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

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

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.

  • 40 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

Match list - 1 (Equations) with list - 2  (Type of processes) and select the correct option

    List 1                                                                            List 2

  Equations                                                                    Type pf process

(1) K_{p}> Q                                                                 (i) Non spontaneous

(2) \Delta G^{o}< RT\ In\ Q                                               (ii) Equilibrium

(3) K_{p}=Q                                                                 (iii) spontaneous and                                                                                                                endothermic

(4) T> \frac{\Delta H}{\Delta S}                                                               (iv) spontaneous

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

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 a spontaneous reaction the \Delta G , equilibrium constant (K) and E^{\circ}_{cell}  will be respectively

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The correct thermodynamic conditions for the spontaneous reaction at all temperatures is:

For a given reaction, \Delta H=35.5kJmol^{-1}\ and\ \Delta S=83.6JK^{-1}mol^{-1}. The reaction is spontaneous at :(Assume that \Delta H \ and\ \Delta S do not vary with tempearature)

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Which of the following statements is correct for a reversible process in a state of equilibrium?

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 which 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 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 be 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 favourable 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 high temp

Spontaneous

-

-

- ve at low temp

Spontaneous

+ ve at 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 play s an important role to decide the spontaneity of a process. A process which 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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