Relation between electron gain enthalpy and electron affinity


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> The correct relation between electron gain enthalpy $(Δ\_\mathrm{eg}H)$ and electron affinity $A\_\mathrm{e}$ at any temperature '$T$' is
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> A) $Δ\_\mathrm{eg}H = -A\_\mathrm{e} - \frac{5}{2}RT$
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> B) $Δ\_\mathrm{eg}H = \frac{A\_\mathrm{e}}{RT}$
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> C) $Δ\_\mathrm{eg}H = \frac{-A\_\mathrm{e}}{RT}$
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> D) $Δ\_\mathrm{eg}H = \frac{-A\_\mathrm{e}}{RT} + \frac{1}{A\_\mathrm{e}^2}$
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> 


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"](https://webbook.nist.gov/chemistry/ion/#EC) 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


$$Δ\_\mathrm{eg}H^⦵ = - A\_\mathrm{e} - \frac{5}{2}RT.$$


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