Gibbs-Helmholtz equation: Worked examples — Physics deep dive

The scenario matrix

Every Gibbs–Helmholtz problem is really one of a handful of shapes. Here is the full grid — each row is a "case class", and later each worked example is tagged with the cell it fills.

#	Case class	What's special	Filled by
A	$Δ H < 0$ (exothermic), $Δ G$ rises with $T$	slope of $Δ G / T$ vs $1/ T$ is negative	Ex 1
B	$Δ H > 0$ (endothermic), $Δ G$ falls with $T$	slope positive	Ex 2
C	Single $G (T)$ point + slope $(\partial G / \partial T)_{P}$	use the $(⋆)$ form, not two points	Ex 3
D	$Δ H \approx 0$ (degenerate / flat)	$Δ G / T$ is constant — the line is horizontal	Ex 4
E	Limiting behaviour: $T \to 0$ and $T \to \infty$	what $G / T$ and $- H / T^{2}$ do at the edges	Ex 5
F	Equilibrium constant $K$ (van 't Hoff face)	plug $Δ G^{\circ} = - R T ln K$	Ex 6
G	Real-world word problem (battery / cell)	$Δ G = - n F E$ , temperature coefficient	Ex 7
H	Exam twist: wrong-form / sign trap	tests if you know $\partial T$ vs $\partial (1/ T)$	Ex 8

The plot below is the mental picture behind rows A, B and D at once: $Δ G / T$ against $1/ T$ is a straight line whose slope is $Δ H$ .

Case A — exothermic, $Δ G$ rises with $T$

Worked example Ex 1 (cell A) — recover a negative

$Δ H$ A reaction has $Δ G^{\circ} = - 20.0$ kJ/mol at $T_{1} = 300$ K and $Δ G^{\circ} = - 18.0$ kJ/mol at $T_{2} = 320$ K. Assume $Δ H$ constant. Find $Δ H^{\circ}$ .

Forecast: $Δ G$ got less negative as we heated — the reaction became less favourable when hot. Cold favours it, which is the fingerprint of releasing heat. So guess $Δ H < 0$ .

Write the two-point form. Why this step? Constant $Δ H$ means $\partial (Δ G / T) / \partial (1/ T) = Δ H$ integrates to a straight line between the two points. $\frac{Δ G _{2}}{T _{2}} - \frac{Δ G _{1}}{T _{1}} = Δ H (\frac{1}{T _{2}} - \frac{1}{T _{1}})$
Left-hand side. Why: it is the change in the quantity $Δ G / T$ that the equation predicts. $\frac{- 18.0}{320} - \frac{- 20.0}{300} = - 0.056250 + 0.066667 = + 0.010417 \frac{kJ}{mol\cdotK}$
The $1/ T$ step. Why: it is the run of our line. $\frac{1}{320} - \frac{1}{300} = - 2.0833 \times 1 0^{- 4} K^{- 1}$
Divide (rise over run = slope = $Δ H$ ). Why: slope of $Δ G / T$ vs $1/ T$ is $Δ H$ . $Δ H = \frac{0.010417}{- 2.0833 \times 1 0 ^{- 4}} = - 50.0 \frac{kJ}{mol}$

Verify: Sign is negative — matches our forecast (exothermic). Units: $(kJ\cdotmol^{- 1} K^{- 1}) / (K^{- 1}) = kJ\cdotmol^{- 1}$ . ✓

Case B — endothermic, $Δ G$ falls with $T$

Case C — one temperature, using the slope

Worked example Ex 3 (cell C) — single

$G (T)$ measurement At $T = 298$ K a system has $G = - 10.0$ kJ and a measured slope $(\partial G / \partial T)_{P} = - 0.100$ kJ/K. Find $H$ .

Forecast: The slope $(\partial G / \partial T)_{P} = - S$ , so $S = + 0.100$ kJ/K here. Then $H = G + T S = - 10 + (298) (0.1) \approx + 20$ kJ — positive, and bigger than $G$ because we add back the entropy term.

Use the raw $(⋆)$ form. Why this step? With only one temperature we have no line to slope — but we do have the slope of $G$ itself, which the identity $H = G - T (\partial G / \partial T)_{P}$ turns straight into $H$ . $H = G - T (\frac{\partial G}{\partial T})_{P}$
Substitute. Why: plug numbers. $H = - 10.0 - (298) (- 0.100) = - 10.0 + 29.8 = + 19.8 kJ$
Cross-check through $S$ . Why: catches a sign slip in the slope term. $S = - (\partial G / \partial T)_{P} = + 0.100$ kJ/K, so $H = G + T S = - 10.0 + (298) (0.100) = + 19.8 kJ$

Verify: Both routes give $+ 19.8$ kJ. ✓ Units: kJ. Matches the forecast.

Case D — the degenerate flat line ( $Δ H \approx 0$ )

Worked example Ex 4 (cell D) — zero enthalpy

A reaction gives $Δ G^{\circ} = - 8.00$ kJ/mol at $T_{1} = 300$ K and $Δ G^{\circ} = - 9.60$ kJ/mol at $T_{2} = 360$ K. Find $Δ H^{\circ}$ and interpret.

Forecast: Watch the ratio $Δ G / T$ . At 300 K it is $- 8/300$ ; at 360 K it is $- 9.6/360$ . Both equal $- 0.02 \overline{6}$ ! If $Δ G / T$ doesn't change, its slope vs $1/ T$ is zero, so $Δ H = 0$ . This is a purely entropy-driven process.

Compute both $Δ G / T$ values. Why: the equation only cares about $Δ G / T$ , not $Δ G$ . $\frac{- 8.00}{300} = - 0.026667, \frac{- 9.60}{360} = - 0.026667$
Take the difference. Why: that difference is $Δ H Δ (1/ T)$ . $Δ (\frac{Δ G}{T}) = 0$
Conclude. Why: $0 = Δ H \cdot (nonzero Δ (1/ T)) \Rightarrow Δ H = 0$ . $Δ H^{\circ} = 0 \frac{kJ}{mol}$

Verify: Since $Δ G = Δ H - T Δ S$ and $Δ H = 0$ , we get $Δ S = - Δ G / T = + 0.026667$ kJ·mol⁻¹·K⁻¹ at both temperatures — a constant entropy, no enthalpy. ✓ On the $s 01$ figure this is the horizontal orange line.

Case E — the limits ( $T \to 0$ , $T \to \infty$ )

Worked example Ex 5 (cell E) — what happens at the edges

Take a fixed exothermic system with $H = - 40$ kJ (treat $H$ as roughly constant). Describe $G / T$ and its slope $- H / T^{2}$ as $T \to 0^{+}$ and as $T \to \infty$ .

Forecast: $- H / T^{2} = + 40/ T^{2}$ . As $T \to 0$ this explodes to $+ \infty$ ; as $T \to \infty$ it dies to $0$ . So the curve of $G / T$ is very steep when cold and nearly flat when hot.

The cold limit $T \to 0^{+}$ . Why this step? $- H / T^{2} = + 40/ T^{2} \to + \infty$ : the slope of $G / T$ vs $T$ becomes infinite. Physically, dividing a finite $G$ by a tiny $T$ makes $G / T$ enormous. Evaluate at $T = 1$ K: $- H / T^{2} = + 40$ kJ/K².
The hot limit $T \to \infty$ . Why: $- H / T^{2} \to 0$ . The $G / T$ curve flattens; the enthalpy's grip on the slope fades because $T^{2}$ in the denominator wins. At $T = 1000$ K: $- H / T^{2} = + 40/1 0^{6} = + 4.0 \times 1 0^{- 5}$ kJ/K².
Sanity on $G / T$ itself. Why: using $G = H - T S$ , $G / T = H / T - S$ . As $T \to \infty$ , $H / T \to 0$ so $G / T \to - S$ ; as $T \to 0^{+}$ , $H / T$ dominates and $G / T \to \pm \infty$ with the sign of $H$ (here $- \infty$ ).

Verify: At $T = 1$ : $- H / T^{2} = 40$ . At $T = 1000$ : $- H / T^{2} = 4 \times 1 0^{- 5}$ . Ratio $= 1 0^{6} = T^{2}$ -ratio. ✓ Consistent with the $1/ T^{2}$ law — no scenario left unshown: cold steep, hot flat.

Case F — the van 't Hoff face (equilibrium $K$ )

Worked example Ex 6 (cell F) — from two

$K$ values to $Δ H^{\circ}$ An equilibrium has $K_{1} = 10.0$ at $T_{1} = 298$ K and $K_{2} = 50.0$ at $T_{2} = 328$ K. Take $R = 8.314$ J·mol⁻¹·K⁻¹. Find $Δ H^{\circ}$ .

Forecast: $K$ grew with temperature, so heating pushes the reaction forward — endothermic, $Δ H^{\circ} > 0$ .

Use the integrated van 't Hoff form (Gibbs–Helmholtz with $Δ G^{\circ} = - R T ln K$ ). Why this step? Substituting makes $Δ G^{\circ} / T = - R ln K$ , and $\partial (Δ G^{\circ} / T) / \partial (1/ T) = Δ H^{\circ}$ becomes $d ln K / d (1/ T) = - Δ H^{\circ} / R$ , which integrates to: $ln \frac{K _{2}}{K _{1}} = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$
Left side. Why: rise of $ln K$ . $ln \frac{50.0}{10.0} = ln 5 = 1.60944$
The $1/ T$ run. Why: denominator. $\frac{1}{328} - \frac{1}{298} = - 3.0704 \times 1 0^{- 4} K^{- 1}$
Solve for $Δ H^{\circ}$ . Why: rearrange step 1. $Δ H^{\circ} = - R \frac{l n ( K _{2} / K _{1} )}{( 1/ T _{2} - 1/ T _{1} )} = - 8.314 \cdot \frac{1.60944}{- 3.0704 \times 1 0 ^{- 4}} = + 4.358 \times 1 0^{4} \frac{J}{mol}$

Verify: $Δ H^{\circ} \approx + 43.6$ kJ/mol, positive — matches the endothermic forecast. Units: J·mol⁻¹K⁻¹·(dimensionless)/K⁻¹ = J·mol⁻¹. ✓

Case G — real-world word problem (a galvanic cell)

Worked example Ex 7 (cell G) — enthalpy of a battery reaction

A galvanic cell drives $n = 2$ moles of electrons. Its cell voltage is $E = 1.100$ V at $298$ K and its temperature coefficient is $(\partial E / \partial T)_{P} = - 5.00 \times 1 0^{- 4}$ V/K. Faraday constant $F = 96485$ C/mol. Find $Δ H$ of the cell reaction.

Forecast: $Δ G = - n F E$ is negative (cell does work), around $- 212$ kJ. The negative temperature coefficient means voltage drops as it heats — so entropy is being lost, and $H$ will be even more negative than $G$ . Guess $Δ H < Δ G < 0$ .

Get $Δ G$ . Why this step? Electrical work links to Gibbs energy: $Δ G = - n F E$ . $Δ G = - (2) (96485) (1.100) = - 212267 J = - 212.27 kJ$
Get the slope $(\partial G / \partial T)_{P}$ . Why: Gibbs–Helmholtz $(⋆)$ needs it. Differentiate $Δ G = - n F E$ : $(\frac{\partial Δ G}{\partial T})_{P} = - n F (\frac{\partial E}{\partial T})_{P} = - (2) (96485) (- 5.00 \times 1 0^{- 4}) = + 96.485 \frac{J}{K}$
Apply the $(⋆)$ form. Why: it turns $G$ + its slope into $H$ directly. $Δ H = Δ G - T (\frac{\partial Δ G}{\partial T})_{P} = - 212267 - (298) (96.485) = - 241000 J$ $Δ H \approx - 241.0 kJ$

Verify: $Δ H = - 241.0$ kJ is more negative than $Δ G = - 212.3$ kJ, exactly as forecast for a falling voltage. Cross-check via entropy: $Δ S = (\partial Δ G / \partial T)_{P} \cdot (- 1) = - 96.485$ J/K, and $Δ H = Δ G + T Δ S = - 212267 + (298) (- 96.485) = - 241000$ J. ✓ Units: J. ✓

Case H — the exam sign trap

Worked example Ex 8 (cell H) — spot and fix the wrong form

A student writes: "Since $\partial (G / T) / \partial T = - H / T^{2}$ , to get $Δ H$ from a plot I plot $Δ G / T$ against $T$ and the slope is $Δ H$ ." At $T = 400$ K a system has $Δ G / T = - 30.0$ J/(mol·K) and the numerically measured slope of $Δ G / T$ vs $T$ there is $- 1.25 \times 1 0^{- 4}$ J·mol⁻¹·K⁻². What did the student get wrong, and what is $Δ H$ ?

Forecast: The slope of $Δ G / T$ vs $T$ is not $Δ H$ — it is $- Δ H / T^{2}$ . So the student's slope must be multiplied by $- T^{2}$ to recover $Δ H$ .

Name the two legitimate forms. Why this step? The trap is confusing $\partial T$ with $\partial (1/ T)$ .

vs $T$ : $\frac{\partial ( Δ G / T )}{\partial T} = - \frac{Δ H}{T ^{2}}$ (slope is $- Δ H / T^{2}$ , not $Δ H$ ).
vs $1/ T$ : $\frac{\partial ( Δ G / T )}{\partial ( 1/ T )} = Δ H$ (slope is $Δ H$ — plot against $1/ T$ , not $T$ ).

Recover $Δ H$ from the vs- $T$ slope. Why: invert the correct relation. $Δ H = - T^{2} (\frac{\partial ( Δ G / T )}{\partial T})_{P} = - (400)^{2} (- 1.25 \times 1 0^{- 4}) = + 20.0 \frac{J}{mol}$

Verify: The student's slope was tiny ( $- 1.25 \times 1 0^{- 4}$ ); if that were " $Δ H$ " it would be absurdly small. Multiplying by $- T^{2} = - 1.6 \times 1 0^{5}$ rescues a sensible $Δ H = + 20.0$ J/mol. ✓ Units: K²·(J·mol⁻¹·K⁻²)=J·mol⁻¹. ✓ See the Enthalpy and Gibbs free energy notes for why $Δ H$ should sit on a normal energy scale.

Active recall

Recall Which cell, which tool?

Given two $Δ G$ values at two $T$ → which formula? ::: two-point integrated form (cells A, B, D) Given one $G$ and $(\partial G / \partial T)_{P}$ → which formula? ::: the $(⋆)$ form $H = G - T (\partial G / \partial T)_{P}$ (cell C) Given two $K$ values → which formula? ::: integrated van 't Hoff $ln (K_{2} / K_{1}) = - \frac{Δ H ^{\circ}}{R} (1/ T_{2} - 1/ T_{1})$ (cell F) $Δ G / T$ unchanged across $T$ means? ::: $Δ H = 0$ , entropy-driven (cell D) As $T \to 0^{+}$ , what does $- H / T^{2}$ do? ::: diverges (steep); as $T \to \infty$ it $\to 0$ (flat) (cell E) Slope of $Δ G / T$ vs $T$ equals? ::: $- Δ H / T^{2}$ , not $Δ H$ (cell H trap)

Connections

Parent: Gibbs–Helmholtz equation — the identity every example uses.
Gibbs free energy — the $Δ G$ we feed in (Ex 1–4, 7, 8).
Enthalpy — the $Δ H$ we extract everywhere.
Entropy — the term that vanishes; reappears in Ex 4 and 7 cross-checks.
van 't Hoff equation — Ex 6 is Gibbs–Helmholtz wearing its $K$ mask.
Kirchhoff's law (thermochemistry) — the fix when the "constant $Δ H$ " assumption of Ex 1, 2 breaks over wide $T$ .
Helmholtz free energy — swap $G \to A$ , $H \to U$ for the constant-volume twin.
Maxwell relations — same family of exact thermodynamic derivatives.