# Relation between electron gain enthalpy and electron affinity

The correct relation between electron gain enthalpy $$(Δ_\mathrm{eg}H)$$ and electron affinity $$A_\mathrm{e}$$ at any temperature '$$T$$' is

A) $$Δ_\mathrm{eg}H = -A_\mathrm{e} - \frac{5}{2}RT$$

B) $$Δ_\mathrm{eg}H = \frac{A_\mathrm{e}}{RT}$$

C) $$Δ_\mathrm{eg}H = \frac{-A_\mathrm{e}}{RT}$$

D) $$Δ_\mathrm{eg}H = \frac{-A_\mathrm{e}}{RT} + \frac{1}{A_\mathrm{e}^2}$$

I have read that electron affinity could be taken as electron gain enthalpy at absolute zero, i.e. 0 K. So, maybe we could use Kirchoff's law for temperature $$T$$, but why is $$C_p$$ taken and not $$C_v$$ or something else, is it from the definition?

You are correct, at absolute zero $$Δ_\mathrm{eg}H^⦵ = - A_\mathrm{e}$$ for the gas-phase act of gaining an electron:

$$\ce{X(g) + e-(g) → X-(g)}$$

$$A_\mathrm{e} = E(\ce{X(g)}) - E(\ce{X-(g)})$$

The term $$5/2RT$$ arises from the so-called "electron convention" when electron is treated as ideal gas with corresponding heat capacity $$C_p$$ from Boltzmann statistics and the element of enthalpy

$$H_T(\ce{e-}) - H_0(\ce{e-}) = \frac{5}{2}RT,$$

resulting in exact relation

$$Δ_\mathrm{eg}H^⦵ = - A_\mathrm{e} - \frac{5}{2}RT.$$