Ionization And Electron Gain Enthalpy MCQ - Practice Questions & Answers

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For electron gain enthalpies of the elements denoted as $\Delta_{\mathrm{eg}} \mathrm{H}$ , the incorrect option is :

$\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Te})<\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{PO})$

$2 . \Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Se})<\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{S})$

$\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{Cl})<\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{F})$

$\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{I})<\Delta_{\mathrm{eg}} \mathrm{H}(\mathrm{At})$

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Concepts Covered - 1

Ionization And Electron Gain Enthalpy

Ionization Energy and Electron Affinity

Ionization energy and electron affinity are defined at absolute zero. At any other temperature, heat capacities for the reactants and the products have to be taken into account.

$\begin{aligned} & \mathrm{M}(\mathrm{g}) \rightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-}(\text {for ionization }) \\ & \mathrm{M}(\mathrm{g})+\mathrm{e}^{-} \rightarrow \mathrm{M}^{-}(\mathrm{g}) \text { (for electron gain) }\end{aligned}$

at temperature, T is

$\Delta_r H^{\ominus}(T)=\Delta_r H^{\ominus}(0)+\int_0^T \Delta_r C_p^{\ominus} d T$

The value of C_p for each species in the above reaction is 5/2 R (C_V = 3/2R)

So, $\Delta_r C_p^{\ominus}=+5 / 2 \mathrm{R}$ (for ionization) $\Delta_I C_p^{\ominus}=-5 / 2 \mathrm{R}$ (for electron gain)

Therefore,

$\Delta_r H^{\ominus}($ ionization enthalpy $)=E_0($ ionization energy $)+5 / 2 \mathrm{RT}$
$\Delta_r H^{\ominus}($ electron gain enthalpy $)=-\mathrm{A}($ electron affinity $)-5 / 2 \mathrm{RT}$