• Written By Pavithra VG
  • Last Modified 25-01-2023

Gibbs Free Energy Formula: Check Solved Example

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Gibbs Free Energy Formula: Gibbs free energy is a phrase used to quantify the largest amount of work done in a thermodynamic system when temperature and pressure remain constant. Gibb’s free energy is represented by the letter G. Joules or Kilojoules are the units of energy. Gibbs free energy is the maximum amount of work that can be collected from a closed system.

According to the second rule of thermodynamics, the entropy of the cosmos increases with every spontaneous change. Can entropy alone predict the reaction’s spontaneity? Gibbs’ free energy determines the reaction’s spontaneity. The meaning of Gibbs free energy, standard Gibbs free energy change, its unit, derivation of Gibbs- Helmholtz equation, conditions of spontaneity, the relationship between free energy and equilibrium constant, and many other topics are covered in the article Gibbs free energy formula.

What is Gibbs Free Energy Formula?

J. Williard Gibbs has introduced the term free energy to predict the direction of spontaneity. Free energy \(\left( {\rm{G}} \right)\) is defined as the amount of energy available for doing useful work under conditions of constant temperature and constant pressure.

\({\rm{G}} = {\rm{H}} – {\rm{TS}}\)

Here, \({\rm{H}}\) is the enthalpy of the system, \({\rm{S}}\) is the entropy of the system, and \({\rm{T}}\) is the temperature of the system on the Kelvin scale.

Free energy (Gibbs free energy) is a state function. Therefore, the change in Gibbs free energy depends only upon the initial and final states of the system and does not depend upon the path by which the change has been carried out. The change in Gibbs free energy is presented by \(\Delta {\rm{G}}{\rm{.}}\)

The Gibbs free energy, \({\rm{G}} = {\rm{H}} – {\rm{TS}}\)

We know that enthalpy, \({\rm{H}} = {\rm{U}} + {\rm{PV}}\)

Therefore, \({\rm{G}} = {\rm{U}} + {\rm{PV}} – {\rm{TS}}\)

The change in Gibbs free energy can be expressed as

\(\Delta {\rm{G}} = \Delta {\rm{U}} + \Delta \left( {{\rm{PV}}} \right) – \Delta \left( {{\rm{TS}}} \right)\)

\(\Delta {\rm{G}} = \Delta {\rm{U}} + {\rm{P}}\Delta {\rm{V}} + {\rm{V}}\Delta {\rm{P}} – {\rm{T}}\Delta {\rm{S}} – {\rm{S}}\Delta {\rm{T}}\)

If the change is carried out at a constant temperature and constant pressure, \(\Delta {\rm{T}} = 0\) and \(\Delta {\rm{P}} = 0.\)

Therefore, \(\Delta {\rm{G}} = \Delta {\rm{U}} + {\rm{P}}\Delta {\rm{V}} – {\rm{T}}\Delta {\rm{S}}\)

Since,

\(\Delta {\rm{H}} = \Delta {\rm{U}} + {\rm{P}}\Delta {\rm{V}}\)

\(\Delta {\rm{G}} = \Delta {\rm{H}} – {\rm{T}}\Delta {\rm{S}}\)

The equation \(\Delta {\rm{G}} = \Delta {\rm{H}} – {\rm{T}}\Delta {\rm{S}}\) is called Gibbs- Helmholtz equation.

What is the Formula of Gibbs Free Energy?

The Gibbs free energy change is represented by \(\Delta {\rm{G}},\) and its unit is Joule or kilojoule.

Dimensionally if we analyze \(\Delta {\rm{G}}\) has the unit of energy, i.e., \({\rm{kJ}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}\) or \({\rm{J}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}\) because both \(\Delta {\rm{H}}\) and \({\rm{T}}\Delta {\rm{S}}\) are energy terms:

\(\Delta {\rm{G}}\left( {{\rm{J}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}} \right) = \Delta {\rm{H}}\left( {{\rm{J}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}} \right) – {\rm{T}}\left( {\rm{K}} \right)\Delta {\rm{S}}\left( {{\rm{J}}\,{{\rm{K}}^{ – 1}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}} \right)\)

Example of Change in Gibbs Free Energy Formula

Melting of ice at room temperature, formation of ammonia \(\left( {{\rm{N}}{{\rm{H}}_3}} \right)\) from nitrogen \(\left( {{{\rm{N}}_2}} \right),\) and hydrogen gas \(\left( {{{\rm{H}}_2}} \right)\) are examples of the spontaneous processes in which Gibbs’s free energy is negative.

Gibbs Free Energy Formula in Electrochemistry and Spontaneity

According to Gibbs- Helmholtz equation

\(\Delta {\rm{G}} = \Delta {\rm{H}} – {\rm{T}}\Delta {\rm{S}}\)

For reaction to be spontaneous \(\Delta {\rm{G}}\) should be negative \(\left( {\Delta {\rm{G}} < 0} \right).\,\Delta {\rm{G}}\) can be negative under the following conditions:

  • \(\Delta {\rm{H}}\) is negative, and \({\rm{T}}\Delta {\rm{S}}\) is positive.
  • Both \(\Delta {\rm{H}}\) and \({\rm{T}}\Delta {\rm{S}}\) are negative. In this case, \(\Delta {\rm{H}}\) favors while \({\rm{T}}\Delta {\rm{S}}\) opposes the spontaneous process. Thus, the process can be spontaneous if \(\Delta {\rm{H}} > {\rm{T}}\Delta {\rm{S}}{\rm{.}}\)
  • Both \(\Delta {\rm{H}}\) and \({\rm{T}}\Delta {\rm{S}}\) are positive. In this case, \({\rm{T}}\Delta {\rm{S}}\) favors the spontaneous process, and \(\Delta {\rm{H}}\) opposes the spontaneous reaction. Thus, the process can be spontaneous if \(\Delta {\rm{H}} < {\rm{T}}\Delta {\rm{S}}{\rm{.}}\)

If\(\Delta {\rm{G}}\) is zero, the process does not occur, or the system is in equilibrium.

Gibbs Free Energy Table

Sign of \(\Delta {\rm{H}}\)Sign of \(\Delta {\rm{S}}\)Magnitude of \({\rm{T}}\)Spontaneity (the Sign of \(\Delta {\rm{G}}\))
\( – \)\( + \)AnySpontaneous, \(\Delta {\rm{G}} = \, – {\rm{ve}}\)
\( + \)\( – \)AnyNon-spontaneous, \(\Delta {\rm{G}} = \, + {\rm{ve}}\)
\( – \)\( – \)LowSpontaneous, \(\Delta {\rm{G}} = \, – {\rm{ve}}\)
\( – \)\( – \)HighNon-spontaneous, \(\Delta {\rm{G}} = \, + {\rm{ve}}\)
\( + \)\( + \)LowNon-spontaneous, \(\Delta {\rm{G}} = \, + {\rm{ve}}\)
\( + \)\( + \)HighSpontaneous, \(\Delta {\rm{G}} = \, – {\rm{ve}}\)

What is Standard Gibbs Free Energy Formula Change?

Standard free energy change of a reaction is defined as the change in the free energy which takes place when the reactants in the standard state \(\left( {1\,{\rm{atm}},\,298\,{\rm{K}}} \right)\) are converted into the products in their standard state.

The Gibbs- Helmholtz equation under standard conditions may be written as:

\(\Delta {{\rm{G}}^{\rm{o}}} = \Delta {{\rm{H}}^{\rm{o}}} – {\rm{T}}\Delta {{\rm{S}}^{\rm{o}}}\)

Here, \(\Delta {{\rm{H}}^{\rm{o}}} = \)Standard enthalpy change

\(\Delta {{\rm{S}}^{\rm{o}}} = \)Standard entropy change

\({\rm{T}} = \) Standard temperature \(\left( {298\,{\rm{K}}} \right)\)

The standard free energy change of a reaction is also calculated as the difference in the standard free energy change of the formation of products and reactants.

\(\Delta {{\rm{G}}^{\rm{o}}} = \sum {{\Delta _{\rm{f}}}{{\rm{G}}^{\rm{o}}}_{\left( {{\rm{products}}} \right)}} – \sum {{\Delta _{\rm{f}}}{{\rm{G}}^{\rm{o}}}_{\left( {{\rm{reactants}}} \right)}} \)

Standard free energy of formation of a substance may be defined as the free energy change which accompanies the formation of one mole of a substance from its constituent elements in its standard state.

What is the Relation Between the Gibbs Free Energy change and Equilibrium Constant?

Free energy is an extensive property, and therefore, its value for a given substance will depend on the concentration of the substance.

Let us consider a general reaction:

\({\rm{A}} + {\rm{B}}⇋{\rm{C}} + {\rm{D}}\)

The Gibbs energy change of reaction, \({\Delta _{\rm{r}}}{\rm{G}}\) is related to the composition of the reaction mixture and standard reaction Gibbs energy change, \({\Delta _{\rm{r}}}{{\rm{G}}^{\rm{o}}}\) as:

\({\Delta _{\rm{r}}}{\rm{G}} = {\Delta _{\rm{r}}}{{\rm{G}}^{\rm{o}}} + {\rm{RT}}\,{\rm{lnQ}}\)

Where \({\Delta _{\rm{r}}}{\rm{G}}\) is the reaction Gibbs energy change at a definite, fixed composition of the mixture.

\({\Delta _{\rm{r}}}{{\rm{G}}^{\rm{o}}}\) is the difference in standard Gibbs energies of formation of the product and the reactants both in their standard states.

\({\rm{Q}}\) is the reaction quotient.

\({\rm{R}}\) is the gas constant having the value of \(8.314\,{\rm{J}}\,{{\rm{K}}^{ – 1}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}.\)

If the species are gases, these concentrations are expressed in partial pressures, and the reaction quotient will be \({{\rm{Q}}_{\rm{P}}},\) and if the species are in solution, the reaction quotient will be expressed in terms of molar concentration as \({{\rm{Q}}_{\rm{C}}}.\)

At equilibrium \({\rm{Q}} = {\rm{K,}}\) called equilibrium constant and \({\Delta _{\rm{r}}}{\rm{G}} = 0,\) because the reaction mixture has no tendency to change in its concentration. Therefore, the above equation becomes:

\(0 = {\Delta _{\rm{r}}}{{\rm{G}}^{\rm{o}}} + {\rm{RT}}\,{\rm{lnK}}\)

\({\Delta _{\rm{r}}}{{\rm{G}}^{\rm{o}}} = \, – {\rm{RT}}\,{\rm{lnK}}\)

\({\Delta _{\rm{r}}}{{\rm{G}}^{\rm{o}}} = \, – 2.303\,{\rm{RT}}\,{\rm{log}}\,{\rm{K}}\)

The above equation may also be written as:

\({\rm{K}} = {{\rm{e}}^{\frac{{ – {\Delta _{\rm{r}}}{{\rm{G}}^{\rm{o}}}}}{{{\rm{RT}}}}}}\)

\({\rm{K}} = {10^{\frac{{ – {\Delta _{\rm{r}}}{{\rm{G}}^{\rm{o}}}}}{{{\rm{2}}.{\rm{303}}{\mkern 1mu} {\rm{RT}}}}}}\)

Gibbs Free Energy Change and Electrical Work Done in a Cell

Gibbs energy is related to the electrical work done in a cell. If \({{\rm{E}}^{\rm{o}}}\) is the emf of the cell, then

\(\Delta {{\rm{G}}^{\rm{o}}} = \, – {\rm{nF}}{{\rm{E}}^{\rm{o}}}\)

Here, \({\rm{n}}\) is the number of moles of electrons transferred in the balanced equation for the reaction occurring in the cell.

\({\rm{F}}\) is Faraday’s constant, which has a value of \({\rm{F}} = 96485\,{\rm{C}}/{\rm{mol}}\)

Gibbs Free Energy Problems/ Numerical

Q.1. For the melting of ice at \(25\,^\circ {\mkern 1mu} {\rm{C}},\) the enthalpy of fusion is \(6.97\,{\rm{KJ}}\,{\rm{mo}}{{\rm{l}}^{ – 1}},\) and the entropy of fusion is \(2.54\,{\rm{J}}{{\rm{K}}^{ – 1}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}.\) Calculate the free energy change and predict whether the melting of ice is spontaneous at this temperature or not.
Ans: \(\Delta {{\rm{H}}_{\left( {{\rm{fusion}}} \right)}} = 6.97\,{\rm{KJ}}\,{\rm{mo}}{{\rm{l}}^{ – 1}} = 6970\,{\rm{J}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}\)
\(\Delta {{\rm{S}}_{\left( {{\rm{fusion}}} \right)}} = 25.4\,{\rm{J}}{{\rm{K}}^{ – 1}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}\)
\({{\rm{T}}_{\rm{f}}} = 25 + 273 = 298\,{\rm{K}}\)
According to Gibbs- Helmholtz equation,
\(\Delta {\rm{G}} = \Delta {\rm{H}} – {\rm{T}}\Delta {\rm{S}}\)
\(\Delta {\rm{G}} = 6970 – \left( {298 \times 25.4} \right)\)
\(\Delta {\rm{G}} = 6970 – 7569.2\)
\(\Delta {\rm{G}} = \, – 599.2\,{\rm{J}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}\)
Since \(\Delta {\rm{G}}\) is negative, therefore, the melting of ice is spontaneous in nature.

Q.2. Calculate \(\Delta {{\rm{G}}^{\rm{o}}}\) for the reaction:
\({\rm{Zn}}\left( {\rm{s}} \right) + {\rm{C}}{{\rm{u}}^{2 + }}\left( {{\rm{aq}}} \right) \to {\rm{Z}}{{\rm{n}}^{2 + }}\left( {{\rm{aq}}} \right) + {\rm{Cu}}\left( {\rm{s}} \right)\)
Given that standard free energy for \({\rm{C}}{{\rm{u}}^{2 + }}\left( {{\rm{aq}}} \right)\) and \({\rm{Z}}{{\rm{n}}^{2 + }}\left( {{\rm{aq}}} \right)\) are \(65.1\,{\rm{KJ}}/{\rm{mol}}\) and \( – 147.2\,{\rm{KJ}}/{\rm{mol}},\) respectively,
Ans: \(\Delta {{\rm{G}}^{\rm{o}}} = {\sum {{{\rm{G}}^{\rm{o}}}} _{\left( {{\rm{products}}} \right)}} – {\sum {{{\rm{G}}^{\rm{o}}}} _{\left( {{\rm{reactants}}} \right)}}\)
\(\Delta {{\rm{G}}^{\rm{o}}} = \left( {{{\rm{G}}^{\rm{o}}}_{{\rm{Z}}{{\rm{n}}^{2 + }}} + {{\rm{G}}^{\rm{o}}}_{{\rm{Cu}}}} \right) – \left( {{{\rm{G}}^{\rm{o}}}_{{\rm{Zn}}} + {{\rm{G}}^{\rm{o}}}_{{\rm{C}}{{\rm{u}}^{2 + }}}} \right)\)
\(\Delta {{\rm{G}}^{\rm{o}}} = \left( { – 147.2 + {\rm{0}}} \right) – \left( {{\rm{0}} + 65.1} \right)\)
\(\Delta {{\rm{G}}^{\rm{o}}} = \, – 212.3\,{\rm{KJ}}/{\rm{mol}}\)

Q.3. Calculate the standard free energy change for the reaction:
\({\rm{Cu}}\left( {\rm{s}} \right) + 2\,{\rm{A}}{{\rm{g}}^{2 + }}\left( {{\rm{aq}}} \right) \to {\rm{C}}{{\rm{u}}^{2 + }}\left( {{\rm{aq}}} \right) + 2\,{\rm{Ag}}\left( {\rm{s}} \right),\,{{\rm{E}}^{\rm{o}}} = 0.46\,{\rm{V}}\)
Ans: \(\Delta {{\rm{G}}^{\rm{o}}} = \, – {\rm{nF}}{{\rm{E}}^{\rm{o}}}\)
Here \({\rm{n}} = 2,\,{\rm{F}} = 96500\,{\rm{C}},\,{{\rm{E}}^{\rm{o}}} = 0.46\,{\rm{V}}\)
\(\Delta {{\rm{G}}^{\rm{o}}} = \, – 2 \times 96500 \times 0.46\)
\(\Delta {{\rm{G}}^{\rm{o}}} = {\mkern 1mu} \, – 88780\,{\rm{CV}}\)
\(\Delta {{\rm{G}}^{\rm{o}}} = {\mkern 1mu} \, – 88780{\mkern 1mu} \,{\rm{J}}\)

Q.4. The equilibrium constant of the reaction:
\({\rm{C}}{{\rm{O}}_2}\left( {\rm{g}} \right) + {{\rm{H}}_2}\left( {\rm{g}} ⇌\right){\rm{CO}}\left( {\rm{g}} \right) + {{\rm{H}}_2}{\rm{O}}\left( {\rm{g}} \right)\)
At \(298\,{\rm{K}}\) is \(73.\) Calculate the value of the standard free energy change. \(\left( {{\rm{R}} = 8.314\,{\rm{J}}\,{\rm{mo}}{{\rm{l}}^{ – 1}}\,{{\rm{K}}^{ – 1}}} \right)\)
Ans: \(\Delta {{\rm{G}}^{\rm{o}}} = \, – 2.303\,{\rm{RT}}\,{\rm{log}}\,{{\rm{K}}_{\rm{C}}}\)
\(\Delta {{\rm{G}}^{\rm{o}}} = \, – 2.303 \times 8.314 \times 298 \times \log \,73\)
\(\Delta {{\rm{G}}^{\rm{o}}} = \, – 2.303 \times 8.314 \times 298 \times 1.8633\)
\(\Delta {{\rm{G}}^{\rm{o}}} = \, – 10631.8\,{\rm{Jmo}}{{\rm{l}}^{ – 1}}\)
\(\Delta {{\rm{G}}^{\rm{o}}} = \, – 10.632\,{\rm{kJmo}}{{\rm{l}}^{ – 1}}\)

Summary

In the article, Gibbs free energy formula you have understood the meaning of Gibbs free energy, standard free energy change, its unit, and formula. The Gibbs- Helmholtz equation can be derived with this knowledge. The spontaneity of the given reaction can be predicted from values of entropy, enthalpy, and temperature. The equilibrium constant can be calculated from the equation \(\Delta {{\rm{G}}^{\rm{o}}} = \, – 2.303\,{\rm{RT}}\,{\rm{log}}\,{{\rm{K}}_{\rm{C}}},\) emf of the cell can be calculated from the equation \(\Delta {{\rm{G}}^{\rm{o}}} = \, – {\rm{nF}}{{\rm{E}}^{\rm{o}}}.\)

FAQs on Gibbs Free Energy Formula

Q.1. Why do we use Gibbs free energy change?
Ans:
Gibbs energy is used to predict the spontaneity of the reaction. For spontaneous reactions, Gibbs free energy change is negative. It is also used to calculate equilibrium constant using equation \(\Delta {{\rm{G}}^{\rm{o}}} = \, – 2.303\,{\rm{RT}}\,{\rm{log}}\,{{\rm{K}}_{\rm{C}}}.\)

Q.2. What happens if Gibbs’s free energy change is positive?
Ans: If Gibbs free energy change is positive, then the reaction is non-spontaneous.

Q.3. Why is Gibbs free energy change negative for the spontaneous process?
Ans: During the spontaneous process, free energy decreases; therefore, Gibbs free energy change is negative.

Q.4. Why is Gibbs free energy called free energy?
Ans: During the reaction, energy is released into the surrounding from the system. Therefore, Gibbs free energy is called free energy.

Q.5. How do you remember Gibbs free energy change?
Ans:
Gibbs free energy is remembered with formula \(\Delta {\rm{G}} = \Delta {\rm{H}} – {\rm{T}}\Delta {\rm{S,}}\) here \(\Delta {\rm{G}}\) is free energy change, \(\Delta {\rm{H}}\) is the change in enthalpy, \({\rm{T}}\) is the temperature, and \(\Delta {\rm{S}}\) is the change in entropy.

Q.6. Can Gibbs free energy change be negative?
Ans: Yes, Gibbs free energy change can be negative for spontaneous reactions.

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