Introduction to Gibbs Free Energy Revision Notes | A Level AQA Chemistry

1. Why Enthalpy Is Not the Whole Story

You already know that exothermic reactions release energy to the surroundings — and for a long time, chemists assumed that was the driving force behind all spontaneous change. If a reaction releases heat, it must be favourable, right?

Here is the problem: some reactions that are clearly endothermic — absorbing heat from the surroundings — still happen spontaneously. Dissolving ammonium nitrate in water is a classic example. The solution turns cold, yet the process occurs without any input of energy. Enthalpy alone cannot explain this.

There must be a second factor at work. That factor is entropy — and understanding it is the key to predicting whether any reaction will actually occur.

The AQA specification is explicit on this point: $\Delta H$, whilst important, is not sufficient to explain feasible change. You need to account for entropy as well — and that is exactly what Gibbs free energy does.

2. Entropy: Measuring Disorder

Picture a box of gas molecules. When you open a valve connecting it to an empty container, the gas spreads out immediately — it never spontaneously contracts back into the original box. The system moves from a more ordered state to a more disordered one. That tendency towards disorder is what entropy quantifies.

The key insight is that nature favours increasing disorder. When a reaction produces more disorder in the system, that tendency supports the reaction occurring — even if the enthalpy change is unfavourable.

You can predict the sign of $\Delta S$ by thinking about the physical state and number of particles involved. The general rules are straightforward once you see the pattern.

Notice that gases have far more entropy than liquids or solids, because gas molecules move freely and occupy a much larger volume of space. Any reaction that produces gas from a solid or liquid will almost always show a significant increase in entropy.

3. Gibbs Free Energy: Combining Enthalpy and Entropy

Now that you have both pieces — enthalpy and entropy — you need a way to combine them into a single quantity that tells you whether a reaction is feasible. That quantity is the Gibbs free energy change, $\Delta G$.

Look at the equation carefully. The $T\Delta S$ term is subtracted from $\Delta H$. This means that at higher temperatures, the entropy term has a greater influence on $\Delta G$. Temperature acts as a weighting factor for entropy — the hotter the system, the more entropy matters.

The feasibility rule is clean and unambiguous: a reaction is feasible when $\Delta G$ is zero or negative ($\Delta G \leq 0$). When $\Delta G$ is positive, the reaction is not feasible under those conditions.

4. Calculating $\Delta G$: Worked Examples

The equation is straightforward to apply, but unit consistency is where most marks are lost. Work through the following example carefully, paying attention to the unit conversion at the start.

Worked example: Calculating $\Delta G$ and Determining Feasibility

For the reaction $\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$, $\Delta H = +178$ kJ mol$^{-1}$ and $\Delta S = +165$ J mol$^{-1}$ K$^{-1}$. Calculate $\Delta G$ at 298 K and determine whether the reaction is feasible at this temperature.

1. $\Delta S = \frac{+165}{1000} = +0.165$ kJ mol$^{-1}$ K$^{-1}$ — The $\Delta H$ value is in kJ mol$^{-1}$, so $\Delta S$ must also be in kJ mol$^{-1}$ K$^{-1}$ for the units to be consistent. This is the most common source of error.
2. $T = 298$ K — The temperature is already given in Kelvin, so no conversion is needed here.
3. $T\Delta S = 298 \times 0.165 = 49.17$ kJ mol$^{-1}$ — Multiply temperature (K) by entropy change (kJ mol$^{-1}$ K$^{-1}$). The Kelvin units cancel, leaving kJ mol$^{-1}$.
4. $\Delta G = +178 - 49.17 = +128.83$ kJ mol$^{-1}$ — The enthalpy term dominates at this temperature because $T$ is relatively low.
5. $\Delta G = +128.83$ kJ mol$^{-1}$ (positive) — Because $\Delta G > 0$, the reaction is not feasible at 298 K. The endothermic enthalpy change is not sufficiently offset by the entropy increase at this temperature.

Answer: $\Delta G = +128.83$ kJ mol$^{-1}$. The reaction is not feasible at 298 K because $\Delta G > 0$.

Notice that even though $\Delta S$ is positive (which favours feasibility), the large positive $\Delta H$ dominates at 298 K. This raises an important question: is there a temperature at which this reaction becomes feasible?

Worked example: Finding the Crossover Temperature

Using the same data ($\Delta H = +178$ kJ mol$^{-1}$, $\Delta S = +0.165$ kJ mol$^{-1}$ K$^{-1}$), calculate the minimum temperature at which the decomposition of calcium carbonate becomes feasible.

1. $0 = \Delta H - T\Delta S$ — At the crossover temperature, $\Delta G$ is exactly zero — the reaction is on the boundary between feasible and non-feasible.
2. $T = \frac{\Delta H}{\Delta S}$ — Rearranging $0 = \Delta H - T\Delta S$ gives $T\Delta S = \Delta H$, so $T = \frac{\Delta H}{\Delta S}$.
3. $T = \frac{+178}{+0.165}$ — Both values must be in consistent units: $\Delta H$ in kJ mol$^{-1}$ and $\Delta S$ in kJ mol$^{-1}$ K$^{-1}$.
4. $T = 1078.8$ K — Dividing 178 by 0.165: $178 \div 0.165 = 1078.8$ K. Above this temperature, $\Delta G$ becomes negative and the reaction is feasible.
5. $T_{\text{crossover}} \approx 1079$ K — The reaction becomes feasible above approximately 1079 K. This is consistent with the industrial use of high temperatures in lime kilns to decompose limestone.

Answer: The crossover temperature is approximately 1079 K. Above this temperature, $\Delta G < 0$ and the reaction is feasible.

5. The Four Combinations of $\Delta H$ and $\Delta S$

The real power of the Gibbs equation comes from analysing how different combinations of $\Delta H$ and $\Delta S$ affect feasibility across a range of temperatures. There are four possible combinations, and each tells a different story.

The two most interesting cases are the middle two rows — where feasibility depends on temperature. When $\Delta H$ and $\Delta S$ have the same sign, there is always a crossover temperature at which $\Delta G$ switches from positive to negative (or vice versa).

The Haber process is a good example of the second row: $\Delta H$ is negative and $\Delta S$ is negative (fewer moles of gas are produced). At low temperatures, $\Delta G$ is negative and the reaction is feasible — but at high temperatures, the $-T\Delta S$ term (which is positive, since $\Delta S$ is negative) eventually makes $\Delta G$ positive. This is one reason why the Haber process uses a compromise temperature rather than the lowest possible one.

6. Exam Technique and Common Pitfalls

AQA questions on Gibbs free energy tend to follow predictable patterns. Knowing what the examiner is looking for — and where marks are most commonly dropped — puts you in a strong position.

Calculation questions almost always award one mark for the correct method (substituting into $\Delta G = \Delta H - T\Delta S$), one mark for correct handling of units (especially the J to kJ conversion for $\Delta S$), and one mark for the correct numerical answer with a feasibility conclusion. Missing the unit conversion costs you the accuracy mark and the conclusion mark in one go.

For 'explain' questions about feasibility, the mark scheme expects you to link the sign of $\Delta G$ explicitly to feasibility. Saying 'the reaction is exothermic' is not sufficient — you must state that $\Delta G$ is negative (or zero) and therefore the reaction is feasible. Use the word 'feasible' in your answer.

One final point worth reinforcing: a negative $\Delta G$ tells you a reaction is thermodynamically feasible, but it says nothing about the rate at which it occurs. A reaction can have a very negative $\Delta G$ and still proceed imperceptibly slowly if the activation energy is high. Thermodynamic feasibility and kinetic feasibility are separate considerations — do not conflate them in an exam answer.