Chemistry Labs

Physical chemistry

Enthalpy, entropy, Gibbs free energy

Combines enthalpy and entropy through Gibbs energy to predict the thermodynamically allowed direction at fixed temperature and pressure.

1 Lower secondary2 Upper secondary3 Undergraduate4 Advanced5 Research

IntuitionIntuition: downhill energy versus dispersal

Heat flow has two drives in disguise. A system that releases energy to its surroundings has a lower final enthalpy, which looks favourable, yet gases still expand against that enthalpy preference because spreading molecules and energy over more arrangements is itself favourable. At constant temperature and pressure the compromise between these two tendencies is the Gibbs energy.

Four sign combinations of ΔH and ΔS produce four qualitative behaviours; the grey zero line locates the crossover temperature.

SchoolSchool: the spontaneity criterion

Definition: Gibbs energy criterion

At constant temperature and pressure a change is thermodynamically allowed in the direction that lowers G: ΔG < 0 forward, ΔG > 0 reverse, ΔG = 0 at equilibrium. “Spontaneous” states the driving force only; it says nothing about how fast the change happens.

ΔG=ΔH−TΔS\Delta G=\Delta H-T\Delta S

Enthalpy H captures heat absorbed at constant pressure, H = U + pV. Entropy S measures how broadly energy and matter are spread among microscopic arrangements; it grows with disorder, mixing, expansion and temperature. The minus sign in −TΔS means that, at high enough temperature, a process that raises the system’s entropy can win over a modestly endothermic enthalpy change.

Sign analysis of ΔG = ΔH − TΔS
ΔHΔSBehaviour
−+ΔG < 0 at all temperatures
+−ΔG > 0 at all temperatures
−−favoured below T* = ΔH/ΔS
++favoured above T* = ΔH/ΔS

Example: Freezing just below 0 °C

At 268 K, freezing water has ΔH = −6.01 kJ mol⁻¹ and ΔS = −22.0 J mol⁻¹ K⁻¹. Estimate ΔG for the freezing process.

Solution

First convert ΔS to kJ mol⁻¹ K⁻¹: −0.0220. Then ΔG = −6.01 − 268 × (−0.0220) = −0.114 kJ mol⁻¹. The small negative value confirms that freezing is only weakly favoured five degrees below the melting point — as it should be, since ΔG = 0 at the melting point itself.

UndergraduateUniversity: standard states and the reaction quotient

The difference ΔrG measured under current conditions is split into a standard part, evaluated at a defined reference (usually 1 bar or 1 mol L⁻¹), and a term containing the reaction quotient Q:

ΔrG=ΔrG∘+RTln⁡Q;ΔrG∘=−RTln⁡K\Delta_r G=\Delta_r G^\circ+RT\ln Q\qquad;\qquad \Delta_r G^\circ=-RT\ln K

Activities, not raw concentrations, enter Q rigorously: gases appear as pᵢ/p°, solutes as cᵢ/c°, pure solids and liquids as 1. The second identity shows why K depends only on temperature for a fixed standard state: K = exp(−ΔrG°/RT). An exothermic reaction therefore normally has a smaller K when heated (van ’t Hoff relation).

Example: Crossover temperature

A reaction has ΔrH° = +58.0 kJ mol⁻¹ and ΔrS° = +176 J mol⁻¹ K⁻¹ (values close to N₂O₄(g) ⇌ 2 NO₂(g)). Estimate the temperature at which ΔrG° changes sign.

Solution

Set ΔrG° = 0: T* = ΔH°/ΔS° = 58 000/176 ≈ 330 K. Above about 330 K, TΔS° outweighs ΔH° and the dissociation becomes spontaneous under standard conditions — consistent with the well-known darkening of N₂O₄/NO₂ mixtures on warming.

AdvancedAdvanced: the thermodynamic structure behind G

Gibbs energy is a Legendre transform of the internal energy, G = U + pV − TS, chosen so its natural variables are (T, p) — the quantities the laboratory controls. Differentiating gives dG = −S dT + V dp for a closed system, so (∂G/∂T)ₚ = −S < 0: Gibbs energies always fall with temperature, faster for high-entropy phases, which is the deep reason vapours win at high T. For composition changes, the chemical potential μᵢ = (∂G/∂nᵢ) governs; chemical equilibrium is the equality of chemical potentials weighted by stoichiometry.

(∂G/∂T)p=−S(∂G/∂p)T=Vμi=(∂G∂ni)T,p(\partial G/\partial T)_p=-S\qquad(\partial G/\partial p)_T=V\qquad\mu_i=\left(\frac{\partial G}{\partial n_i}\right)_{T,p}

References