# Entropy Calculation for JEE

> Entropy calculations for JEE Chemistry: Delta S for heating, expansion and phase change, Delta S of surroundings, spontaneity test and Trouton's rule.

- Canonical URL: https://prepelephant.com/topics/jee/chemistry/entropy-calculation-jee
- Exam / course: JEE · Subject: Chemistry
- Publisher: PrepElephant (https://prepelephant.com) — Prepared and reviewed by the PrepElephant Academic Review Team
- First published: 2026-10-02
- Last updated: 2026-10-02
- How to cite: "Entropy Calculation for JEE", PrepElephant, https://prepelephant.com/topics/jee/chemistry/entropy-calculation-jee

## 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.