Equilibrium as the Minimum of Gibbs Energy

Drop a beaker of N₂O₄ into an empty flask and watch it turn brown. Some N₂O₄ converts to NO₂, but the reaction stops well before every molecule has reacted. Why does it halt exactly there, and not sooner or later? The answer is not that the reaction "runs out of energy." It is that the Gibbs energy of the reacting mixture, plotted against the extent of reaction, has a minimum. The system rolls downhill in $G$ until it reaches the bottom of the valley, then stops. That single geometric idea is the foundation of all chemical equilibrium theory. This article builds that picture from the ground up. You will see how $G$ depends on the extent of reaction $\xi$ . Then you will see why the slope of that curve is the reaction Gibbs energy. Finally you will see why the minimum forces $\Delta_rG = 0$ , setting up the equilibrium constant.

Key takeaways

$G(\xi)$ for a reacting mixture is a smooth curve with a single minimum between pure reactants and pure products.
The slope of that curve is the reaction Gibbs energy, $\Delta_rG = (\partial G/\partial \xi)_{T,p}$ .
Equilibrium is defined as the point where the slope is zero: $\Delta_rG = 0$ , not where reactants are "used up."
Away from equilibrium, $\Delta_rG < 0$ drives the reaction forward and $\Delta_rG > 0$ drives it backward.
Mixing entropy is what curves $G(\xi)$ and guarantees an interior minimum, so real reactions never go 100% to completion.
Key idea: chemical equilibrium is the composition at which the reaction mixture sits at the bottom of its own Gibbs-energy valley, where $(\partial G/\partial \xi)_{T,p}=0$ .

The core insight

At constant temperature and pressure, a reacting system evolves spontaneously toward lower total Gibbs energy; it stops changing composition precisely when $G$ can decrease no further, and that stationary point—not complete conversion—defines chemical equilibrium.

The physical picture

Consider a reaction run in a closed vessel held at constant $T$ and $p$ , the standard laboratory condition. As the reaction proceeds, we track its progress with the extent of reaction, $\xi$ (units: mol), defined so that the amount of each species changes as $n_J = n_{J,0} + \nu_J \xi$ , with $\nu_J$ the stoichiometric number (negative for reactants, positive for products). At every value of $\xi$ the mixture has some total Gibbs energy $G(\xi)$ , computed by summing $n_J \mu_J$ over all species, where $\mu_J$ is the chemical potential of species $J$ in the mixture. As $\xi$ increases from zero (pure reactants) toward its maximum value (pure products), two things happen. First, the intrinsic energy content shifts as reactant bonds break and product bonds form. Second—and this is the subtle, often-missed part—the entropy of mixing reactants with products contributes a large negative term to $G$ at intermediate $\xi$ , since a partially reacted mixture is always more disordered than either pure end. The mixing term is what gives $G(\xi)$ its characteristic bowl shape. Even for a reaction that is thermodynamically very favorable (strongly negative standard Gibbs energy), the curve still bends back upward before reaching pure products, because forcing the last trace of reactant to disappear would require an infinitely negative mixing entropy contribution for the vanishing reactant. This is precisely why virtually no reaction goes to 100% completion: the mixing entropy penalty for eliminating one component entirely is thermodynamically infinite in magnitude, so the minimum of $G$ almost always lies strictly inside the interval, at both reactants and products present in some ratio. The slope of this curve, $(\partial G/\partial \xi)_{T,p}$ , is exactly the reaction Gibbs energy $\Delta_rG$ introduced when chemical potentials were first defined. Where the slope is negative, increasing $\xi$ (running the reaction forward) lowers $G$ , so the forward reaction is spontaneous. Where the slope is positive, decreasing $\xi$ (running the reverse reaction) lowers $G$ . Only where the slope is exactly zero does neither direction lower $G$ — that is equilibrium.

Figure

Gibbs energy versus extent of reaction, showing the minimum at ξ_eq

Notice the two flanking regions in the figure. To the left of the minimum, the tangent slope is negative: $\Delta_rG < 0$ , and the forward reaction is spontaneous. To the right, the tangent is positive: $\Delta_rG > 0$ , and the reverse reaction is spontaneous. Exactly at $\xi_{eq}$ the tangent is horizontal. A second view helps connect this to what's measured in a lab: composition versus time. The system doesn't oscillate around the minimum or overshoot it — under fixed $T$ and $p$ with no external driving, it approaches $\xi_{eq}$ monotonically and stays there.

Figure

Composition approaching equilibrium in time, mirroring the descent on the G(ξ) curve

The core relations

\xi = \frac{n_J - n_{J,0}}{\nu_J}, \qquad dn_J = \nu_J\, d\xi

where: $\xi$ is the extent of reaction (mol), $\nu_J$ is positive for products and negative for reactants.

\Delta_rG \equiv \left(\frac{\partial G}{\partial \xi}\right)_{T,p} = \sum_J \nu_J \mu_J

where: $\mu_J$ is the chemical potential of species $J$ in the reaction mixture at the current composition.

dG = -S\,dT + V\,dp + \Delta_rG\, d\xi

where: at constant $T$ and $p$ , $dG = \Delta_rG\, d\xi$ , so the sign of $\Delta_rG$ fixes the spontaneous direction of change in $\xi$ .

\text{Equilibrium condition:}\qquad \left(\frac{\partial G}{\partial \xi}\right)_{T,p} = \Delta_rG = 0

where: this defines $\xi_{eq}$ , the composition at the bottom of the $G(\xi)$ valley.

Derivation AWhy the equilibrium condition is exactly Δ_rG = 0

Goal. Show, from the second law applied at constant $T$ and $p$ , that a closed reacting system evolves until $\Delta_rG = 0$ , and that this is a genuine minimum of $G$ , not just a stationary point.

Step 1: Start from the fundamental criterion for spontaneity. For any process at constant temperature and pressure with no non-expansion work, the second law guarantees

dG \le 0

with equality only at equilibrium. This is the constant- $T$ ,- $p$ specialization of $dG = dH - TdS \le 0$ , itself derived from $dS_{\text{total}} = dS_{\text{sys}} + dS_{\text{surr}} \ge 0$ .

Step 2: Express $dG$ for the reacting mixture in terms of $\xi$ . The total Gibbs energy of the mixture is $G = \sum_J n_J \mu_J$ . Holding $T$ and $p$ fixed and using $dn_J = \nu_J\, d\xi$ ,

dG = \sum_J \mu_J\, dn_J = \left(\sum_J \nu_J \mu_J\right) d\xi

Define the bracketed quantity as the reaction Gibbs energy:

\Delta_rG \equiv \sum_J \nu_J \mu_J = \left(\frac{\partial G}{\partial \xi}\right)_{T,p}

Step 3: Combine the spontaneity criterion with this slope. Since $dG = \Delta_rG\, d\xi \le 0$ for any spontaneous infinitesimal step, and $d\xi$ can be either sign depending on which direction is spontaneous, consistency requires:

\Delta_rG < 0 \Rightarrow d\xi > 0 \text{ (forward is spontaneous)}, \qquad \Delta_rG > 0 \Rightarrow d\xi < 0 \text{ (reverse is spontaneous)}

In both cases the reaction moves so as to make $G$ decrease. The only way for no direction of change to lower $G$ — i.e., for the system to have truly stopped evolving — is

\Delta_rG = 0

Step 4: Confirm this stationary point is a minimum, not a maximum or inflection. Write $G(\xi)$ as the sum of an "ideal" linear term (from bond energies, roughly linear in $\xi$ ) and a mixing-entropy term. For an ideal mixture the mixing contribution is

G_{\text{mix}} = RT\sum_J n_J \ln x_J

where $x_J$ is the mole fraction of $J$ . As $\xi \to 0$ or $\xi \to \xi_{\max}$ (a pure end), one mole fraction $x_J \to 0$ , and $n_J \ln x_J \to 0$ but

\frac{d}{d\xi}\left(n_J \ln x_J\right) \to -\infty \text{ or } +\infty

at the respective end, because $\ln x_J$ diverges even though $n_J \ln x_J$ itself vanishes. This means the slope $\Delta_rG$ is driven to $-\infty$ at the reactant-only end and to $+\infty$ at the product-only end. A curve whose slope sweeps continuously from $-\infty$ to $+\infty$ must cross zero, and because it is monotonically increasing in slope (the mixing term is convex), that crossing is unique — a single interior minimum.

Step 5: State the boxed result.

\boxed{\left(\frac{\partial G}{\partial \xi}\right)_{T,p} = \Delta_rG = 0 \ \text{at equilibrium, and this stationary point is always a minimum of } G(\xi)}

Worked examples

Example 1 — Locating equilibrium qualitatively from Δ_rG data (foundational). A gas-phase isomerization A(g) ⇌ B(g) is studied at 500 K and 1 bar in a rigid vessel initially containing 1.000 mol of pure A. Measurements of the reaction Gibbs energy at several extents of reaction give:

$\xi$ / mol	$\Delta_rG$ / kJ mol⁻¹
0.10	−8.4
0.30	−2.1
0.50	+0.3
0.55	+1.0

Step 1 — interpret the sign. At $\xi = 0.10$ and $0.30$ , $\Delta_rG$ is negative, so the forward reaction (A → B) is still spontaneous; $\xi$ must increase further. Step 2 — bracket the root. Between $\xi = 0.30$ ( $\Delta_rG = -2.1$ kJ mol⁻¹) and $\xi = 0.50$ ( $\Delta_rG = +0.3$ kJ mol⁻¹) the sign flips. Equilibrium lies in this interval. Step 3 — linear interpolation for an estimate. Treating $\Delta_rG$ as locally linear in $\xi$ ,

\xi_{eq} \approx 0.30 + (0.50-0.30)\times\frac{0-(-2.1)}{0.3-(-2.1)} = 0.30 + 0.20 \times \frac{2.1}{2.4} = 0.30+0.175 = 0.475\ \text{mol}

Step 4 — sanity check. At $\xi = 0.50$ , $\Delta_rG = +0.3$ kJ mol⁻¹, only slightly positive, consistent with $\xi_{eq}$ sitting just below 0.50 mol, close to our estimate of 0.475 mol. The mixture at equilibrium contains roughly 0.525 mol A and 0.475 mol B — not pure B, exactly as the mixing-entropy argument predicts.

Example 2 — Estimating an equilibrium constant from the curvature picture (applied, harder). For the same reaction A(g) ⇌ B(g), suppose the standard reaction Gibbs energy at 500 K is

\Delta_rG^{\ominus} = -1.50\ \text{kJ mol}^{-1}

and both gases behave ideally at 1 bar total pressure. We want the equilibrium mole fraction of B, using the relation (foreshadowed here, developed fully in the equilibrium-constant treatment) that for an ideal-gas reaction

\Delta_rG = \Delta_rG^{\ominus} + RT\ln Q, \qquad Q = \frac{p_B/p^\ominus}{p_A/p^\ominus} = \frac{x_B}{x_A}

with total pressure held at $p^\ominus = 1$ bar so partial pressures equal mole fractions numerically. Step 1 — apply the equilibrium condition. At equilibrium $\Delta_rG = 0$ , so

0 = \Delta_rG^{\ominus} + RT\ln K, \qquad \ln K = -\frac{\Delta_rG^{\ominus}}{RT}

Step 2 — plug in numbers. With $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ and $T = 500\ \text{K}$ ,

RT = 8.314 \times 500 = 4157\ \text{J mol}^{-1} = 4.157\ \text{kJ mol}^{-1}

\ln K = -\frac{(-1.50\ \text{kJ mol}^{-1})}{4.157\ \text{kJ mol}^{-1}} = 0.3608

K = e^{0.3608} = 1.434

Step 3 — convert $K$ to composition. Since $K = x_B/x_A$ and $x_A + x_B = 1$ ,

x_B = \frac{K}{1+K} = \frac{1.434}{2.434} = 0.589, \qquad x_A = 1 - 0.589 = 0.411

Step 4 — connect back to $G(\xi)$ . Starting from 1.000 mol pure A, this means $\xi_{eq} = 0.589$ mol. Because $\Delta_rG^{\ominus}$ is only mildly negative (−1.50 kJ mol⁻¹, much smaller in magnitude than $RT = 4.16$ kJ mol⁻¹), the equilibrium sits well short of complete conversion — mixing entropy has pulled the minimum strongly toward the interior, even though the reaction is "favorable" by the sign of $\Delta_rG^{\ominus}$ . Sanity check. $K > 1$ correctly signals more product than reactant at equilibrium, consistent with $\Delta_rG^{\ominus} < 0$ ; and $x_B = 0.589$ is neither 0 nor 1, exactly as required by the interior-minimum argument in Step 4 of the derivation.

Limiting cases & common traps

Situation	What actually happens	Common mistake
$\Delta_rG^{\ominus}$ very negative	$K \gg 1$ ; reaction goes nearly (never exactly) to completion	Believing $\Delta_rG^{\ominus} < 0$ means "the reaction finishes"
$\Delta_rG^{\ominus} = 0$	$K = 1$ ; equal partial pressures/mole fractions of reactants and products at equilibrium	Thinking $\Delta_rG^{\ominus}=0$ means "no reaction occurs"
Mixture already at $\xi_{eq}$	$\Delta_rG = 0$ ; adding heat or catalyst doesn't shift $\xi$ (catalyst only speeds approach)	Confusing $\Delta_rG = 0$ (mixture condition) with $\Delta_rG^{\ominus}=0$ (a fixed, composition-independent number)
Reaction quotient $Q < K$	$\Delta_rG < 0$ , forward spontaneous	Assuming direction is fixed by $\Delta_rG^{\ominus}$ alone, ignoring current composition
Pure-reactant or pure-product limit	$\Delta_rG \to \mp\infty$ from mixing entropy	Treating $G(\xi)$ as a straight line with no curvature

The chemist behind it

The modern language of "reaction Gibbs energy" and the extent-of-reaction formalism used throughout this article was systematized by the Belgian thermodynamicist Théophile de Donder in the 1920s, building directly on the chemical potential concept that Josiah Willard Gibbs had introduced half a century earlier. De Donder's great contribution was to define $\Delta_rG$ (he called it the "affinity," with an opposite sign convention) as a single derivative, $-\partial G/\partial \xi$ , turning equilibrium into a clean calculus problem: find where the slope vanishes. This reframing linked the Gibbs energy surface directly to the direction and eventual halting point of any chemical change, exactly the picture developed above.

Origins · 1 minThéophile de Donder and how this was discovered

Théophile de Donder

1872–1957 · Belgian physicist who founded the modern thermodynamics of irreversible chemical reactions and introduced the extent of reaction

Ask the tutor: Want to see how this same G(ξ) minimum leads directly to the equilibrium constant expression K = ∏(activities)^ν, including how pressure and concentration standard states enter?

Is Δ_rG = 0 the same statement as ΔG = 0 for a general process?

Yes, in essence: it is the specialization of the general equilibrium criterion $dG=0$ (at constant $T,p$ ) to a reacting mixture parametrized by extent of reaction $\xi$ . Because $dG = \Delta_rG\,d\xi$ at constant $T$ and $p$ , saying "no further spontaneous change is possible" is identical to saying $\Delta_rG=0$ . The reaction-specific notation just makes the single relevant coordinate, $\xi$ , explicit instead of leaving it buried inside $G$ .

Does Δ_rG = 0 mean the reaction quotient Q equals the equilibrium constant K?

Yes. Using $\Delta_rG = \Delta_rG^{\ominus} + RT\ln Q$ , setting $\Delta_rG=0$ gives $\Delta_rG^{\ominus} = -RT\ln K$ where $K$ is defined as the value of $Q$ at equilibrium. This is exactly the bridge used in Example 2 above, and it is the relation that the dedicated equilibrium-constant treatment develops in full, including nonideal corrections via activities.

Can Δ_rG ever be exactly zero away from equilibrium, just by coincidence?

In principle yes, momentarily, if the system passes through that exact composition while still evolving under some external driving (for example, if $T$ or $p$ is being changed simultaneously). But under strictly constant $T$ and $p$ with no external work, $\Delta_rG = 0$ is a stable fixed point: the system that reaches it stays there, since any small excursion either direction restores a nonzero slope pushing it back. That stability is what makes it a genuine equilibrium rather than a fleeting crossing.

How does this connect to Le Chatelier's principle?

Le Chatelier's principle describes how $\xi_{eq}$ shifts when $T$ , $p$ , or composition is perturbed. In the $G(\xi)$ picture, changing conditions reshapes or shifts the curve, and the system relaxes to the new minimum. For example, adding more reactant increases the reactant-rich mixing-entropy penalty, which shifts the zero-slope point toward higher $\xi$ — more product forms, exactly the qualitative shift Le Chatelier's principle predicts, now derived rather than asserted.