Temperature Dependence of the Gibbs Energy

Ice melts spontaneously above 0 °C but not below it. The chemical identity of water has not changed. What changed is temperature, and temperature reshapes the Gibbs energy landscape that decides spontaneity. The Gibbs energy is the thermodynamic compass for constant-pressure processes: reactions run toward lower $G$ . But $G$ is not a fixed number for a substance — it slides with temperature, and it slides differently for gases, liquids, and solids. Understanding that slope is the key to predicting phase transitions, reaction spontaneity at high or low temperature, and eventually the van 't Hoff equation for equilibrium constants. This article derives the fundamental relation between $G$ , $T$ , and entropy from the Gibbs energy's natural variables. You will see how the sign and steepness of that slope come directly from $S$ . Then you will see why gases' free energy falls faster with heating than solids' does. Finally you will use these ideas to locate transition temperatures and explain why some reactions flip from spontaneous to nonspontaneous as you heat them.

Key takeaways

Starting from $dG = -SdT + VdP$ , holding pressure constant gives $(\partial G/\partial T)_P = -S$ .
Since $S \geq 0$ always, $G$ decreases as temperature rises, at constant pressure and composition.
Gases have the largest entropy, so their $G$ falls fastest with $T$ ; solids fall slowest.
A plot of $G$ versus $T$ is concave down everywhere, because $(\partial^2 G/\partial T^2)_P = -C_P/T < 0$ .
Crossing points on $G$ – $T$ diagrams for different phases mark equilibrium transition temperatures.
Key idea: the slope of Gibbs energy against temperature is minus the entropy, so higher-entropy phases lose free-energy advantage fastest as temperature climbs, which is exactly why melting and boiling happen where they do.

The core insight

Because $dG = -SdT + VdP$ is an exact expression for the Gibbs energy in its natural variables, holding pressure fixed strips away the volume term and leaves $(\partial G/\partial T)_P = -S$ — a compact, exact statement that entropy is literally the negative slope of free energy against temperature, and since entropy is always positive, every substance's Gibbs energy must decrease as it warms.

The physical picture

Recall that $G$ has natural variables $T$ and $P$ : the total differential $dG = -SdT + VdP$ follows directly from combining the first and second laws with the definition $G = H - TS$ (see Natural Variables of the Thermodynamic Potentials). At constant pressure, $dP = 0$ , and the relation collapses to a single-variable statement about how $G$ responds to heating. Physically, entropy measures the density of accessible microstates — how "spread out" energy and configurations are. A gas has enormously more positional and momentum freedom than a liquid, which in turn has more than a rigid crystal. So $S_{gas} \gg S_{liquid} > S_{solid}$ for the same substance. Since the slope of $G$ against $T$ is $-S$ , the gas phase's Gibbs energy plunges downward with temperature far more steeply than the solid's does. This differential steepness is not a minor detail — it is the entire reason phase transitions exist at fixed, reproducible temperatures. Think of $G(T)$ for each phase as a descending curve. At low temperature the solid curve lies lowest (most stable, smallest $G$ ), because at $T=0$ enthalpy dominates and rigid, low-energy structures win. As $T$ rises, the solid curve descends slowly, the liquid curve descends faster, and the gas curve descends fastest of all. Eventually the liquid curve dips below the solid curve — that crossing is the melting point. Further heating brings the gas curve below the liquid curve — that crossing is the boiling point. Everything in phase behavior traces back to these differing slopes, each one just $-S$ for that phase.

Figure

Gibbs energy of three phases as functions of temperature, showing crossings at

T_{fus}

and

T_{vap}

A second useful picture is the curvature: because $(\partial^2 G/\partial T^2)_P = -(C_P/T)$ , and heat capacities are positive, every $G$ – $T$ curve bends downward (concave), never upward. This guarantees exactly one crossing between two phase curves near each transition, not multiple oscillating intersections.

Figure

Concavity of a single G-T curve, showing that the slope steepens (more negative) with temperature

The core relations

\left(\frac{\partial G}{\partial T}\right)_P = -S

where: $S$ is the entropy of the system at that temperature and pressure, always $S \geq 0$ .

dG = -S\,dT + V\,dP

where: this is the total differential of $G$ in its natural variables $T$ and $P$ ; setting $dP=0$ recovers the boxed relation above.

\left(\frac{\partial^2 G}{\partial T^2}\right)_P = -\left(\frac{\partial S}{\partial T}\right)_P = -\frac{C_P}{T}

where: $C_P$ is the constant-pressure heat capacity, always positive for stable phases, so $G$ is concave in $T$ .

\Delta_r G(T_2) - \Delta_r G(T_1) = -\int_{T_1}^{T_2} \Delta_r S \, dT

where: $\Delta_r S$ is the reaction entropy, integrated over the temperature interval at constant $P$ .

Derivation AFrom the fundamental equation to (∂G/∂T)_P = −S

Goal. Show rigorously that the Gibbs energy's temperature derivative at constant pressure equals minus the entropy, starting only from the first and second laws and the definition of $G$ .

Step 1. Begin with the combined statement of the first and second laws for a closed system doing only $PV$ work, under reversible conditions:

dU = T\,dS - P\,dV

This is exact because $U$ is a state function; $T\,dS$ replaces reversible heat and $-P\,dV$ replaces reversible work.

Step 2. Define the Gibbs energy as $G \equiv H - TS = U + PV - TS$ . Take its total differential using the product rule on each product term:

dG = dU + P\,dV + V\,dP - T\,dS - S\,dT

Step 3. Substitute $dU = T\,dS - P\,dV$ from Step 1 into the expression from Step 2:

dG = (T\,dS - P\,dV) + P\,dV + V\,dP - T\,dS - S\,dT

The $T\,dS$ terms cancel exactly, and the $P\,dV$ terms cancel exactly, leaving:

dG = V\,dP - S\,dT

This is the fundamental equation for $G$ , exact and valid for any closed system in equilibrium doing only expansion work.

Step 4. Because $G$ is a state function with natural variables $T$ and $P$ , its total differential must also be expressible as:

dG = \left(\frac{\partial G}{\partial T}\right)_P dT + \left(\frac{\partial G}{\partial P}\right)_T dP

Step 5. Compare term by term with the result of Step 3. The coefficient of $dP$ gives $(\partial G/\partial P)_T = V$ , and the coefficient of $dT$ gives directly:

\left(\frac{\partial G}{\partial T}\right)_P = -S

Step 6. Since entropy is a positive quantity for any system with accessible microstates (third law: $S \to 0$ only as $T \to 0$ for a perfect crystal), the derivative $(\partial G/\partial T)_P$ is always zero or negative. $G$ never increases with temperature at constant pressure and fixed composition.

\boxed{\left(\frac{\partial G}{\partial T}\right)_P = -S \leq 0}

Worked examples

Example 1 — Foundational: estimating ΔG for heating liquid water at constant pressure. Liquid water at 298 K, 1 bar has molar entropy $S_m = 69.9\ \text{J K}^{-1}\text{mol}^{-1}$ . Assume $S_m$ is roughly constant over a small interval (a good approximation since $C_P$ changes slowly). Estimate the change in molar Gibbs energy when the temperature is raised from 298 K to 308 K at constant pressure. Step 1 — set up the integral form of the relation, treating $S$ as constant over the small 10 K interval:

\Delta G_m \approx -S_m \, \Delta T

Step 2 — insert numbers:

\Delta G_m \approx -(69.9\ \text{J K}^{-1}\text{mol}^{-1})(10\ \text{K}) = -699\ \text{J mol}^{-1}

So $G_m$ drops by about 0.70 kJ mol⁻¹ for this 10 K rise. Sanity check: the sign is negative, consistent with $(\partial G/\partial T)_P = -S \leq 0$ ; the magnitude (under 1 kJ mol⁻¹) is small compared to typical bond or reaction energies (hundreds of kJ mol⁻¹), which is reasonable since we heated by only 10 K.

Example 2 — Applied: locating where a solid-to-liquid transition becomes spontaneous. Consider a hypothetical simple molecular solid A(s) melting to A(l). At 1 bar, thermodynamic tables give:

\Delta_{fus}H^\circ = 6.0\ \text{kJ mol}^{-1}, \qquad \Delta_{fus}S^\circ = 22.0\ \text{J K}^{-1}\text{mol}^{-1}

both roughly temperature-independent near the transition. Find the equilibrium melting temperature, and confirm using the $(\partial G/\partial T)_P = -S$ relation why melting becomes favorable above that temperature. Step 1 — at the equilibrium melting point, solid and liquid have equal Gibbs energy, so $\Delta_{fus}G = 0$ :

\Delta_{fus}G = \Delta_{fus}H - T\Delta_{fus}S = 0

Step 2 — solve for $T$ :

T_{fus} = \frac{\Delta_{fus}H}{\Delta_{fus}S} = \frac{6000\ \text{J mol}^{-1}}{22.0\ \text{J K}^{-1}\text{mol}^{-1}} = 272.7\ \text{K}

Step 3 — check the direction of change above and below $T_{fus}$ . We know

\left(\frac{\partial (\Delta_{fus}G)}{\partial T}\right)_P = -\Delta_{fus}S = -22.0\ \text{J K}^{-1}\text{mol}^{-1}

This is negative, so $\Delta_{fus}G$ decreases as $T$ increases. Below 272.7 K, $\Delta_{fus}G > 0$ (melting nonspontaneous, solid favored); above 272.7 K, $\Delta_{fus}G < 0$ (melting spontaneous, liquid favored). Sanity check: 272.7 K is close to 273.15 K, the real melting point of ice, and the entropy of fusion of water (22.0 J K⁻¹ mol⁻¹ is close to the real value, 22.0 J K⁻¹ mol⁻¹) is realistic for a hydrogen-bonded molecular solid. The signs and trend match physical intuition: melting is entropy-driven and becomes favorable only once $T\Delta S$ overtakes $\Delta H$ .

Limiting cases & common traps

Situation	What happens to $(\partial G/\partial T)_P$	Common mistake
$T \to 0$ K (third law)	$S \to 0$ for a perfect crystal, so slope $\to 0$ ; $G$ – $T$ curve flattens	Assuming $G$ keeps falling linearly all the way to 0 K
Gas phase vs. condensed phase	Gas has much larger $S$ , so much steeper negative slope	Forgetting that different phases have different slopes, not just different intercepts
Phase transition point	Two phases' $G$ curves cross; $\Delta G = 0$ there	Thinking $G$ itself is zero at a transition, rather than the difference between phases
Increasing $P$ at fixed $T$	Slope $-S$ is essentially unaffected for condensed phases, more affected for gases	Confusing $(\partial G/\partial T)_P$ with $(\partial G/\partial P)_T = V$
Reaction with $\Delta_r S < 0$	$\Delta_r G$ increases with $T$	Assuming heating always helps spontaneity — it only helps if $\Delta S > 0$

The chemist behind it

The formalism connecting $G$ , $S$ , and $T$ traces back to Josiah Willard Gibbs, whose 1870s papers "On the Equilibrium of Heterogeneous Substances" laid out the full apparatus of chemical thermodynamics: the free energy function, chemical potential, and the phase rule. Gibbs worked largely in isolation at Yale, publishing in a relatively obscure journal (the Transactions of the Connecticut Academy), and his work was slow to be appreciated in America. It took European scientists — notably Maxwell, who built a plaster model of Gibbs's thermodynamic surface for water, and later Ostwald and van der Waals, who translated and publicized his papers — to bring his ideas into mainstream physical chemistry. The relation $(\partial G/\partial T)_P = -S$ is one of the "Maxwell-relation-adjacent" results falling directly out of Gibbs's exact differential for $G$ , and it underlies nearly every temperature-dependent equilibrium calculation used today, from melting points to the van 't Hoff analysis of equilibrium constants.

Origins · 1 minJosiah Willard Gibbs and how this was discovered

Josiah Willard Gibbs

1839–1903 · American mathematical physicist who founded chemical thermodynamics and defined the Gibbs free energy

Ask the tutor: How does the Gibbs–Helmholtz equation repackage this same relation in terms of enthalpy instead of entropy?

Why does (∂G/∂T)_P equal −S and not −S plus some other term

The result follows exactly from combining the first and second laws into $dG = -S\,dT + V\,dP$ , with no approximations. Comparing this total differential to the general form $dG = (\partial G/\partial T)_P dT + (\partial G/\partial P)_T dP$ term by term shows the $dT$ -coefficient must be $-S$ precisely. There is no missing term because the derivation only used exact state-function properties: the definition $G = H - TS$ and the combined law $dU = TdS - PdV$ . This is why the relation holds for any pure substance or mixture at equilibrium, not just as an approximation.

Does (∂G/∂T)_P = −S mean entropy is temperature-independent

No. The relation gives the instantaneous slope of $G$ against $T$ at a given temperature; $S$ itself generally changes with $T$ , growing as $C_P/T$ per unit temperature via $(\partial S/\partial T)_P = C_P/T$ . That is why the $G$ – $T$ curve is not just a straight line but bends downward (concave), getting steeper as $T$ rises, since $S$ itself grows with temperature for most substances away from phase transitions.

How does this relation connect to the van 't Hoff equation

The van 't Hoff equation describes how the equilibrium constant $K$ varies with temperature, and it is derived by applying temperature-differentiation to $\Delta_rG^\circ = -RT\ln K$ together with the Gibbs–Helmholtz equation, which itself is built from $(\partial G/\partial T)_P = -S$ combined with $G = H + T(\partial G/\partial T)_P$ . In short, this relation is the foundational brick; Gibbs–Helmholtz repackages it in terms of enthalpy, and van 't Hoff applies that repackaging to equilibrium constants.

Can (∂G/∂T)_P ever be positive for a real substance

No, not for a substance obeying the third law of thermodynamics, since entropy is non-negative for all stable phases at any temperature above absolute zero. At exactly $T = 0$ K, a perfect crystal has $S = 0$ , so the slope is momentarily zero, but it cannot go positive. Claims of a positive slope usually indicate confusion between a pure substance's $G$ and a reaction's $\Delta_rG$ , which behaves differently depending on the sign of $\Delta_rS$ .

Why is the G-T curve concave rather than a straight line

Because the second derivative $(\partial^2G/\partial T^2)_P$ equals $-C_P/T$ , and heat capacities $C_P$ are positive for any stable phase. A negative second derivative means the curve bends downward everywhere it is defined, so the slope (which is $-S$ ) becomes increasingly negative as temperature rises. This curvature is what guarantees a clean, single crossing point between two phases' $G$ - $T$ curves at each phase transition, rather than multiple intersections.