)CI(g) DeltaH_(7) = - 314 kJ mol^(-1)`
`DeltaH = (1)/(2) DeltaH_(1) - DeltaH_(2) +DeltaH_(3) - DeltaH_(4) - DeltaH_(5) - (1)/(2)DeltaH_(6) +DeltaH_(7)`
`=- 718 kJ mol^(-1)`