Gibbs Energy as the Criterion of Spontaneity

Ice melts in a warm room, but not because the melting itself lowers the room's entropy budget in an obvious way. The universe's entropy still rises, hidden in the warming of the surroundings. Gibbs energy repackages that awkward two-part bookkeeping — system entropy plus surroundings entropy — into one system-only quantity you can compute from a beaker of chemicals. This matters because almost every chemistry experiment runs at constant temperature and pressure, not constant energy or volume. The second law, stated with total entropy, is true but clumsy for a chemist standing at a bench with a thermometer and a barometer. Gibbs energy converts that universal law into a local, practical test. This article builds GG from the second law, derives why dG≤0\mathrm{d}G \le 0 at constant TT and pp, and shows exactly what "equilibrium" means in that language. Two full numerical examples follow. You will finish able to predict spontaneity from tabulated data alone, without ever mentioning the surroundings explicitly.

Key takeaways
  • At constant TT and pp, a process is spontaneous only if dG≤0\mathrm{d}G \le 0; equilibrium is the state where dG=0\mathrm{d}G = 0.
  • GG is built to absorb the surroundings' entropy change into a system-only function: G=H−TSG = H - TS.
  • The criterion follows directly from the second law, dSuniv≥0\mathrm{d}S_{\text{univ}} \ge 0, applied to a system exchanging heat with surroundings at fixed TT, pp.
  • ΔG<0\Delta G < 0 means spontaneous forward reaction; ΔG>0\Delta G > 0 means the reverse is spontaneous; ΔG=0\Delta G = 0 means equilibrium, not "no reaction."
  • Spontaneity is a statement about direction and driving force, not about speed — a large negative ΔG\Delta G can still describe a reaction that takes years.
  • Key idea: dG≤0\mathrm{d}G \le 0 at constant T,pT,p is the second law wearing a system-only disguise, valid because −ΔG/T-\Delta G/T equals the total entropy change of universe.
The core insight

Gibbs energy converts the universal, but surroundings-dependent, criterion dSuniv≥0\mathrm{d}S_{\text{univ}} \ge 0 into a criterion computable entirely from system properties, dG≤0\mathrm{d}G \le 0 at constant TT and pp, because at those fixed conditions ΔGsys\Delta G_{\text{sys}} is exactly −T-T times the total entropy change of the universe.

The physical picture

Consider a beaker sitting open to the atmosphere, thermostatted in a water bath. Both pressure and temperature are pinned by the surroundings — this is the constant-TT, constant-pp world where essentially all wet chemistry, biochemistry, and industrial synthesis happens. The second law says a process is spontaneous only if it increases the total entropy of the universe: system plus surroundings. The trouble is that "surroundings entropy" is inconvenient. It depends on how much heat leaks out and at what temperature the bath absorbs it. Josiah Willard Gibbs's insight, formalized in the 1870s, was to notice that at constant TT and pp the surroundings' entropy change can be written using only system quantities: heat leaving the system at temperature TT becomes −qp/T-q_p/T entering the surroundings, and qpq_p at constant pressure is just ΔH\Delta H of the system. Substituting this back into the total entropy inequality produces a new function, G=H−TSG = H - TS, whose sign change alone tells you whether the universe's entropy went up. Physically, GG is a compromise between two competing tendencies: systems want to lower their enthalpy (settle into stronger bonds, more stable arrangements) and simultaneously want to raise their own entropy (spread energy and matter more widely). At constant TT and pp, ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S weighs these two drives on a common footing, with temperature setting the exchange rate between them.

Figure
Gibbs energy as the balance of enthalpy and entropy driving forces
ΔH (bond, lattice energy) −TΔS (disorder term) ΔG = ΔH − TΔS Fulcrum: constant T, p ΔG less than 0 spontaneous ΔG greater than 0 non-spontaneous

A second picture worth holding in mind is how GG behaves as a reaction proceeds from pure reactants to pure products. Plotting GG against the extent of reaction ξ\xi gives a curve with a minimum — not necessarily at ξ=1\xi = 1 (complete conversion), but wherever the slope ∂G/∂ξ\partial G/\partial \xi first hits zero. That minimum, not "all reactants consumed," is chemical equilibrium.

Figure
Gibbs energy versus extent of reaction, showing the equilibrium minimum
extent of reaction ξ G ξ_eq dG/dξ = 0 pure reactants pure products

The core relations

G≡H−TSG \equiv H - TS

where: HH is enthalpy, TT is absolute temperature, SS is entropy, all of the system.

dSuniv=dSsys+dSsurr≥0\mathrm{d}S_{\text{univ}} = \mathrm{d}S_{\text{sys}} + \mathrm{d}S_{\text{surr}} \ge 0

where: the second law demands total entropy never decreases for any real process.

ΔGsys=−T ΔSuniv(at constant T,p)\Delta G_{\text{sys}} = -T\,\Delta S_{\text{univ}} \quad (\text{at constant } T, p)

where: this identity is the bridge — it converts the universal criterion into a system-only one.

dG≤0(constant T,p)\mathrm{d}G \le 0 \quad (\text{constant } T, p)

where: equality holds only at equilibrium; strict inequality marks a spontaneous direction.

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S

where: the working formula for computing spontaneity from tabulated thermodynamic data.

The core relations

Derivation AFrom the second law to dG ≤ 0

Goal. Show that at constant temperature TT and pressure pp, the second law's universal criterion dSuniv≥0\mathrm{d}S_{\text{univ}} \ge 0 is exactly equivalent to dG≤0\mathrm{d}G \le 0, using only system properties.

Step 1. Start from the second law. For any process, the Clausius statement of the second law requires

dSsys+dSsurr≥0.\mathrm{d}S_{\text{sys}} + \mathrm{d}S_{\text{surr}} \ge 0 .

Equality holds only for a reversible process; strict inequality marks irreversibility, i.e., a spontaneous real process.

Step 2. Express the surroundings' entropy change. The surroundings are taken to be a large thermal reservoir at fixed temperature TT, so any heat dqsurr\mathrm{d}q_{\text{surr}} they absorb changes their entropy reversibly:

dSsurr=dqsurrT=−dqsysT.\mathrm{d}S_{\text{surr}} = \frac{\mathrm{d}q_{\text{surr}}}{T} = \frac{-\mathrm{d}q_{\text{sys}}}{T} .

Heat leaving the system enters the surroundings, so the sign flips.

Step 3. Bring in constant pressure. At constant pressure with only pVpV work, the heat absorbed by the system equals its enthalpy change,

dqsys,p=dH.\mathrm{d}q_{\text{sys},p} = \mathrm{d}H .

This is the standard result from the first law applied at constant pp — a prerequisite result, not re-derived here.

Step 4. Substitute back into the second law. Combining Steps 1–3,

dSsys−dHT≥0.\mathrm{d}S_{\text{sys}} - \frac{\mathrm{d}H}{T} \ge 0 .

Multiply through by −T-T (a negative number, so the inequality reverses):

dH−T dSsys≤0.\mathrm{d}H - T\,\mathrm{d}S_{\text{sys}} \le 0 .

Step 5. Recognize the left side as dG. At constant TT, the differential of G=H−TSG = H - TS is

dG=dH−T dS−S dT=dH−T dS(dT=0).\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S - S\,\mathrm{d}T = \mathrm{d}H - T\,\mathrm{d}S \quad (\mathrm{d}T = 0).

This is precisely the left-hand side of Step 4.

Step 6. Conclude.

dG≤0at constant T,p\boxed{\mathrm{d}G \le 0 \quad \text{at constant } T, p}

Equality corresponds to a reversible change or to equilibrium; strict inequality corresponds to a spontaneous irreversible process. No surroundings variable appears — the entire criterion is now internal to the system.

Worked examples

Example 1 — Foundational: is ice melting at 263 K and 298 K spontaneous? For the fusion of ice, H2O(s)→H2O(l)\mathrm{H_2O(s)} \rightarrow \mathrm{H_2O(l)}, standard data at 273 K give

ΔHfus=+6.01 kJ mol−1,ΔSfus=+22.0 J K−1mol−1.\Delta H_{\text{fus}} = +6.01\ \text{kJ mol}^{-1}, \qquad \Delta S_{\text{fus}} = +22.0\ \text{J K}^{-1}\text{mol}^{-1}.

(These are consistent: ΔHfus/Tfus=6010 J mol−1/273 K=22.0 J K−1mol−1\Delta H_{\text{fus}}/T_{\text{fus}} = 6010\ \text{J mol}^{-1} / 273\ \text{K} = 22.0\ \text{J K}^{-1}\text{mol}^{-1}, confirming equilibrium exactly at the melting point.) Treat ΔH\Delta H and ΔS\Delta S as roughly temperature-independent over a narrow range near 273 K, and evaluate ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta S at two temperatures. At T=263T = 263 K (−10 °C, supercooled liquid region / normal freezing conditions):

ΔG=6010 J mol−1−(263 K)(22.0 J K−1mol−1)=6010−5786=+224 J mol−1.\Delta G = 6010\ \text{J mol}^{-1} - (263\ \text{K})(22.0\ \text{J K}^{-1}\text{mol}^{-1}) = 6010 - 5786 = +224\ \text{J mol}^{-1}.

ΔG>0\Delta G > 0: melting is non-spontaneous at 263 K; the reverse process, freezing, is spontaneous. This matches everyday experience — water freezes below 0 °C. At T=298T = 298 K (25 °C):

ΔG=6010 J mol−1−(298 K)(22.0 J K−1mol−1)=6010−6556=−546 J mol−1.\Delta G = 6010\ \text{J mol}^{-1} - (298\ \text{K})(22.0\ \text{J K}^{-1}\text{mol}^{-1}) = 6010 - 6556 = -546\ \text{J mol}^{-1}.

ΔG<0\Delta G < 0: melting is spontaneous at room temperature — ice melts, as observed. At exactly T=273.0T = 273.0 K,

ΔG=6010−(273.0)(22.0)=6010−6006=+4 J mol−1≈0,\Delta G = 6010 - (273.0)(22.0) = 6010 - 6006 = +4\ \text{J mol}^{-1} \approx 0,

the small residual reflecting rounding in the tabulated ΔS\Delta S; physically this is the coexistence point, ΔG=0\Delta G = 0, solid and liquid in equilibrium. The sign flip between 263 K and 298 K, straddling 273 K, is exactly the expected physical behavior — a strong sanity check.

Example 2 — Applied: does calcium carbonate decompose spontaneously at 800 °C and 1 bar? Consider CaCO3(s)→CaO(s)+CO2(g)\mathrm{CaCO_3(s)} \rightarrow \mathrm{CaO(s)} + \mathrm{CO_2(g)}, important in cement manufacture (calcination of limestone). Standard thermodynamic data at 298 K:

ΔH∘=+178.3 kJ mol−1,ΔS∘=+160.6 J K−1mol−1.\Delta H^{\circ} = +178.3\ \text{kJ mol}^{-1}, \qquad \Delta S^{\circ} = +160.6\ \text{J K}^{-1}\text{mol}^{-1}.

Step 1: check at 298 K.

ΔG298∘=178,300 J mol−1−(298 K)(160.6 J K−1mol−1)\Delta G^{\circ}_{298} = 178{,}300\ \text{J mol}^{-1} - (298\ \text{K})(160.6\ \text{J K}^{-1}\text{mol}^{-1})
=178,300−47,859=+130,441 J mol−1≈+130.4 kJ mol−1.= 178{,}300 - 47{,}859 = +130{,}441\ \text{J mol}^{-1} \approx +130.4\ \text{kJ mol}^{-1}.

Strongly positive: limestone does not decompose spontaneously at room temperature, consistent with limestone buildings surviving centuries.

Step 2: check at 800 °C = 1073 K, the industrial calcination temperature, assuming ΔH∘\Delta H^{\circ} and ΔS∘\Delta S^{\circ} stay roughly constant (a simplification revisited when temperature-dependence of GG is treated later):

ΔG1073∘=178,300 J mol−1−(1073 K)(160.6 J K−1mol−1)\Delta G^{\circ}_{1073} = 178{,}300\ \text{J mol}^{-1} - (1073\ \text{K})(160.6\ \text{J K}^{-1}\text{mol}^{-1})
=178,300−172,363=+5,937 J mol−1≈+5.9 kJ mol−1.= 178{,}300 - 172{,}363 = +5{,}937\ \text{J mol}^{-1} \approx +5.9\ \text{kJ mol}^{-1}.

Still slightly positive at standard pressure of CO₂ (1 bar) — decomposition is not yet spontaneous under standard conditions, though the driving force has shrunk by a factor of 22.

Step 3: find the temperature where ΔG° = 0 (the "onset" temperature under standard conditions):

T=ΔH∘ΔS\circo=178,300 J mol−1160.6 J K−1mol−1=1110 K=837 ∘C.T = \frac{\Delta H^{\circ}}{\Delta S^{\circo}} = \frac{178{,}300\ \text{J mol}^{-1}}{160.6\ \text{J K}^{-1}\text{mol}^{-1}} = 1110\ \text{K} = 837\ ^{\circ}\text{C}.

Above roughly 837 °C, ΔG∘\Delta G^{\circ} turns negative and calcination becomes spontaneous at standard 1 bar CO₂ pressure — matching the well-known industrial kiln temperatures (900–1000 °C, used partly for reaction rate and partly to push well past this threshold, and because real kilns keep CO₂ partial pressure below 1 bar, lowering the threshold further via the reaction quotient). The 837 °C figure is a benchmark every cement chemistry course reproduces.

Limiting cases & common traps

SituationWhat ΔG\Delta G tells youCommon trap
ΔG<0\Delta G < 0Forward process spontaneous as writtenAssuming it happens fast — spontaneity says nothing about rate
ΔG>0\Delta G > 0Reverse process spontaneousConcluding the forward reaction "cannot occur at all," ignoring that a nonzero reverse population still exists at equilibrium
ΔG=0\Delta G = 0System at equilibriumThinking no reaction occurs — forward and reverse rates are equal but nonzero
ΔH<0,ΔS>0\Delta H < 0, \Delta S > 0Spontaneous at all TTForgetting this is the only quadrant with no temperature dependence of sign
ΔH>0,ΔS<0\Delta H > 0, \Delta S < 0Non-spontaneous at all TT—
Constant T,VT,V instead of T,pT,pUse A=U−TSA = U - TS (Helmholtz), not GGApplying dG≤0\mathrm{d}G \le 0 to a rigid, insulated bomb calorimeter step

The chemist behind it

Josiah Willard Gibbs, working largely in isolation at Yale in the 1870s, published "On the Equilibrium of Heterogeneous Substances" — a paper so dense and so far ahead of experimental chemistry that it took decades for chemists to fully absorb it. He introduced the free energy function, chemical potential, and the phase rule in one extended memoir, essentially inventing chemical thermodynamics as a predictive discipline rather than a bookkeeping exercise for engines.

Origins · 1 minJosiah Willard Gibbs and how this was discovered
JW
Josiah Willard Gibbs
1839–1903 · American mathematical physicist who founded chemical thermodynamics with a single monumental 1876 memoir

Ask the tutor: How does ΔG connect to the equilibrium constant K, and why does that relationship change with temperature?

Why isn't ΔS of the system alone the criterion for spontaneity?

The second law requires only that total entropy — system plus surroundings — never decreases. A system's own entropy can decrease during a spontaneous process, as when water freezes and becomes more ordered, provided the surroundings' entropy rises by more. Gibbs energy is useful precisely because it folds that surroundings term into a system-only quantity, so you never have to track the bath separately.

Does ΔG = 0 mean the reaction has stopped?

No. At equilibrium, forward and reverse reactions continue at equal rates, so there is no net change, but molecules keep converting in both directions constantly. ΔG=0\Delta G = 0 is a statement about the balance of driving forces at that specific composition, not a statement that molecular motion or reaction events have ceased.

Why does the criterion require both constant T and constant p?

The derivation substitutes qp=ΔHq_p = \Delta H, valid only when pressure is constant, and assumes the surroundings act as a reservoir fixing TT. Change either constraint and GG stops being the natural function: at constant TT and VV, the analogous role is played by the Helmholtz energy A=U−TSA = U - TS, whose criterion is dA≤0\mathrm{d}A \le 0.

How does this connect to equilibrium constants and reaction Gibbs energy?

This article establishes that dG≤0\mathrm{d}G \le 0 governs spontaneity at fixed T,pT,p. A closely related quantity, the reaction Gibbs energy ΔrG\Delta_r G, tracks how GG changes with extent of reaction and links directly to the equilibrium constant KK via ΔrG∘=−RTln⁡K\Delta_r G^{\circ} = -RT\ln K. The temperature dependence of GG itself, needed to extend today's examples to arbitrary temperatures rigorously, is treated separately.

Is Gibbs energy the same as "free energy" and is it always the Gibbs free energy people mean?

"Free energy" historically referred to either Gibbs energy GG (constant T,pT,p) or Helmholtz energy AA (constant T,VT,V); modern IUPAC usage prefers the unambiguous names Gibbs energy and Helmholtz energy. In solution and biochemistry, where reactions run open to the atmosphere at fixed pressure, "free energy" essentially always means Gibbs energy unless stated otherwise.

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