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Gibbs Free Energy Formula

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  • Last Updated : 01 Jun, 2022
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Gibbs free energy, also known as Gibbs energy, Gibbs function, or free enthalpy in thermodynamics, is the thermodynamic potential. It’s used to figure out how much work a thermodynamic system can do at constant pressure and temperature. Let’s look at the concept of Gibbs Free Energy Formula.

What is Gibbs Free Energy?

To calculate changes in entropy and enthalpy values, Willard Gibbs developed the Gibbs energy (G) function. 

The amount of energy created, as measured by a decrease in the thermodynamic parameter known as Gibbs free energy, is the greatest work done.

Furthermore, when a system enters chemical equilibrium at constant temperature and pressure, the Gibbs energy is the thermodynamic potential that is minimized. We may also establish whether a reaction is preferred or disfavored using Gibbs’ formula.

Gibbs free energy is a term that describes how much work may be done in a thermodynamic system when the temperature and pressure remain constant. G is used to represent Gibb’s free energy. The SI units of Gibb’s free energy are Joules or Kilojoules. Gibbs free energy is the most work a closed system can collect.

Gibbs Free Energy Formula

Gibbs’ free energy formula is as follows:



  • ΔG = Gibbs Free Energy,
  • ΔH = Enthalpy Change,
  • T = Temperature (Kelvin),
  • ΔS = Entropy change.

Derivation of Gibbs Free Energy Formula


G = H – TS              …..(1)

We have the enthalpy,

H = U + PV

Put value of H in Equation 1, we get

G = U + PV – TS

Gibbs free energy changes can be written as

ΔG = ΔU + Δ(PV) – Δ(TS)



ΔT=0 and ΔP=0 if the change is done at the same temperature and pressure.





G = ΔH – TΔS

Gibbs Free Energy Formula Example – Gibbs free energy is negative in spontaneous processes such as ice melting at ambient temperature, ammonia (NH3) synthesis from nitrogen (N2), and hydrogen gas (H2).

Gibbs Free Energy Formula in Spontaneity and Electrochemistry

The Gibbs-Helmholtz equation states that,


ΔG must be negative (ΔG < 0) for the response to be spontaneous. The following conditions can cause ΔG to be negative:

  • TΔS is positive whereas ΔH is negative.
  • TΔS and ΔH both have a negative value. TΔS opposes the spontaneous process in this example, while ΔH promotes it. As a result, if ΔH > TΔS is true, the process can be spontaneous.
  • ΔH and TΔS are both affirmative. TΔS prefers the spontaneous reaction in this scenario, while ΔH opposes it.  As a result, if ΔH < TΔS is present, the procedure can be spontaneous.

The process does not take place if ΔG is 0, or the system is in equilibrium.

Sample Questions

Question 1: What is the significance of Gibbs’s free energy change?


The reaction’s spontaneity is predicted using Gibbs energy. Gibbs free energy changes are negative for spontaneous reactions. It’s also utilized in the equation ΔGo=-2.303RTlogKC to get the equilibrium constant.

Question 2: What makes Gibbs’s free energy so special?

Answer :

Because, The Energy is released into the environment by the system during the reaction. Gibbs free energy is hence referred to as free energy.

Question 3: The enthalpy of fusion for melting ice at 25°C is 7.12 KJmol-1 and the entropy of fusion is 2.54 JK-1mol-1. Calculate the free energy change and determine if ice melting at this temperature is spontaneous or not.


Given : ΔH = 7.12 KJmol-1 = 7120 Jmol-1, ΔS = 3.11 JK-1mol-1, T = 45°C = 45 + 273 = 318 K



ΔG = 7120 – (318 × 3.11)

= 7120 – 988.98

= 6131 Jmol-1

Because ΔG is positive, ice melting is not a spontaneous in nature.

Question 4: Calculate ΔGo for the following reaction:

Zn(s) + Cu2+(aq)  →  Zn2+(aq) + Cu(s)

Cu2+(aq) has standard free energy of 56.19 KJ/mol, while Zn2+(aq) has standard free energy of -137.1 KJ/mol.


ΔGo = ∑Go(product) – ∑Go(reactants)

∴ ΔGo = (-137.1 + 0) – (56.19 + 0)

= -137.1 – 56.19

= -193.29 J/mol

Question 5: When the Enthalpy change is 3.12 kJ/mol at a temperature of 32 degrees C and the Change of Entropy is 2.55 J/K/mol then calculate the Gibbs free energy.


Given : ΔH = 3.12 KJmol-1 = 3120 Jmol-1, ΔS = 2.55 JK-1mol-1, T = 32°C = 32 + 273 = 305 K



∴ ΔG = 3120 – (305 × 2.55)

= 3120 – 777.75

= 2342 J/mol

Question 6: Calculate the Enthalpy Change when the Gibbs free energy is 3217.2 J/mol at 451 K and the Change of Entropy is 1.82 J/K/mol.


Given : ΔG = 3217.2 Jmol-1, ΔS = 1.82 JK-1mol-1, T = 451 K


ΔG = ΔH – TΔS  

∴ ΔH = ΔG + TΔS

= 3217.2 + (451 × 1.82)

= 3217.2 + 820.82 

= 4038 J/mol

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