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Gibbs Free Energy of Reaction - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

Quick Facts

  • 23 Questions around this concept.

Solve by difficulty

The values of \Delta H \ and \ \Delta S for the reaction, C_{\left ( graphite \right )}+CO_{2}\ _{\left ( g \right )}\rightarrow 2CO_{\left ( g \right )}  are 170 kJ and 170JK^{-1}, respectively. The reaction will be spontaneous at

Given

E_{Cl_{2}/Cl^{-1}}^{0}=1.36V,E_{Cr^{3 +}/Cr}^{0}=-0.74V

E_{Cr_{2}O_{7}^{2-}/Cr^{3+}}^{0}=1.33V,E_{MnO_{4}^{-}/Mn^{2+}}^{0}=1.1V

Among the following, the strongest reducing agent is :

Statement I:-  There will be no effect of catalyst on Gibbs energy
Statement II:- Activation energy increase by Catalyst

Choose Correct option.

Concepts Covered - 1

Feasibility and Gibbs Free Energy of Reaction

Let n faraday charge be involved in a cell generating an emf E, then the magnitude of the work done by the cell will be calculated as:
|Work| = Charge × Potential = (nF) \times (E)

Now, the maximum work that can be extracted from the cell is equal to the decrease in the Gibb's free energy.

So, it can be said that

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

The negative sign is incorporated to include the spontaineity relation between E and Gibb's free Energy

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}}

Thus, for a spontaneous cell reaction, 

E > 0 and \DeltaG < 0

 

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Feasibility and Gibbs Free Energy of Reaction

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