# Gibbs Free Energy Formula

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:

Î”G = Î”H – TÎ”Swhere,

- Î”G = Gibbs Free Energy,
- Î”H = Enthalpy Change,
- T = Temperature (Kelvin),
- Î”S = Entropy change.

### Derivation of Gibbs Free Energy Formula

Since,

G = H – TS…..(1)We have the enthalpy,

H = U + PVPut value of H in Equation 1, we get

G = U + PV – TSGibbs free energy changes can be written as

Î”G = Î”U + Î”(PV) – Î”(TS)or

Î”G = Î”U + Î”PV + PÎ”V – Î”TS – TÎ”SÎ”T=0 and Î”P=0 if the change is done at the same temperature and pressure.

Î”G = Î”U + PÎ”V – TÎ”SHere,

Î”H = Î”U + PÎ”Vor

G = Î”H – TÎ”S

**Gibbs Free Energy Formula Example –** Gibbs free energy is negative in spontaneous processes such as ice melting at ambient temperature, ammonia (NH_{3}) synthesis from nitrogen (N_{2}), and hydrogen gas (H_{2}).

### Gibbs Free Energy Formula in Spontaneity and Electrochemistry

The Gibbs-Helmholtz equation states that,

Î”G = Î”H – TÎ”SÎ”G must be negative (Î”G < 0) for the response to be spontaneous. The following conditions can cause Î”G to be negative:

- TÎ”S is positive whereas Î”H is negative.
- TÎ”S and Î”H both have a negative value. TÎ”S opposes the spontaneous process in this example, while Î”H promotes it. As a result, if Î”H > TÎ”S is true, the process can be spontaneous.
- Î”H and TÎ”S are both affirmative. TÎ”S prefers the spontaneous reaction in this scenario, while Î”H opposes it. As a result, if Î”H < TÎ”S is present, the procedure can be spontaneous.
The process does not take place if Î”G is 0, or the system is in equilibrium.

### Sample Questions

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

**Answer:**

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 Î”G

^{o}=-2.303RTlogK_{C}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 25Â°C is 7.12 KJmol ^{-1} and the entropy of fusion is 2.54 JK^{-1}mol^{-1}. Calculate the free energy change and determine if ice melting at this temperature is spontaneous or not.**

**Answer:**

Given : Î”H = 7.12 KJmol

^{-1 }= 7120 Jmol^{-1}, Î”S = 3.11 JK^{-1}mol^{-1}, T = 45Â°C = 45 + 273 = 318 KSince,

Î”G = Î”H – TÎ”S

Î”G = 7120 – (318 Ã— 3.11)

= 7120 – 988.98

= 6131 Jmol^{-1}Because Î”G is positive, ice melting is not a spontaneous in nature.

**Question 4: Calculate Î”G ^{o} for the following reaction:**

**Zn(s) + Cu ^{2+}(aq) â†’ Zn^{2+}(aq) + Cu(s)**

**Cu ^{2+}(aq) has standard free energy of 56.19 KJ/mol, while Zn^{2+}(aq) has standard free energy of -137.1 KJ/mol.**

**Answer**:

Î”G

^{o}= âˆ‘G^{o}_{(product)}– âˆ‘G^{o}_{(reactants)}âˆ´ Î”G

^{o}= (-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.**

**Answer:**

Given : Î”H = 3.12 KJmol

^{-1}= 3120 Jmol^{-1}, Î”S = 2.55 JK^{-1}mol^{-1}, T = 32Â°C = 32 + 273 = 305 KSince,

Î”G = Î”H – TÎ”S

âˆ´ Î”G = 3120 – (305 Ã— 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.**

**Answer:**

Given : Î”G = 3217.2 Jmol

^{-1}, Î”S = 1.82 JK^{-1}mol^{-1}, T = 451 KSince,

Î”G = Î”H – TÎ”S

âˆ´ Î”H = Î”G + TÎ”S

= 3217.2 + (451 Ã— 1.82)

= 3217.2 + 820.82

= 4038 J/mol

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