Gibbs free energy

Chemistry· Unit 17: Further chemical energetics· 15 min read

1. Definition and the Core Gibbs Equation★★☆☆☆⏱ 4 min

📘 Definition

Gibbs Free Energy Change

\Delta G

A thermodynamic quantity that combines enthalpy change (ΔH) and entropy change (ΔS) to determine reaction spontaneity at constant temperature and pressure.

Example:

ΔG is the key value used to predict whether a reaction will occur spontaneously.

Delta G = \Delta H - T\Delta S

Standard Gibbs free energy change ((\Delta G^\ominus)) is measured under standard conditions: 1 atm pressure, 1 mol dm⁻³ concentration, usually 298 K. (\Delta G^\ominus) can also be calculated from standard Gibbs free energies of formation ((\Delta G^\ominus_f)) using the formula: (\Delta G^\ominus = \sum\Delta G^\ominus_f(\text{products}) - \sum\Delta G^\ominus_f(\text{reactants})).

📐 Worked Example

Given (\Delta H = -20) kJ mol⁻¹, (\Delta S = -50) J K⁻¹ mol⁻¹, calculate (\Delta G) at 298 K.

1
Convert ΔS to kJ to match the units of ΔH:
2
$\Delta S = \frac{-50}{1000} = -0.050 \text{ kJ K}^{-1} \text{mol}^{-1}$
3
Substitute values into the Gibbs equation:
4
$\Delta G = (-20) - (298 \times -0.050) = -20 + 14.9$
5
Calculate the final result:
6
$\Delta G = -5.1 \text{ kJ mol}^{-1}$

Exam tip:

Always check units of ΔH and ΔS match! Convert between J and kJ if needed, this is the most commonly tested mistake.

2. ΔG and Reaction Spontaneity★★☆☆☆⏱ 5 min

The sign of ΔG directly tells us if a reaction is thermodynamically spontaneous (feasible) at a given temperature:

If (\Delta G < 0): reaction is spontaneous (feasible) in the forward direction
If (\Delta G = 0): reaction is at equilibrium, no net change
If (\Delta G > 0): reaction is non-spontaneous in the forward direction (spontaneous in reverse)

📐 Worked Example

Predict the temperature range where a reaction with (\Delta H = +120) kJ mol⁻¹ and (\Delta S = +400) J K⁻¹ mol⁻¹ is spontaneous.

1
A reaction is spontaneous when (\Delta G < 0). Convert ΔS to kJ:
2
$\Delta S = 0.400 \text{ kJ K}^{-1} \text{mol}^{-1}$
3
Rearrange the inequality to solve for (T):
4
$\Delta H - T\Delta S < 0 \implies T > \frac{\Delta H}{\Delta S}$
5
Substitute values:
6
$T > \frac{120}{0.400} = 300 \text{ K}$
7
Conclusion: The reaction is spontaneous at all temperatures above 300 K.

3. Calculating ΔG⊖ from Formation Values★★★☆☆⏱ 4 min

📘 Definition

Standard Gibbs Free Energy of Formation

\Delta G^\ominus_f

The Gibbs free energy change when 1 mole of a compound is formed from its elements in their standard states. (\Delta G^\ominus_f = 0) for any element in its standard state.

Example:

(\Delta G^\ominus_f(O_2(g)) = 0), (\Delta G^\ominus_f(CO_2(g)) = -394) kJ mol⁻¹

To calculate the standard Gibbs free energy change for a full reaction, you use the same sum of products minus sum of reactants rule used for enthalpy change calculations:

\Delta G^\ominus = \sum n\Delta G^\ominus_f(\text{products}) - \sum m\Delta G^\ominus_f(\text{reactants})

📐 Worked Example

Calculate (\Delta G^\ominus) for: (C_2H_4(g) + 3O_2(g) \rightarrow 2CO_2(g) + 2H_2O(l)). Given: (\Delta G^\ominus_f(C_2H_4(g)) = +68), (\Delta G^\ominus_f(CO_2(g)) = -394), (\Delta G^\ominus_f(H_2O(l)) = -237) kJ mol⁻¹.

1
Write the expression for ΔG⊖, remembering (\Delta G^\ominus_f(O_2(g)) = 0):
2
$\Delta G^\ominus = [2\Delta G^\ominus_f(CO_2) + 2\Delta G^\ominus_f(H_2O)] - [\Delta G^\ominus_f(C_2H_4)]$
3
Substitute the values:
4
$\Delta G^\ominus = [(2 \times -394) + (2 \times -237)] - 68$
5
Calculate the result:
6
$\Delta G^\ominus = -1262 - 68 = -1330 \text{ kJ mol}^{-1}$

4. ΔG⊖ and the Equilibrium Constant★★★☆☆A2 only⏱ 4 min

The standard Gibbs free energy change is directly related to the equilibrium constant (K) by the relationship:

\Delta G^\ominus = -RT \ln K

Where (R = 8.31) J K⁻¹ mol⁻¹, and (T) is absolute temperature. This relationship lets us predict the position of equilibrium from ΔG⊖:

If (\Delta G^\ominus < 0): (K > 1), products are favoured at equilibrium
If (\Delta G^\ominus = 0): (K = 1), equal amounts of products and reactants
If (\Delta G^\ominus > 0): (K < 1), reactants are favoured at equilibrium

📐 Worked Example

Calculate (K) for the combustion of ethene at 298 K, given (\Delta G^\ominus = -1330) kJ mol⁻¹.

1
Convert ΔG⊖ to J to match the units of (R):
2
$\Delta G^\ominus = -1330 \times 1000 = -1330000 \text{ J mol}^{-1}$
3
Rearrange to solve for (\ln K):
4
$\ln K = -\frac{\Delta G^\ominus}{RT} = -\frac{-1330000}{8.31 \times 298} \approx 537$
5
Exponentiate to get (K):
6
$K = e^{537} \approx 10^{233}$
7
This very large value makes sense: combustion of ethene goes almost to completion.

5. Common Pitfalls

Wrong move:

Forgetting to convert ΔS from J to kJ before substituting into the Gibbs equation

Why:

ΔH is almost always given in kJ mol⁻¹, so leaving ΔS in J gives a ΔG that is 1000× the correct value

Correct move:

Always check units first, convert ΔS to kJ K⁻¹ mol⁻¹ to match ΔH's units

Wrong move:

Assuming that a negative ΔH always means the reaction is spontaneous

Why:

ΔG depends on both ΔH and the (TΔS) term. If ΔS is negative enough, even exothermic reactions can be non-spontaneous at high temperatures

Correct move:

Always use the full Gibbs equation to determine spontaneity, never rely on ΔH alone

Wrong move:

Confusing spontaneity (ΔG sign) with reaction rate

Why:

Students often assume a negative ΔG means the reaction will happen quickly

Correct move:

Remember ΔG only describes thermodynamic feasibility, reaction rate depends on activation energy, not ΔG

Wrong move:

Using ΔG⊖ to predict spontaneity for non-standard concentration conditions

Why:

ΔG⊖ is only defined for standard conditions (1 M concentration). ΔG for non-standard conditions is (\Delta G = \Delta G^\ominus + RT \ln Q)

Correct move:

Only use ΔG⊖ for spontaneity under standard conditions, or when relating to the equilibrium constant

6. Quick Reference Cheatsheet

Concept	Formula	Key Note
Core Gibbs equation	(\Delta G = \Delta H - T\Delta S)	Check units match (J/kJ)
ΔG from formation values	(\Delta G^\ominus = \sum \Delta G^\ominus_f(products) - \sum \Delta G^\ominus_f(reactants))	(\Delta G^\ominus_f(element) = 0)
Spontaneous forward		(\Delta G < 0)
At equilibrium		(\Delta G = 0)
Non-spontaneous forward		(\Delta G > 0)
ΔG⊖ and K	(\Delta G^\ominus = -RT \ln K)	R = 8.31 J K⁻¹ mol⁻¹
Products favoured		(\Delta G^\ominus < 0 \implies K > 1)
Reactants favoured		(\Delta G^\ominus > 0 \implies K < 1)

7. Frequently Asked

Does a negative ΔG mean the reaction happens instantly?

No. ΔG only describes thermodynamic feasibility (whether the reaction can occur spontaneously), not reaction rate. A spontaneous reaction can be kinetically very slow (e.g. diamond converting to graphite).

Why do endothermic reactions ever occur spontaneously?

If entropy change (ΔS) is large and positive, the $T Δ S$ term can make $Δ G = Δ H - T Δ S$ negative even when ΔH is positive, especially at high temperatures.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 2
Calculate ΔG from ΔH and ΔS
2023 · 1
Predict spontaneity from ΔG sign
2021 · 4
Relate ΔG⊖ to equilibrium constant

Going deeper

syllabusCIE 9701 2025-2026 SyllabusSee page 42 for thermodynamics outcomes

What's Next

Gibbs free energy is the foundation of chemical thermodynamics for A-level chemistry, linking energetics, equilibrium and redox chemistry together. You will use ΔG to explain why reactions proceed in a given direction, and to calculate equilibrium constants from thermodynamic data, which is a common high-weightage topic in A2 Paper 4. This concept also underpins more advanced topics like electrode potentials, where you will use ΔG to calculate cell potential and predict spontaneous redox reactions. Mastering ΔG calculations and spontaneity rules is critical for high scores in energetics questions.