Entropy Change

Quantifies the change in disorder: ΔS = q_rev/T for a process, or products minus reactants of tabulated standard entropies.

\Delta S = \frac{q_{\text{rev}}}{T} \quad ; \quad \Delta S^{\circ}_{\text{rxn}} = \sum n S^{\circ}(\text{products}) - \sum m S^{\circ}(\text{reactants})

ΔS = q_rev/T ; ΔS°rxn = ΣnS°(products) − ΣmS°(reactants)

What it means

Entropy is nature's bookkeeping for disorder, and its change measures how irreversibly a process spreads energy. The defining relation divides reversible heat by absolute temperature: the same joule of heat creates more entropy at low temperature than at high, because it represents a larger relative disturbance. For reactions, tabulated standard molar entropies let you compute ΔS°rxn as products minus reactants — more gas moles on the product side almost always means a positive ΔS. When to use it: pairing with ΔH in the Gibbs equation to judge spontaneity, or assessing whether a process increases or decreases the universe's total entropy.

Variables

Symbol	Name	Unit	Meaning
$ΔS$	Entropy change	J/(mol·K)	The change in entropy; positive means more disorder.
$q_rev$	Reversible heat	J	Heat exchanged along a reversible (equilibrium) path.
$T$	Temperature	K	Absolute temperature at which the heat is transferred — kelvin required.
$S°$	Standard molar entropy	J/(mol·K)	Tabulated standard molar entropy of a species.

Assumptions

The heat transfer occurs reversibly (a quasi-static, equilibrium path) for ΔS = q_rev/T.
Standard entropies S° refer to 25°C and 1 atm; stoichimetric coefficients weight each term.
Entropy is a state function, so the path between states does not matter.

Worked example

Problem

Find the entropy change when 1.00 mol of ice melts at 273 K (ΔHfus = 6.01 kJ/mol).

Given:

1.00 mol of ice melting at its melting point, T = 273 K
Heat absorbed: q = ΔHfus = 6.01 kJ/mol = 6010 J/mol

Melting at the melting point is a reversible isothermal process, so ΔS = q_rev / T.
Convert the heat to joules: 6.01 kJ/mol = 6010 J/mol (positive — heat absorbed).
Substitute: ΔS = 6010 J/mol / 273 K.
Divide: 6010 / 273 = 22.018 J/(mol·K), which rounds to 22.0 J/(mol·K).

ΔS°fus = 22.0 J/(mol·K) — melting increases disorder

Melting absorbs heat reversibly at 273 K and raises the system's entropy by 22.0 J per kelvin per mole — liquid water is more disordered than ice.