Back to the on-screen lesson ·
The Helmholtz and Gibbs free energies, $\Delta G = \Delta H - T\Delta S$ as the test of which way a process runs, crossover temperatures, and the maximum work of electrolysis and fuel cells.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to derive the Gibbs criterion from the second law, decide with $\Delta G = \Delta H - T\Delta S$ whether a process runs, find the temperature at which it reverses, and find the maximum work and voltage of an electrochemical cell.
You know that the entropy of an isolated system never falls, that heat $Q$ into surroundings at temperature $T$ raises their entropy by $Q/T$, and that enthalpy $H = U + PV$ is the heat exchanged at constant pressure. This lesson combines them into the single quantity that decides what happens in a system held at the temperature and pressure of its surroundings.
| Term | What it means |
|---|---|
| Helmholtz free energy | $F = U - TS$; at fixed $T$ and $V$ it never increases, and $-\Delta F$ is the most work obtainable. |
| Gibbs free energy | $G = H - TS = U + PV - TS$; at fixed $T$ and $P$ it never increases. |
| Spontaneous | Able to happen by itself; at fixed $T$ and $P$, a process with $\Delta G < 0$. |
| Maximum non-expansion work | $-\Delta G$: the most electrical or other useful work a process at fixed $T$ and $P$ can deliver. |
| Crossover temperature | $T = \Delta H/\Delta S$, where $\Delta G$ changes sign and a process reverses direction. |
| Endothermic | Absorbing heat from the surroundings, $\Delta H > 0$. |
| Exothermic | Releasing heat to the surroundings, $\Delta H < 0$. |
| Faraday constant | $F = 96\,485$ C/mol, the charge of a mole of electrons. |
Most processes happen in contact with surroundings that hold the temperature — and often the pressure — fixed. The second law still applies to the system and surroundings together, but the surroundings are awkward to track. The free energies do it for us.
Suppose a system at fixed temperature $T$ and pressure $P$ absorbs heat $Q = \Delta H$ from its surroundings. The surroundings' entropy falls by $\Delta H/T$, so the total entropy change is $\Delta S_{\text{total}} = \Delta S - \Delta H/T$. The second law requires this to be positive, and multiplying by $-T$:
$$\Delta G = \Delta H - T\Delta S \leq 0, \qquad G = H - TS.$$
So at fixed temperature and pressure a process runs by itself exactly when it lowers the Gibbs free energy. Two contributions compete: releasing heat ($\Delta H < 0$) helps, because it raises the surroundings' entropy; raising the system's own entropy ($\Delta S > 0$) helps, weighted by the temperature. When both help or both hurt the answer is obvious; when they disagree, the temperature decides, and the process changes direction at $T = \Delta H/\Delta S$.
The free energy also measures useful work. At fixed $T$ and $P$ the most non-expansion work — electrical work, say — a process can deliver is $-\Delta G$; at fixed $T$ and volume the most total work is $-\Delta F$, with the Helmholtz free energy $F = U - TS$. The name free means available: energy that can be turned into work, as opposed to the $TS$ part that must stay behind as heat.
Another way: picture
Think of $\Delta H$ as a bill and $T\Delta S$ as a subsidy. A process that releases heat is paid to happen; one that raises entropy is subsidized, and the subsidy grows with the temperature. $\Delta G$ is what is left after the subsidy. Melting ice costs $6$ kJ per mole but earns a subsidy of $22$ J/K times the temperature; above $273$ K the subsidy covers the bill and the ice melts.
Another way: steps
Checks. The four sign combinations give four behaviors: $\Delta H < 0$ and $\Delta S > 0$ runs at every temperature; $\Delta H > 0$ and $\Delta S < 0$ never runs; the mixed cases run above or below the crossover. At the melting or boiling point of a substance $\Delta G = 0$ exactly, so $\Delta S = \Delta H/T$, which is the $L/T$ of the entropy lesson. And $|\Delta G|$ for a reaction releasing heat is less than $|\Delta H|$ when $\Delta S < 0$: some of the energy must leave as heat to pay the entropy.
Splitting a mole of liquid water into hydrogen and oxygen at $298$ K has $\Delta H = +285.8$ kJ and $\Delta S = +163.2$ J/K. Its Gibbs free energy change is $\Delta G = 285.8 - 298 \times 0.1632 = 237.2$ kJ. That is the least electrical work that must be supplied; the other $48.6$ kJ can be drawn from the surroundings as heat, because the products have more entropy than the water.
Two electrons move per molecule, so the least voltage is $\mathcal{E} = \Delta G/(2F) = 237\,200/(2 \times 96\,485) = 1.23$ V. Real electrolyzers run at about $1.8$ V, the excess lost as heat in the cell.
Run the reaction backward in a fuel cell and hydrogen and oxygen combine to give at most $237.2$ kJ of electrical work per mole, while releasing $285.8$ kJ of enthalpy. The limit $237.2/285.8 = 83$ percent is not a Carnot limit: no temperature difference is involved. It is the fraction of the reaction's energy that is free, the rest being the $T\Delta S$ that must leave as heat because the water formed has less entropy than the gases.
The Gibbs free energy is the chemist's tool; the Helmholtz free energy $F = U - TS$ is the statistical physicist's. For a system held at fixed temperature and volume, $F$ can be computed directly from the list of the system's energy levels, through the partition function of the next unit: $F = -kT\ln Z$. From $F$ everything else follows by differentiation: $S = -\partial F/\partial T$, $P = -\partial F/\partial V$, and $\mu = \partial F/\partial N$.
That is why $F$ matters even for problems nobody would run at fixed volume. It is the natural meeting point of the two halves of the subject: thermodynamics says what to do with $F$, and statistical mechanics says how to calculate it from the microscopic physics.
Every living cell runs on free energy. The hydrolysis of ATP, the molecule cells use as an energy currency, has $\Delta G$ of about $-30.5$ kJ/mol under standard conditions and around $-50$ kJ/mol under the concentrations inside a working cell. Cells couple that drop to reactions that would not run on their own, such as building proteins, so that the combined $\Delta G$ is negative.
A person uses and remakes roughly their own body weight in ATP every day. The underlying source is the oxidation of glucose, for which $\Delta G = -2870$ kJ/mol; a cell captures about $30$ ATP per glucose, $30 \times 50 = 1500$ kJ, roughly half the free energy available. The rest leaves as heat, which is why a resting body gives off about $100$ W. Life does not violate the second law by building order; it pays for that order with free energy drawn from food and ultimately from sunlight.
Portland cement, the binder in concrete, begins with limestone heated until it decomposes: $\text{CaCO}_3 \to \text{CaO} + \text{CO}_2$. The reaction absorbs $178.3$ kJ per mole and raises the entropy by $160.6$ J/K, mostly because it releases a gas.
At room temperature, $\Delta G = 178.3 - 298 \times 0.1606 = 130.4$ kJ/mol, strongly positive: limestone is stable, which is why marble statues last. The crossover is at $178\,300/160.6 = 1110$ K, about $840$ °C, and kilns run at about $900$ °C to drive the reaction quickly with carbon dioxide at atmospheric pressure above it. At $1200$ K, $\Delta G = 178.3 - 1200 \times 0.1606 = -14.4$ kJ/mol.
The same numbers explain why cement making is responsible for roughly $8$ percent of global carbon dioxide emissions. Each mole of lime releases a mole of $\text{CO}_2$ from the rock itself, about $0.79$ kg for every kilogram of lime, before counting the fuel burned to reach the crossover temperature. No change of fuel can remove the first part, which is why the industry is studying carbon capture at the kiln.
A Toyota Mirai stores $5.6$ kg of hydrogen at $700$ atmospheres. Its fuel cells can turn at most the Gibbs free energy of the reaction into electricity: $237.2$ kJ per mole, and $5.6$ kg is $5600/2.016 = 2778$ mol, so the ceiling is $2778 \times 237.2 = 659\,000$ kJ, or $183$ kWh.
The enthalpy released is $2778 \times 285.8 = 794\,000$ kJ, or $221$ kWh, and the $38$ kWh difference must leave as heat whatever the fuel cell is made of. Real stacks deliver about $60$ percent of the Gibbs limit, around $110$ kWh, and the car's EPA-rated range of about $400$ miles corresponds to roughly $0.27$ kWh per mile.
The comparison with an engine is the point. A hydrogen engine burning the same tank would first turn the $221$ kWh into heat and then lose the Carnot share, delivering perhaps $75$ kWh at the wheels. The fuel cell's advantage is that it never passes the energy through heat at all, so its limit is set by entropy alone rather than by entropy and a temperature difference.
A common rule of thumb says that reactions giving off heat happen by themselves and reactions absorbing heat do not. It fails constantly. An instant cold pack works because ammonium nitrate dissolves in water by itself while absorbing heat; ice melts on a warm day while absorbing heat; water evaporates from a glass while absorbing heat. In each, the system's entropy rises by enough that $T\Delta S$ outweighs $\Delta H$ and $\Delta G$ is negative.
The opposite trap is to think a process with $\Delta G < 0$ must happen quickly. The free energy says only whether a process can run by itself, not how fast. A diamond has a higher Gibbs free energy than graphite at room conditions, so it can turn into graphite — but the rate is so slow that diamonds last billions of years. Hydrogen and oxygen can sit mixed for years until a spark lets them react. How fast is a question for kinetics, and it is separate from which way.
For melting ice, $\Delta H = 6010$ J/mol and $\Delta S = 22.0$ J/(mol K). Write the free energy change.
$\Delta G = \Delta H - T\Delta S = 6010 - 22.0T$
At fixed temperature and pressure its sign decides.
Evaluate it at $-10$ °C.
$\Delta G = 6010 - 22.0 \times 263 = 6010 - 5786 = +224\ \text{J/mol}$
Positive: ice does not melt.
Evaluate it at $+10$ °C.
$\Delta G = 6010 - 22.0 \times 283 = 6010 - 6226 = -216\ \text{J/mol}$
Negative: ice melts.
Find the crossover.
$T = \dfrac{\Delta H}{\Delta S} = \dfrac{6010}{22.0} = 273\ \text{K}$
The melting point, where solid and liquid are in balance.
Say which term wins on each side.
$T < 273\ \text{K}: \ \Delta H \text{ wins}; \qquad T > 273\ \text{K}: \ T\Delta S \text{ wins}$
Melting costs energy but gains entropy, and the entropy term grows with temperature.
For splitting liquid water at $298$ K, $\Delta H = +285.8$ kJ/mol and $\Delta S = +163.2$ J/(mol K). Convert the entropy to kJ/K.
$\Delta S = 0.1632\ \text{kJ/(mol K)}$
The units must match.
Find the entropy term.
$T\Delta S = 298 \times 0.1632 = 48.6\ \text{kJ/mol}$
Heat that can come from the surroundings.
Find the Gibbs free energy change.
$\Delta G = 285.8 - 48.6 = 237.2\ \text{kJ/mol}$
The least work that must be supplied electrically.
Count the electrons transferred per molecule.
$\text{H}_2\text{O} \to \text{H}_2 + \tfrac{1}{2}\text{O}_2: \quad n = 2$
Two hydrogen ions each take an electron.
Find the least voltage.
$\mathcal{E} = \dfrac{\Delta G}{nF} = \dfrac{237\,200}{2 \times 96\,485} = 1.23\ \text{V}$
Work per unit charge.
Compare with a real electrolyzer.
$\dfrac{1.23}{1.8} \approx 0.68$
Industrial cells run near $1.8$ V; the rest is lost to resistance and to driving the reactions quickly.
In a fuel cell, $\text{H}_2 + \tfrac{1}{2}\text{O}_2 \to \text{H}_2\text{O}$ at $298$ K. Write its enthalpy and entropy changes.
$\Delta H = -285.8\ \text{kJ/mol}, \qquad \Delta S = -0.1632\ \text{kJ/(mol K)}$
The reverse of splitting water.
Find the Gibbs free energy change.
$\Delta G = -285.8 - 298 \times (-0.1632) = -285.8 + 48.6 = -237.2\ \text{kJ/mol}$
Negative: the reaction runs by itself.
Find the most electrical work per mole.
$W_{\max} = -\Delta G = 237.2\ \text{kJ}$
At fixed $T$ and $P$ the drop in $G$ is the most non-expansion work.
Find the heat that must be released.
$Q = T\Delta S = 298 \times (-0.1632) = -48.6\ \text{kJ}$
The water formed has less entropy than the gases, so this much heat must leave to raise the surroundings' entropy.
Check the energy balance.
$237.2 + 48.6 = 285.8\ \text{kJ} = -\Delta H$
Work plus heat equals the enthalpy released.
Find the efficiency limit.
$e_{\max} = \dfrac{-\Delta G}{-\Delta H} = \dfrac{237.2}{285.8} = 0.83$
Not a Carnot limit: no temperature difference is involved.
Compare with burning the hydrogen in an engine.
$e_{\text{engine}} \approx 0.35$
Burning turns the chemical energy into heat first, and the Carnot limit then takes its share; the fuel cell skips that step.
Find the cell voltage.
$\mathcal{E} = \dfrac{237\,200}{2 \times 96\,485} = 1.23\ \text{V}$
Real cells deliver about $0.7$ V under load, so stacks of hundreds of cells are used.
Set the free energy change to zero.
$\Delta H - T\Delta S = 0 \quad\Rightarrow\quad T = \dfrac{\Delta H}{\Delta S}$
The crossover temperature.
Put both in joules.
$T = \dfrac{178\,300}{160.6}$
Kilojoules to joules.
Divide the two numbers.
Dissolving ammonium nitrate in water at $298$ K absorbs heat: $\Delta H = +20$ kJ/mol, with $\Delta S = +105$ J/(mol K). Does it dissolve by itself, and why?
Complete the worked solution: for splitting liquid water into hydrogen and oxygen, $\Delta H = +285.8$ kJ and $\Delta S = +0.1632$ kJ/K per mole of reaction. Find $\Delta G$ for $4$ moles at $410$ K.
Multiply the enthalpy change by the number of moles.
$n\Delta H = 4 \times 285.8 =$ a kJ
Enthalpy is extensive.
Multiply the entropy change by the moles and the temperature.
$nT\Delta S = 4 \times 410 \times 0.1632 =$ b kJ
The entropy term, in kJ.
Subtract the entropy term from the enthalpy term.
$\Delta G = n\Delta H - nT\Delta S =$ c kJ
$G = H - TS$ at fixed temperature.
Match each set of conditions to the quantity a system drives to an extreme in equilibrium, for example in a lab at $285$ K.
| entropy $S$ is largest | Helmholtz free energy $F$ is smallest | Gibbs free energy $G$ is smallest | enthalpy $H$ is smallest | |
|---|---|---|---|---|
| isolated: fixed energy, volume and particles | ||||
| fixed temperature and volume | ||||
| fixed temperature and pressure | ||||
| fixed entropy and pressure |
For melting ice, $\Delta H = 6010$ J/mol and $\Delta S = 22.0$ J/(mol K). For $2$ mol, fill in $T\Delta S$ and $\Delta G$, in joules, at each temperature.
| $T\Delta S$ (J) | $\Delta G$ (J) | |
|---|---|---|
| 263 K | ||
| 273 K | ||
| 283 K |
For decomposing baking soda, $\Delta H = +135.6$ kJ and $\Delta S = +0.3344$ kJ/K. What is $\Delta G$ at $260$ K, in kJ?
Answer: kJ
For calcium carbonate decomposing, $\Delta H = +178.3$ kJ and $\Delta S = +160.6$ J/K per mole at one atmosphere, both nearly independent of temperature. Above what temperature, in kelvin, does it run by itself?
Answer: K
A fuel-cell car carries $5$ kg of hydrogen. Combining hydrogen with oxygen into liquid water releases $\Delta H = -285.8$ kJ/mol and has $\Delta G = -237.2$ kJ/mol at $298$ K. What is the most electrical energy, in kWh, the fuel cell could deliver from the tank?
Answer: kWh
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For melting ice, $\Delta H = 6010$ J/mol and $\Delta S = 22.0$ J/(mol K). For $6$ mol, fill in $T\Delta S$ and $\Delta G$, in joules, at each temperature.
| $T\Delta S$ (J) | $\Delta G$ (J) | |
|---|---|---|
| 263 K | ||
| 273 K | ||
| 283 K |
You can decide which way a process runs at a given temperature and how much useful work it can give. Explain to someone how an instant cold pack can absorb heat and still work.
15. Your turn: for decomposing calcium carbonate, $\Delta H = +178.3$ kJ and $\Delta S = +160.6$ J/K. Above what temperature does it run?, step 3
$T \approx 1110\ \text{K} \approx 837\ ^{\circ}\text{C}$
Lime kilns run near $900$ °C, just above the crossover.