Gibbs free energy change, ΔG

Q: What is the master equation used to calculate the standard Gibbs Free Energy change (ΔG⊖)?

\(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\)

Q: What common unit discrepancy must be resolved before substituting values into the Gibbs equation?

\(\Delta H\) is usually given in \(\text{kJ}\text{ mol}^{-1}\) whereas \(\Delta S\) is given in \(\text{J}\text{ K}^{-1}\text{ mol}^{-1}\); you must convert \(\Delta S\) to \(\text{kJ}\text{ K}^{-1}\text{ mol}^{-1}\) by dividing by 1000 (or convert \(\Delta H\) to Joules).

Q: What does the sign of \(\Delta G\) indicate about the thermodynamic feasibility of a chemical reaction?

A reaction is thermodynamically feasible (spontaneous) if \(\Delta G 0\), and at equilibrium if \(\Delta G = 0\).

Chemistry (9701) · Cambridge International A Level · 5 min read · Updated: 28 Aug 2026

🔬 Gibbs Free Energy Change, \(\Delta G^{\ominus}\): The Ultimate Decision Maker in Chemistry

Welcome to one of the most important concepts in Physical Chemistry! In your AS Level studies, you learned about Enthalpy change (\(\Delta H\))—whether a reaction gives out heat or takes it in. You then learned about Entropy change (\(\Delta S\))—how disordered the system becomes.

But here's the catch: a reaction can be exothermic (\(\Delta H < 0\)) but still not occur, or it can be endothermic but occur spontaneously! Why?

We need one grand master quantity that combines both energy (\(\Delta H\)) and disorder (\(\Delta S\)). This is where the Gibbs Free Energy Change (\(\Delta G\)) comes in. It determines the true feasibility (or spontaneity) of a reaction.

Section 1: The Components of Feasibility (\(\Delta H\) and \(\Delta S\))

Before diving into \(\Delta G\), let's quickly recap the two factors that influence whether a reaction "wants" to happen.

Enthalpy Change (\(\Delta H\))

Enthalpy is all about energy transfer. Systems naturally prefer to lose energy (become more stable).

Exothermic Reaction: \(\Delta H\) is negative (-). This contributes favorably to feasibility.
Endothermic Reaction: \(\Delta H\) is positive (+). This works against feasibility.

Entropy Change (\(\Delta S\))

Entropy measures the dispersal of energy and matter (disorder). Systems naturally tend towards maximum disorder.
Analogy: Imagine your bedroom. It takes energy to keep it tidy (low entropy); it takes no effort for it to become messy (high entropy)!

Increase in disorder: \(\Delta S\) is positive (+). This contributes favorably to feasibility.
Decrease in disorder: \(\Delta S\) is negative (-). This works against feasibility.

We calculate \(\Delta S^{\ominus}\) for a reaction in the same way we calculate \(\Delta H^{\ominus}\):
\(\Delta S^{\ominus} = \Sigma S^{\ominus}\ (\text{products}) - \Sigma S^{\ominus}\ (\text{reactants})\)

Quick Review: For a reaction to be spontaneous at all temperatures, we want \(\Delta H\) to be negative and \(\Delta S\) to be positive. When they conflict, \(\Delta G\) settles the outcome.

Section 2: The Gibbs Equation: Combining Energy and Disorder

Definition and the Master Equation (LO 23.4.1)

The Gibbs Free Energy Change (\(\Delta G\)) indicates the thermodynamic feasibility of a reaction or process under standard conditions.

The key equation that links enthalpy and entropy is the Gibbs Equation:
\(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\)

Where:

\(\Delta G^{\ominus}\) is the Standard Gibbs Free Energy Change (\(\text{J}\,\text{mol}^{-1}\) or \(\text{kJ}\,\text{mol}^{-1}\)).
\(\Delta H^{\ominus}\) is the Standard Enthalpy Change (\(\text{J}\,\text{mol}^{-1}\) or \(\text{kJ}\,\text{mol}^{-1}\)).
\(T\) is the absolute temperature (\(\text{K}\), Kelvin).
\(\Delta S^{\ominus}\) is the Standard Entropy Change (\(\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}\)).

⚠ Accessibility Alert: Units are CRUCIAL! ⚠

This is the number one mistake students make in \(\Delta G\) calculations.
\(\Delta H\) is almost always given in kilojoules (\(\text{kJ}\,\text{mol}^{-1}\)), but \(\Delta S\) is almost always given in joules per Kelvin (\(\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}\)).
You must make them consistent before subtracting!
Memory Aid: Convert \(\Delta S\) into \(\text{kJ}\,\text{K}^{-1}\,\text{mol}^{-1}\) by dividing by \(1000\) OR convert \(\Delta H\) into \(\text{J}\,\text{mol}^{-1}\) by multiplying by \(1000\).

Remember: Temperature (\(T\)) must be in Kelvin, and all energy terms (\(\Delta G\), \(\Delta H\), \(T\Delta S\)) must share the exact same energy units. Standard temperature is \(298\,\text{K}\).

Section 3: Predicting Feasibility using \(\Delta G\) (LO 23.4.3)

The Feasibility Rule

The sign of \(\Delta G\) tells us whether a process is thermodynamically feasible under given conditions:

If \(\Delta G < 0\) (Negative): The process is Feasible (Spontaneous).
If \(\Delta G > 0\) (Positive): The process is Not Feasible (Non-spontaneous). It will only proceed if energy is continuously supplied.
If \(\Delta G = 0\): The system is in a state of dynamic Equilibrium (e.g., during a phase change at its boiling point).

Thermodynamic Feasibility vs. Kinetic Inertness

A reaction with \(\Delta G^{\ominus} < 0\) is thermodynamically feasible, but it might not happen at a measurable rate at room temperature if it has a very high activation energy (\(E_a\)). In such cases, the reaction is said to be kinetically inert or kinetically stable (for example, diamond converting to graphite, or the combustion of methane in air without a spark).

Section 4: Calculations and Temperature Effects (LO 23.4.2 & 23.4.4)

Step-by-Step Calculation Guide (LO 23.4.2)

To calculate \(\Delta G^{\ominus}\) for a reaction, follow these steps:

Find \(\Delta H^{\ominus}\): Obtain the standard enthalpy change of the reaction (usually in \(\text{kJ}\,\text{mol}^{-1}\)).
Find \(\Delta S^{\ominus}\): Calculate or obtain the standard entropy change of the reaction (usually in \(\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}\)).
Check Units: Convert \(\Delta S^{\ominus}\) to \(\text{kJ}\,\text{K}^{-1}\,\text{mol}^{-1}\) (divide by \(1000\)).
Determine \(T\): Identify the temperature in Kelvin (e.g., \(298\,\text{K}\) for standard conditions).
Calculate the \(T\Delta S^{\ominus}\) term: Multiply \(T\) by the converted \(\Delta S^{\ominus}\).
Calculate \(\Delta G^{\ominus}\): Substitute into \(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\).

Example Calculation Check:
If \(\Delta H^{\ominus} = -100\,\text{kJ}\,\text{mol}^{-1}\) and \(\Delta S^{\ominus} = +50\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1}\) at \(T = 300\,\text{K}\):
1. Convert \(\Delta S^{\ominus}\): \(50\,\text{J}\,\text{K}^{-1}\,\text{mol}^{-1} = 0.050\,\text{kJ}\,\text{K}^{-1}\,\text{mol}^{-1}\)
2. Calculate \(T\Delta S^{\ominus}\): \(300\,\text{K} \times 0.050\,\text{kJ}\,\text{K}^{-1}\,\text{mol}^{-1} = 15\,\text{kJ}\,\text{mol}^{-1}\)
3. Calculate \(\Delta G^{\ominus}\): \(\Delta G^{\ominus} = (-100) - (15) = -115\,\text{kJ}\,\text{mol}^{-1}\)
Result: Since \(\Delta G^{\ominus} < 0\), the reaction is feasible at \(300\,\text{K}\).

Predicting the Effect of Temperature (LO 23.4.4)

The feasibility of a reaction often depends heavily on the temperature \(T\), because \(T\) scales the entropy term (\(T\Delta S\)).

The Four Combinations of \(\Delta H\) and \(\Delta S\)

Case	\(\Delta H\)	\(\Delta S\)	\(\Delta G = \Delta H - T\Delta S\)	Temperature Dependence
1	Negative (-)	Positive (+)	Always Negative (\(\Delta G < 0\))	Feasible at ALL temperatures.
2	Positive (+)	Negative (-)	Always Positive (\(\Delta G > 0\))	Not feasible at ANY temperature.
3	Negative (-)	Negative (-)	Negative at low \(T\), Positive at high \(T\)	Feasible only below critical temperature \(T_{\text{crit}}\).
4	Positive (+)	Positive (+)	Positive at low \(T\), Negative at high \(T\)	Feasible only above critical temperature \(T_{\text{crit}}\).

Calculating the Critical Temperature (\(T_{\text{crit}}\))

Feasibility switches at the transition temperature where \(\Delta G = 0\):
\(0 = \Delta H^{\ominus} - T_{\text{crit}}\Delta S^{\ominus}\)
Rearranging gives:
\(T_{\text{crit}} = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}\)

Note: Remember to use consistent units (e.g., both \(\Delta H^{\ominus}\) and \(\Delta S^{\ominus}\) in Joules) when calculating \(T_{\text{crit}}\)!

Case 3 (\(\Delta H < 0, \Delta S < 0\)): Feasible when \(T < T_{\text{crit}}\).
Case 4 (\(\Delta H > 0, \Delta S > 0\)): Feasible when \(T > T_{\text{crit}}\).

Example: The reason water boils at \(100\,^{\circ}\text{C}\) (\(373\,\text{K}\)) under standard pressure is because \(373\,\text{K}\) is the critical temperature (\(T_{\text{crit}}\)) where \(\Delta G = 0\) for \(\text{H}_2\text{O}(\text{l}) \to \text{H}_2\text{O}(\text{g})\). Above \(373\,\text{K}\), \(\Delta G\) is negative and vaporization becomes feasible.

KEY TAKEAWAY: Gibbs Free Energy combines enthalpy and entropy into one single criterion for feasibility. If \(\Delta G < 0\), the process is thermodynamically feasible. Temperature determines feasibility when \(\Delta H\) and \(\Delta S\) have the same sign.

References

Cambridge International AS & A Level Chemistry 9701 syllabus for 2025, 2026 and 2027 — Cambridge Assessment International Education
Gibbs free energy change, ΔG – Study Notes & Past Paper Style Questions (Cambridge International A Level Chemistry 9701) — Tutopiya

Quick check

Can you answer these now?

Open each question to check the key ideas from this chapter.

What is the master equation used to calculate the standard Gibbs Free Energy change (ΔG⊖)?

\(\Delta G^{\ominus} = \Delta H^{\ominus} - T\Delta S^{\ominus}\)

What common unit discrepancy must be resolved before substituting values into the Gibbs equation?

\(\Delta H\) is usually given in \(\text{kJ}\text{ mol}^{-1}\) whereas \(\Delta S\) is given in \(\text{J}\text{ K}^{-1}\text{ mol}^{-1}\); you must convert \(\Delta S\) to \(\text{kJ}\text{ K}^{-1}\text{ mol}^{-1}\) by dividing by 1000 (or convert \(\Delta H\) to Joules).

What does the sign of \(\Delta G\) indicate about the thermodynamic feasibility of a chemical reaction?

A reaction is thermodynamically feasible (spontaneous) if \(\Delta G < 0\), not feasible if \(\Delta G > 0\), and at equilibrium if \(\Delta G = 0\).

Why might a reaction with a negative \(\Delta G^{\ominus}\) still fail to occur at room temperature?

The reaction has a high activation energy (\(E_a\)), making it kinetically inert (kinetically stable) despite being thermodynamically feasible.

Under what condition is a reaction feasible at all temperatures?

When the reaction is exothermic (\(\Delta H < 0\)) and disorder increases (\(\Delta S > 0\)).

How does temperature affect feasibility when both \(\Delta H\) and \(\Delta S\) are positive?

The reaction becomes feasible only at high temperatures above the critical temperature (\(T > T_{\text{crit}}\)), where the \(-T\Delta S\) term outweighs \(\Delta H\).

How is the critical temperature (\(T_{\text{crit}}\)) at which a reaction becomes feasible calculated?

\(T_{\text{crit}} = \frac{\Delta H^{\ominus}}{\Delta S^{\ominus}}\) (calculated at \(\Delta G = 0\)).

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