Gibbs-Helmholtz equation: Exercises — Physics deep dive

Before we start, one shared picture. We will plot the quantity $Δ G / T$ (the free-energy change of a reaction, divided by the absolute temperature $T$ ) against $1/ T$ (one divided by temperature — a "coldness" axis). The parent note proved the slope of that straight line is $Δ H$ , the enthalpy change. Hold this graph in mind for the analysis problems.

Level 1 — Recognition

Goal: recognise which form of the equation you are looking at, and what each symbol means.

L1.1 Which of these is the Gibbs–Helmholtz equation written so that a straight-line plot gives $Δ H$ as the slope? (a) $(\frac{\partial ( G / T )}{\partial T})_{P} = - \frac{H}{T ^{2}}$ (b) $(\frac{\partial ( Δ G / T )}{\partial ( 1/ T )})_{P} = Δ H$ (c) $(\frac{\partial G}{\partial T})_{P} = - S$

Recall Solution L1.1

The answer is (b).

What each is. (a) is the temperature form; its slope is not constant, so it does not give a straight line. (c) is not Gibbs–Helmholtz at all — it is the definition $(\partial G / \partial T)_{P} = - S$ , the entropy relation.
Why (b). If you put $x = 1/ T$ on the horizontal axis and $y = Δ G / T$ on the vertical axis, then $\frac{d y}{d x} = Δ H$ . A derivative that equals a constant is the slope of a straight line. So plotting $Δ G / T$ vs $1/ T$ and measuring the slope hands you $Δ H$ directly.

L1.2 In the equation $(\partial (G / T) / \partial T)_{P} = - H / T^{2}$ , what does the subscript $P$ mean, and why is it there?

Recall Solution L1.2

Meaning. The subscript $P$ means "hold pressure constant" while you change temperature and take the derivative. A partial derivative $\partial / \partial T$ only makes sense once you say what is being kept fixed.
Why here. Gibbs energy $G$ naturally depends on both $T$ and $P$ (from $d G = - S d T + V d P$ ). If pressure were allowed to drift, the extra $V d P$ term would sneak in and the clean result would break. Constant $P$ kills that term, leaving only the temperature effect.

L1.3 Match each symbol to its plain-English name: $G$ , $H$ , $S$ , $A$ , $U$ .

Recall Solution L1.3

$G$ = Gibbs free energy (Gibbs free energy) — the "will it happen at constant $T, P$ ?" quantity.
$H$ = enthalpy (Enthalpy) — heat content at constant pressure.
$S$ = entropy (Entropy) — spread/disorder measure.
$A$ = Helmholtz free energy (Helmholtz free energy) — the constant-volume cousin of $G$ .
$U$ = internal energy — total microscopic energy, the constant-volume cousin of $H$ .

Level 2 — Application

Goal: put numbers in and turn the crank carefully.

L2.1 A reaction has $Δ G^{\circ} = - 30.0$ kJ/mol at $T_{1} = 280$ K and $Δ G^{\circ} = - 27.0$ kJ/mol at $T_{2} = 310$ K. Assuming $Δ H$ is roughly constant, find $Δ H$ .

Recall Solution L2.1

What we use, and why it integrates. The $1/ T$ form says the rate of change of $Δ G / T$ with respect to $1/ T$ equals $Δ H$ : $\frac{d ( Δ G / T )}{d ( 1/ T )} = Δ H .$ Why we can integrate. If $Δ H$ is (assumed) constant, this is exactly "a quantity whose slope is a fixed number" — a straight line. Multiply both sides by the little step $d (1/ T)$ and add up (integrate) from state 1 to state 2: $\int_{1}^{2} d (\frac{Δ G}{T}) = \int_{1}^{2} Δ H d (\frac{1}{T}) = Δ H \int_{1}^{2} d (\frac{1}{T}) .$ The left integral is just the total change of $Δ G / T$ ; the right is $Δ H$ times the total change of $1/ T$ . So $\frac{Δ G _{2}}{T _{2}} - \frac{Δ G _{1}}{T _{1}} = Δ H (\frac{1}{T _{2}} - \frac{1}{T _{1}}) .$ Left side (units kJ·mol⁻¹·K⁻¹): $\frac{- 27.0}{310} - \frac{- 30.0}{280} = - 0.087097 + 0.107143 = + 0.020045.$ The $1/ T$ difference (units K⁻¹): $\frac{1}{310} - \frac{1}{280} = 0.0032258 - 0.0035714 = - 3.4562 \times 1 0^{- 4} .$ Divide: $Δ H = \frac{0.020045}{- 3.4562 \times 1 0 ^{- 4}} = - 58.0 kJ/mol .$ Sign check. $Δ G$ became less negative as $T$ rose → the reaction is less favourable when hot → it is exothermic, so $Δ H < 0$ . ✓

L2.2 At $T = 298$ K a system has $G = - 12.0$ kJ and slope $(\partial G / \partial T)_{P} = - 0.15$ kJ/K. Find $H$ and $S$ .

Recall Solution L2.2

Where the formula comes from. Start from the definition $G = H - T S$ . We also know (from $d G = - S d T + V d P$ at constant $P$ ) that $(\partial G / \partial T)_{P} = - S$ , i.e. $S = - (\partial G / \partial T)_{P}$ . Substitute this $S$ into the definition: $G = H - T = S [- (\frac{\partial G}{\partial T})_{P}] = H + T (\frac{\partial G}{\partial T})_{P} .$ Rearranging for $H$ gives the " $(⋆)$ form": $H = G - T\left(\frac{\partial G}{\partial T}\right)_P. \tag{$\star$}$ So it is not a new law — it is $G = H - T S$ with $S$ rewritten as a slope of $G$ . Entropy. $S = - (\partial G / \partial T)_{P} = - (- 0.15) = + 0.15$ kJ/K. Enthalpy via $(⋆)$ : $H = - 12.0 - 298 (- 0.15) = - 12.0 + 44.7 = + 32.7 kJ .$ Cross-check with $H = G + T S = - 12.0 + 298 (0.15) = - 12.0 + 44.7 = + 32.7$ kJ. ✓

L2.3 Convert between forms: if $(\partial (G / T) / \partial T)_{P} = - H / T^{2}$ , show the step that turns it into $(\partial (G / T) / \partial (1/ T))_{P} = H$ .

Recall Solution L2.3

What we do. Change variable from $T$ to $β \equiv 1/ T$ . Why. We want the slope to be constant so a plot is a straight line. How. Since $β = 1/ T = T^{- 1}$ , differentiate: $d β = - T^{- 2} d T$ , i.e. $d T = - T^{2} d β$ . Then by the chain rule $\frac{\partial ( G / T )}{\partial ( 1/ T )} = \frac{\partial ( G / T )}{\partial T} \cdot \frac{d T}{d β} = (- \frac{H}{T ^{2}}) (- T^{2}) = H .$ The two minus signs and the $T^{2}$ all cancel, leaving simply $H$ .

Level 3 — Analysis

Goal: reason about slopes, signs, and every case — hot, cold, exo-, endo-.

L3.1 You plot $Δ G / T$ (vertical) against $1/ T$ (horizontal) and get a straight line with positive slope. Is the reaction exothermic or endothermic? What is the sign of $Δ H$ ?

Recall Solution L3.1

Reasoning. The slope equals $Δ H$ (the $1/ T$ form). Positive slope ⇒ $Δ H > 0$ ⇒ endothermic (absorbs heat). Look at figure s01: the amber line has positive slope, and the arrow marks "slope $= Δ H$ ". Physical cross-check. For endothermic reactions, raising $T$ (moving left toward smaller $1/ T$ ) makes $Δ G$ more negative — heat helps them. Consistent.

L3.2 Sketch/describe all four sign combinations of $(Δ H, Δ S)$ and say, for each, how $Δ G$ behaves as temperature rises. (Use $Δ G = Δ H - T Δ S$ .)

Recall Solution L3.2

Since $Δ G = Δ H - T Δ S$ , the slope of $Δ G$ vs $T$ is $- Δ S$ . Four cases:

$Δ H < 0, Δ S > 0$ : $Δ G < 0$ at all $T$ and gets more negative hot. Always spontaneous.
$Δ H > 0, Δ S < 0$ : $Δ G > 0$ at all $T$ , more positive hot. Never spontaneous.
$Δ H < 0, Δ S < 0$ : spontaneous only when cold ( $T < Δ H /Δ S$ ); the $- T Δ S$ term eventually wins.
$Δ H > 0, Δ S > 0$ : spontaneous only when hot ( $T > Δ H /Δ S$ ). See figure s02 for the four lines and their crossovers.

L3.3 (degenerate case). What happens to the plot $Δ G / T$ vs $1/ T$ if $Δ H = 0$ (a "thermoneutral" reaction)? And what if $Δ S = 0$ ?

Recall Solution L3.3

$Δ H = 0$ : the slope is zero → the line is flat (horizontal). Then $Δ G / T = - Δ S$ = constant, so $Δ G$ scales linearly with $T$ and equilibrium never shifts with temperature (van 't Hoff: $d ln K / d (1/ T) = 0$ , $K$ is $T$ -independent).
$Δ S = 0$ : now $Δ G = Δ H$ (constant), so $Δ G / T = Δ H / T = Δ H \cdot (1/ T)$ . Plotting vs $1/ T$ gives a straight line through the origin with slope $Δ H$ — the cleanest possible case.

Level 4 — Synthesis

Goal: combine Gibbs–Helmholtz with other laws.

L4.1 Starting from $Δ G^{\circ} = - R T ln K$ (where $R = 8.314$ J·mol⁻¹·K⁻¹ is the gas constant and $K$ is the equilibrium constant), derive the van 't Hoff equation using Gibbs–Helmholtz.

Recall Solution L4.1

Divide by $T$ : $\frac{Δ G ^{\circ}}{T} = - R ln K .$ Apply the $1/ T$ form $\frac{\partial ( Δ G ^{\circ} / T )}{\partial ( 1/ T )} = Δ H^{\circ}$ : $\frac{\partial ( - R l n K )}{\partial ( 1/ T )} = Δ H^{\circ} .$ Why the partial becomes an ordinary derivative. At the fixed standard pressure of the standard state, the only variable left that $Δ G^{\circ} / T$ (hence $ln K$ ) still depends on is temperature. When a quantity depends on just one variable, the partial derivative $\partial / \partial (1/ T)$ and the ordinary derivative $d / d (1/ T)$ are the same thing. So we may write $d ln K / d (1/ T)$ . Since $R$ is a constant, pull it out: $- R \frac{d l n K}{d ( 1/ T )} = Δ H^{\circ} \Rightarrow \frac{d ln K}{d ( 1/ T )} = - \frac{Δ H ^{\circ}}{R} .$ What it says: a plot of $ln K$ vs $1/ T$ has slope $- Δ H^{\circ} / R$ . The van 't Hoff equation is Gibbs–Helmholtz wearing an equilibrium-constant costume.

L4.2 For a reaction $ln K = 4.00$ at $T_{1} = 300$ K and $ln K = 2.50$ at $T_{2} = 350$ K. Find $Δ H^{\circ}$ (assume constant). Use $R = 8.314$ J·mol⁻¹·K⁻¹.

Recall Solution L4.2

Integrated van 't Hoff: $ln K_{2} - ln K_{1} = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) .$ Numbers. $ln K_{2} - ln K_{1} = 2.50 - 4.00 = - 1.50$ . $\frac{1}{350} - \frac{1}{300} = 0.00285714 - 0.00333333 = - 4.7619 \times 1 0^{- 4}$ K⁻¹. Solve for $Δ H^{\circ}$ : $Δ H^{\circ} = \frac{( l n K _{2} - l n K _{1} ) R}{- ( 1/ T _{2} - 1/ T _{1} )} = \frac{( - 1.50 ) ( 8.314 )}{- ( - 4.7619 \times 1 0 ^{- 4} )} = \frac{- 12.471}{4.7619 \times 1 0 ^{- 4}} = - 26190 J/mol .$ So $Δ H^{\circ} \approx - 26.2$ kJ/mol. Sign check: $K$ dropped as $T$ rose → hotter is less favourable → exothermic, $Δ H^{\circ} < 0$ . ✓

L4.3 Write the Helmholtz (constant-volume) analogue and explain, symbol by symbol, why $U$ replaces $H$ .

Recall Solution L4.3

The twin: $(\frac{\partial ( A / T )}{\partial ( 1/ T )})_{V} = U .$ Why. $G = H - T S$ came from constant-pressure life, where $H$ is the natural energy. At constant volume we use Helmholtz free energy $A = U - T S$ instead — same algebra with $H \to U$ and constant- $P$ → constant- $V$ . The derivation of the parent note is identical after that swap, because $d A = - S d T - P d V$ gives $(\partial A / \partial T)_{V} = - S$ , mirroring $(\partial G / \partial T)_{P} = - S$ .

Level 5 — Mastery

Goal: build a correction, handle the case the easy formula ignores.

L5.1 The two-point formula assumes $Δ H$ is constant. In reality $Δ H$ drifts with temperature via Kirchhoff's law (thermochemistry): $\frac{d Δ H}{d T} = Δ C_{p}$ . Suppose $Δ C_{p} = - 40$ J·mol⁻¹·K⁻¹ is constant, and $Δ H (300 K) = - 58.0$ kJ/mol. Find $Δ H$ at $310$ K.

Recall Solution L5.1

Kirchhoff, integrated with constant $Δ C_{p}$ : $Δ H (T_{2}) = Δ H (T_{1}) + Δ C_{p} (T_{2} - T_{1}) .$ Numbers. $Δ H (310) = - 58000 + (- 40) (310 - 300) = - 58000 - 400 = - 58400$ J/mol $= - 58.4$ kJ/mol. Reading it: a negative $Δ C_{p}$ makes the reaction slightly more exothermic as it warms — the easy constant- $Δ H$ assumption was off by $0.4$ kJ/mol over $10$ K, small but real.

L5.2 (Synthesis + Mastery). Derive the full integrated Gibbs–Helmholtz result when $Δ H (T) = Δ H_{0} + Δ C_{p} (T - T_{0})$ (constant $Δ C_{p}$ ), giving $Δ G (T) / T$ as a function of $1/ T$ .

Recall Solution L5.2

Start from $\frac{d ( Δ G / T )}{d T} = - \frac{Δ H ( T )}{T ^{2}}$ and substitute $Δ H (T) = (Δ H_{0} - Δ C_{p} T_{0}) + Δ C_{p} T$ . Write $a \equiv Δ H_{0} - Δ C_{p} T_{0}$ so $Δ H = a + Δ C_{p} T$ . Integrate term by term from $T_{0}$ to $T$ : $\frac{Δ G ( T )}{T} - \frac{Δ G ( T _{0} )}{T _{0}} = - \int_{T_{0}}^{T} \frac{a + Δ C _{p} T ^{'}}{T ^{'2}} d T^{'} = - a (- \frac{1}{T ^{'}})_{T_{0}}^{T} - Δ C_{p} ln T^{'}_{T_{0}}^{T} .$ Result: $\frac{Δ G ( T )}{T} = \frac{Δ G ( T _{0} )}{T _{0}} + a (\frac{1}{T} - \frac{1}{T _{0}}) - Δ C_{p} ln \frac{T}{T _{0}},$ with $a = Δ H_{0} - Δ C_{p} T_{0}$ . Meaning. The first correction term is linear in $1/ T$ (the constant- $Δ H$ line); the $ln (T / T_{0})$ term is the curvature Kirchhoff introduces. If $Δ C_{p} = 0$ , both extra pieces vanish and you recover the straight two-point line. This is the honest formula for wide temperature ranges.

L5.3 (numeric). Using the full formula from L5.2 with $Δ G (T_{0}) = - 30.0$ kJ/mol at $T_{0} = 280$ K, $Δ H_{0} = - 58.0$ kJ/mol (the enthalpy change at $T_{0}$ ), and constant $Δ C_{p} = - 40$ J·mol⁻¹·K⁻¹, find $Δ G$ at $T = 310$ K.

Recall Solution L5.3

Work in J/mol throughout (never mix kJ and J). Step 1 — the constant piece $a$ . $a = Δ H_{0} - Δ C_{p} T_{0} = - 58000 - (- 40) (280) = - 58000 + 11200 = - 46800$ . Step 2 — evaluate the three terms of the L5.2 formula for $Δ G (310) /310$ . $\frac{Δ G ( 310 )}{310} = \frac{- 30000}{280} + (- 46800) (\frac{1}{310} - \frac{1}{280}) - (- 40) ln \frac{310}{280} .$

First term: $\frac{- 30000}{280} = - 107.1429$ .
Second term: $\frac{1}{310} - \frac{1}{280} = 0.0032258 - 0.0035714 = - 3.4562 \times 1 0^{- 4}$ ; times $- 46800 = + 16.175$ .
Third term: $- (- 40) ln (310/280) = 40 ln (1.107143) = 40 (0.101783) = 4.0713$ . Step 3 — sum and multiply by $T$ . Sum $= - 107.1429 + 16.175 + 4.0713 = - 86.897$ J·mol⁻¹·K⁻¹. $Δ G (310) = - 86.897 \times 310 = - 26938 J/mol \approx - 26.9 kJ/mol .$ Sanity: $Δ G$ rose from $- 30.0$ to $- 26.9$ kJ/mol as $T$ increased — an exothermic reaction losing favour when hot. ✓

Connections

Gibbs free energy — the quantity every problem differentiates.
Enthalpy — the prize each exercise extracts.
Entropy — the term that cancels; watch it in L3.2.
van 't Hoff equation — L4 is entirely this bridge.
Kirchhoff's law (thermochemistry) — powers the L5 corrections.
Helmholtz free energy — the constant-volume twin in L4.3.
Maxwell relations — same identity family behind $(\partial G / \partial T)_{P} = - S$ .