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Gibbs Energy Change And Criteria For Equilibrium - Practice Questions & MCQ

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

Quick Facts

  • Spontaneity Criteria With Gibbs Energy (G) is considered one the most difficult concept.

  • Gibbs Energy And Change In Gibbs Energy is considered one of the most asked concept.

  • 45 Questions around this concept.

Solve by difficulty

For a particular reversible reaction at temperature  T,\Delta H\: and \: \Delta S\: were found to be both +Ve.

If T_{e} is the temperature at equilibrium, the reaction would be spontaneous when

Standard entropy of \mathrm{ X_{2},Y_{2}\; and\; XY_{3}}  are 60, 40 and 50 J K-1 mol-1, respectively. For the reaction,

\mathrm{\frac{1}{2} X_{2}+\frac{3}{2} Y_{2}\rightarrow XY_{3},\; \Delta H=-30\, kJ} to be at equilibrium, the temperature will be

In a fuel cell methanol is used as fuel and oxygen gas is used as an oxidizer. The reaction is

CH_{3}OH_{\left ( l \right )}+\frac{3}{2}O_{2\left ( g \right )}\rightarrow CO_{2\left ( g \right )}+2H_{2}O_{\left ( l \right )}

At 298 K standard Gibb’s energies of formation for CH_{3}OH_{\left ( l \right )},H_{2}O_{\left ( l \right )}\: and\: CO_{2\left ( g \right )}

are –166.2, –237.2 and –394.4 kJ mol-1 respectively. If standard enthalpy of combustion of methanol is –726 kJ mol-1 , efficiency of the fuel cell will be

The entropy change \mathrm{(\Delta S)} for a chemical reaction is given by:
\mathrm{\Delta S=\sum n_{i} \cdot S_{i, \text { products }}-\sum n_{j} \cdot S_{j, \text { reactants }}}

Calculate the entropy change for the reaction:

\mathrm{2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightarrow 2 \mathrm{SO}_{3}(g)}
Given the standard molar entropy \mathrm{\left(S^{\circ}\right) values: }
S^{\circ}\left(\mathrm{SO}_{2}\right)=248.15 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}
S^{\circ}\left(\mathrm{O}_{2}\right)=205.15 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}
$ $S^{\circ}\left(\mathrm{SO}_{3}\right)=256.2 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}

Consider a substance that undergoes a phase transition from a solid to a liquid at its melting point. The melting point of the substance is \mathrm{100^{\circ}C}. The molar enthalpy of fusion \mathrm{\left(\Delta H_{\text {fusion }}\right)} is 10 kJ/mol, and the molar entropy of fusion \mathrm{\left(\Delta S_{\text {fusion }}\right)} is 30 J/(mol K).Calculate the change in Gibbs free energy \mathrm{(\Delta G)} for the phase transition at its melting point.

Consider a chemical reaction that takes place at 298 K and standard pressure:

\mathrm{A(g)+2 B(g) \rightarrow C(g)+D(g)}

The standard molar enthalpy change \mathrm{(\Delta H^{\circ})} for the reaction is -400 kJ/mol. The standard molar entropy change \mathrm{(\Delta S^{\circ})} for the reaction is 600 J/(mol·K).
Calculate the standard Gibbs free energy change \mathrm{(\Delta G^{\circ})} for the reaction.

 

For the reactions,

C+O_{2}\rightarrow CO_2 \; ;           \Delta H=-393\, J

2Zn+O_2\rightarrow 2ZnO\, ;    \Delta H=-412\, J

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If an endothermic reaction is non­-spontaneous at the freezing point of water and becomes feasible at its boiling point, then

For the reaction:

\mathrm{2 \mathrm{NO}(g)+\mathrm{O}_2(g) \rightarrow 2 \mathrm{NO}_2(g)}

Given that the standard Gibbs free energy change\mathrm{\Delta G^{\circ}} is -150 kJ/mol at 25°C and the standard entropy change \mathrm{\Delta S^{\circ}}is 150 J/(mol·K), determine if the reaction is spontaneous at 25°C.

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Which of the following is not correct?

Concepts Covered - 3

Gibbs Energy And Change In Gibbs Energy

It was introduced in order to relate H, S and to explain spontaneity. According to J. Willard Gibb's Free energy of a system is defined as the maximum amount of energy available to a system during a process that can be converted into useful work. 

or 

It is the thermodynamic quantity specially characterizing the system, the decrease in whose value during a process is equal to the useful work done by the system. 

 

It is denoted by G and it is given mathematically as follows: 

\mathrm{G = H-TS}

Here,

H = Enthalpy

T = Absolute Temperature

S = Entropy

Also, we learnt that 

\mathrm{H = E + PV}

\mathrm{G = E + PV - TS}

 

Therefore, Free energy change at constant temperature and pressure is given as:

\\ \mathrm{\Delta G = \Delta E + P\Delta V - T\Delta S} \\\\ \mathrm{As\ \Delta H = \Delta E + P\Delta V} \\\\ \mathrm{So,\ \Delta G = \Delta H - T\Delta S}

At standard condition that is, 298 K and 1 atm pressure

\\ \mathrm{\ \Delta G^o = \Delta H^o - T\Delta S^o }

It is called Gibbs equation and it is used to explain criterion of spontaneity, driving force etc.

It is a state function and an extensive property. 

 

Gibb's Free Energy change for a Reaction

For a general reaction, it can be given as follows: 

\\ \mathrm{pA + qB \rightarrow rC+sD}\\ \\ \begin{array}{l}{\Delta \mathrm{G}^{\circ}=\sum \Delta \mathrm{G}_{\mathrm{P}}^{\circ}-\sum \Delta \mathrm{G}_{\mathrm{R}}^{\circ}}\\ \\ {=\left[\left(\mathrm{r} \sum \mathrm{G}_{\mathrm{C}}^{\circ}+\mathrm{s} \sum \Delta \mathrm{G}_{\mathrm{D}}^{\circ}\right)-\left(\mathrm{p} . \sum \Delta \mathrm{G}_{\mathrm{A}}^{\circ}+\mathrm{q} \sum \Delta \mathrm{G}_{\mathrm{B}}^{\circ})]\right.\right.}\end{array}

This requires the exact same treatment as \mathrm{\Delta H} or \mathrm{\Delta S}

 

Gibb's Free Energy Change for small changes in a Reversible process

\mathrm{G=H-TS}

\mathrm{dG=dH-TdS-SdT}\ \ \ \ \rightarrow (1)

Now, 

\mathrm{dH=dE + PdV + VdP}\ \ \ \ \rightarrow (2)

Using equations (1) and (2), we can write

\mathrm{dG=dE + PdV + VdP-TdS - SdT}\ \ \ \ \rightarrow (3)

Now, 

\mathrm{dE= dq +dw \ ; \ dq= TdS\ ; dw = -PdV}

Putting these values in the above expression (3), we have 

\mathrm{dG = VdP -SdT}

Note: Remember this important formula for small changes in dG values 

Spontaneity Criteria With Gibbs Energy (G)

and Criteria of Spontaneity

Suppose we consider a system that is not isolated from its surroundings then for such a system is given as:

\\ \mathrm{\ \Delta S_{total} = \ \Delta S_{system} +\ \Delta S_{surrounding} }

If we consider that qp amount of heat is given by the system to the surroundings at constant temperature and constant pressure then 

 \left(q_{\mathrm{p}}\right)_{\text { surroundings }}=-\left(\mathrm{q}_{\mathrm{p}}\right)_{\text { system }}=-\Delta \mathrm{H}_{\text { system }}
 

\Delta \mathrm{S}_{\text { surroundings }}=\frac{\left(\mathrm{q})_{\mathrm{p} \text { surroundings }}\right.}{\mathrm{T}}=\frac{-\Delta \mathrm{H}_{\text { system }}}{\mathrm{T}} \dots(ii)

From equation (i) and (ii)

\Delta \mathrm{S}_{\text { total }}=\Delta \mathrm{S}_{\text { system }}-\frac{\Delta \mathrm{H}_{\text { system }}}{\mathrm{T}}

Or

\mathrm{T\Delta \mathrm{S}_{\text { total }}=T\Delta \mathrm{S}_{\text { system }}-\Delta \mathrm{H}_{\text { system }}}

-\mathrm{T\Delta \mathrm{S}_{\text { total }}=\Delta \mathrm{H}_{\text { system }}-T\Delta \mathrm{S}_{\text { system }}}

As according to Gibb-Helmholtz equation,

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

So, \\ \mathrm{\ \Delta G_{system} = \Delta H_{system} - T\Delta S_{system}}

\\ \mathrm{\ \Delta G_{system} =- T\Delta S_{total}}

As for the spontaneous process

\\ \mathrm{\Delta S_{total} > 0}

Hence  \\ \mathrm{\Delta G = -ve}

Thus for a spontaneous process must be positive.

Or must be negative.

 

Case I.  Suppose both energy and entropy factors oppose a process that is,

\mathrm{= (+ ve) - (-ve) = +ve}

Thus, is positive for a non-spontaneous process.

 

  Case II.  Suppose both tendencies are equal in magnitude but opposite, that is,

\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=0

Thus, the process is said to be at equilibrium.

 

Case III.  Suppose entropy and energy, both factors are favorable for a process, that is,

\mathrm{= (-ve)-(+ve) = -ve}

Thus, this process is spontaneous at every temperature.


 

Remark

-

+

Always -ve

Spontaneous

+

-

Always +ve

Non-spontaneous

+

+

+ ve at low temp

Non-spontaneous

- ve at a high temp

Spontaneous

-

-

- ve at low temp

Spontaneous

+ ve at a high temp

Non-spontaneous

 

  • \mathrm{\Delta G=negative},  Spontaneous process

  • \mathrm{\Delta G=positive}, Non-spontaneous process

  • , Process in equilibrium

 

  • In exergonic reaction \mathrm{\Delta G=negative}

  • In endergonic reaction \mathrm{\Delta G=positive}

  • Temperature also plays an important role to decide the spontaneity of a process. A process that is not spontaneous at low temperature can become spontaneous at high temperature and vice-versa.


 

Gibbs Energy At Equilibrium

\begin{array}{l}{\text { Relationship between } \Delta \mathbf{G}^{\circ} \text { and Equilibrium }} {\text { constant }\left(\mathbf{K}_{\mathrm{eq}}\right)}\end{array}

\begin{array}{l}{\text { for a reversible reaction }} \\ {\qquad \mathrm{P}+\mathrm{Q} \rightleftharpoons \mathrm{R}+\mathrm{S}}\end{array}

\begin{array}{l}{\Delta \mathrm{G}, \Delta \mathrm{G}^{\circ} \text { and Reaction Quotient (Q) are related as }} \\ {\text { follows }} \\ {\qquad \Delta \mathrm{G}=\Delta \mathrm{G}^{\circ}+\mathrm{RT} \log _{\mathrm{e}} \mathrm{Q}} \\ {\text { as at equilibrium } \Delta \mathrm{G}=0}\end{array}

\begin{array}{l}{\mathrm{Q}=\mathrm{Keq}} \\ {0=\Delta \mathrm{G}^{\circ}+\mathrm{R} \mathrm{T} \log _{\mathrm{e}} \mathrm{Keq}} \\ {\Delta \mathrm{G}^{\circ}=-\mathrm{RT} \log _{\mathrm{e}} \mathrm{Keq}} \\ {\Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log _{10} \mathrm{Keq}}\end{array}

 

Relationship between \Delta G or \Delta G^o with E or E^o:-

Free energy change \Delta G in an electrochemical cell can be related to electrical work done (E) in cell as follows 

\Delta G = -nFE

when we use standard conditions than 

\Delta G^o = -nFE^o

Here E^o= standard E.M.F of the cell

n = No. of moles of e- transferred 

F = Faraday's constant

 

Study it with Videos

Gibbs Energy And Change In Gibbs Energy
Spontaneity Criteria With Gibbs Energy (G)
Gibbs Energy At Equilibrium

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Books

Reference Books

Gibbs Energy And Change In Gibbs Energy

Chemistry Part I Textbook for Class XI

Page No. : 184

Line : 15

Spontaneity Criteria With Gibbs Energy (G)

Chemistry Part I Textbook for Class XI

Page No. : 184

Line : 1

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