Entropy Calculation for JEE
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Direct answer
One mole of ice melting at 0°C absorbs about 6 kJ and gains 22 J K^-1 of entropy — that division, delta-S = delta-H(transition)/T, is the first of the three entropy recipes JEE tests. The second covers heating an ideal gas: delta-S = n Cp ln(T2/T1) at constant pressure, and the third covers expansion: delta-S = nR ln(V2/V1) for an isothermal change, or nR ln(p1/p2) in pressure form. Spontaneity is judged by the total entropy of system plus surroundings, delta-S(total) = delta-S(system) + delta-S(surroundings) > 0, with equilibrium where it equals zero; at constant temperature and pressure the surroundings contribute delta-S(surroundings) = −delta-H(system)/T.
What you must remember
- Phase transitions: delta-S = delta-H(trans)/T — fusion of ice gives 6000/273 = 22 J K^-1 mol^-1; vaporisation of water gives 40,660/373 = 109 J K^-1 mol^-1.
- Trouton's rule: most normal liquids vaporise with delta-S(vap) near 85 J K^-1 mol^-1 at their boiling points; water overshoots to 109 because hydrogen bonding orders the liquid.
- State function discipline: entropy change is path-independent, so compute it along any reversible route between the same states, even for an irreversible real process.
- Surroundings term: delta-S(surroundings) = −delta-H(system)/T at constant T, p — the piece students most often forget when asked about "the universe".
- Third law and standard entropies: a perfect crystal has S = 0 at 0 K; standard molar entropies rank gas above liquid above solid (N2 gas 191.6, graphite 5.7, diamond 2.4 J K^-1 mol^-1).
- Sign instincts: gas moles increasing makes delta-S positive; dissolving a gas in a liquid makes it negative; mixing similar liquids positive.
Two calculations that cover the syllabus
Heat 2 mol of an ideal gas (Cp = 29 J K^-1 mol^-1) from 300 K to 600 K, then expand it isothermally to double its volume. Heating: delta-S = nCp ln(T2/T1) = 2 × 29 × ln 2 = 58 × 0.693 = 40.2 J K^-1. Expansion: delta-S = nR ln(V2/V1) = 2 × 8.314 × ln 2 = 11.5 J K^-1. Total 51.7 J K^-1 — and note both numbers came from the same ln 2, which is why exam papers love factor-of-two changes.
Now the equilibrium insight wrapped in a phase change. Vaporising water at exactly 373 K: delta-S(system) = +109 J K^-1 mol^-1, delta-S(surroundings) = −109 J K^-1 mol^-1, total zero — the boiling point is where the two precisely cancel, which is the thermodynamic definition of a normal boiling point. Heat the water a few degrees more and the surroundings' loss shrinks below the system's gain, the total turns positive, and vaporisation turns spontaneous. Entropy arithmetic quietly becomes a phase-diagram argument.
Where students slip
Units cause more losses than concepts: delta-H arrives in kilojoules while entropy lives in joules per kelvin, so every division by T needs a × 1000 — the single most common numerical error in this chapter. The second slip is the surroundings' sign: an exothermic reaction heats the surroundings, so delta-S(surroundings) is positive, and writing it as delta-H/T instead of −delta-H/T flips the spontaneity verdict. The third is expecting system entropy to rise in every spontaneous process — freezing water below 0°C has delta-S(system) negative yet spontaneous, because the surroundings' gain outweighs it. JEE Advanced adds the reversible-path requirement: compute along a reversible route even when the actual change is free expansion, since S is a state function and the value is identical.
Frequently asked questions
How do you calculate entropy change for a phase transition?
As delta-S = delta-H(transition)/T at the transition temperature, because temperature stays constant while the enthalpy change is absorbed reversibly.
What is delta-S for the surroundings of an exothermic reaction?
Positive and equal to −delta-H(system)/T at constant temperature and pressure — heat released to the surroundings disorder them.
Why does water violate Trouton's rule?
Its boiling entropy is about 109 J K^-1 mol^-1 against the typical 85, because hydrogen bonding makes liquid water more ordered than ordinary liquids, so vaporisation disorderises it more.
When is the total entropy change zero?
At equilibrium, where delta-S(system) exactly cancels delta-S(surroundings) — as in vaporisation at the normal boiling point.
How is entropy change found for an irreversible process?
Compute delta-S along any convenient reversible path between the same initial and final states, because entropy is a state function independent of path.