Gibbs Energy Formula: Check Solved Example

Gibbs Energy Formula: Gibbs 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 energy is represented by the letter G. Joules or Kilojoules are the units of energy. Gibbs 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’ energy determines the reaction’s spontaneity. The meaning of Gibbs energy, standard Gibbs energy change, its unit, derivation of Gibbs- Helmholtz equation, conditions of spontaneity, the relationship between energy and equilibrium constant, and many other topics are covered in the article Gibbs energy formula.

What is Gibbs Energy Formula?

J. Williard Gibbs has introduced the term energy to predict the direction of spontaneity. 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.

energy (Gibbs energy) is a state function. Therefore, the change in Gibbs 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 energy is presented by \(\Delta {\rm{G}}{\rm{.}}\)

The Gibbs 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 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 Energy?

The Gibbs 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 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 energy is negative.

Gibbs 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 Energy Table

What is Standard Gibbs Energy Formula Change?

Standard energy change of a reaction is defined as the change in the 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 energy change of a reaction is also calculated as the difference in the standard 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 energy of formation of a substance may be defined as the 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 Energy change and Equilibrium Constant?

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 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 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 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 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 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 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 energy formula you have understood the meaning of Gibbs energy, standard 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 Energy Formula

Q.1. Why do we use Gibbs energy change?
Ans: Gibbs energy is used to predict the spontaneity of the reaction. For spontaneous reactions, Gibbs 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 energy change is positive?
Ans: If Gibbs energy change is positive, then the reaction is non-spontaneous.

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

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

Q.5. How do you remember Gibbs energy change?
Ans: Gibbs energy is remembered with formula \(\Delta {\rm{G}} = \Delta {\rm{H}} – {\rm{T}}\Delta {\rm{S,}}\) here \(\Delta {\rm{G}}\) is 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 energy change be negative?
Ans: Yes, Gibbs energy change can be negative for spontaneous reactions.

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