Chapter 18. Chemical Thermodynamics

# Gibbs Free Energy

Jessie A. Key

### Learning Objectives

- To gain an understanding of Gibbs free energy.
- To understand the relationship between the sign of Gibbs free energy change and the spontaneity of a process.
- To be able to determine Gibbs free energy using standard free energies of formation.

Figure 18.4. J. Willard Gibbs

A portrait of J. Willard Gibbs^{[1]}

J. Willard Gibbs (1839-1903) proposed a single state function to determine spontaneity:

*G = H – TS*

where *H* is the enthalpy of the system, *S* is the entropy of the system, and *G* is **Gibbs free energy**.

The change in Gibbs free energy, Δ*G*, is the maximum amount of free energy available to do useful work. For an isothermal process, it can be expressed as:

Δ*G = *Δ*H – T*Δ*S *or at standard conditions:* *Δ*G*⁰* = *Δ*H*⁰ *–** T*Δ*S*⁰

This single term, Gibbs free energy (*G*), allows us to avoid calculating the entropy of the surroundings. It is really just a simplification of our previous method of estimating spontaneity:

Δ*S*_{universe} = Δ*S*_{sys} + Δ*S*_{surr}

= + ( )

Multiply both sides of the equation by –*T*:

–*T*Δ*S*_{universe}= + ( )

–*T*Δ*S*_{universe} = Δ*H*_{sys} – *TΔS*_{sys}

Therefore Δ*G = *–*T*Δ*S*_{universe}

As a result of this relationship, the sign of Gibbs free energy provides information on the spontaneity of a given reaction:

If Δ*G *> 0, the reaction is nonspontaneous in the direction written.

If Δ*G *= 0, the reaction is in a state of equilibrium.

If Δ*G *< 0, the reaction is spontaneous in the direction written.

The significance of the sign of a change in Gibbs free energy parallels the relationship of terms from the equilibrium chapter: the reaction quotient, *Q*, and the equilibrium constant, *K*.

If *Q *> *K*, the reaction is nonspontaneous in the direction written.

If *Q *=* K*, the reaction is in a state of equilibrium.

If *Q *< *K*, the reaction is spontaneous in the direction written.

**Example 4**

Calculate Δ*G*⁰ for a reaction where Δ*H*⁰ is equal to 36.2 kJ and Δ*S*⁰ is equal to 123 J/K at 298 K. Is this a spontaneous reaction?

Solution

Δ*G*⁰* = *Δ*H*⁰ *– T*Δ*S*⁰

Δ*G*⁰* = *36.2 kJ* –* (298 K x 123 J/K)

Δ*G*⁰* = *-0.4 kJ

Therefore the reaction is spontaneous because Δ*G*⁰ is negative.

## Determining Δ*G*⁰ from Standard Free Energy of Formation

The standard Gibbs free energy change, Δ*G*⁰, for a reaction can be calculated from the standard free energies of formation, Δ*G*⁰* _{f} *.

Δ*G*⁰* _{f} *= ∑

*n*Δ

*G*⁰

*(products) – ∑*

_{f}*m*Δ

*G*⁰

*(reactants)*

_{f}where *n* and *m* are the coefficients in the balanced chemical equation of the reaction.

Standard free energies of formation values are listed in the appendix, “Standard Thermodynamic Quantities for Chemical Substances at 25°C.”

**Example 5**

Calculate the standard free energy change for the following reaction, using standard free energies of formation:

5 C(s) + 2 SO_{2}(g) → CS_{2}(g) + 4 CO(g)

Is this a spontaneous reaction?

Solution

Δ*G*⁰= ∑*n*Δ*G*⁰* _{f}*(products) – ∑

*m*Δ

*G*⁰

*(reactants)*

_{f}Δ*G*⁰ = [(4 x −137.2 kJ/mol) + (67.1 kJ/mol)] – [(5 x 0 kJ/mol) + (2 x −300.1 kJ/mol)]

Δ*G*⁰= (-481.7 kJ/mol) – (-600.2 kJ/mol)

Δ*G*⁰= 118.5 kJ/mol

Δ*G*⁰ has a positive value so this is not a spontaneous process.

### Key Takeaways

- The change in Gibbs free energy (Δ
*G*) is the maximum amount of free energy available to do useful work. - If Δ
*G*> 0, the reaction is nonspontaneous in the direction written. If Δ*G*=0, the reaction is in a state of equilibrium. If Δ*G*< 0, the reaction is spontaneous in the direction written. - The standard Gibbs free energy change,
*G*⁰, for a reaction can be calculated from the standard free energies of formation, Δ*G*⁰._{f}

- Gibbs Josiah Willard/Public Domain ↵