OCR A Level Chemistry A: Enthalpy and entropy (H432) – Revision Notes

These notes condense 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. They cover all six learning outcomes, 5.2.2(a) to (f), which sit in Module 5 (Physical chemistry and transition elements) of this full A level. The specification’s assessment overview places Module 5 in the Periodic table, elements and physical chemistry component and the Unified chemistry component.

For full explanations and fully worked examples, use the Enthalpy and entropy study guide. When you are ready, try the Enthalpy and entropy practice questions. Track the whole course on the OCR chemistry hub and the printable checklist, and find gaps fast with the free diagnostics.

5.2.2(a): Entropy in one paragraph

Entropy, S: a measure of the dispersal of energy in a system. The more disordered the system, the greater its entropy. Units: J mol⁻¹ K⁻¹. Standard entropies are quoted at 298 K and are positive for pure substances. Elements have non-zero S values, unlike their ∆fH values, which are zero.

5.2.2(b): Which has more entropy?

(i) States. For one substance: solid < liquid < gas.

State	Particle picture	Entropy
Solid	Vibrate about fixed lattice positions	Low
Liquid	Move past each other, no fixed positions	Higher
Gas	Fast, random motion through the whole volume	Much higher

(ii) Gas moles. Count gaseous molecules on each side.

• Products have more gas molecules: ∆S positive.
• Products have fewer gas molecules: ∆S negative.
• Equal numbers: ∆S small; use data to decide.

Ignore solids and liquids in the count, but only after you have checked every state symbol.

5.2.2(c): Calculating ∆S

∆S = ΣS(products) − ΣS(reactants)

Method in steps:

1. Write each S value next to its species in the equation.
2. Multiply by the balancing number.
3. Add the products; add the reactants.
4. Subtract: products minus reactants.
5. Give a sign and units, J mol⁻¹ K⁻¹.

Related quantities. If ∆S is given and one S value is missing, write the full expression with the unknown as a letter, then rearrange. Divide by its balancing number at the end.

Worked reminder: a missing entropy

For 2C(graphite) + O₂(g) → 2CO(g), ∆S = +178.8 J mol⁻¹ K⁻¹. Use S(C(graphite)) = 5.7 and S(O₂(g)) = 205.2 J mol⁻¹ K⁻¹.

• +178.8 = 2S(CO) − [2(5.7) + 205.2] = 2S(CO) − 216.6
• 2S(CO) = 178.8 + 216.6 = 395.4
• S(CO) = 197.7 J mol⁻¹ K⁻¹

Sense check: 1 mol of gas becomes 2 mol, so a positive ∆S fits.

5.2.2(d) and (e): Feasibility and ∆G

∆G = ∆H − T∆S. A process is feasible when ∆G is negative.

Symbol	Meaning	Usual units
∆G	Free energy change	kJ mol⁻¹
∆H	Enthalpy change of the system	kJ mol⁻¹
T	Temperature	K (never °C)
∆S	Entropy change of the system	J mol⁻¹ K⁻¹, so divide by 1000 to use with kJ
T at ∆G = 0	Threshold temperature	T = ∆H ÷ ∆S, both in kJ units

Feasibility depends on two terms: ∆H, and T∆S. A negative ∆H helps. A positive ∆S helps more and more as T rises.

∆H	∆S	∆G against T	Result
−	+	always below zero	Feasible at all T
+	−	always above zero	Never feasible
−	−	rises as T rises	Feasible below the threshold T
+	+	falls as T rises	Feasible above the threshold T

Graph reminder. ∆G against T is a straight line (if ∆H and ∆S stay roughly constant): intercept ∆H, gradient −∆S.

Method in steps for ∆G:

1. Find ∆H (given, or from ∆fH data with products minus reactants).
2. Find ∆S, then change it to kJ mol⁻¹ K⁻¹ (÷ 1000).
3. Make sure T is in kelvin, not °C.
4. Put the values into ∆G = ∆H − T∆S. Watch the sign when ∆S is negative.
5. State the conclusion: negative means feasible.

Method for a threshold temperature:

1. Set ∆G = 0, so T = ∆H ÷ ∆S.
2. Use the four-case table to decide whether the process is feasible above or below this T.

Worked reminder: heating zinc carbonate

ZnCO₃(s) → ZnO(s) + CO₂(g), ∆H = +68.8 kJ mol⁻¹. Use S values (J mol⁻¹ K⁻¹): ZnCO₃(s) 82.4, ZnO(s) 43.7, CO₂(g) 213.8.

• ∆S = (43.7 + 213.8) − 82.4 = +175.1 J mol⁻¹ K⁻¹ (a solid releases 1 mol of gas)
• ∆G at 298 K = 68.8 − 298 × 0.1751 = 68.8 − 52.2 = +16.6 kJ mol⁻¹: not feasible
• Threshold T = 68.8 ÷ 0.1751 = 393 K
• At 600 K, ∆G = 68.8 − 105.1 = −36.3 kJ mol⁻¹: feasible

Both ∆H and ∆S are positive, so the decomposition is feasible only above about 393 K.

5.2.2(f): Kinetic limits

• ∆G tells you whether a reaction can happen, not how fast.
• A reaction with negative ∆G but a large activation energy can be too slow to observe. Examples: charcoal or graphite in air; fuels such as methane that sit in air until ignited.
• A catalyst changes the rate only. ∆H, ∆S and ∆G stay the same.
• When estimating an outcome, give the ∆G verdict, then comment on rate.

Must-know distinctions

• S versus ∆S: S belongs to one substance; ∆S belongs to a reaction.
• Feasible versus fast: thermodynamics versus kinetics.
• J versus kJ: entropy data are in J; enthalpy data are in kJ.
• Threshold temperature versus “reaction starts”: at the threshold, ∆G = 0. Feasibility begins on one side of it.
• ∆fH(element) = 0, S(element) ≠ 0: elements count in entropy sums.

Quick self-test

1. Which has the greater entropy, 1 mol of liquid water or 1 mol of steam at the same temperature? Explain.
2. Predict the sign of ∆S for 3C₂H₂(g) → C₆H₆(l), with a reason.
3. Predict the sign of ∆S for 2KNO₃(s) → 2KNO₂(s) + O₂(g), with a reason.
4. Calculate ∆S for H₂(g) + Cl₂(g) → 2HCl(g). S values: H₂ 130.7, Cl₂ 223.1, HCl 186.9 J mol⁻¹ K⁻¹.
5. Calculate ∆S for 2O₃(g) → 3O₂(g). S values: O₃ 238.9, O₂ 205.2 J mol⁻¹ K⁻¹.
6. A reaction has ∆H = −96 kJ mol⁻¹ and ∆S = −204 J mol⁻¹ K⁻¹. Find ∆G at 298 K and the temperature range over which it is feasible.
7. A reaction has ∆H = +66 kJ mol⁻¹ and ∆S = +172 J mol⁻¹ K⁻¹. Is it feasible at 298 K? Find the minimum temperature for feasibility.
8. A process has ∆G = 0 at 520 K and ∆S = +135 J mol⁻¹ K⁻¹. Find ∆H.
9. Which sign combination of ∆H and ∆S is never feasible?
10. Hydrogen and oxygen can be stored mixed at room temperature without reacting, although ∆G for forming water is negative. Explain.
11. Does adding a catalyst change ∆G? Explain.

Answers

1. Steam. Gas particles move randomly through the whole volume, so energy is far more dispersed than in the liquid.
2. Negative: 3 mol of gas become a liquid, so disorder falls.
3. Positive: 1 mol of gas forms from solids only.
4. ∆S = 2(186.9) − (130.7 + 223.1) = 373.8 − 353.8 = +20.0 J mol⁻¹ K⁻¹ (small, as gas moles do not change)
5. ∆S = 3(205.2) − 2(238.9) = 615.6 − 477.8 = +137.8 J mol⁻¹ K⁻¹
6. ∆G = −96 − 298 × (−0.204) = −96 + 60.8 = −35.2 kJ mol⁻¹ (feasible). T = −96 ÷ −0.204 = 471 K, so feasible below about 471 K.
7. ∆G = 66 − 298 × 0.172 = +14.7 kJ mol⁻¹, so not feasible. Minimum T = 66 ÷ 0.172 = 384 K.
8. ∆H = T∆S = 520 × 0.135 = +70.2 kJ mol⁻¹
9. ∆H positive and ∆S negative: both terms make ∆G positive at every temperature.
10. Few collisions at room temperature have energy equal to or above the large activation energy, so the reaction is too slow to notice. A spark provides that energy and starts it.
11. No. A catalyst lowers the activation energy and changes the rate, but ∆H, ∆S and so ∆G are unchanged.

Where marks are usually lost

• Substituting ∆S in J mol⁻¹ K⁻¹ alongside ∆H in kJ mol⁻¹.
• Using 25 instead of 298 for T.
• Missing O₂(g), H₂(g) or another element from the entropy sum.
• Not multiplying S values by balancing numbers.
• Writing reactants minus products for ∆S.
• Dropping the sign of ∆S, so −T∆S has the wrong sign.
• Saying “feasible above” when ∆H and ∆S are both negative.
• Stating that a negative ∆G means a reaction is fast.
• Explaining entropy as “chaos” with no reference to how energy or particles are dispersed.

Next steps

• Full explanations: Enthalpy and entropy study guide
• Exam-style practice: Enthalpy and entropy practice questions
• ∆H from data: Enthalpy changes revision notes
• Course map: OCR chemistry hub and checklist
• Check your level: free diagnostics

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.