The Gibbs Energy

Imagine a beaker sitting open on a bench, held at constant temperature and constant atmospheric pressure. Almost every reaction a chemist actually runs happens exactly like this. The question is simple to ask but subtle to answer: how do we predict, from thermodynamic data alone, whether a reaction will proceed? Entropy alone cannot answer this, because the surroundings also gain or lose entropy as heat flows in or out. J. Willard Gibbs solved this in the 1870s by defining a single function that bundles enthalpy and entropy together. That function, the Gibbs energy $G$ , becomes the natural potential for constant-temperature, constant-pressure change. This article builds $G = H - TS$ from first principles. You will see why it is the right variable for benchtop chemistry. Then you will see how it connects to the Helmholtz energy. Finally you will see two full numerical examples showing $\Delta G$ decide spontaneity.

Key takeaways

$G = H - TS$ combines enthalpy and entropy into one state function with units of energy.
At constant $T$ and $p$ , spontaneous processes have $dG \le 0$ ; equilibrium is $dG = 0$ .
$G$ is the constant- $T$ , $p$ analogue of the Helmholtz energy $A$ , which governs constant- $T$ , $V$ processes.
$\Delta G$ measures the maximum non-expansion work extractable from a process (foreshadowing later material).
Standard Gibbs energies of formation let you calculate $\Delta_rG^\circ$ for any reaction from tabulated data.
Key idea: $G$ is the thermodynamic potential that lets a closed system's entropy-maximization criterion be rewritten purely in terms of the system's own enthalpy and entropy, at fixed $T$ and $p$ .

The core insight

Because real chemistry happens at constant temperature and pressure, not constant energy or volume, we need a state function whose sign alone tells us the direction of spontaneous change without our having to separately track the entropy of the surroundings; $G = H - TS$ is precisely that function, built so that $\Delta G = \Delta H - T\Delta S$ folds the surroundings' entropy change into the system's own accounting.

The physical picture

Spontaneity is fundamentally governed by the second law: for an isolated system, entropy increases until equilibrium, $dS_{\text{univ}} \ge 0$ . But a beaker on a bench is not isolated. It exchanges heat with the surroundings to stay at constant $T$ , and it exchanges volume work with the atmosphere to stay at constant $p$ . Tracking the total entropy of system plus surroundings is inconvenient — we want a criterion built entirely from properties of the system. Consider a system in thermal contact with surroundings at temperature $T$ , undergoing a process at constant pressure $p$ . Heat released by the system, $-dq$ , flows into the surroundings reversibly at the same $T$ , so the surroundings' entropy changes by $dS_{\text{surr}} = -dq_p/T$ . At constant pressure with only expansion work, $dq_p = dH$ . The second law requires

dS_{\text{sys}} + dS_{\text{surr}} \ge 0

dS_{\text{sys}} - \frac{dH}{T} \ge 0

Multiplying through by $-T$ (which flips the inequality since $T > 0$ ) gives $dH - T\,dS_{\text{sys}} \le 0$ . This combination of purely-system quantities, $H - TS$ , is defined as the Gibbs energy $G$ . Its differential decrease at constant $T$ and $p$ is exactly the criterion for spontaneous change, expressed without ever looking at the surroundings directly. This mirrors how the Helmholtz energy $A = U - TS$ was built for constant $T$ and $V$ , using $dq_V = dU$ instead of $dq_p = dH$ . Enthalpy replaces internal energy because pressure, not volume, is now the controlled mechanical variable. The pattern is general: whichever pair of variables you hold fixed, a Legendre transform of $U$ produces the natural potential for that pair.

Figure

Constructing G from H, S, and T at constant pressure

Figure

G versus reaction progress: spontaneity and equilibrium

The core relations

G \equiv H - TS

where: $G$ is Gibbs energy, $H$ enthalpy, $T$ absolute temperature, $S$ entropy — all state functions of the system.

dG = dH - T\,dS - S\,dT

where: this is the exact differential of $G$ , valid for any process, reversible or not.

\Delta G = \Delta H - T\Delta S \quad (\text{constant } T)

where: the practical working formula for isothermal processes, using system $\Delta H$ and $\Delta S$ only.

dG \le 0 \quad (\text{constant } T, p, \text{ no non-expansion work})

where: equality holds at equilibrium; strict inequality signals spontaneous change in the forward direction.

G = A + pV

where: $A = U - TS$ is the Helmholtz energy; $G$ and $A$ are related exactly as $H$ and $U$ are, by $+pV$ .

The core relations

Derivation AFrom the Clausius inequality to dG ≤ 0

Goal. Show that for a closed system at constant temperature and pressure, exchanging only expansion work with its surroundings, spontaneous change corresponds to $dG \le 0$ , using only system-level quantities.

Step 1. Start from the Clausius inequality. For any process, the entropy change of the system obeys

dS \ge \frac{dq}{T}

where $dq$ is the heat absorbed by the system from surroundings at temperature $T$ . Equality holds for a reversible process; the inequality holds for an irreversible (spontaneous) one.

Step 2. Specialize to constant pressure, expansion work only. If the only work is $p\,dV$ against a constant external pressure equal to the system pressure, the first law gives $dU = dq - p\,dV$ , so $dq = dU + p\,dV$ . At constant pressure this is $dq_p = d(U + pV) = dH$ . Substituting,

dS \ge \frac{dH}{T}

Step 3. Rearrange to isolate a system-only inequality. Multiply both sides by $T > 0$ (inequality direction unchanged) and rearrange:

T\,dS \ge dH

dH - T\,dS \le 0

Step 4. Recognize the combination as a differential of G. At constant $T$ , $d(TS) = T\,dS$ , so $dH - T\,dS = dH - d(TS) = d(H - TS)$ . Since $G \equiv H - TS$ , this is simply $dG$ . Hence

dG \le 0 \qquad (\text{constant } T, p)

Step 5. Interpret the equality case. When the process is reversible, the system is infinitesimally close to equilibrium at every stage, and $dG = 0$ . This is the condition satisfied at a genuine chemical equilibrium at fixed $T$ and $p$ : the Gibbs energy sits at a minimum with respect to the extent of reaction.

\boxed{dG \le 0 \text{ (spontaneous)}, \quad dG = 0 \text{ (equilibrium)} \quad \text{at constant } T, p}

Worked examples

Example 1 — Foundational: melting of ice at two temperatures. For the process $\text{H}_2\text{O}(s) \rightarrow \text{H}_2\text{O}(l)$ at 1 bar, take $\Delta H_{\text{fus}} = +6.01\ \text{kJ mol}^{-1}$ and $\Delta S_{\text{fus}} = +22.0\ \text{J K}^{-1}\text{mol}^{-1}$ (both roughly constant over the small temperature range considered). At $T = 263\ \text{K}$ ( $-10^\circ$ C):

\Delta G = \Delta H - T\Delta S = 6010\ \text{J mol}^{-1} - (263\ \text{K})(22.0\ \text{J K}^{-1}\text{mol}^{-1})

\Delta G = 6010 - 5786 = +224\ \text{J mol}^{-1}

Positive $\Delta G$ : melting is non-spontaneous at $-10^\circ$ C. Water below 0°C stays frozen — consistent with everyday experience. At $T = 283\ \text{K}$ ( $+10^\circ$ C):

\Delta G = 6010 - (283)(22.0) = 6010 - 6226 = -216\ \text{J mol}^{-1}

Negative $\Delta G$ : melting is spontaneous at $+10^\circ$ C, matching observation. At $T = \Delta H/\Delta S = 6010/22.0 = 273\ \text{K}$ , $\Delta G = 0$ exactly — the two phases coexist at equilibrium, correctly reproducing the normal melting point of ice, 0°C.

Example 2 — Applied: is a reaction spontaneous under standard conditions, and how does temperature shift the answer? Consider the synthesis of ammonia,

\text{N}_2(g) + 3\text{H}_2(g) \rightarrow 2\text{NH}_3(g)

Standard formation data at 298 K: $\Delta_fH^\circ(\text{NH}_3, g) = -46.1\ \text{kJ mol}^{-1}$ ; $S^\circ(\text{N}_2) = 191.6$ , $S^\circ(\text{H}_2) = 130.7$ , $S^\circ(\text{NH}_3) = 192.5\ \text{J K}^{-1}\text{mol}^{-1}$ (all elements in standard states have $\Delta_fH^\circ = 0$ ). Step 1: reaction enthalpy.

\Delta_rH^\circ = 2(-46.1) - [0 + 3(0)] = -92.2\ \text{kJ mol}^{-1}

Step 2: reaction entropy.

\Delta_rS^\circ = 2(192.5) - [191.6 + 3(130.7)] = 385.0 - 583.7 = -198.7\ \text{J K}^{-1}\text{mol}^{-1}

The large negative $\Delta_rS^\circ$ reflects four moles of gas becoming two — a substantial loss of translational and configurational entropy, so the surroundings must absorb entropy via heat release for this to be favorable. Step 3: Gibbs energy at 298 K.

\Delta_rG^\circ = \Delta_rH^\circ - T\Delta_rS^\circ = -92200\ \text{J mol}^{-1} - (298\ \text{K})(-198.7\ \text{J K}^{-1}\text{mol}^{-1})

\Delta_rG^\circ = -92200 + 59212 = -32988\ \text{J mol}^{-1} \approx -33.0\ \text{kJ mol}^{-1}

Negative: ammonia synthesis is spontaneous at 298 K under standard conditions (this matches the tabulated $\Delta_fG^\circ(\text{NH}_3) \approx -16.5\ \text{kJ mol}^{-1}$ per mole NH $_3$ , i.e. $-33.0\ \text{kJ}$ for 2 mol, confirming internal consistency). Step 4: repeat at industrial operating temperature, 700 K, assuming $\Delta H^\circ$ and $\Delta S^\circ$ stay roughly constant:

\Delta_rG^\circ(700\ \text{K}) = -92200 - (700)(-198.7) = -92200 + 139090 = +46890\ \text{J mol}^{-1}

At 700 K the reaction is non-spontaneous under standard conditions ( $\Delta_rG^\circ \approx +46.9\ \text{kJ mol}^{-1}$ ). This is exactly why the Haber process runs at high pressure — to shift the equilibrium position back toward products despite the unfavorable high-temperature $\Delta G^\circ$ , trading thermodynamic favorability for faster kinetics.

Limiting cases & common traps

Situation	What to remember
Constant $T$ , $V$ (not $p$ )	Use Helmholtz energy $A = U - TS$ , not $G$ ; $dA \le 0$ is the correct criterion there.
$\Delta H < 0$ , $\Delta S < 0$	Spontaneity is temperature-dependent; low $T$ favors reaction, as in ammonia synthesis.
$\Delta H > 0$ , $\Delta S > 0$	Spontaneous only above a crossover temperature $T^* = \Delta H/\Delta S$ .
$\Delta H < 0$ , $\Delta S > 0$	Spontaneous at all temperatures.
$\Delta H > 0$ , $\Delta S < 0$	Non-spontaneous at all temperatures.
$\Delta_rG^\circ$ vs $\Delta_rG$	$\Delta_rG^\circ$ is for pure standard states; actual $\Delta_rG$ depends on composition via $\Delta_rG = \Delta_rG^\circ + RT\ln Q$ .

The chemist behind it

Josiah Willard Gibbs worked in relative isolation at Yale, publishing his monumental "On the Equilibrium of Heterogeneous Substances" between 1876 and 1878 in the little-read Transactions of the Connecticut Academy of Sciences. In it he laid out chemical potential, phase rule, and the free energy function that now bears his name, essentially inventing chemical thermodynamics as a mathematical discipline before European chemists like Ostwald and van't Hoff translated and popularized his work.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American mathematical physicist who founded chemical thermodynamics and statistical mechanics, working almost entirely from New Haven, Connecticut

Ask the tutor: How does ΔG connect to the equilibrium constant K, and why does ΔG° = −RT ln K?

Why do we need G at all if we already have entropy and the second law?

The second law, as stated, requires tracking the entropy of the entire universe — system plus surroundings — which is inconvenient because we rarely have direct access to the surroundings' entropy change. Gibbs energy repackages that requirement using only system properties ( $H$ , $S$ , $T$ ), valid specifically at constant temperature and pressure. It turns a universal but impractical criterion into a practical, tabulable one for benchtop and industrial chemistry.

How is G different from the Helmholtz energy A?

Both are Legendre transforms of internal energy $U$ built to match different sets of controlled variables. $A = U - TS$ is the natural potential at constant temperature and volume, useful for rigid, sealed containers. $G = H - TS = A + pV$ is the natural potential at constant temperature and pressure, the far more common laboratory condition. Their relationship, $G = A + pV$ , exactly parallels $H = U + pV$ .

Does ΔG being negative mean the reaction will happen quickly?

No. $\Delta G$ is purely a thermodynamic quantity describing whether products are favored over reactants at equilibrium; it says nothing about the rate. A reaction can have a very negative $\Delta G$ yet proceed immeasurably slowly because of a large activation energy — the classic example is diamond converting to graphite. Kinetics and thermodynamics are independent axes of analysis.

What exactly is Q in the equation ΔG = ΔG° + RT ln Q?

$Q$ is the reaction quotient, built from activities or partial pressures of products over reactants raised to their stoichiometric powers, evaluated at the current (not necessarily equilibrium) composition. This relation, covered in depth when connecting Gibbs energy to the equilibrium constant, shows that $\Delta G^\circ$ alone only describes standard-state conditions; real mixtures need this correction term.

Why does G also foreshadow non-expansion work?

It can be shown that $\Delta G$ equals the maximum non-expansion (useful) work obtainable from a process at constant $T$ and $p$ — electrical work from a battery, for instance. This result, covered under the maximum non-expansion work theorem, is what makes $\Delta G$ so central to electrochemistry and biochemical energetics, not just to predicting reaction direction.

The Gibbs Energy

The physical picture

The core relations

The core relations

Worked examples

Limiting cases & common traps

The chemist behind it

Before this