AHL: Gibbs free energy and reaction spontaneity

IB Chemistry HL· AHL R1.2 Gibbs free energy· 25 min read

1. Gibbs Free Energy and the Gibbs Equation★★☆☆☆⏱ 8 min

📘 Definition

Gibbs free energy (G)

ΔG = change in Gibbs free energy

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

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

📐 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.

2. Predicting Spontaneity from ΔG Sign★★★☆☆⏱ 7 min

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

If $Δ G < 0$ : reaction is spontaneous (proceeds forward as written)
If $Δ G = 0$ : reaction is at dynamic equilibrium (no net change)
If $Δ G > 0$ : 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 $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ , 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: $Δ S = - 0.199$ 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)

3. Calculating Standard ΔG from ΔGf°★★★☆☆⏱ 5 min

📘 Definition

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: $C H_{4} (g) + 2 O_{2} (g) \to C O_{2} (g) + 2 H_{2} O (l)$ . Given ΔGf° values: $C H_{4} (g) = - 50.7$ kJ mol⁻¹, $C O_{2} (g) = - 394.4$ kJ mol⁻¹, $H_{2} O (l) = - 237.1$ 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.

4. ΔG° and the Equilibrium Constant K★★★★☆⏱ 5 min

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 $Δ G ° < 0$ , $K > 1$ : products are favored at equilibrium
If $Δ G ° = 0$ , $K = 1$ : products and reactants are equally favored
If $Δ G ° > 0$ , $K < 1$ : 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.

5. Common Pitfalls

Wrong move:

Forgetting to convert ΔS from J K⁻¹ to kJ K⁻¹ when calculating ΔG

Why:

ΔH is almost always given in kJ, so mismatched units produce a ΔG with the wrong magnitude, and often the wrong sign

Correct move:

Always check units before substitution: divide ΔS by 1000 to convert from J to kJ

Wrong move:

Claiming a non-spontaneous reaction (ΔG > 0) can never occur

Why:

ΔG only predicts spontaneity under the given conditions. Non-spontaneous reactions can be driven by external energy input

Correct move:

Only state the reaction is non-spontaneous under the specified temperature and pressure conditions

Wrong move:

Confusing ΔG and ΔG° when relating to the equilibrium constant

Why:

Students often mix up the meaning of the two values and incorrectly state ΔG = -RT ln K

Correct move:

Remember: at equilibrium, $Δ G = 0$ , and $Δ G^{\circ} = - R T ln K$

Wrong move:

Assuming all exothermic reactions (ΔH < 0) are always spontaneous

Why:

Spontaneity depends on both ΔH, ΔS and temperature. A very negative ΔS can make even an exothermic reaction non-spontaneous at high temperature

Correct move:

Always calculate $Δ G = Δ H - T Δ S$ to confirm spontaneity, never rely on ΔH alone

6. Quick Reference 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	$Δ G^{\circ} = - R T ln K$
Standard ΔG calculation		$Δ G^{\circ} = \sum Δ G_{f}^{\circ} (p r o d) - \sum Δ G_{f}^{\circ} (r e a c t)$

7. Frequently Asked

How do I know if a reaction is spontaneous at all temperatures?

A reaction is spontaneous at all temperatures if $ ΔH < 0 $(e x o t h er mi c) an d$ ΔS > 0 $(in cr e a se in e n t r o p y) . T hi s m e an s$ ΔG = ΔH - TΔS $w i l l a l w a y s b e n e g a t i v e, r e g a r d l esso f t h e v a l u eo f$ T$.

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
ΔG calculation from ΔH and ΔS
2023 · 1
Identify spontaneous reaction conditions
2024 · 2
Relate ΔG° to equilibrium constant K

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.