Gibbs Free Energy (AQA A-Level Chemistry): Revision Notes

5.1.7 Gibbs Free Energy

What is Gibbs Free Energy?

The Gibbs free energy change ($\Delta G$) combines the effects of enthalpy ($\Delta H$) and entropy ($\Delta S$) to predict the feasibility of a chemical reaction.

This is determined by the equation:

$$\Delta G = \Delta H - T\Delta S$$

where:

• $\Delta G$: Gibbs free energy change (kJ mol$⁻¹$)
• $\Delta H$: Enthalpy change of the reaction (kJ mol$⁻¹$)
• $T$: Temperature (in Kelvin, K)
• $\Delta S$: Entropy change of the reaction (J K$⁻¹$ mol$⁻¹$)

For a reaction to be feasible, the value of $\Delta G$ must be zero or negative.

This means that the reaction is either spontaneous ($\Delta G < 0$) or at equilibrium ($\Delta G = 0$).

Balancing Entropy and Enthalpy

The feasibility of a reaction depends on both:

• Enthalpy ($\Delta H$): Reactions that release energy ($\Delta H < 0$) favour feasibility.
• Entropy ($\Delta S$): Reactions that increase disorder ($\Delta S > 0$) also favour feasibility.

The relationship between these two factors, along with temperature, dictates whether $\Delta G$ will be negative or positive.

Calculating Gibbs Free Energy

To calculate $\Delta G$, use the given values for $\Delta H$, $T$, and $\Delta S$ in the equation:

$$\Delta G = \Delta H - T\Delta S$$

Remember to convert $\Delta S$ into kJ (if it's in J) by dividing by 1000 to match units with $\Delta H$

Calculating $ΔS$ for vaporisation of Water

Determining Feasibility at Different Temperatures

By rearranging the Gibbs free energy equation, you can calculate the temperature at which $\Delta G = 0$, which is the threshold for reaction feasibility:

$$T = \frac{\Delta H}{\Delta S}$$

Plotting $\Delta G$ Versus Temperature

Graphing $\Delta G$ against $T$ can help visualise how temperature impacts reaction feasibility:

• A negative slope in the graph of $\Delta G$ versus $T$ indicates that the reaction may become feasible as temperature increases.
• By analysing the intercept, you can determine the temperature at which $\Delta G = 0$, confirming the temperature at which the reaction becomes feasible.