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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

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  • 23 Questions around this concept.

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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

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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 :

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

Choose Correct option.

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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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Gibbs Free Energy of Reaction Current Topic

Oxidation Number
Electrochemistry
Redox Reactions
Electrochemical Series
Displacement Reaction
Faraday’s Laws of Electrolysis
Galvanic Cells
Salt Bridge
Standard Hydrogen Electrode
Gibbs Free Energy of Reaction

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