What is entropy change, ΔS, in the context of Chemical Energetics for Cambridge International A Levels?

Entropy change, ΔS, is a measure of the disorder or randomness in a system. In the context of Chemical Energetics for Cambridge International A Levels, it refers to the change in entropy when a chemical reaction occurs. It helps in predicting the spontaneity of reactions, where a positive ΔS indicates an increase in disorder and is often associated with spontaneous processes.

How is entropy change, ΔS, calculated in a chemical reaction?

Entropy change, ΔS, in a chemical reaction is calculated using the formula ΔS = ΣS(products) - ΣS(reactants), where ΣS represents the sum of the standard molar entropies of the products and reactants. This calculation helps determine whether a reaction leads to an increase or decrease in disorder.

Why is entropy change, ΔS, important in determining the spontaneity of a reaction in A-Level Chemistry?

Entropy change, ΔS, is crucial in determining the spontaneity of a reaction because it is a component of the Gibbs free energy equation, ΔG = ΔH - TΔS. A negative ΔG indicates a spontaneous reaction. Therefore, a positive ΔS can contribute to a negative ΔG, especially at higher temperatures, making the reaction more likely to occur spontaneously.

Can entropy change, ΔS, be negative, and what does it imply in a chemical process?

Yes, entropy change, ΔS, can be negative, which implies that the system becomes more ordered during the chemical process. This is often seen in reactions where gases condense into liquids or solids, or when a reaction results in fewer gas molecules. A negative ΔS does not necessarily mean the reaction is non-spontaneous; it must be considered alongside enthalpy change and temperature in the Gibbs free energy equation.

How does temperature affect entropy change, ΔS, in chemical reactions studied in A-Level Chemistry?

Temperature affects entropy change, ΔS, because it influences the degree of disorder in a system. Generally, as temperature increases, the entropy of a system also increases due to greater molecular motion. In the context of the Gibbs free energy equation, the impact of ΔS on reaction spontaneity becomes more significant at higher temperatures, potentially favoring reactions with positive ΔS.

Why is "Quoting ΔS in kJ K⁻¹ mol⁻¹ instead of J K⁻¹ mol⁻¹." a common mistake in Entropy change, ΔS?

Why it happens: Carrying over enthalpy units. How to avoid it: Entropy is in J K⁻¹ mol⁻¹; only convert to kJ when combining with ΔH for ΔG. Source: 9701 Examiner Reports 2022-2024

Why is "Setting S° of an element to zero." a common mistake in Entropy change, ΔS?

Why it happens: Confusing with enthalpy of formation. How to avoid it: Elements have a real, non-zero standard entropy.

Why is "Predicting ΔS sign from whether the reaction is exo/endothermic." a common mistake in Entropy change, ΔS?

Why it happens: Conflating entropy with enthalpy. How to avoid it: Judge ΔS from the change in disorder (especially moles of gas).

Chemical Energetics(A-Level Physical Chemistry)Cambridge International A Level Chemistry 9701

Entropy change, ΔS

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Entropy Change, ΔS — Cambridge International AS & A Level Chemistry 9701 (2025-2027 syllabus)

Entropy as a measure of disorder: predicting the sign of ΔS, calculating ΔS° from standard entropies, and seeing why some endothermic changes happen anyway.

What you’ll learn

Mapped to the Cambridge International A Level 9701 syllabus (2025-2027).

23.3 — Define entropy as a measure of the disorder of a system.
23.3 — Predict the sign of the entropy change for a given physical or chemical change.
23.3 — Calculate the standard entropy change ΔS° from standard molar entropies.

Entropy and predicting its sign

More ways to arrange particles = more entropy; gases are the most disordered.

Entropy (S) measures the disorder of a system — strictly, the number of ways the particles and their energy can be arranged. The more disordered (more 'spread out') a system is, the higher its entropy.

Key order of disorder:

$S(\text{solid}) < S(\text{liquid}) < S(\text{gas})$

Entropy also increases with temperature (particles move and vibrate more).

Predicting the sign of ΔS — look at how disorder changes:

Positive ΔS (more disorder): solid/liquid → gas; a reaction producing more moles of gas; a solid dissolving.
Negative ΔS (less disorder): gas → liquid/solid; a reaction producing fewer moles of gas.

The single most useful clue in a chemical reaction is the change in the number of moles of gas.

S(solid) < S(liquid) < S(gas); rises with temperature.
Gas produced / solid dissolving → ΔS positive.
Fewer moles of gas → ΔS negative.

See the full worked example for entropy change, δs →

Calculating ΔS°

Products minus reactants, using standard molar entropies in J K⁻¹ mol⁻¹.

Standard molar entropies (S°) are tabulated for elements and compounds. Calculate the standard entropy change as products minus reactants, multiplying each by its coefficient:

$\Delta S^{\circ} = \sum S^{\circ}(\text{products}) - \sum S^{\circ}(\text{reactants})$

Worked. For $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$ , with S°: N₂ 192, H₂ 131, NH₃ 192 (J K⁻¹ mol⁻¹): $\Delta S^{\circ} = 2(192) - [192 + 3(131)] = 384 - 585 = -201\,\text{J K}^{-1}\,\text{mol}^{-1}$

The negative value makes sense: 4 moles of gas become 2 — disorder decreases.

Note the units are J K⁻¹ mol⁻¹ (not kJ). You only convert to kJ when you combine ΔS with ΔH to find ΔG.

ΔS° = ΣS°(products) − ΣS°(reactants).
Multiply each S° by its coefficient.
Units: J K⁻¹ mol⁻¹.

See the full worked example for entropy change, δs →

Why endothermic changes can be spontaneous

A large positive ΔS can outweigh a positive ΔH — a preview of ΔG.

Some endothermic processes happen readily — for example, ammonium nitrate dissolving in water (which gets cold) or thermal decomposition of carbonates at high temperature. How?

Spontaneity is governed by Gibbs free energy, $\Delta G = \Delta H - T\Delta S$ , not by ΔH alone. Dissolving and decomposition both have a large positive ΔS (an ordered solid becomes dispersed ions, or a gas is released). The favourable $-T\Delta S$ term can be large enough to outweigh a positive ΔH, giving ΔG < 0.

This is why entropy matters: it explains spontaneous endothermic changes that enthalpy alone cannot.

Endothermic ≠ impossible.
Dissolving / gas release → large positive ΔS.
Positive ΔS can make ΔG negative despite positive ΔH.

See the full worked example for entropy change, δs →

How it’s examined

Usually a short Paper 4 item: predict the sign of ΔS (justified by gas moles or state changes), then a ΔS° calculation, often feeding into a ΔG calculation in the next part. Common errors: quoting ΔS in kJ, setting an element's S° to zero, and predicting the sign from enthalpy instead of disorder.

Sources: Cambridge International AS & A Level Chemistry 9701 syllabus (2025-2027), section 23.3; 9701 Examiner Reports 2022-2024; 9701/41 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-20.

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Step-by-step worked examples — Entropy change, ΔS

Step-by-step solutions to past-paper-style questions on entropy change, δs, written exactly the way a tutor would explain them at the board.

1Sign of ΔS for state changes (2 marks)
Getting started• entropy sign
Question
State the sign of ΔS for (a) ice melting and (b) steam condensing.
Step-by-step solution
1. Step 1
  (a) Melting solid → liquid: more disorder → ΔS positive.
2. Step 2
  (b) Gas → liquid: less disorder → ΔS negative.
Answer
(a) ΔS positive; (b) ΔS negative.
Examiner tip
Disorder increases solid < liquid < gas; more disorder = positive ΔS.
2Sign of ΔS for a reaction (2 marks)
Getting started• entropy sign
Question
Predict the sign of ΔS for $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$ and explain.
Step-by-step solution
1. Step 1
  A gas (CO₂) is produced from a solid.
2. Step 2
  Gas has much greater disorder → ΔS positive.
Answer
ΔS is positive — a gas is formed from a solid.
Examiner tip
Look at the change in moles of GAS — gas formed → positive ΔS.
3Calculating ΔS° (3 marks)
Building confidence• calculation
Question
Calculate ΔS° for $\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$ . S°: CaCO₃ 93, CaO 40, CO₂ 214 (J K⁻¹ mol⁻¹).
Step-by-step solution
1. Step 1
  ΔS° = ΣS°(products) − ΣS°(reactants).
  $\Delta S^{\circ} = (40 + 214) - 93$
2. Step 2
  Evaluate.
  $\Delta S^{\circ} = +161\,\text{J K}^{-1}\,\text{mol}^{-1}$
Answer
$\Delta S^{\circ} = +161\,\text{J K}^{-1}\,\text{mol}^{-1}$ .
Examiner tip
Entropy units are J K⁻¹ mol⁻¹ (note J, not kJ).
4ΔS° for the Haber reaction (3 marks)
Building confidence• calculation
Question
Calculate ΔS° for $\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)$ . S°: N₂ 192, H₂ 131, NH₃ 192 (J K⁻¹ mol⁻¹). Comment on the sign.
Step-by-step solution
1. Step 1
  Products − reactants.
  $\Delta S^{\circ} = 2(192) - [192 + 3(131)]$
2. Step 2
  Evaluate.
  $\Delta S^{\circ} = 384 - 585 = -201\,\text{J K}^{-1}\,\text{mol}^{-1}$
3. Step 3
  Negative — 4 mol gas → 2 mol gas (less disorder).
Answer
$\Delta S^{\circ} = -201\,\text{J K}^{-1}\,\text{mol}^{-1}$ (gas moles fall 4 → 2).
5Entropy of dissolving (4 marks)
Stretch• dissolving
Question
Predict and explain the sign of ΔS when an ionic solid dissolves in water, and note one factor that can reduce it. (4)
Step-by-step solution
1. Step 1
  The ordered lattice breaks up and ions disperse through the solution → big increase in disorder.
2. Step 2
  So ΔS is usually positive.
3. Step 3
  However, the ions order surrounding water molecules (hydration shells), which decreases entropy.
4. Step 4
  For small, highly charged ions this ordering is large and can make ΔS less positive (even negative).
Answer
Usually ΔS positive (lattice → dispersed ions); but ordering of water by small/high-charge ions reduces it.
Examiner tip
Mention the competing effect: dispersal increases entropy, hydration shells decrease it.
6Why endothermic reactions can occur (4 marks)
Stretch• spontaneity
Question
Ammonium nitrate dissolving in water is endothermic, yet it happens spontaneously at room temperature. Explain in terms of entropy. (4)
Step-by-step solution
1. Step 1
  Dissolving has a large positive ΔS (solid lattice → dispersed aqueous ions).
2. Step 2
  Spontaneity depends on ΔG = ΔH − TΔS, not on ΔH alone.
3. Step 3
  The positive ΔS makes −TΔS large and negative.
4. Step 4
  This can outweigh the positive ΔH, giving ΔG < 0 → spontaneous despite being endothermic.
Answer
The large positive entropy change makes −TΔS outweigh the positive ΔH, so ΔG < 0 (entropy-driven).
Examiner tip
Endothermic ≠ impossible — a big positive ΔS can drive the reaction (forward look to ΔG).

Key Formulae — Entropy change, ΔS

The formulae you need to memorise for entropy change, δs on the Cambridge International A Level 9701 paper, with every variable defined in plain English and a note on when to use it.

Standard entropy change
$\Delta S^{\circ} = \sum S^{\circ}(\text{products}) - \sum S^{\circ}(\text{reactants})$
$S^{\circ}$
standard molar entropy / J K⁻¹ mol⁻¹
When to use
Calculating ΔS° from data-book entropies (multiply each by its coefficient).

Key Definitions and Keywords — Entropy change, ΔS

Definitions to memorise and the exact keywords mark schemes credit for entropy change, δs answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9701 sitting.

Entropy (S)
Examiner keyword
A measure of the disorder (number of ways energy/particles can be arranged) of a system; units J K⁻¹ mol⁻¹.
Order of entropy
Examiner keyword
Solid < liquid < gas; entropy increases with temperature and with the number of gas molecules.
Standard molar entropy (S°)
Examiner keyword
The entropy of one mole of a substance under standard conditions; unlike enthalpy, elements have a non-zero S°.

Common Mistakes and Misconceptions — Entropy change, ΔS

The traps other students keep falling into on entropy change, δs questions — taken from recent Cambridge International A Level 9701 examiner reports and mark schemes — and how to avoid them.

✕Quoting ΔS in kJ K⁻¹ mol⁻¹ instead of J K⁻¹ mol⁻¹.
9701 Examiner Reports 2022-2024
Why it happens
Carrying over enthalpy units.
How to avoid it
Entropy is in J K⁻¹ mol⁻¹; only convert to kJ when combining with ΔH for ΔG.
✕Setting S° of an element to zero.
Why it happens
Confusing with enthalpy of formation.
How to avoid it
Elements have a real, non-zero standard entropy.
✕Predicting ΔS sign from whether the reaction is exo/endothermic.
Why it happens
Conflating entropy with enthalpy.
How to avoid it
Judge ΔS from the change in disorder (especially moles of gas).

Practice questions

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Entropy change, ΔS — frequently asked questions

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Entropy change, ΔS

Short Study Notes in the form of Slides

Short Study Notes — Entropy change, ΔS

Detailed Notes

What you’ll learn

How it’s examined

1Sign of ΔS for state changes (2 marks)

2Sign of ΔS for a reaction (2 marks)

3Calculating ΔS° (3 marks)

4ΔS° for the Haber reaction (3 marks)

5Entropy of dissolving (4 marks)

6Why endothermic reactions can occur (4 marks)

Standard entropy change

Entropy (S)

Order of entropy

Standard molar entropy (S°)

✕Quoting ΔS in kJ K⁻¹ mol⁻¹ instead of J K⁻¹ mol⁻¹.

✕Setting S° of an element to zero.

✕Predicting ΔS sign from whether the reaction is exo/endothermic.

Practice questions

Past paper style quiz

Assess your understanding

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Entropy change, ΔS — frequently asked questions