The amount of energy when million atoms of iodine are completely converted into $${I^ - }$$ ions in the vapour state according to the equation, $${I_{\left( g \right)}} + {e^ - } \to I_{\left( g \right)}^ - $$    is $$4.9 \times {10^{ - 13}}\,J.$$   What would be the electron gain enthalpy of iodine in terms of $$kJ\,mo{l^{ - 1}}$$  and $$eV$$  per atom?

Solution :
The electron gain enthalpy of iodine is equal to the energy released when 1 mole of iodine atoms in vapour state is converted into $${I^ - }$$ ions
$$\eqalign{ & {\Delta _{eg}}H = \frac{{4.9 \times {{10}^{ - 13}} \times 6.023 \times {{10}^{23}}}}{{{{10}^6}}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 29.5 \times {10^4}\,J\,mo{l^{ - 1}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 295\,kJ\,mo{l^{ - 1}} \cr} $$
Thus, electron gain enthalpy of iodine $$ = - 295\,kJ\,mo{l^{ - 1}}$$
We know that 1 $$eV$$  per atom $$ = 96.49\,kJ\,mo{l^{ - 1}}$$
∴ Electron gain enthalpy of iodine in $$eV$$  per atom $$ = - \frac{{295}}{{96.49}} = - 3.06\,eV$$     per atom