# AHL: Gibbs free energy and reaction spontaneity

> IB Chemistry HL · IB Chemistry HL R1: What drives chemical reactions?
> Source: https://www.owlsprep.com/study/ib-chemistry-hl-u4-ahl-gibbs-free-energy-and/

This subtopic explains how enthalpy and entropy combine to give Gibbs free energy, and how to use ΔG to predict whether a reaction is spontaneous at any given temperature. You will also connect ΔG° to the equilibrium constant.

**Prerequisites:** [Enthalpy change and Hess's law](https://www.owlsprep.com/study/ib-chemistry-hl-r1-enthalpy-and-hess-law/); [Entropy and entropy changes](https://www.owlsprep.com/study/ib-chemistry-hl-r1-entropy-and-entropy-changes/)

## Learning objectives

- Relate Gibbs free energy change to reaction spontaneity at constant T and P
- Calculate ΔG from ΔH, ΔS and standard Gibbs free energies of formation
- Connect standard Gibbs free energy change to the equilibrium constant K

## Gibbs Free Energy and the Gibbs Equation

**Gibbs free energy (G)** — A thermodynamic potential that combines enthalpy (total energy change) and entropy (disorder change) to predict reaction spontaneity at constant temperature and pressure.

*Notation:* ΔG = change in Gibbs free energy

*Example:* Melting of ice at 1 atm and 25°C has ΔG < 0, so it is spontaneous.

At constant temperature and pressure, the fundamental relationship between ΔG, ΔH and ΔS is:

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

> **tip**
>
> Always check units before calculation: ΔH is typically in kJ mol⁻¹, while ΔS is usually in J K⁻¹ mol⁻¹. Convert ΔS to kJ by dividing by 1000 to get consistent units.

**Worked example:** Calculate ΔG for a reaction with ΔH = -100 kJ mol⁻¹ and ΔS = -200 J K⁻¹ mol⁻¹ at 300 K, then state if the reaction is spontaneous.

1. Convert ΔS to kJ to match ΔH units:
2. $$\Delta S = -200 \text{ J K}^{-1} \text{mol}^{-1} = -0.2 \text{ kJ K}^{-1} \text{mol}^{-1}$$
3. Substitute values into the Gibbs equation:
4. $$\Delta G = (-100) - (300 \times -0.2) = -100 + 60 = -40 \text{ kJ mol}^{-1}$$
5. Since ΔG = -40 kJ mol⁻¹ < 0, the reaction is spontaneous at 300 K.

## Predicting Spontaneity from ΔG Sign

The sign of ΔG directly indicates spontaneity at constant temperature and pressure:

- If $ΔG < 0$: reaction is **spontaneous** (proceeds forward as written)
- If $ΔG = 0$: reaction is at **dynamic equilibrium** (no net change)
- If $ΔG > 0$: reaction is **non-spontaneous** as written (reverse is spontaneous)

Different combinations of ΔH and ΔS signs lead to different temperature dependence of spontaneity, summarized in the table below:

| ΔH sign | ΔS sign | Spontaneous when? | Low T ΔG sign | High T ΔG sign |
| --- | --- | --- | --- | --- |
| - | + | All temperatures | - | - |
| + | - | No temperatures | + | + |
| - | - | Low temperatures | - | + |
| + | + | High temperatures | + | - |

**Worked example:** For the Haber process $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$, ΔH = -92 kJ mol⁻¹ and ΔS = -199 J K⁻¹ mol⁻¹. Determine if the reaction is spontaneous at 25°C (298 K) and 500°C (773 K).

1. Convert ΔS to kJ: $ΔS = -0.199$ kJ K⁻¹ mol⁻¹
2. Calculate ΔG at 298 K:
3. $$\Delta G = -92 - (298 \times -0.199) = -32.7 \text{ kJ mol}^{-1}$$
4. ΔG < 0, so spontaneous at 298 K (25°C)
5. Calculate ΔG at 773 K:
6. $$\Delta G = -92 - (773 \times -0.199) = +61.8 \text{ kJ mol}^{-1}$$
7. ΔG > 0, so non-spontaneous at 773 K (500°C)

> **Exam tip**
>
> Always confirm your conclusion matches the ΔH/ΔS combination rule: for negative ΔH and negative ΔS, spontaneity only at low temperature, which matches our result here.

## Calculating Standard ΔG from ΔGf°

**Standard Gibbs free energy of formation (ΔGf°)** — The Gibbs free energy change for forming 1 mole of a compound from its elements in their standard states. ΔGf° = 0 for elements in their standard state.

Just like standard enthalpy change, we can calculate standard ΔG for a reaction using ΔGf° values:

$$\Delta G^\circ = \sum \Delta G^\circ_f (\text{products}) - \sum \Delta G^\circ_f (\text{reactants})$$

**Worked example:** Calculate ΔG° for combustion of methane: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(l)$. Given ΔGf° values: $CH_4(g) = -50.7$ kJ mol⁻¹, $CO_2(g) = -394.4$ kJ mol⁻¹, $H_2O(l) = -237.1$ kJ mol⁻¹, $O_2(g) = 0$ kJ mol⁻¹.

1. Calculate sum of ΔGf° for products:
2. $$\sum \Delta G^\circ_f (products) = -394.4 + 2(-237.1) = -868.6 \text{ kJ mol}^{-1}$$
3. Calculate sum of ΔGf° for reactants:
4. $$\sum \Delta G^\circ_f (reactants) = -50.7 + 2(0) = -50.7 \text{ kJ mol}^{-1}$$
5. Calculate ΔG°:
6. $$\Delta G^\circ = -868.6 - (-50.7) = -817.9 \text{ kJ mol}^{-1}$$
7. ΔG° is negative, so the reaction is spontaneous under standard conditions.

## ΔG° and the Equilibrium Constant K

Standard Gibbs free energy change is directly related to the equilibrium constant, which describes how far the reaction proceeds at equilibrium:

$$\Delta G^\circ = -RT \ln K$$

- R = 8.31 J K⁻¹ mol⁻¹ (gas constant), T = absolute temperature in Kelvin
- If $ΔG° < 0$, $K > 1$: products are favored at equilibrium
- If $ΔG° = 0$, $K = 1$: products and reactants are equally favored
- If $ΔG° > 0$, $K < 1$: reactants are favored at equilibrium

**Worked example:** Given ΔG° = -32.7 kJ mol⁻¹ for the Haber process at 298 K, calculate K and comment on the equilibrium position.

1. Convert ΔG° to J to match R units: ΔG° = -32700 J mol⁻¹
2. Rearrange to solve for ln K:
3. $$\ln K = -\frac{\Delta G^\circ}{RT} = -\frac{(-32700)}{(8.31)(298)} \approx 13.2$$
4. Calculate K by exponentiating both sides:
5. $$K = e^{13.2} \approx 5.4 \times 10^5$$
6. $K >> 1$, so products are heavily favored at equilibrium at 298 K.

> **Exam tip**
>
> Remember that ΔG° = -RT ln K applies to standard Gibbs change, not the actual ΔG. At equilibrium, actual ΔG = 0, not ΔG°.

## Common pitfalls

- **Wrong:** Forgetting to convert ΔS from J K⁻¹ to kJ K⁻¹ when calculating ΔG
  - Why it fails: ΔH is almost always given in kJ, so mismatched units produce a ΔG with the wrong magnitude, and often the wrong sign
  - Correct: Always check units before substitution: divide ΔS by 1000 to convert from J to kJ
- **Wrong:** Claiming a non-spontaneous reaction (ΔG > 0) can never occur
  - Why it fails: ΔG only predicts spontaneity under the given conditions. Non-spontaneous reactions can be driven by external energy input
  - Correct: Only state the reaction is non-spontaneous under the specified temperature and pressure conditions
- **Wrong:** Confusing ΔG and ΔG° when relating to the equilibrium constant
  - Why it fails: Students often mix up the meaning of the two values and incorrectly state ΔG = -RT ln K
  - Correct: Remember: at equilibrium, $\Delta G = 0$, and $\Delta G^\circ = -RT \ln K$
- **Wrong:** Assuming all exothermic reactions (ΔH < 0) are always spontaneous
  - Why it fails: Spontaneity depends on both ΔH, ΔS and temperature. A very negative ΔS can make even an exothermic reaction non-spontaneous at high temperature
  - Correct: Always calculate $\Delta G = \Delta H - T\Delta S$ to confirm spontaneity, never rely on ΔH alone

## Cheatsheet

| ΔH/ΔS Combination | Spontaneity | ΔG Sign | K Relationship |
| --- | --- | --- | --- |
| ΔH -, ΔS + | All temperatures | Always - | Always K > 1 |
| ΔH +, ΔS - | No temperatures | Always + | Always K < 1 |
| ΔH -, ΔS - | Only low T | Low T: -, High T: + | K > 1 at low T |
| ΔH +, ΔS + | Only high T | Low T: +, High T: - | K > 1 at high T |
| At equilibrium | No net change | ΔG = 0 | $\Delta G^\circ = -RT \ln K$ |
| Standard ΔG calculation |  | $\Delta G^\circ = \sum \Delta G_f^\circ (prod) - \sum \Delta G_f^\circ (react)$ |  |

## What's next

Mastering Gibbs free energy and spontaneity is the foundation of all thermodynamic reasoning in chemistry, connecting thermodynamics to equilibrium, electrochemistry, acid-base chemistry and solubility. This concept explains why some reactions proceed spontaneously while others require energy input, and how temperature changes affect the direction of reaction. It is a heavily tested topic in both IB Chemistry Paper 1 (multiple choice) and Paper 2 (extended calculation questions), so regular practice of calculations is essential. Below are related topics to study next to build on this knowledge.

- [R2: How much / how fast / how far?](https://www.owlsprep.com/study/ib-chemistry-hl-u5-overview/)
- [Stoichiometric relationships](https://www.owlsprep.com/study/ib-chemistry-hl-u5-stoichiometric-relationships/)
- [Limiting and excess reactants](https://www.owlsprep.com/study/ib-chemistry-hl-u5-limiting-and-excess-reactants/)

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