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

To calculate the proton affinity of \( NH_3(g) \), we need to determine the enthalpy change (\( \Delta H \)) for the following reaction:

\[ NH_3(g) + H^+(g) \rightarrow NH_4^+(g) \]

Let's denote this enthalpy change as \( X \).

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