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

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: 

$\mathrm{G}=\mathrm{H}-\mathrm{TS}$

Here,

H = Enthalpy

T = Absolute Temperature

S = Entropy

Also, we learnt that 

$H=E+P V$

$\mathrm{G}=\mathrm{E}+\mathrm{PV}-\mathrm{TS}$

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

$
\Delta \mathrm{G}=\Delta \mathrm{E}+\mathrm{P} \Delta \mathrm{~V}-\mathrm{T} \Delta \mathrm{~S}
$

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

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

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

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: 

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

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

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

$\mathrm{G}=\mathrm{H}-\mathrm{TS}$

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

Now, 

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

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

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

Now, 

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

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

$\mathrm{dG}=\mathrm{VdP}-\mathrm{SdT}$

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

$\Delta \mathrm{G}$ and Criteria of Spontaneity

Suppose we consider a system that is not isolated from its surroundings then for such a system $\Delta \mathrm{H}$ is given as:

$\Delta \mathrm{S}_{\text {total }}=\Delta \mathrm{S}_{\text {system }}+\Delta \mathrm{S}_{\text {surrounding }}$

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

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

$\Delta \mathrm{S}_{\text {surroundings }}=\frac{(\mathrm{q})_{\mathrm{p} \text { surroundings }}}{\mathrm{T}}=\frac{-\Delta \mathrm{H}_{\text {system }}}{\mathrm{T}} \ldots(i i)$

From equation (i) and (ii)

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

Or

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

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

As according to Gibb-Helmholtz equation,

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

So, $\Delta \mathrm{G}_{\text {system }}=\Delta \mathrm{H}_{\text {system }}-\mathrm{T} \Delta \mathrm{S}_{\text {system }}$

$\Delta \mathrm{G}_{\text {system }}=-T \Delta \mathrm{~S}_{\text {total }}$

As for the spontaneous process

$\Delta S_{\text {total }}>0$

Hence  $\Delta \mathrm{G}=-\mathrm{ve}$

Thus for a spontaneous process $\mathrm{T} \Delta \mathrm{S}_{\text {total }}$ must be positive.

Or $\Delta \mathrm{G}$ must be negative.

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

$\Delta \mathrm{H}=+\mathrm{ve}$ and $\mathrm{T} \Delta \mathrm{S}=-\mathrm{ve}$

$\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=(+\mathrm{ve})-(-\mathrm{ve})=+\mathrm{ve}$ 

Thus, $\Delta G$ is positive for a non-spontaneous process.

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

$\begin{aligned} \Delta \mathrm{H} & =+\mathrm{ve} \text { and } \mathrm{T} \Delta \mathrm{S}=+\mathrm{ve} \\ \Delta \mathrm{H} & =\mathrm{T} \Delta \mathrm{S} \\ \Delta \mathrm{G} & =\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=0\end{aligned}$

Thus, the process is said to be at equilibrium.

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

$\begin{aligned} & \Delta \mathrm{H}=-\mathrm{ve} \text { and } \mathrm{T} \Delta \mathrm{S}=+\mathrm{ve} \\ & \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}=(-\mathrm{ve})-(+\mathrm{ve})=-\mathrm{ve}\end{aligned}$

Thus, this process is spontaneous at every temperature.

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

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

• $\Delta \mathrm{G}=$ 0 , Process in equilibrium

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

• In endergonic reaction $\Delta \mathrm{G}=$ 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.

Relationship between $\Delta \mathbf{G}^{\circ}$ and Equilibrium constant $\left(\mathbf{K}_{\mathrm{eq}}\right)$

for a reversible reaction

$
\mathrm{P}+\mathrm{Q} \rightleftharpoons \mathrm{R}+\mathrm{S}
$
 

$\Delta \mathrm{G}, \Delta \mathrm{G}^{\circ}$ and Reaction Quotient (Q) are related as follows

$
\Delta \mathrm{G}=\Delta \mathrm{G}^{\circ}+\mathrm{RT} \log _{\mathrm{e}} \mathrm{Q}
$

as at equilibrium $\Delta \mathrm{G}=0$

$\begin{aligned} & \mathrm{Q}=\mathrm{Keq} \\ & 0=\Delta \mathrm{G}^{\circ}+\mathrm{RT} \log _{\mathrm{e}} \mathrm{Keq} \\ & \Delta \mathrm{G}^{\circ}=-\mathrm{RT} \log _{\mathrm{e}} \mathrm{Keq} \\ & \Delta \mathrm{G}^{\circ}=-2.303 \mathrm{RT} \log _{10} \mathrm{Keq}\end{aligned}$

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 

$\Delta G=-n F E$

when we use standard conditions than 

$\Delta G^o=-n F E^o$

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

n = No. of moles of e- transferred 

F = Faraday's constant