Gibbs Free Energy Calculations Revision Notes | A Level AQA Chemistry
To calculate ΔG, convert ΔH and ΔS to the same units, then use ΔG = ΔH − TΔS with T in Kelvin. Free AQA notes with 3 worked examples and 13 questions answered.
Updated 5 Sept 2026 · Revise Right Team
Gibbs Free Energy Calculations is part of Thermodynamics (A-level only) in A Level AQA Chemistry.
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What this lesson covers
To calculate ΔG, convert ΔH and ΔS to the same units, then use ΔG = ΔH − TΔS with T in Kelvin. Free AQA notes with 3 worked examples and 13 questions answered.
Calculate entropy changes for reactions using standard absolute entropy values
Apply the relationship $\Delta G = \Delta H - T\Delta S$ to determine how Gibbs free energy varies with temperature
Determine the temperature at which a reaction becomes feasible using $\Delta G = \Delta H - T\Delta S$
Evaluate whether a reaction is spontaneous based on the sign of $\Delta G$
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Use this lesson to stay inside Thermodynamics (A-level only) while connecting the detail back to the wider Physical chemistry topic.
Gibbs Free Energy ($\Delta G$): A thermodynamic quantity that combines enthalpy and entropy to predict whether a reaction is feasible. Calculated using $\Delta G = \Delta H - T\Delta S$. A reaction is spontaneous when $\Delta G < 0$.
Standard Entropy ($S^\circ$): The absolute entropy of a substance under standard conditions (298 K, 1 atm). Measured in J K$^{-1}$ mol$^{-1}$. Used to calculate $\Delta S^\circ_{\text{reaction}} = \Sigma S^\circ_{\text{products}} - \Sigma S^\circ_{\text{reactants}}$.
Feasibility Temperature: The temperature at which $\Delta G = 0$, marking the crossover point between spontaneous and non-spontaneous. Calculated as $T = \frac{\Delta H}{\Delta S}$ when $\Delta H$ and $\Delta S$ have the same sign.
Entropy Change ($\Delta S$): The change in disorder of a system during a reaction. Measured in J K$^{-1}$ mol$^{-1}$. Positive when disorder increases (e.g. solid to gas); negative when disorder decreases.
Revision notes
Read the core explanations from Gibbs Free Energy Calculations before testing yourself from memory.
1. Why Enthalpy Alone Cannot Predict Feasibility
You already know that exothermic reactions release energy — but here is the problem: some endothermic reactions happen spontaneously too. Dissolving ammonium nitrate in water is endothermic, yet it proceeds without any input of energy. Enthalpy alone cannot explain this.
The missing piece is entropy — the measure of disorder in a system. When ammonium nitrate dissolves, the ions spread out into solution, massively increasing disorder. That increase in entropy drives the process forward even though energy is absorbed.
Gibbs free energy, $\Delta G$, combines both enthalpy and entropy into a single quantity that tells you whether a reaction is thermodynamically feasible at a given temperature. This is the central idea of this lesson.
2. Calculating Entropy Changes from Absolute Entropy Values
Unlike enthalpy, entropy has an absolute scale — every substance has a positive standard absolute entropy value, $S^\circ$, measured in J K$^{-1}$ mol$^{-1}$, though the entropy change $\Delta S^\circ$ for a reaction can be negative. You calculate the entropy change for a reaction the same way you use Hess's Law for enthalpy: products minus reactants.
Stoichiometric coefficients matter here. If two moles of a product are formed, you multiply its $S^\circ$ value by two before summing. Forgetting this is one of the most common errors in entropy calculations.
Worked example: Calculating $\Delta S^\circ$ for the Combustion of Methane
Calculate the standard entropy change for the combustion of methane: $\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(l)}$
Standard entropy values: $S^\circ[\text{CH}_4\text{(g)}] = 186$ J K$^{-1}$ mol$^{-1}$; $S^\circ[\text{O}_2\text{(g)}] = 205$ J K$^{-1}$ mol$^{-1}$; $S^\circ[\text{CO}_2\text{(g)}] = 214$ J K$^{-1}$ mol$^{-1}$; $S^\circ[\text{H}_2\text{O(l)}] = 70$ J K$^{-1}$ mol$^{-1}$
$\Sigma S^\circ_{\text{products}} = S^\circ[\text{CO}_2] + 2 \times S^\circ[\text{H}_2\text{O}]$
$= 214 + (2 \times 70)$
$= 214 + 140 = 354$ J K$^{-1}$ mol$^{-1}$ — Multiply each value by its stoichiometric coefficient. CO$_2$ has coefficient 1; H$_2$O has coefficient 2.
$\Sigma S^\circ_{\text{reactants}} = S^\circ[\text{CH}_4] + 2 \times S^\circ[\text{O}_2]$
$= 186 + (2 \times 205)$
$= 186 + 410 = 596$ J K$^{-1}$ mol$^{-1}$ — O$_2$ has coefficient 2, so its entropy is doubled before adding.
$\Delta S^\circ = 354 - 596 = -242$ J K$^{-1}$ mol$^{-1}$ — The negative value makes sense: gases are converted partly to liquid water, reducing disorder.
Reactants: 3 mol gas. Products: 1 mol gas + 2 mol liquid. Fewer gas molecules means less disorder. — A negative $\Delta S$ is expected whenever the number of gas-phase moles decreases across the reaction.
With $\Delta S$ in hand, you now have everything needed to calculate Gibbs free energy. This is where the real power of the topic emerges — because $\Delta G$ links entropy, enthalpy, and temperature into one decisive test of feasibility.
3. The Gibbs Free Energy Equation and What It Tells You
One critical unit trap lurks here. $\Delta H$ is typically given in kJ mol$^{-1}$, while $\Delta S$ is in J K$^{-1}$ mol$^{-1}$. Before substituting into $\Delta G = \Delta H - T\Delta S$, you must convert both quantities to the same unit — either both in J or both in kJ. Mixing units is the single most common error in this calculation.
Think of the equation as a tug-of-war between enthalpy and entropy. The $T\Delta S$ term grows with temperature, so at high enough temperatures, a large positive $T\Delta S$ can overcome a positive $\Delta H$ and make $\Delta G$ negative. Temperature is the lever that shifts the balance.
$\Delta H$
$\Delta S$
Effect on $\Delta G$
Feasibility
Negative (exothermic)
Positive
$\Delta G$ always negative
Feasible at all temperatures
Negative (exothermic)
Negative
$\Delta G$ negative at low $T$, positive at high $T$
Feasible only below a crossover temperature
Positive (endothermic)
Positive
$\Delta G$ positive at low $T$, negative at high $T$
Feasible only above a crossover temperature
Positive (endothermic)
Negative
$\Delta G$ always positive
Never feasible at any temperature
This table is worth memorising. The two interesting cases — where feasibility depends on temperature — are the ones examiners focus on most. Both involve $\Delta H$ and $\Delta S$ having the same sign, which creates a crossover temperature.
4. How $\Delta G$ Varies with Temperature
Because $T$ appears explicitly in $\Delta G = \Delta H - T\Delta S$, Gibbs free energy changes linearly with temperature. If you plotted $\Delta G$ against $T$, you would get a straight line with gradient $-\Delta S$ and intercept $\Delta H$.
Notice that the gradient of each line is $-\Delta S$. A positive $\Delta S$ gives a negative gradient — $\Delta G$ falls as temperature rises. A negative $\Delta S$ gives a positive gradient — $\Delta G$ rises as temperature rises. This is a direct consequence of the $-T\Delta S$ term in the equation.
The crossover point — where the line crosses $\Delta G = 0$ — is the feasibility temperature. Below this temperature, one feasibility condition holds; above it, the other does. Identifying this crossover is a core exam skill.
5. Finding the Temperature at Which a Reaction Becomes Feasible
When $\Delta H$ and $\Delta S$ have the same sign, there is a specific temperature at which $\Delta G$ crosses zero. At this exact temperature, the reaction is at the boundary between feasible and non-feasible. Setting $\Delta G = 0$ in the equation gives you the crossover temperature directly.
After finding the crossover temperature, you must state what happens on either side of it. For a reaction with positive $\Delta H$ and positive $\Delta S$, the reaction is feasible above the crossover temperature. For negative $\Delta H$ and negative $\Delta S$, it is feasible below the crossover temperature. Examiners expect this interpretation — a bare number without a conclusion will not earn full marks.
Worked example: Finding the Feasibility Temperature for the Thermal Decomposition of Calcium Carbonate
Calcium carbonate decomposes according to: $\text{CaCO}_3\text{(s)} \rightarrow \text{CaO(s)} + \text{CO}_2\text{(g)}$
Given: $\Delta H = +178$ kJ mol$^{-1}$ and $\Delta S = +165$ J K$^{-1}$ mol$^{-1}$.
Calculate the minimum temperature at which this reaction becomes feasible.
$\Delta H = +178 \text{ kJ mol}^{-1} = +178{,}000 \text{ J mol}^{-1}$ — Both $\Delta H$ and $\Delta S$ must be in consistent units before dividing. Converting $\Delta H$ to J mol$^{-1}$ matches the units of $\Delta S$ in J K$^{-1}$ mol$^{-1}$.
$\Delta H = +178{,}000$ J mol$^{-1}$ (positive); $\Delta S = +165$ J K$^{-1}$ mol$^{-1}$ (positive) — Both have the same sign, so a crossover temperature exists. The reaction will be feasible above this temperature.
$0 = \Delta H - T\Delta S$
$T = \frac{\Delta H}{\Delta S}$ — At the crossover point, $\Delta G = 0$ exactly. Rearranging gives the feasibility temperature formula.
$T = \frac{178{,}000}{165}$ — Numerator in J mol$^{-1}$, denominator in J K$^{-1}$ mol$^{-1}$. The mol$^{-1}$ cancels, leaving K.
$T = \frac{178{,}000}{165} = 1079$ K (to 4 significant figures) — Verify: $178{,}000 \div 165$. $165 \times 1000 = 165{,}000$; remainder $= 13{,}000$; $165 \times 78 = 12{,}870$; remainder $\approx 130$; so $T \approx 1079$ K. This is consistent.
The reaction becomes feasible above approximately 1079 K. — Above 1079 K, $T\Delta S > \Delta H$, so $\Delta G < 0$. Below 1079 K, $\Delta G > 0$ and the reaction is not feasible. This matches the industrial reality that CaCO$_3$ decomposition requires very high temperatures.
Answer: The reaction becomes feasible above approximately 1079 K.
Notice how the unit analysis works out cleanly: J mol$^{-1}$ divided by J K$^{-1}$ mol$^{-1}$ gives K. Checking units at this step is a reliable way to catch errors before they propagate into your final answer.
6. Full $\Delta G$ Calculation: Bringing It All Together
In many exam questions, you will need to calculate $\Delta S$ from absolute entropy values first, then substitute into $\Delta G = \Delta H - T\Delta S$ to find $\Delta G$ at a specific temperature. This multi-stage approach is where marks are most often dropped — usually through unit inconsistency or forgetting to apply stoichiometric coefficients.
Worked example: Full $\Delta G$ Calculation for the Haber Process Reaction
For the reaction $\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightarrow 2\text{NH}_3\text{(g)}$:
Standard entropy values: $S^\circ[\text{N}_2\text{(g)}] = 192$ J K$^{-1}$ mol$^{-1}$; $S^\circ[\text{H}_2\text{(g)}] = 131$ J K$^{-1}$ mol$^{-1}$; $S^\circ[\text{NH}_3\text{(g)}] = 193$ J K$^{-1}$ mol$^{-1}$.
$\Delta H^\circ = -92$ kJ mol$^{-1}$.
(a) Calculate $\Delta S^\circ$ for the reaction.
(b) Calculate $\Delta G^\circ$ at 298 K.
(c) Determine the temperature above which the reaction is no longer feasible.
$\Sigma S^\circ_{\text{products}} = 2 \times S^\circ[\text{NH}_3] = 2 \times 193 = 386$ J K$^{-1}$ mol$^{-1}$ — Two moles of NH$_3$ are produced, so multiply by 2.
$\Sigma S^\circ_{\text{reactants}} = S^\circ[\text{N}_2] + 3 \times S^\circ[\text{H}_2]$
$= 192 + (3 \times 131)$
$= 192 + 393 = 585$ J K$^{-1}$ mol$^{-1}$ — Three moles of H$_2$ are consumed, so multiply by 3. One mole of N$_2$ needs no multiplier.
$\Delta S^\circ = 386 - 585 = -199$ J K$^{-1}$ mol$^{-1}$ — Negative $\Delta S$ makes sense: 4 moles of gas become 2 moles of gas, so disorder decreases.
$\Delta H = -92 \text{ kJ mol}^{-1} = -92{,}000 \text{ J mol}^{-1}$ — Units must match before substituting into $\Delta G = \Delta H - T\Delta S$.
$\Delta G = \Delta H - T\Delta S$
$\Delta G = -92{,}000 - (298 \times -199)$
$= -92{,}000 - (-59{,}302)$
$= -92{,}000 + 59{,}302$
$= -32{,}698$ J mol$^{-1}$
$\approx -32.7$ kJ mol$^{-1}$ — At 298 K, $\Delta G$ is negative, so the reaction is feasible. Note: $298 \times 199 = 298 \times 200 - 298 = 59{,}600 - 298 = 59{,}302$.
$T = \frac{\Delta H}{\Delta S} = \frac{-92{,}000}{-199} = 462$ K (to 3 significant figures) — Both $\Delta H$ and $\Delta S$ are negative, so the negatives cancel. Above 462 K, $\Delta G$ becomes positive and the reaction is no longer feasible. Verify: $199 \times 462 = 199 \times 400 + 199 \times 62 = 79{,}600 + 12{,}338 = 91{,}938 \approx 92{,}000$. Consistent.
The reaction is feasible below approximately 462 K. Above 462 K, $\Delta G > 0$ and the reaction is not thermodynamically feasible. — This explains why the Haber process uses a compromise temperature: higher temperatures speed up the rate but reduce thermodynamic feasibility.
Answer: (a) $\Delta S^\circ = -199$ J K$^{-1}$ mol$^{-1}$; (b) $\Delta G^\circ \approx -32.7$ kJ mol$^{-1}$ at 298 K (feasible); (c) The reaction is no longer feasible above approximately 462 K.
7. Exam Technique and Common Mistakes
The most mark-losing errors in this topic are predictable and avoidable. The unit mismatch between $\Delta H$ (kJ mol$^{-1}$) and $\Delta S$ (J K$^{-1}$ mol$^{-1}$) catches students out repeatedly. Always convert $\Delta H$ to J mol$^{-1}$ before substituting — or convert $\Delta S$ to kJ K$^{-1}$ mol$^{-1}$. Either approach works, but you must be consistent throughout the calculation.
A second common error is forgetting to interpret the crossover temperature. Calculating $T = \Delta H / \Delta S$ and stopping there will typically cost you a mark. You must state whether the reaction is feasible above or below that temperature, and link this back to the signs of $\Delta H$ and $\Delta S$.
Mastering Gibbs free energy calculations means holding three quantities — $\Delta H$, $\Delta S$, and $T$ — in mind simultaneously and understanding how they interact. The equation $\Delta G = \Delta H - T\Delta S$ is the unifying tool that connects entropy calculations, enthalpy data, and temperature dependence into a single, exam-ready framework.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-05.
How do you calculate entropy change for a reaction?
To calculate the entropy change for a reaction, use ΔS = ΣS(products) − ΣS(reactants), multiplying each standard absolute entropy by its stoichiometric coefficient. For the combustion of methane, the products total 354 and the reactants 596 J K⁻¹ mol⁻¹, so ΔS = −242 J K⁻¹ mol⁻¹.
How do you calculate Gibbs free energy change?
To calculate ΔG, convert ΔH and ΔS to the same units, then substitute into ΔG = ΔH − TΔS with T in Kelvin. For the Haber process at 298 K, ΔG = −92,000 − (298 × −199) = −32,698 J mol⁻¹, about −32.7 kJ mol⁻¹, so the reaction is feasible.
Can entropy be negative?
Absolute entropy cannot be negative: every substance has a positive standard absolute entropy value, S°, in J K⁻¹ mol⁻¹. The entropy change for a reaction, ΔS°, can be negative, for example when the number of moles of gas decreases.
How do you find the temperature at which a reaction becomes feasible?
To find the temperature at which a reaction becomes feasible, set ΔG = 0 and rearrange to T = ΔH ÷ ΔS, using consistent units. For the decomposition of calcium carbonate, T = 178,000 ÷ 165 = 1079 K, so the reaction is feasible at temperatures higher than about 1079 K.
What does a graph of ΔG against temperature show?
A graph of ΔG against temperature is a straight line with gradient −ΔS and intercept ΔH, because ΔG = ΔH − TΔS. A positive ΔS gives a negative gradient, so ΔG falls as temperature rises, and the point where the line crosses ΔG = 0 is the feasibility temperature.
Why is the entropy change negative for the Haber process?
The entropy change for N₂ + 3 H₂ → 2 NH₃ is negative, −199 J K⁻¹ mol⁻¹, because 4 moles of gas become 2 moles of gas, so disorder decreases. With ΔH = −92 kJ mol⁻¹, the reaction is feasible only at temperatures lower than about 462 K.
What is the entropy change for the combustion of methane?
The entropy change for the combustion of methane is −242 J K⁻¹ mol⁻¹. Products: 214 + 2 × 70 = 354; reactants: 186 + 2 × 205 = 596; so ΔS = 354 − 596 = −242. The negative value makes sense because 3 mol of gas become 1 mol of gas and 2 mol of liquid.
At what temperature does calcium carbonate decomposition become feasible?
Calcium carbonate decomposition becomes feasible at temperatures higher than approximately 1079 K. With ΔH = +178,000 J mol⁻¹ and ΔS = +165 J K⁻¹ mol⁻¹, setting ΔG = 0 gives T = ΔH ÷ ΔS = 178,000 ÷ 165 = 1079 K, which matches the very high temperatures needed industrially.
What is ΔG for the Haber process at 298 K?
ΔG for the Haber process at 298 K is about −32.7 kJ mol⁻¹, so it is feasible. ΔS = 386 − 585 = −199 J K⁻¹ mol⁻¹ and ΔH = −92,000 J mol⁻¹, so ΔG = −92,000 − (298 × −199) = −32,698 J mol⁻¹.
At what temperature does the Haber process stop being feasible?
The Haber process reaction is no longer feasible at temperatures higher than approximately 462 K. Both ΔH (−92,000 J mol⁻¹) and ΔS (−199 J K⁻¹ mol⁻¹) are negative, so T = ΔH ÷ ΔS = 462 K; at higher temperatures, ΔG becomes positive.
What does ΔG = 0 mean?
ΔG = 0 means the reaction is at the crossover point between feasible and non-feasible. Setting ΔG = 0 in ΔG = ΔH − TΔS gives the crossover temperature, T = ΔH ÷ ΔS, which only exists when ΔH and ΔS have the same sign.
What are common errors in Gibbs free energy calculations?
The most common error in Gibbs free energy calculations is mixing units, since ΔH is in kJ mol⁻¹ and ΔS is in J K⁻¹ mol⁻¹; both must be converted to J or both to kJ. Another is stopping at T = ΔH ÷ ΔS without stating whether the reaction is feasible at higher or lower temperatures.
Why must stoichiometric coefficients be used in entropy calculations?
Stoichiometric coefficients must be used in entropy calculations because each S° value is per mole of substance. If two moles of a product form, its S° is multiplied by two before summing; for the Haber process, 2 × 193 = 386 for NH₃ and 3 × 131 = 393 for H₂.