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So far in this course, we have focused on enthalpy changes to understand whether reactions release or absorb energy. But enthalpy alone does not determine whether a reaction will actually occur. Some endothermic reactions happen spontaneously, while some exothermic reactions do not. To understand why, we need the concept of entropy. This lesson covers what entropy is, how to calculate entropy changes, the second law of thermodynamics, and how entropy explains processes that enthalpy alone cannot.
Entropy (symbol S) is a measure of the disorder or dispersal of energy in a system. More precisely, it quantifies the number of ways that energy can be distributed among the particles in a system.
A system with high entropy has many possible arrangements of its particles and energy — it is highly disordered. A system with low entropy has few possible arrangements — it is highly ordered.
The units of entropy are J K⁻¹ mol⁻¹ (note: joules, not kilojoules).
Important unit warning: Entropy is in J K⁻¹ mol⁻¹, but enthalpy is in kJ mol⁻¹. When using both in the Gibbs equation (ΔG = ΔH − TΔS), you must convert ΔS to kJ K⁻¹ mol⁻¹ by dividing by 1000, or convert ΔH to J mol⁻¹ by multiplying by 1000.
Every substance has a standard molar entropy (S°), which is the entropy of one mole of the substance under standard conditions (298 K, 100 kPa).
| Substance | S° / J K⁻¹ mol⁻¹ | State | Notes |
|---|---|---|---|
| C(s, diamond) | 2.4 | Solid | Very ordered crystal |
| C(s, graphite) | 5.7 | Solid | Layers can slide |
| Fe(s) | 27.3 | Solid | Metallic solid |
| NaCl(s) | 72.1 | Solid | Ordered ionic lattice |
| H₂O(s) | 48.0 | Solid | Ice |
| H₂O(l) | 69.9 | Liquid | More disordered than ice |
| C₂H₅OH(l) | 160.7 | Liquid | Large molecule |
| H₂O(g) | 188.7 | Gas | Much more disordered |
| H₂(g) | 130.6 | Gas | Small molecule |
| N₂(g) | 191.6 | Gas | Slightly larger than H₂ |
| O₂(g) | 205.0 | Gas | Larger still |
| CO₂(g) | 213.6 | Gas | Three atoms — more ways to vibrate |
| C₂H₆(g) | 229.5 | Gas | Larger molecule with many vibrational modes |
| NH₃(g) | 192.3 | Gas | Four atoms |
Key observations:
You can often predict whether ΔS is positive or negative by considering the states and number of moles of gas:
| Change | Effect on entropy | Sign of ΔS |
|---|---|---|
| Solid → liquid (melting) | Increased disorder | Positive |
| Liquid → gas (boiling) | Large increase in disorder | Positive (large) |
| Solid → gas (sublimation) | Very large increase in disorder | Positive (very large) |
| Fewer moles of gas → more moles of gas | More gas particles = more disorder | Positive |
| More moles of gas → fewer moles of gas | Less gas particles = less disorder | Negative |
| Dissolving a solid in water | Ions dispersed in solution | Usually positive |
| Gas dissolving in liquid | Gas loses freedom | Negative |
The standard entropy change for a reaction is calculated using:
ΔS°reaction = Σ S°(products) − Σ S°(reactants)
This is similar to the Hess's law formula for enthalpies of formation, but uses absolute entropy values (not entropy changes).
CaCO₃(s) → CaO(s) + CO₂(g)
Given:
ΔS° = [39.7 + 213.6] − [92.9]
ΔS° = 253.3 − 92.9 = +160.4 J K⁻¹ mol⁻¹
This is positive, as expected: a gas is produced from a solid, greatly increasing disorder.
N₂(g) + 3H₂(g) → 2NH₃(g)
Given:
ΔS° = [2 × 192.3] − [191.6 + 3 × 130.6]
ΔS° = 384.6 − [191.6 + 391.8]
ΔS° = 384.6 − 583.4 = −198.8 J K⁻¹ mol⁻¹
This is negative because 4 moles of gas become 2 moles of gas — a decrease in disorder.
C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)
Given:
ΔS° = [2(213.6) + 3(69.9)] − [160.7 + 3(205.0)]
ΔS° = [427.2 + 209.7] − [160.7 + 615.0]
ΔS° = 636.9 − 775.7 = −138.8 J K⁻¹ mol⁻¹
Although this reaction is strongly exothermic, the entropy change is actually negative. There are 3 moles of gas on the left but only 2 moles on the right (with 3 moles of liquid), so the system becomes more ordered. This reaction is still feasible because the large negative ΔH dominates at room temperature.
The second law states that the total entropy of the universe (system + surroundings) increases in any spontaneous process:
ΔS(total) = ΔS(system) + ΔS(surroundings) > 0 for a spontaneous process.
When an exothermic reaction occurs, energy is released to the surroundings, which increases the entropy of the surroundings:
ΔS(surroundings) = −ΔH / T
This is why many exothermic reactions are spontaneous even when ΔS(system) is negative — the increase in entropy of the surroundings more than compensates.
The dissolving of ammonium nitrate in water is endothermic (ΔH = +25.7 kJ mol⁻¹) yet it happens spontaneously. How?
At room temperature, TΔS > ΔH, so the process is feasible. This is formally quantified using Gibbs free energy in the next lesson.
| Mistake | Correction |
|---|---|
| Saying entropy is "randomness" | More precise: entropy is the number of ways to distribute energy among particles |
| Forgetting units are J K⁻¹ mol⁻¹, not kJ | Always check units before using in ΔG = ΔH − TΔS |
| Confusing S° with ΔS° | S° is the absolute entropy of a substance; ΔS° is the change for a reaction |
| Assuming negative ΔS means the reaction cannot occur | It can still occur if ΔH is sufficiently negative (exothermic) |
| Forgetting to multiply S° by the stoichiometric coefficient | Use 2 × S° if there are 2 moles of that substance in the equation |
Edexcel 9CH0 specification Topic 13 introduces standard entropy S° as a measure of disorder, requires the calculation of ΔS° = ΣS°(products) − ΣS°(reactants) for chemical and physical processes, and demands the prediction of the sign of ΔS° based on phase changes and changes in moles of gas (refer to the official specification document for exact wording). Entropy precedes Gibbs free energy in the spec sequence and provides the thermodynamic justification for why some endothermic processes (NH₄NO₃ dissolution) proceed spontaneously. Entropy questions appear on Paper 1 as quantitative items (ΔS calculation, sign prediction) and on Paper 3 as synoptic items combining entropy with equilibrium, Gibbs free energy, and group 2 chemistry. Standard entropies are tabulated in J K⁻¹ mol⁻¹ (note: not kJ); careful unit conversion is required when entropy is combined with enthalpy in Gibbs calculations.
Question (8 marks):
Standard entropies (J K⁻¹ mol⁻¹): N₂(g) 192; H₂(g) 131; NH₃(g) 193.
(a) Calculate ΔS° for the Haber synthesis: N₂(g) + 3H₂(g) → 2NH₃(g). (3)
(b) Predict the sign of ΔS° (without calculation) for: (i) H₂O(l) → H₂O(g); (ii) NaCl(s) → Na⁺(aq) + Cl⁻(aq); (iii) 2NO₂(g) → N₂O₄(g). Justify each prediction. (3)
(c) Explain why standard entropies of gases are typically much larger than those of liquids, which are larger than those of solids. (2)
Solution with mark scheme:
(a) ΔS° = ΣS°(products) − ΣS°(reactants) = 2(193) − [192 + 3(131)] = 386 − [192 + 393] = 386 − 585 = −199 J K⁻¹ mol⁻¹.
M1 — correct equation. M1 — correct substitution including stoichiometric coefficients (×2 for NH₃, ×3 for H₂). A1 — answer with correct units (J K⁻¹ mol⁻¹) and sign.
(b) (i) B1 — ΔS° positive: liquid → gas; molecular freedom increases dramatically; gas has many more accessible microstates.
(ii) B1 — ΔS° positive: 1 mole of solid → 2 moles of (hydrated) ions in solution; greater disorder despite ordering of water around each ion.
(iii) B1 — ΔS° negative: 2 moles of gas → 1 mole of gas; reduction in moles of gas is the dominant entropy criterion.
(c) B1 — gases have much greater translational, rotational and vibrational freedom than liquids; liquids have greater motional freedom than rigid solids.
B1 — therefore the number of accessible microstates is gas ≫ liquid > solid; entropy ∝ ln(microstates).
Total: 8 marks (M2 A1 B5).
Question (6 marks):
Standard entropies (J K⁻¹ mol⁻¹): CaCO₃(s) 92.9; CaO(s) 39.7; CO₂(g) 213.7.
(a) Calculate ΔS° for: CaCO₃(s) → CaO(s) + CO₂(g). (3)
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