The big idea: Entropy, given the symbol S, measures how dispersed (spread out) the matter and energy of a system are.
The more ways the particles and their energy can be arranged, the higher the entropy.
A tidy, ordered arrangement has few ways to arrange it → low entropy. A jumbled, spread-out arrangement has many ways → high entropy.
Two things get dispersed: Entropy goes up when either is spread out more:
- Matter — particles free to move into more positions (a gas filling a room vs a solid block). - Energy — the same total energy shared over more particles and more ways of moving (vibrating, rotating, translating).
So melting, boiling, dissolving and especially making more gas all tend to raise entropy.
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ΔS is the change in entropy: positive (ΔS > 0) means the system becomes more dispersed; negative (ΔS < 0) means it becomes more ordered.
You can usually predict the sign without any data by asking: does the change spread the particles and energy out more, or pack them in more?
ΔS is POSITIVE (+)
- solid → liquid → gas (melting, boiling, subliming)
- a solid or liquid dissolves
- moles of gas increase (strongest effect)
ΔS is NEGATIVE (−)
- gas → liquid → solid (condensing, freezing)
- moles of gas decrease
- two species combine into fewer particles
Gas moles dominate
- count gas moles on each side first
- more gas on the right → ΔS > 0
- no change in gas moles → look at state/solution
Gas moles are the deciding factor: Gases have far higher entropy than liquids or solids, so the change in the number of moles of gas usually decides the sign of ΔS.
Compare moles of gas on each side of the balanced equation. If gas moles rise, ΔS is positive; if they fall, ΔS is negative.
Worked example — predicting the sign of ΔS
Predict the sign of ΔS for the decomposition of dinitrogen tetroxide:
N2O4(g) → 2NO2(g)
Solution
- Count the moles of gas on each side:
- Gas moles increase (1 → 2), so the particles and their energy become more dispersed.
- More gas means more ways to arrange matter and energy, so:
Final answer
ΔS is positive — the number of moles of gas increases from 1 to 2.
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The standard molar entropy, S°, is the entropy of one mole of a substance under standard conditions (100 kPa, a stated temperature, usually 298 K).
Its units are J K⁻¹ mol⁻¹ — note the J (joules, not kJ), unlike enthalpy values.
Key facts about S°: - S° values are always positive for any substance above 0 K. - For the same substance, gas > liquid > solid — gases have the most dispersed matter and energy. - A perfect crystal at 0 K has S = 0 — the particles are in one perfectly ordered arrangement with no thermal motion (the only zero-entropy state).
Watch the units: S° is in J K⁻¹ mol⁻¹, but enthalpy (ΔH) is usually in kJ mol⁻¹.
When you later combine them (in 4.4.2), you must convert so both are in the same energy unit — a very common slip.
Worked example — ranking entropy by state
Place these three substances in order of increasing standard molar entropy and explain your choice: diamond (a solid), liquid bromine, and chlorine gas.
Solution
- Recall the rule: for comparable substances, gas > liquid > solid because gases have the most dispersed matter and energy.
- Diamond is a rigid solid (ordered lattice) → lowest S°. Liquid bromine has more freedom → middle. Chlorine gas particles are spread out → highest S°.
- So the order of increasing S° is:
Final answer
diamond (s) < Br2(l) < Cl2(g) — gas has the highest entropy, the solid the lowest.
To get the standard entropy change of a reaction, add up the S° of the products and subtract the S° of the reactants — each multiplied by its stoichiometric coefficient.
- standard entropy change of the reaction (J K⁻¹ mol⁻¹)
- standard molar entropy of each product (J K⁻¹ mol⁻¹)
- standard molar entropy of each reactant (J K⁻¹ mol⁻¹)
- sum, with each S° multiplied by its stoichiometric coefficient
How this is tested: R1.4 is HL-only content, so it appears on Paper 2 (short-answer / extended response) and Paper 1A (MCQ).
- Paper 1A: predict the sign of ΔS from an equation, or pick the correct ΔS° expression. - Paper 2: a part-marks ΔS° calculation from a small S° table, then often a comment on the sign.
The classic traps: ignoring the coefficients, and mixing J (entropy) with kJ (enthalpy).
Three easy marks: (1) Write ΔS° = ΣS°(products) − ΣS°(reactants) first. (2) Multiply each S° by its coefficient. (3) Keep the unit J K⁻¹ mol⁻¹ and check the sign matches the gas-mole change.
IB-style question — combustion of methanol (a)
Methanol burns according to:
2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)
Using the standard molar entropies below, (a) calculate ΔS° for this reaction.
S°: CH3OH(l) = 127, O2(g) = 205, CO2(g) = 214, H2O(g) = 189 (all in J K⁻¹ mol⁻¹).
Solution
- Formula first — write the relationship:
- Sum the products, each × its coefficient:
- Sum the reactants, each × its coefficient:
- Subtract — keep the unit:
Final answer
ΔS° = +315 J K⁻¹ mol⁻¹.
IB-style question — combustion of methanol (b)
(b) Explain, in terms of the moles of gas, why the sign of ΔS° you calculated is reasonable. [2]
How to score the marks
- State the change in moles of gas across the equation.
- Link more moles of gas to greater dispersal of matter and energy, hence a positive ΔS° — consistent with the +315 J K⁻¹ mol⁻¹ found.
Final answer
Moles of gas rise from 3 to 6, so matter and energy become more dispersed; ΔS° is therefore positive, agreeing with the calculated +315 J K⁻¹ mol⁻¹.