Gibbs Free Energy of Reaction MCQ - Practice Questions & Answers

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Feasibility and Gibbs Free Energy of Reaction is considered one the most difficult concept.
23 Questions around this concept.

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Feasibility and Gibbs Free Energy of Reaction

Let n faraday charge be taken out of a cell of emf E, then work done by the cell will be calculated as:
Work = Charge × Potential
Work done by the cell is equal to the decrease in the free energy.

$-\Delta \mathrm{G}=\mathrm{nFE}$

Similarly, maximum obtainable work from the cell at standard condition will be:

$\mathrm{W}_{\max }=\mathrm{nF} \mathrm{E}_{\mathrm{cell}}^{0} \quad \text { where } \mathrm{E}_{\mathrm{cell}}^{0}=\text { standard emf of standard cell potential }$

$\mathrm{-\Delta G^{\circ}=n F E_{\text {cell }}^{0}}$

Variation of E.M.F. (E_cell) with Temperature
The temperature coefficient of e.m.f. of the cell is written as:

$\Delta \mathrm{G}=\Delta \mathrm{H}+\mathrm{T}\left(\frac{\mathrm{d}(\Delta \mathrm{G})}{\mathrm{dT}}\right)_{\mathrm{Const.} \text {Pressure }}$

$\Rightarrow \quad-\mathrm{nFE}_{\mathrm{cell}}=\Delta \mathrm{H}+\mathrm{T}\left(\frac{\mathrm{d}\left(-\mathrm{nFE}_{\mathrm{cell}}\right)}{\mathrm{dT}}\right)_{\mathrm{P}} \quad\quad\quad\left[\because \Delta \mathrm{G}=-\mathrm{nFE}_{\mathrm{cell}}\right]$

$\Rightarrow \quad\left(\frac{\mathrm{dE}_{\mathrm{cell}}}{\mathrm{dT}}\right)_{\mathrm{P}}=\frac{\Delta \mathrm{H}}{\mathrm{nFT}}+\frac{\mathrm{E}_{\mathrm{cell}}}{\mathrm{T}}$

Enthalpy Change: From above equation, we have:

$\Delta \mathrm{H}=\Delta \mathrm{G}-\mathrm{T}\left(\frac{\mathrm{d}}{\mathrm{dT}}(\Delta \mathrm{G})\right)_{\mathrm{P}}$

Thus, enthalpy(ΔH) is given as:

$\Delta \mathrm{H}=-\mathrm{nF}\left[\mathrm{E}_{\mathrm{cell}}-\mathrm{T}\left(\frac{\mathrm{dE}_{\mathrm{cell}}}{\mathrm{dT}}\right)_{\mathrm{P}}\right]$

Entropy change: For entropy change we have:

$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}$

$\Delta \mathrm{S}=-\left(\frac{\mathrm{d}}{\mathrm{dT}}(\Delta \mathrm{G})\right)_{\mathrm{P}}=\mathrm{nF}\left(\frac{\mathrm{dE}_{\mathrm{cell}}}{\mathrm{dT}}\right)_{\mathrm{P}}$