The Reaction Gibbs Energy

Mix hydrogen and nitrogen at 400°C in a rigid vessel. The mixture inches toward ammonia without any external push. Plot the Gibbs energy of the whole reacting mixture against how far the reaction has run. You get a curve, not a straight line. The slope of that curve at any point is the reaction Gibbs energy, $\Delta_rG$ . It is the single number that tells you which way a reaction wants to go right now. This quantity matters because spontaneity is not a fixed property of "products vs reactants." It depends on composition, and $\Delta_rG$ captures that dependence exactly. It is the bridge between the static idea of chemical potential and the dynamic question of how far a reaction proceeds before it stops. This article builds $\Delta_rG$ from first principles. You will see how it emerges as a derivative of $G$ with respect to extent of reaction $\xi$ . You will see how it collapses into a sum of chemical potentials weighted by stoichiometric numbers. Two worked examples show the idea in action. One is foundational, and one is applied to a real industrial equilibrium.

Key takeaways

$\Delta_rG$ is defined as the slope of the $G$ versus $\xi$ curve at constant $T$ and $p$ .
Algebraically, $\Delta_rG = \sum_i \nu_i \mu_i$ , a weighted sum of chemical potentials.
The sign of $\Delta_rG$ tells you the spontaneous direction of reaction at that composition.
$\Delta_rG$ changes continuously as $\xi$ changes, unlike a single fixed "reaction energy."
Equilibrium is reached exactly where the slope is zero, $\Delta_rG = 0$ .
Key idea: $\Delta_rG$ is not a constant — it is the instantaneous slope of $G(\xi)$ , telling you which way and how strongly the reaction is pushed at the current composition.

The core insight

The reaction Gibbs energy $\Delta_rG$ is the partial derivative of the total Gibbs energy of a reacting mixture with respect to the extent of reaction, at fixed temperature and pressure; because $G$ is built from chemical potentials, this derivative reduces exactly to $\sum_i \nu_i \mu_i$ , turning an abstract slope into a computable, composition-dependent quantity that vanishes precisely at equilibrium.

The physical picture

Consider a general reaction written with stoichiometric numbers $\nu_i$ , positive for products and negative for reactants,

0 = \sum_i \nu_i \, \mathrm{A}_i .

As the reaction advances by an infinitesimal extent $d\xi$ , the amount of each species changes by $dn_i = \nu_i \, d\xi$ . This is the definition of extent of reaction from the prerequisite topic: one variable, $\xi$ , tracks the entire composition of the mixture as it moves from pure reactants ( $\xi=0$ ) toward products. Now hold $T$ and $p$ fixed and ask how the total Gibbs energy of the mixture, $G(T,p,\xi)$ , responds to that advance. Because $G$ is a well-defined thermodynamic function of the amounts of every species present, and every amount is tied to $\xi$ through the stoichiometry, $G$ becomes a function of the single variable $\xi$ once $T$ and $p$ are fixed. Plotting $G$ against $\xi$ gives a curve — generally not a straight line — that starts at the Gibbs energy of the reactants, dips through a minimum somewhere in between, and rises again toward the Gibbs energy of the pure products (unless the reaction goes to completion, in which case the minimum sits at the boundary). The reaction Gibbs energy is the tangent slope of that curve at any given point:

\Delta_rG \equiv \left(\frac{\partial G}{\partial \xi}\right)_{T,p}.

Where the slope is negative, increasing $\xi$ lowers $G$ , so the forward reaction is spontaneous. Where the slope is positive, decreasing $\xi$ lowers $G$ , so the reverse reaction is spontaneous. Where the slope is exactly zero, the mixture sits at the minimum of the curve — perturbing $\xi$ in either direction can only raise $G$ , which is forbidden at constant $T,p$ . That is the equilibrium composition, the subject the next topic develops in full. Because $G$ is built additively from the chemical potentials of the species present, $G = \sum_i n_i \mu_i$ , and because $dn_i = \nu_i\,d\xi$ , the slope collapses algebraically into a sum of chemical potentials weighted by stoichiometric numbers. This is the content of the core relation below: an abstract graphical slope becomes a concrete, calculable combination of the same $\mu_i$ you met when studying the chemical potential itself.

Figure

Gibbs energy versus extent of reaction, with the tangent slope at three compositions

A second view makes the composition-dependence explicit for a simple isomerization A → B, where $\Delta_rG$ can be written in terms of the mole fraction of product using the chemical potentials of an ideal mixture.

Figure

Δ_rG for A to B versus mole fraction of B, showing the sign change at equilibrium

The core relations

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

where $G$ is the total Gibbs energy of the reacting mixture and $\xi$ is the extent of reaction at fixed $T,p$ .

\Delta_rG = \sum_i \nu_i \mu_i

where $\nu_i$ is the stoichiometric number of species $i$ (positive for products, negative for reactants) and $\mu_i$ is its chemical potential in the mixture.

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

where this is the fundamental equation for a reacting system, showing $\Delta_rG$ as the natural conjugate variable to $\xi$ .

\Delta_rG < 0 \;(\text{forward spontaneous}), \quad \Delta_rG = 0 \;(\text{equilibrium}), \quad \Delta_rG > 0 \;(\text{reverse spontaneous})

where the three cases classify the direction of spontaneous change at the current composition.

Derivation AFrom the slope of G(ξ) to a sum of chemical potentials

Goal. Show that the abstract derivative $(\partial G/\partial \xi)_{T,p}$ reduces to $\sum_i \nu_i \mu_i$ , turning a graphical slope into a calculable quantity built from chemical potentials.

Step 1. Write G in terms of chemical potentials. For any mixture, the total Gibbs energy is the sum of partial molar Gibbs energies weighted by amount,

G = \sum_i n_i \mu_i .

This is simply the definition of chemical potential as the partial molar Gibbs energy, carried over from the prerequisite topic.

Step 2. Introduce the extent of reaction. For the reaction $0 = \sum_i \nu_i \mathrm{A}_i$ , each amount changes with $\xi$ according to

n_i(\xi) = n_i(0) + \nu_i \xi ,

so that

dn_i = \nu_i \, d\xi

for every species simultaneously — this is the one-variable bookkeeping that extent of reaction was built to provide.

Step 3. Differentiate G with respect to ξ at constant T, p. Take the total differential of $G$ at fixed $T$ and $p$ , so that only the $n_i$ vary:

dG = \sum_i \mu_i \, dn_i \quad (\text{at constant } T, p).

This follows from the general Gibbs energy differential $dG = -S\,dT + V\,dp + \sum_i \mu_i\, dn_i$ , dropping the $dT$ and $dp$ terms since $T,p$ are held fixed.

Step 4. Substitute the stoichiometric relation. Insert $dn_i = \nu_i\, d\xi$ from Step 2:

dG = \sum_i \mu_i \, \nu_i \, d\xi = \left(\sum_i \nu_i \mu_i\right) d\xi .

Step 5. Identify the derivative. Dividing through by $d\xi$ at constant $T,p$ gives exactly the definition of $\Delta_rG$ from the core relations:

\left(\frac{\partial G}{\partial \xi}\right)_{T,p} = \sum_i \nu_i \mu_i .

Step 6. Interpret the sign convention. Because products carry positive $\nu_i$ and reactants carry negative $\nu_i$ , this sum is literally "chemical potential of products minus chemical potential of reactants," each weighted by how many moles the balanced equation assigns them. A negative value means the products currently sit lower in Gibbs energy per unit advancement than the reactants, so the forward direction is downhill.

\boxed{\Delta_rG = \left(\frac{\partial G}{\partial \xi}\right)_{T,p} = \sum_i \nu_i \mu_i}

Worked examples

Example 1 — Foundational: sign and magnitude of Δ_rG for a simple gas-phase reaction. Consider the isomerization of gaseous cis-2-butene to trans-2-butene at 400 K,

\text{cis-2-butene(g)} \rightarrow \text{trans-2-butene(g)}.

Suppose at a particular composition the chemical potentials, computed from standard data and the actual partial pressures present, are

\mu_{\text{cis}} = -170.0\ \text{kJ mol}^{-1}, \qquad \mu_{\text{trans}} = -172.3\ \text{kJ mol}^{-1}.

Here $\nu_{\text{cis}} = -1$ and $\nu_{\text{trans}} = +1$ , since cis is reactant and trans is product. Apply the core relation:

\Delta_rG = \nu_{\text{trans}}\mu_{\text{trans}} + \nu_{\text{cis}}\mu_{\text{cis}}

\Delta_rG = (+1)(-172.3\ \text{kJ mol}^{-1}) + (-1)(-170.0\ \text{kJ mol}^{-1})

\Delta_rG = -172.3 + 170.0 = -2.3\ \text{kJ mol}^{-1}.

Sanity check. The result is negative, so at this composition the forward reaction — cis converting to trans — is spontaneous. That agrees with the chemical fact that trans-2-butene is the thermodynamically more stable isomer (less steric strain), so $\mu_{\text{trans}} < \mu_{\text{cis}}$ makes physical sense, and the magnitude, a few kJ per mole, is typical for a mild isomerization rather than a highly exergonic combustion.

Example 2 — Applied: Δ_rG at a specific composition during ammonia synthesis. Take the Haber–Bosch reaction at $T = 700\ \text{K}$ ,

\mathrm{N_2(g)} + 3\,\mathrm{H_2(g)} \rightarrow 2\,\mathrm{NH_3(g)},

so $\nu_{\mathrm{N_2}} = -1$ , $\nu_{\mathrm{H_2}} = -3$ , $\nu_{\mathrm{NH_3}} = +2$ . At a snapshot of the reaction mixture, suppose partial pressures are $p_{\mathrm{N_2}} = 40\ \text{bar}$ , $p_{\mathrm{H_2}} = 120\ \text{bar}$ , $p_{\mathrm{NH_3}} = 10\ \text{bar}$ , all relative to the standard pressure $p^{\ominus}=1\ \text{bar}$ , and treat the gases as ideal so each chemical potential is

\mu_i = \mu_i^{\ominus}(T) + RT\ln\!\left(\frac{p_i}{p^{\ominus}}\right).

Suppose the standard chemical potentials at 700 K (derived from tabulated standard Gibbs energies of formation, extrapolated to 700 K for this illustration) are

\mu_{\mathrm{N_2}}^{\ominus} = 0, \qquad \mu_{\mathrm{H_2}}^{\ominus} = 0, \qquad \mu_{\mathrm{NH_3}}^{\ominus} = -32.7\ \text{kJ mol}^{-1}.

(Elements in their reference state are assigned zero standard chemical potential of formation, as usual.) First compute each actual chemical potential using $RT = (8.314\ \text{J K}^{-1}\text{mol}^{-1})(700\ \text{K}) = 5819.8\ \text{J mol}^{-1} = 5.820\ \text{kJ mol}^{-1}$ .

\mu_{\mathrm{N_2}} = 0 + 5.820\ln(40) = 5.820 \times 3.689 = 21.47\ \text{kJ mol}^{-1}

\mu_{\mathrm{H_2}} = 0 + 5.820\ln(120) = 5.820 \times 4.787 = 27.86\ \text{kJ mol}^{-1}

\mu_{\mathrm{NH_3}} = -32.7 + 5.820\ln(10) = -32.7 + 5.820 \times 2.303 = -32.7 + 13.40 = -19.30\ \text{kJ mol}^{-1}

Now assemble $\Delta_rG = \sum_i \nu_i \mu_i$ :

\Delta_rG = 2\mu_{\mathrm{NH_3}} - \mu_{\mathrm{N_2}} - 3\mu_{\mathrm{H_2}}

\Delta_rG = 2(-19.30) - 21.47 - 3(27.86)

\Delta_rG = -38.60 - 21.47 - 83.58 = -143.65\ \text{kJ mol}^{-1}.

Sanity check. $\Delta_rG$ is strongly negative, meaning at this particular composition (excess N₂ and H₂, only a little NH₃ formed so far) the forward reaction is powerfully downhill, consistent with the industrial practice of continuously feeding N₂/H₂ and removing NH₃ to keep driving the reaction forward. As NH₃ accumulates, its $\ln(p_i)$ term grows and $\mu_{\mathrm{NH_3}}$ rises, while consuming N₂ and H₂ lowers their $\mu_i$ ; the sum climbs toward zero as the mixture approaches the equilibrium composition, exactly the flattening of the $G$ -versus- $\xi$ curve pictured above.

Limiting cases & common traps

Situation	What happens to Δ_rG	Common mistake
Reaction just started, mostly pure reactants	Often strongly negative (large driving force)	Assuming Δ_rG is constant throughout the reaction
Near equilibrium composition	Approaches zero from either side	Confusing "Δ_rG = 0" with "no reactants or products left"
Reaction essentially complete	Can be strongly positive if pushed further	Forgetting Δ_rG changes sign as ξ crosses ξ_eq
Pure, unmixed reactants or products at the ξ boundary	Slope can be very large in magnitude (mixing terms diverge logarithmically)	Ignoring the entropy-of-mixing contribution to μ_i near the pure limits
Reaction at its standard state (all activities = 1)	$\Delta_rG$ equals $\Delta_rG^{\ominus}$ , a single fixed number	Using $\Delta_rG^{\ominus}$ interchangeably with $\Delta_rG$ at arbitrary composition

The chemist behind it

The formalism of extent of reaction and the reaction Gibbs energy as a derivative of $G$ traces to the Belgian chemist Théophile de Donder, who in the 1920s introduced the "affinity" of a reaction, essentially $-\Delta_rG$ , as the thermodynamic force driving chemical change forward. De Donder's insight was to treat a chemical reaction the way mechanics treats a displaced spring: a single generalized coordinate ( $\xi$ , analogous to displacement) paired with a single generalized force (the affinity, or $-\Delta_rG$ ), so that the machinery of equilibrium thermodynamics — energy minimization, response to constraints, stability — could be applied to reacting systems with the same rigor used for mechanical or electrical systems. This affinity formalism became the backbone of the Brussels school of nonequilibrium thermodynamics, later extended by Ilya Prigogine to reactions running away from equilibrium, but its foundation is exactly the $G$ -versus- $\xi$ slope developed here.

Origins · 1 minThéophile de Donder and how this was discovered

Théophile de Donder

1872-1957 · Belgian mathematician and physical chemist who defined chemical affinity as the driving force conjugate to the extent of reaction

Ask the tutor: How does Δ_rG connect to the equilibrium constant K, and what does Δ_rG° actually mean physically?

Is Δ_rG the same thing as ΔG for the reaction?

Yes, provided both refer to the same quantity evaluated at a specific composition, though notation varies by textbook. Older texts sometimes write plain $\Delta G$ for what modern IUPAC convention calls $\Delta_rG$ , the subscript "r" flagging that this is specifically the reaction Gibbs energy, the slope with respect to extent of reaction, rather than a generic Gibbs energy change between two arbitrary states. The subscript matters because $\Delta_rG$ is composition-dependent, unlike a single "reaction energy" you might expect from a simpler picture.

How is Δ_rG related to the equilibrium constant K?

At equilibrium, $\Delta_rG = 0$ , and separately, the standard reaction Gibbs energy relates to $K$ through $\Delta_rG^{\ominus} = -RT\ln K$ . These are two different statements: the first says the slope vanishes at the equilibrium composition, the second says the standard-state constant $\Delta_rG^{\ominus}$ (evaluated with every species at unit activity, not at the equilibrium composition) determines how far $K$ sits from 1. The full connection is developed in the standard-reaction-Gibbs-energy topic.

Can Δ_rG be measured directly in a lab, or is it purely computed from tables?

It can be obtained experimentally, most directly through electrochemical cells, where $\Delta_rG = -nFE_{\text{cell}}$ links the reaction Gibbs energy to a measured cell potential $E_{\text{cell}}$ at a given composition. It can also be reconstructed from partial pressures or concentrations combined with tabulated standard chemical potentials, as done in the worked ammonia example above. Both routes ultimately rest on the same definition, the slope of $G$ against $\xi$ .

Why does Δ_rG depend on composition instead of being fixed by the balanced equation alone?

Because each chemical potential $\mu_i = \mu_i^{\ominus} + RT\ln a_i$ depends on the activity $a_i$ of that species in the current mixture, not just on its identity. As the reaction proceeds, activities of reactants fall and activities of products rise, continuously shifting every $\mu_i$ and hence the weighted sum $\sum_i \nu_i \mu_i$ . Only the standard-state value $\Delta_rG^{\ominus}$ , with every $a_i$ fixed at 1, is a single fixed number independent of how far the reaction has run.

Does a negative Δ_rG mean the reaction happens quickly?

No — sign and magnitude of $\Delta_rG$ tell you thermodynamic direction and driving force, not rate. A reaction can have a large negative $\Delta_rG$ and still proceed immeasurably slowly because of a high activation barrier, a kinetic question entirely separate from thermodynamics. Diamond converting to graphite at room temperature is the classic example: strongly favorable in $\Delta_rG$ , yet effectively frozen in practice.