Equilibrium conditions for closed systems

You already know the rule from school: a reaction runs while $\Delta G<0$ and stops once $\Delta G=0$ . That rule is qualitative — it never tells you why $\Delta G$ stops changing, or what "the reaction" even means once you have a mixture of reactants and products sitting together in one flask. This article makes the rule exact. For a closed system at constant temperature and pressure, the single criterion $dG\le0$ governs everything: which way a reaction runs, when it stops, and even when a liquid stops evaporating into its own vapour. The tool that makes this precise is the extent of reaction $\xi$ , a single number that tracks how far a reaction has proceeded, and its partner the reaction Gibbs energy $\Delta_r G$ , literally the slope of a $G$ versus $\xi$ curve. Where that slope is zero, the system is at equilibrium — and from that one condition falls out the equilibrium constant $K$ itself, derived rather than assumed.

The core insight

At constant $T,P$ a closed system evolves toward the composition that minimises its Gibbs energy $G$ , so equilibrium — for a reaction or for matter moving between phases — is exactly the state where the slope $\Delta_r G=\sum_i\nu_i\mu_i$ or the chemical potential difference $\mu_i^{(\alpha)}-\mu_i^{(\beta)}$ vanishes.

The core idea

The single criterion. The Second Law, specialised to constant $T$ and $P$ (the everyday lab condition — open beaker, atmospheric pressure, thermostatted bath), collapses to one inequality:

dG \le 0

with the equality $dG=0$ marking equilibrium: the state beyond which no further spontaneous change is possible. (At constant $S,V$ the analogous statement is $dU\le0$ ; at constant $T,V$ it is $dA\le0$ — but $dG\le0$ is the one chemists reach for almost always, because constant $T,P$ is how real experiments are run.)

Tracking composition with one variable. A reaction mixture has many mole numbers $n_i$ , but for a single reaction $\sum_i\nu_i A_i=0$ they are not independent — they all move together, tied to one extent of reaction $\xi$ through $dn_i=\nu_i\,d\xi$ (products carry positive $\nu_i$ , reactants negative). Substituting into $dG=\sum_i\mu_i\,dn_i$ collapses the whole sum to a single slope:

dG\Big|_{T,P} = \left(\sum_i \nu_i\mu_i\right)d\xi \equiv \Delta_r G\,d\xi

where: $\mu_i$ is the chemical potential of species $i$ , $\nu_i$ its stoichiometric coefficient, and $\Delta_r G\equiv\sum_i\nu_i\mu_i=(\partial G/\partial\xi)_{T,P}$ is the reaction Gibbs energy — literally the gradient of the $G$ -versus- $\xi$ curve at the current composition.

Reading the sign. Since $d\xi>0$ always means "reaction moves forward," the criterion $dG\le0$ forces: if $\Delta_r G<0$ the reaction runs forward; if $\Delta_r G>0$ it runs in reverse; and equilibrium is the unique point where the slope is exactly zero,

\Delta_r G=\sum_i\nu_i\mu_i=0 \quad\text{at equilibrium.}

This is why the $G$ versus $\xi$ curve is bowl-shaped: it falls while $\Delta_r G<0$ , flattens to zero at the bottom, and would rise (violating $dG\le0$ ) if the reaction overshot — so the system simply cannot pass the minimum.

The same logic for phases, no reaction needed. If species $i$ can migrate between two phases $\alpha$ and $\beta$ inside a sealed container (liquid and vapour, say), moving $dn_i^{(\alpha)}$ moles from $\beta$ into $\alpha$ changes $G$ by $(\mu_i^{(\alpha)}-\mu_i^{(\beta)})\,dn_i^{(\alpha)}$ . The same $dG\le0$ argument then gives phase equilibrium as

\mu_i^{(\alpha)} = \mu_i^{(\beta)}

Both results — $\sum\nu_i\mu_i=0$ for a reaction, $\mu_i^{(\alpha)}=\mu_i^{(\beta)}$ for phase transfer — are the same principle wearing two hats: at constant $T,P$ , equilibrium is the state where $G$ is stationary with respect to every variable the system is free to adjust.

Figure

Gibbs energy plotted against extent of reaction forming a bowl shape with the slope negative on the left, zero at the minimum marked equilibrium, and positive on the right

Derivation AFrom the equilibrium condition to the equilibrium constant

The goal is to show that the single microscopic condition $\Delta_r G=\sum_i\nu_i\mu_i=0$ , applied to an ideal-gas reaction, is not an independent rule but the exact origin of the familiar bridge equation $\Delta_r G^\circ=-RT\ln K$ that connects tabulated thermodynamic data to the equilibrium constant used throughout equilibrium chemistry.

Step 1. Take the gas-phase reaction $\text{N}_2\text{O}_4(g)\rightleftharpoons2\text{NO}_2(g)$ , so $\nu_{N_2O_4}=-1$ and $\nu_{NO_2}=+2$ . Each species has an ideal-gas chemical potential $\mu_i=\mu_i^\circ+RT\ln(p_i/p^\circ)$ , where $\mu_i^\circ$ is its standard chemical potential and $p_i/p^\circ$ its partial pressure relative to the standard pressure.

Step 2. Substitute into $\Delta_r G=\sum_i\nu_i\mu_i=2\mu_{NO_2}-\mu_{N_2O_4}$ and expand each $\mu_i$ : $\Delta_r G=2\left[\mu_{NO_2}^\circ+RT\ln(p_{NO_2}/p^\circ)\right]-\left[\mu_{N_2O_4}^\circ+RT\ln(p_{N_2O_4}/p^\circ)\right]$ .

Step 3. Collect the standard-state terms separately from the logarithmic terms: $\Delta_r G=\left(2\mu_{NO_2}^\circ-\mu_{N_2O_4}^\circ\right)+RT\left[2\ln(p_{NO_2}/p^\circ)-\ln(p_{N_2O_4}/p^\circ)\right]$ . The first bracket is by definition the standard reaction Gibbs energy $\Delta_r G^\circ$ , and the second bracket is $RT\ln\left[(p_{NO_2}/p^\circ)^2/(p_{N_2O_4}/p^\circ)\right]=RT\ln Q$ , where $Q$ is the reaction quotient built from the instantaneous pressures. So generally $\Delta_r G=\Delta_r G^\circ+RT\ln Q$ .

Step 4. Impose the equilibrium condition. At equilibrium $\Delta_r G=0$ and $Q$ takes its special equilibrium value, called $K$ : $0=\Delta_r G^\circ+RT\ln K$ , which rearranges to

\Delta_r G^\circ = -RT\ln K

This is the exact bridge between thermodynamics ( $\Delta_r G^\circ$ , computed from tabulated $\mu_i^\circ$ or $\Delta_f G^\circ$ values) and the equilibrium constant $K$ — derived here from $\sum\nu_i\mu_i=0$ , not asserted as a separate postulate. Verified: for any reaction with ideal-gas or ideal-solution species, $\Delta_r G=\Delta_r G^\circ+RT\ln Q$ reduces at $Q=K$ to $\Delta_r G^\circ=-RT\ln K$ ; the algebra depends only on $\mu_i=\mu_i^\circ+RT\ln(\text{activity}_i)$ .

Worked example

A closed vessel at 350 K holds liquid benzene in equilibrium with its vapour. A small amount of liquid evaporates spontaneously while the system is not yet at equilibrium. What does this say about the sign of $\mu^{(l)}-\mu^{(v)}$ , and how does that quantity change as evaporation proceeds?

Step 1. Set up the extent-of-reaction bookkeeping for the phase transfer: moving $d\xi$ moles from liquid to vapour gives $dn^{(v)}=+d\xi$ and $dn^{(l)}=-d\xi$ , so $dG=\mu^{(v)}dn^{(v)}+\mu^{(l)}dn^{(l)}=(\mu^{(v)}-\mu^{(l)})\,d\xi$ .

Step 2. Apply the spontaneity condition. Evaporation is observed, meaning $d\xi>0$ actually occurs, and it must satisfy $dG\le0$ . Combined, this forces $\mu^{(v)}-\mu^{(l)}\le0$ , i.e. $\mu^{(l)}>\mu^{(v)}$ — the liquid's chemical potential is higher, matching the general rule that matter flows from high $\mu$ to low $\mu$ .

Step 3. Track how the gap closes. As evaporation proceeds in the sealed, fixed-volume vessel, the vapour's partial pressure rises, so $\mu^{(v)}=\mu^{\circ(v)}+RT\ln(p/p^\circ)$ climbs, while $\mu^{(l)}$ (pure liquid, essentially pressure-insensitive) stays flat. The gap $\mu^{(l)}-\mu^{(v)}$ therefore shrinks toward zero.

Answer: evaporation stops exactly when $\mu^{(l)}=\mu^{(v)}$ — the vessel has reached the saturation vapour pressure of benzene at 350 K, a dynamic, self-limiting endpoint driven purely by chemical-potential equalisation rather than a static "reaction has stopped" picture.

Competitive stretch

Suppose at 350 K the standard reaction Gibbs energy for $\text{N}_2\text{O}_4\rightleftharpoons2\text{NO}_2$ were $\Delta_r G^\circ=+5.0\ \text{kJ mol}^{-1}$ . Using $\Delta_r G^\circ=-RT\ln K$ with $R=8.314\ \text{J mol}^{-1}\text{K}^{-1}$ : $\ln K=-\Delta_r G^\circ/RT=-5000/(8.314\times350)=-1.72$ , so $K=e^{-1.72}\approx0.18$ . A positive $\Delta_r G^\circ$ correctly gives $K<1$ : at standard pressures the reaction as written is unfavourable, favouring $\text{N}_2\text{O}_4$ over $\text{NO}_2$ at equilibrium — exactly what the sign of $\Delta_r G^\circ$ predicts.

Common traps

Confusing $\Delta_r G$ (the instantaneous slope $\partial G/\partial\xi$ , changes as the reaction proceeds) with $\Delta_r G^\circ$ (a fixed number at standard conditions) — only at equilibrium is $\Delta_r G=0$ , and $Q=K$ only at equilibrium too.
Reading $\Delta_r G<0$ as "the reaction goes to completion" — it only means $\xi$ is currently increasing; the reaction still halts once $\Delta_r G$ climbs back to zero as $Q$ rises toward $K$ .
Applying $dG\le0$ outside constant $T,P$ — for constant $S,V$ use $dU\le0$ , for constant $T,V$ use $dA\le0$ .
Flipping the sign convention in $\Delta_r G=\sum_i\nu_i\mu_i$ by treating a reactant's $\nu_i$ as positive instead of negative.

Ask the tutor: if the jump from $\Delta_r G=\sum\nu_i\mu_i=0$ to $\Delta_r G^\circ=-RT\ln K$ felt fast, ask the AI tutor to redo the derivation slowly for a reaction of your choosing.

Why is dG less than or equal to zero the criterion only at constant T and P, and not always?

The Second Law in its most basic form is about entropy, $dS\ge0$ for an isolated system. Converting that into a statement about $G=H-TS$ requires holding $T$ and $P$ fixed while the system exchanges heat and work with its surroundings — that is precisely how $dG\le0$ gets derived. Hold different variables constant instead (say $S$ and $V$ , or $T$ and $V$ ) and the bookkeeping produces a different potential — $dU\le0$ or $dA\le0$ respectively — because a different combination of variables is now the natural one to track. Chemists default to $dG\le0$ because most reactions happen open to the atmosphere at roughly constant pressure and temperature, not because it is the only valid form.

Does delta_r G equal zero mean the reaction has stopped happening at the molecular level?

No — equilibrium is dynamic, not static. Forward and reverse reactions both continue at the molecular level, and at equilibrium their rates have simply become equal, so there is no further net change in composition. The thermodynamic statement $\Delta_r G=0$ only tells you the macroscopic slope of $G$ versus $\xi$ is flat; it says nothing about molecules stopping. This is exactly the benzene example: liquid molecules keep evaporating and vapour molecules keep condensing even after the vessel reaches saturation vapour pressure — the two rates have just become equal.

How is the extent of reaction xi different from just tracking one species concentration?

Tracking a single species tells you only that species' story, and different species change by different amounts set by their stoichiometric coefficients — two moles of NO2 form for every one mole of N2O4 consumed, for instance. The extent of reaction xi is the one variable behind all of them: every $dn_i=\nu_i\,d\xi$ moves in lockstep with the same $d\xi$ . That is precisely what lets the whole sum $\sum_i\mu_i\,dn_i$ collapse into a single slope $\Delta_r G\,d\xi$ , turning a multi-variable problem into a one-variable one.

Equilibrium conditions for closed systems

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads