1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.