OCR A Level Chemistry A: Enthalpy and entropy (H432)

This study guide teaches section 5.2.2, Enthalpy and entropy, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. The section sits in Module 5, Physical chemistry and transition elements, and has six lettered learning outcomes, 5.2.2(a) to (f), and each one is taught below with worked examples. H432 is a single full A level. Its assessment overview shows that Module 5 is assessed in the Periodic table, elements and physical chemistry component and in the Unified chemistry component. No PAG is linked to this section.

Every H432 topic is linked from the OCR chemistry hub; tick off outcomes on the printable checklist. To find your weak spots first, try the free diagnostics. You will use ∆H values throughout, so revisit Enthalpy changes if Hess cycles feel rusty.

Outcomes in 5.2.2

Outcome	What you must be able to do	Specification guidance
5.2.2(a)	Explain that entropy measures the dispersal of energy in a system, and is greater the more disordered the system	HSW1: the entropy model explains thermodynamic stability
5.2.2(b)	Explain differences in entropy (i) between solids, liquids and gases, (ii) for a reaction that changes the number of gaseous molecules	–
5.2.2(c)	Calculate ∆S of a system, and related quantities, from the entropies of reactants and products	M2.2, M2.3, M2.4
5.2.2(d)	Explain that feasibility depends on ∆H and on T∆S of the system	HSW1, 5, 6
5.2.2(e)	Explain and calculate ∆G = ∆H − T∆S; a process is feasible when ∆G is negative	M0.0, M2.2, M2.3, M2.4; HSW5
5.2.2(f)	Explain the limits of ∆G predictions in terms of kinetics	M0.3; HSW6

5.2.2(a): What entropy measures

Entropy, S, is a measure of the dispersal of energy in a system. The more ways the particles and their energy can be arranged, the more disordered the system and the larger its entropy. Left to itself, energy tends to spread out: a gas fills its container and a hot object warms its surroundings.

Key facts:

• Entropy has units of J mol⁻¹ K⁻¹ (per mole, per kelvin).
• Tables give standard entropies for each substance at 298 K. For pure substances these values are positive.
• Elements in their standard states have non-zero entropies. This is different from ∆fH, where elements are zero. Leaving O₂(g) out of an entropy sum is a common slip.

5.2.2(b): Comparing entropies

(i) Solids, liquids and gases

For the same substance, S(solid) < S(liquid) < S(gas).

• In a solid, particles vibrate about fixed positions in a regular lattice. There are few ways to arrange them and their energy, so entropy is low.
• In a liquid, particles move past each other with no fixed positions. Disorder is greater.
• In a gas, particles move fast and randomly through the whole container. The energy is spread most widely, so entropy is far higher.

Typical values, in J mol⁻¹ K⁻¹: graphite 5.7 and magnesium oxide 26.9 (both solids), liquid bromine 152.2, and bromine gas 245.5. The jump on boiling (152.2 to 245.5) is much larger than the differences between most solids.

(ii) Changing the number of gaseous molecules

Because gases dominate the entropy of a system, the quickest prediction of the sign of ∆S comes from counting gas moles on each side of the equation.

• More gas molecules in the products: ∆S is positive.
• Fewer gas molecules in the products: ∆S is negative.
• Same number of gas molecules: ∆S is small, and its sign needs data.

Worked example 1: predicting the sign of ∆S

Predict the sign of ∆S for each change.

1. 2O₃(g) → 3O₂(g): 2 mol of gas become 3 mol, so ∆S is positive.
2. 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(l): 9 mol of gas become 4 mol (water is liquid), so ∆S is negative.
3. Br₂(l) → Br₂(g): liquid to gas, so ∆S is positive.
4. MgCO₃(s) → MgO(s) + CO₂(g): no gas becomes 1 mol of gas, so ∆S is positive.

Read the state symbols first: H₂O(g) in example 2 would change the gas count.

5.2.2(c): Calculating ∆S

The entropy change of a system is

∆S = ΣS(products) − ΣS(reactants)

Multiply each entropy by the balancing number in the equation. Give ∆S in J mol⁻¹ K⁻¹ with a sign.

Worked example 2: a decomposition

Calculate ∆S for MgCO₃(s) → MgO(s) + CO₂(g). Use these values (J mol⁻¹ K⁻¹): MgCO₃(s) 65.7, MgO(s) 26.9, CO₂(g) 213.8.

1. ΣS(products) = 26.9 + 213.8 = 240.7
2. ΣS(reactants) = 65.7
3. ∆S = 240.7 − 65.7 = +175.0 J mol⁻¹ K⁻¹

Worked example 3: using balancing numbers

Iron is oxidised to iron(III) oxide: 4Fe(s) + 3O₂(g) → 2Fe₂O₃(s). Calculate ∆S. Use these values (J mol⁻¹ K⁻¹): Fe(s) 27.3, O₂(g) 205.2, Fe₂O₃(s) 87.4.

1. ΣS(products) = 2(87.4) = 174.8
2. ΣS(reactants) = 4(27.3) + 3(205.2) = 109.2 + 615.6 = 724.8
3. ∆S = 174.8 − 724.8 = −550.0 J mol⁻¹ K⁻¹

All 3 mol of gas are used up and only solids form, so a large negative value makes sense.

Worked example 4: a related quantity

For 2Na(s) + Cl₂(g) → 2NaCl(s), ∆S = −181.5 J mol⁻¹ K⁻¹. Given S(Na(s)) = 51.3 and S(Cl₂(g)) = 223.1 J mol⁻¹ K⁻¹, find S(NaCl(s)).

1. Write the expression: −181.5 = 2S(NaCl) − [2(51.3) + 223.1]
2. ΣS(reactants) = 102.6 + 223.1 = 325.7
3. 2S(NaCl) = −181.5 + 325.7 = 144.2
4. S(NaCl) = 72.1 J mol⁻¹ K⁻¹

5.2.2(d): Feasibility depends on ∆H and T∆S

A feasible process is one that is thermodynamically possible: it can happen without outside work being done to drive it. Two factors in the system decide this:

• ∆H: a negative (exothermic) enthalpy change favours feasibility.
• T∆S: a positive entropy change favours feasibility, and its effect grows with temperature because ∆S is multiplied by T.

This is why some endothermic changes happen on their own. Ice melts above 273 K and many salts dissolve with cooling, because the entropy gain, multiplied by a high enough temperature, outweighs the unfavourable ∆H. The specification links this to HSW1, 5 and 6: you use entropy, enthalpy and temperature together to predict feasibility.

∆H	∆S	Feasible?
negative	positive	at all temperatures
positive	negative	at no temperature
negative	negative	only below a certain temperature
positive	positive	only above a certain temperature

5.2.2(e): The Gibbs equation

∆G = ∆H − T∆S

• ∆G is the free energy change, in kJ mol⁻¹.
• T is the temperature in kelvin.
• Convert ∆S from J mol⁻¹ K⁻¹ to kJ mol⁻¹ K⁻¹ by dividing by 1000 before you substitute (M0.0).
• A process is feasible when ∆G is negative.

At the temperature where a process just becomes feasible, ∆G = 0, so T = ∆H ÷ ∆S (with ∆S in kJ mol⁻¹ K⁻¹).

Worked example 5: one reaction, two temperatures

For MgCO₃(s) → MgO(s) + CO₂(g), ∆H = +101 kJ mol⁻¹ and ∆S = +175.0 J mol⁻¹ K⁻¹ (worked example 2).

1. ∆S = 175.0 ÷ 1000 = 0.1750 kJ mol⁻¹ K⁻¹
2. At 298 K: ∆G = 101 − 298 × 0.1750 = 101 − 52.15 = +48.9 kJ mol⁻¹, so not feasible.
3. At 800 K: ∆G = 101 − 800 × 0.1750 = 101 − 140 = −39.0 kJ mol⁻¹, so feasible.
4. Threshold: T = 101 ÷ 0.1750 = 577 K. Above about 577 K the decomposition is feasible.

Worked example 6: finding ∆H first

Iron(III) oxide can be reduced by hydrogen: Fe₂O₃(s) + 3H₂(g) → 2Fe(s) + 3H₂O(g). Use these values: ∆fH (kJ mol⁻¹): Fe₂O₃(s) −824.2, H₂O(g) −241.8. S (J mol⁻¹ K⁻¹): Fe₂O₃(s) 87.4, H₂(g) 130.7, Fe(s) 27.3, H₂O(g) 188.8.

1. ∆H = 3(−241.8) − (−824.2) = −725.4 + 824.2 = +98.8 kJ mol⁻¹
2. ∆S = [2(27.3) + 3(188.8)] − [87.4 + 3(130.7)] = 621.0 − 479.5 = +141.5 J mol⁻¹ K⁻¹
3. At 298 K: ∆G = 98.8 − 298 × 0.1415 = +56.6 kJ mol⁻¹ (not feasible)
4. Minimum temperature: T = 98.8 ÷ 0.1415 = 698 K

Worked example 7: an upper temperature limit

For CO(g) + Cl₂(g) → COCl₂(g), ∆H = −108.6 kJ mol⁻¹ and ∆S = −137.3 J mol⁻¹ K⁻¹.

1. At 298 K: ∆G = −108.6 − 298 × (−0.1373) = −108.6 + 40.9 = −67.7 kJ mol⁻¹ (feasible).
2. ∆G = 0 when T = −108.6 ÷ −0.1373 = 791 K.
3. Above 791 K, the positive −T∆S term outweighs the negative ∆H, so ∆G is positive. At 900 K, ∆G = −108.6 + 123.6 = +15.0 kJ mol⁻¹.

So this reaction is feasible only below about 791 K. Check the direction in the table above before writing “above” or “below”.

Reading a ∆G against T graph

Because ∆G = ∆H − T∆S, a plot of ∆G (y) against T (x) is a straight line, assuming ∆H and ∆S stay roughly constant. The intercept at T = 0 is ∆H and the gradient is −∆S. The line crosses ∆G = 0 at the threshold temperature.

5.2.2(f): Limits of ∆G predictions: kinetics

A negative ∆G shows only that a reaction is thermodynamically possible. The rate is a separate question that ∆G cannot answer. If the activation energy is high, a feasible reaction may be so slow at room temperature that nothing is seen.

Worked example 8: a pencil lead in air

For C(graphite) + O₂(g) → CO₂(g), ∆H = −393.5 kJ mol⁻¹. Use these entropies (J mol⁻¹ K⁻¹): C(graphite) 5.7, O₂(g) 205.2, CO₂(g) 213.8.

1. ∆S = 213.8 − (5.7 + 205.2) = +2.9 J mol⁻¹ K⁻¹ (1 mol of gas on each side, so the change is small)
2. ∆G at 298 K = −393.5 − 298 × 0.0029 = −393.5 − 0.86 = −394.4 kJ mol⁻¹

The reaction is strongly feasible, yet a graphite pencil lead or a lump of charcoal sits in air for years unchanged. Strong covalent bonds in graphite and the O=O bond must start to break before products form, so the activation energy is large and the rate at room temperature is negligible.

Other examples: methane and oxygen do not react until a spark supplies energy, and paper in air stays unburnt until it is lit. In each case ∆G is negative but the rate at room temperature is tiny. This is HSW6: you must weigh entropy-based feasibility against rate. A catalyst can speed up a feasible reaction, but it does not change ∆H, ∆S or ∆G, so it cannot make an infeasible reaction feasible.

When you are asked to estimate an outcome (M0.3), state both parts: the ∆G conclusion, then whether a high activation energy might stop the reaction being observed.

Common errors

• Combining ∆S in J mol⁻¹ K⁻¹ directly with ∆H in kJ mol⁻¹: the T∆S term comes out 1000 times too big.
• Substituting a Celsius temperature, such as 25, where T must be in kelvin (298).
• Leaving out elements, such as O₂(g) or H₂(g), from the entropy sum.
• Forgetting balancing numbers when summing S values.
• Writing “above” for a threshold temperature when ∆H and ∆S are both negative.
• Saying a reaction “will happen quickly” because ∆G is negative.
• Losing the sign of −T∆S when ∆S is negative.

Next steps

Condense this with the Enthalpy and entropy revision notes, then test yourself with the Enthalpy and entropy practice questions. The same feasibility-versus-rate argument returns in 5.2.3, where electrode potentials predict feasibility; the oxidation-number skills you need there are in Redox.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 5.2.2 Enthalpy and entropy, learning outcomes (a) to (f). Published by OCR.