The Gibbs–Helmholtz Equation

A chemist measures $\Delta G^\circ$ for a reaction at 298 K, but the reactor runs at 350 K. Rerunning the whole free-energy calculation seems necessary. It is not. The Gibbs–Helmholtz equation lets you predict how $\Delta G$ shifts with temperature using only $\Delta H$ , a much easier quantity to measure or estimate calorimetrically. It converts a two-variable problem, tracking both energy and disorder, into a one-variable problem governed by enthalpy alone. This article derives the equation from the fundamental properties of $G$ . It shows why the combination $G/T$ , not $G$ itself, obeys the clean derivative law. It closes with two worked numerical examples and a preview of how this equation becomes the van't Hoff equation for equilibrium constants.

Key takeaways

The Gibbs–Helmholtz equation states $\left(\partial(G/T)/\partial T\right)_p = -H/T^2$ .
It follows directly from $G = H - TS$ and the relation $(\partial G/\partial T)_p = -S$ .
The trick is dividing by $T$ first: this eliminates $S$ from the derivative entirely.
Applied to a reaction, it gives $\left(\partial(\Delta G/T)/\partial T\right)_p = -\Delta H/T^2$ .
Integrating over a temperature range, assuming $\Delta H$ is constant, predicts $\Delta G$ at any $T$ from data at one reference temperature.
Key idea: Because $S$ cancels algebraically, the temperature dependence of $\Delta G/T$ is controlled purely by $\Delta H$ , making enthalpy data alone sufficient to extrapolate free energies across temperature.

The core insight

Dividing the Gibbs energy by temperature before differentiating causes the entropy term to vanish algebraically, leaving a derivative that depends only on enthalpy — this is the entire content and utility of the Gibbs–Helmholtz equation.

The physical picture

Start from the familiar temperature dependence of Gibbs energy at constant pressure, established from the fundamental equation $dG = -S\,dT + V\,dp$ :

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

This is useful, but it leaves $S$ sitting in the answer. Entropy is often the harder quantity to pin down experimentally compared with enthalpy, which comes straight from calorimetry. The Gibbs–Helmholtz equation is a rewriting of exactly this same fact, but packaged so that only $H$ appears, never $S$ . The key move is to look not at $G$ itself, but at the ratio $G/T$ . This ratio appears constantly in equilibrium and electrochemical work because $\Delta G^\circ/T$ is directly proportional to $-R\ln K$ (foreshadowing the van't Hoff equation). When you differentiate $G/T$ with respect to $T$ at constant $p$ , the quotient rule produces two terms: one containing $(\partial G/\partial T)_p = -S$ , and another containing $G/T^2$ . Because $G = H - TS$ , these two terms combine so that every occurrence of $S$ cancels exactly, leaving a clean result in $H$ and $T$ alone. This cancellation is not a coincidence — it is guaranteed by the definition of $G$ , and it is the entire reason chemists prefer $G/T$ over $G$ when studying temperature effects. Physically, this tells you something important: a reaction's free energy changes with temperature for two competing reasons — the entropy term $-S\,dT$ pulls in one direction, while the changing weight of the $TS$ term inside $G$ pulls in another. The Gibbs–Helmholtz form shows that after these two effects are combined honestly, the net temperature sensitivity of $G/T$ is set only by how much heat the process absorbs or releases, i.e., by $H$ .

Figure

Slope of G/T versus 1/T reveals ΔH

Figure

Enthalpy and entropy contributions to G versus T

The core relations

\left(\frac{\partial (G/T)}{\partial T}\right)_p = -\frac{H}{T^2}

where: $G$ is Gibbs energy, $H$ is enthalpy, $T$ is absolute temperature, derivative taken at constant pressure.

\left(\frac{\partial (\Delta G/T)}{\partial T}\right)_p = -\frac{\Delta H}{T^2}

where: applies the same relation to reaction quantities, $\Delta G$ and $\Delta H$ for a process at fixed composition/pressure.

\frac{\Delta G(T_2)}{T_2} - \frac{\Delta G(T_1)}{T_1} = \Delta H \left(\frac{1}{T_2} - \frac{1}{T_1}\right)

where: integrated form assuming $\Delta H$ is independent of temperature over $T_1$ to $T_2$ .

Derivation AFrom G = H − TS to the Gibbs–Helmholtz equation

Goal. Show that $\left(\partial(G/T)/\partial T\right)_p = -H/T^2$ , starting only from the definition of $G$ and its known first derivative.

Step 1. State the known derivative. From the fundamental equation for Gibbs energy, $dG = -S\,dT + V\,dp$ , holding pressure constant gives:

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

This was established when studying the temperature dependence of $G$ itself; it is the starting point here, not a new assumption.

Step 2. Differentiate the ratio G/T using the quotient rule. Treat $G/T$ as a function of $T$ at fixed $p$ :

\left(\frac{\partial (G/T)}{\partial T}\right)_p = \frac{1}{T}\left(\frac{\partial G}{\partial T}\right)_p - \frac{G}{T^2}

This is just calculus — the derivative of a quotient $f/g$ where $g = T$ .

Step 3. Substitute the known derivative from Step 1. Replace $(\partial G/\partial T)_p$ with $-S$ :

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

Step 4. Substitute the definition G = H − TS. Write $G = H - TS$ , so $G/T^2 = H/T^2 - S/T$ :

\left(\frac{\partial (G/T)}{\partial T}\right)_p = -\frac{S}{T} - \left(\frac{H}{T^2} - \frac{S}{T}\right)

Step 5. Cancel the entropy terms. The $-S/T$ from Step 3 and the $+S/T$ generated in Step 4 cancel exactly:

\left(\frac{\partial (G/T)}{\partial T}\right)_p = -\frac{S}{T} + \frac{S}{T} - \frac{H}{T^2} = -\frac{H}{T^2}

This cancellation is exact and general — it holds for any system obeying $G = H - TS$ , i.e., always.

Step 6. Apply to a process (reaction). Since $\Delta G = G_{products} - G_{reactants}$ and each side individually obeys the relation above, subtracting gives the same law for the difference:

\boxed{\left(\frac{\partial (\Delta G/T)}{\partial T}\right)_p = -\frac{\Delta H}{T^2}}

Step 7. Integrate for practical use. If $\Delta H$ is treated as constant over the temperature interval of interest (a good approximation when $\Delta C_p \approx 0$ or the range is modest), integrate both sides from $T_1$ to $T_2$ :

\int_{T_1}^{T_2} d\left(\frac{\Delta G}{T}\right) = -\Delta H \int_{T_1}^{T_2} \frac{dT}{T^2}

\frac{\Delta G(T_2)}{T_2} - \frac{\Delta G(T_1)}{T_1} = \Delta H\left(\frac{1}{T_2} - \frac{1}{T_1}\right)

This boxed integrated form is the workhorse equation used to extrapolate free energy data across temperature.

Worked examples

Example 1 — Foundational: extrapolating ΔG for ammonia synthesis. For $\mathrm{N_2(g) + 3H_2(g) \rightarrow 2NH_3(g)}$ at 298 K, standard tables give:

\Delta H^\circ = -92.4\ \text{kJ mol}^{-1}, \qquad \Delta G^\circ_{298} = -32.9\ \text{kJ mol}^{-1}

Find $\Delta G^\circ$ at 500 K, assuming $\Delta H^\circ$ is constant. Step 1. Compute $\Delta G^\circ_{298}/T_1$ :

\frac{\Delta G^\circ_{298}}{298\ \text{K}} = \frac{-32{,}900\ \text{J mol}^{-1}}{298\ \text{K}} = -110.4\ \text{J mol}^{-1}\text{K}^{-1}

Step 2. Apply the integrated Gibbs–Helmholtz relation:

\frac{\Delta G^\circ_{500}}{500} = \frac{\Delta G^\circ_{298}}{298} + \Delta H^\circ\left(\frac{1}{500} - \frac{1}{298}\right)

Step 3. Compute the bracket:

\frac{1}{500} - \frac{1}{298} = 0.002000 - 0.003356 = -0.001356\ \text{K}^{-1}

Step 4. Multiply by $\Delta H^\circ$ :

(-92{,}400\ \text{J mol}^{-1})(-0.001356\ \text{K}^{-1}) = 125.3\ \text{J mol}^{-1}\text{K}^{-1}

Step 5. Sum and solve for $\Delta G^\circ_{500}$ :

\frac{\Delta G^\circ_{500}}{500} = -110.4 + 125.3 = 14.9\ \text{J mol}^{-1}\text{K}^{-1}

\Delta G^\circ_{500} = 500 \times 14.9 = 7450\ \text{J mol}^{-1} \approx +7.5\ \text{kJ mol}^{-1}

Sanity check. $\Delta G^\circ$ flips sign from negative to positive as temperature rises — consistent with Le Chatelier: ammonia synthesis is exothermic with negative $\Delta S^\circ$ (4 mol gas → 2 mol gas), so raising $T$ should disfavor products. The sign change near 450–500 K matches known industrial behavior (high pressure is needed at operating temperatures near 700 K to compensate).

Example 2 — Applied: comparing predictions using ΔH vs. naive constant-ΔG assumption. A biochemist studies a protein-folding equilibrium with $\Delta H^\circ = -180\ \text{kJ mol}^{-1}$ and $\Delta G^\circ_{310} = -8.0\ \text{kJ mol}^{-1}$ at physiological temperature (310 K, 37 °C). Estimate $\Delta G^\circ$ at 315 K (mild fever) and compare with the (wrong) assumption that $\Delta G^\circ$ is temperature-independent. Step 1. Compute $\Delta G^\circ_{310}/310$ :

\frac{-8000\ \text{J mol}^{-1}}{310\ \text{K}} = -25.81\ \text{J mol}^{-1}\text{K}^{-1}

Step 2. Compute the bracket term:

\frac{1}{315} - \frac{1}{310} = 0.0031746 - 0.0032258 = -5.122\times10^{-5}\ \text{K}^{-1}

Step 3. Multiply by $\Delta H^\circ$ :

(-180{,}000\ \text{J mol}^{-1})(-5.122\times10^{-5}\ \text{K}^{-1}) = 9.220\ \text{J mol}^{-1}\text{K}^{-1}

Step 4. Add:

\frac{\Delta G^\circ_{315}}{315} = -25.81 + 9.22 = -16.59\ \text{J mol}^{-1}\text{K}^{-1}

\Delta G^\circ_{315} = 315 \times (-16.59) = -5226\ \text{J mol}^{-1} \approx -5.2\ \text{kJ mol}^{-1}

Comparison. A naive assumption of constant $\Delta G^\circ = -8.0\ \text{kJ mol}^{-1}$ would badly overstate stability. The Gibbs–Helmholtz correction shows folding becomes noticeably less favorable, a swing of about 2.8 kJ/mol over just 5 K — because the large negative $\Delta H^\circ$ makes $\Delta G/T$ very sensitive to $1/T$ . This size of shift is exactly why fever-range temperature changes can perturb protein stability biologically.

Limiting cases & common traps

Situation	What happens	Why
$\Delta H = 0$	$\Delta G/T$ is constant with $T$	Only entropy drives the temperature dependence, and $\Delta G = -T\Delta S$ scales linearly
$\Delta C_p \neq 0$ over a wide range	Integrated formula becomes inaccurate	Assumption of constant $\Delta H$ breaks down; need Kirchhoff's law correction
Small $T_2 - T_1$	Linear approximation in $1/T$ is excellent	Higher-order terms in Taylor expansion are negligible
Confusing $G$ and $G/T$ derivatives	Wrong sign or magnitude	$(\partial G/\partial T)_p = -S$ , NOT $-H/T^2$ ; the $H/T^2$ form applies only to $G/T$

The chemist behind it

The equation bears two names because it synthesizes work from both scientists on the relationship between heat and available (free) energy in chemical and physical processes during the 1870s–1880s. Josiah Willard Gibbs, working largely in isolation at Yale, developed the rigorous thermodynamic potential $G$ (originally his "available energy") and the mathematical apparatus of chemical thermodynamics almost single-handedly, publishing in the relatively obscure Transactions of the Connecticut Academy. Hermann von Helmholtz, already a towering figure in German physics and physiology, independently developed the closely related concept of free energy for physical systems and pushed the equation toward practical use in galvanic cells and chemical affinity. Their combined names reflect how a single deep insight, that $G/T$ rather than $G$ has this simple structure, emerged from two very different scientific traditions converging on the same thermodynamic truth.

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 transform into the van't Hoff equation for predicting equilibrium constants at different temperatures?

Why divide by T instead of just using the plain derivative of G

The plain derivative, $(\partial G/\partial T)_p = -S$ , is correct but requires knowing entropy, which is harder to measure directly than enthalpy. Dividing by $T$ before differentiating is a deliberate algebraic trick: it makes the entropy terms cancel exactly, leaving a result that depends only on $H$ and $T$ . Since $\Delta H$ is readily available from calorimetry or bond energies, this reformulation turns an entropy-dependent problem into an enthalpy-only problem, which is far more convenient experimentally.

Does the Gibbs–Helmholtz equation assume constant ΔH, or is that only for the integrated form

The differential form, $\left(\partial(\Delta G/T)/\partial T\right)_p = -\Delta H/T^2$ , is exact at every temperature with no approximation. The approximation enters only when you integrate this over a finite range and want a simple closed-form answer. There, treating $\Delta H$ as constant (ignoring $\Delta C_p$ ) is the standard simplification, valid when the temperature range is modest or heat capacities of reactants and products are similar.

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

Since $\Delta G^\circ = -RT\ln K$ , dividing by $T$ gives $\Delta G^\circ/T = -R\ln K$ . Substituting this into the Gibbs–Helmholtz relation directly yields $\left(\partial \ln K/\partial T\right)_p = \Delta H^\circ/(RT^2)$ , which is precisely the van't Hoff equation. In this sense, the van't Hoff equation is simply the Gibbs–Helmholtz equation rewritten in terms of the equilibrium constant instead of the free energy itself.

Can this equation be applied to non-reaction processes, like phase transitions

Yes. The derivation never assumed a chemical reaction specifically; it only used $G = H - TS$ and $(\partial G/\partial T)_p = -S$ , both general thermodynamic facts. For a phase transition such as melting or vaporization, the same equation applies with $\Delta H_{fus}$ or $\Delta H_{vap}$ in place of a reaction enthalpy, letting you predict how the free energy difference between phases shifts with temperature near the transition point.

What goes wrong if I ignore the ΔCp correction over a very wide temperature range

If $\Delta C_p \neq 0$ , then $\Delta H$ itself changes with temperature according to Kirchhoff's law, $d(\Delta H)/dT = \Delta C_p$ . Using the constant- $\Delta H$ integrated formula over a wide range then introduces systematic error, typically underestimating or overestimating $\Delta G$ at the far end of the interval. For precise work spanning hundreds of kelvin, chemists use a temperature-dependent $\Delta H(T)$ inside the integral rather than treating it as constant.