Relation between electron gain enthalpy and electron affinity

Question

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?

Answer 1

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

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