# Gibbs free energy

> Chemistry · CIE A-Level 9701
> Source: https://www.owlsprep.com/study/cie-9701-u17-gibbs-free-energy/

This module covers the definition of Gibbs free energy, the core Gibbs equation, the link between ΔG and reaction spontaneity, and common calculations of ΔG for chemical reactions.

**Prerequisites:** [Enthalpy and enthalpy change](https://www.owlsprep.com/study/cie-9701-u17-enthalpy-change/); [Entropy and entropy change](https://www.owlsprep.com/study/cie-9701-u17-entropy/)

## Learning objectives

- Define Gibbs free energy change (ΔG) and standard Gibbs free energy of formation
- Use the Gibbs equation ΔG = ΔH - TΔS to calculate ΔG
- Predict reaction spontaneity from the sign of ΔG
- Calculate ΔG⊖ from standard Gibbs free energy of formation values
- Relate ΔG⊖ to the equilibrium constant K

## Definition and the Core Gibbs Equation

**Gibbs Free Energy Change** — A thermodynamic quantity that combines enthalpy change (ΔH) and entropy change (ΔS) to determine reaction spontaneity at constant temperature and pressure.

*Notation:* \(\Delta G\)

*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.

## ΔG and Reaction Spontaneity

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)

> **info**
>
> ΔG only describes thermodynamic feasibility, it does *not* tell you anything about reaction rate. A feasible reaction can still take years to proceed if it has a very high activation energy.

**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.

## Calculating ΔG⊖ from Formation Values

**Standard Gibbs Free Energy of Formation** — 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.

*Notation:* \(\Delta G^\ominus_f\)

*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. $$\Delta G^\ominus = -1262 - 68 = -1330 \text{ kJ mol}^{-1}$$

## ΔG⊖ and the Equilibrium Constant

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.

## Common pitfalls

- **Wrong:** Forgetting to convert ΔS from J to kJ before substituting into the Gibbs equation
  - Why it fails: ΔH is almost always given in kJ mol⁻¹, so leaving ΔS in J gives a ΔG that is 1000× the correct value
  - Correct: Always check units first, convert ΔS to kJ K⁻¹ mol⁻¹ to match ΔH's units
- **Wrong:** Assuming that a negative ΔH always means the reaction is spontaneous
  - Why it fails: 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: Always use the full Gibbs equation to determine spontaneity, never rely on ΔH alone
- **Wrong:** Confusing spontaneity (ΔG sign) with reaction rate
  - Why it fails: Students often assume a negative ΔG means the reaction will happen quickly
  - Correct: Remember ΔG only describes thermodynamic feasibility, reaction rate depends on activation energy, not ΔG
- **Wrong:** Using ΔG⊖ to predict spontaneity for non-standard concentration conditions
  - Why it fails: 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: Only use ΔG⊖ for spontaneity under standard conditions, or when relating to the equilibrium constant

## 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 |  | \(\Delta G^\ominus > 0 \implies K < 1\) |

## 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.

- [Entropy and entropy change](https://www.owlsprep.com/study/cie-9701-u17-entropy/)
- [Further chemical equilibria](https://www.owlsprep.com/study/cie-9701-u18-overview/)
- [Acid-base equilibria](https://www.owlsprep.com/study/cie-9701-u18-acid-base-equilibria/)

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