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23.3 Entropy change, ΔS

Syllabus
9701–2028–2029
Topic
23.3
Level
A2

Learning objectives

23.3.1Terms: entropy, S, as the number of possible• Define: entropy, S, as the number of possible arrangements of the particles and their energy in a given system23.3.2Predict/explain the sign of the entropy• Predict/explain: the sign of the entropy changes that occur:- (a) during a change in state, e.g. melting, boiling and dissolving (and their reverse)- (b) during a temperature change- (c) during a reaction in which there is a change in the number of gaseous molecules23.3.3The entropy change for a reaction, ΔS• Calculate the entropy change for a reaction, ΔS, given the standard entropies, S ⦵ , of the reactants and products, ΔS ⦵ = ΣS ⦵: (products) – ΣS ⦵ (reactants) (use of ΔS ⦵ = ΔSsurr + ΔSsys is not required)

Entropy counts possible arrangements of particles and energy

Entropy, S, is the number of possible arrangements of the particles and their energy in a given system. More possible arrangements mean higher entropy.

The arrangements include where particles can be and how the system's energy can be distributed. A gas therefore usually has higher entropy than the same amount of the same substance as a liquid or solid because its particles and energy have many more possible arrangements.

S(gas)>S(liquid)>S(solid)for the same substance under comparable conditionsS(\mathrm{gas})>S(\mathrm{liquid})>S(\mathrm{solid})\quad\text{for the same substance under comparable conditions}

Use the syllabus definition rather than treating entropy as everyday 'messiness'. Entropy is a property of the stated system, so identify its particles, state and conditions before comparing arrangements.

Predict the sign of ΔS from the change in accessible arrangements

Change in the system Sign of ΔS Particle-and-energy reason
solid → liquid; liquid → gas + particles gain freedom and more arrangements become possible
gas → liquid; liquid → solid − particles become more constrained
solid solute → dilute solution + solute particles become dispersed through the solvent
crystallisation from solution − dispersed particles enter an ordered lattice
temperature increases without a phase change + energy can be distributed among more accessible energy states
temperature decreases − fewer energy arrangements are accessible
reaction forms more gaseous molecules + gas particles have more positional and energy arrangements
reaction forms fewer gaseous molecules − gas-particle arrangements decrease

Count gaseous molecules using the balanced-equation coefficients and ignore solid, liquid and aqueous coefficients for this specific gas-count test. For N₂(g) + 3H₂(g) → 2NH₃(g), four gaseous molecules become two, so ΔS is predicted to be negative.

Melting and boiling have positive ΔS; freezing and condensing have negative ΔS. CaCO₃(s) → CaO(s) + CO₂(g) has positive ΔS because gas is produced from solids.

If the number of gaseous molecules is unchanged, gas count alone gives no sign. A qualitative prediction is not a numerical calculation, and a positive ΔS alone does not establish feasibility; Gibbs free energy is treated in 23.4.

Calculate ΔS° with coefficients and the correct physical states

ΔS∘=∑νS∘(products)−∑νS∘(reactants)\Delta S^\circ=\sum \nu S^\circ(\mathrm{products})-\sum \nu S^\circ(\mathrm{reactants})

Use the standard molar entropy for each substance in its stated physical state. Multiply every S° value by its stoichiometric coefficient, total the product side and reactant side separately, then subtract reactants from products.

For 2Mg(s) + O₂(g) → 2MgO(s):

Substance S° / J K⁻¹ mol⁻¹ Coefficient Contribution
Mg(s) 32.60 2 65.20
O₂(g) 205.0 1 205.0
MgO(s) 38.20 2 76.40

ΔS∘=(2×38.20)−[(2×32.60)+205.0]=−193.8 J K−1 mol−1\Delta S^\circ=(2\times38.20)-[(2\times32.60)+205.0]=-193.8\ \mathrm{J\ K^{-1}\ mol^{-1}}

The negative sign agrees with the qualitative change: one mole of gas is consumed and only solids remain. Keep entropy in J K⁻¹ mol⁻¹; do not convert it to kJ until a later equation explicitly requires matching kJ units.

The tabulated quantities are S°, not ΔS° values for individual substances. Do not reverse products minus reactants, omit coefficients, use the wrong physical state, or introduce ΔSsurr; the latter is explicitly not required here.