The Third Law of Thermodynamics

Cool any pure, perfectly ordered crystal toward absolute zero. Its entropy drains away. In the limit $T \to 0\ \text{K}$ , the entropy approaches a single universal constant. This constant is taken to be zero. This is the Third Law of Thermodynamics. It is what lets chemists assign absolute entropies to substances, rather than only entropy differences. Without the Third Law, entropy would be as relative as potential energy. It would always be defined up to an arbitrary constant. With the Third Law, calorimetric measurements matter differently. Measurements from near 0 K up to 298 K give a real number, $S^\circ_{298}$ . This number appears in every thermodynamic table. You use it to predict whether a reaction runs spontaneously. This article builds the statistical justification for the law. It derives how absolute entropies are reconstructed from heat-capacity data. It works two full numerical examples. It closes with the common traps: residual entropy, glasses, and why "zero" is a convention with real physical teeth.

Key takeaways

The Third Law states that the entropy of a perfect crystal tends to zero as $T \to 0\ \text{K}$ .
"Perfect crystal" means one unique, non-degenerate microstate — so $S = k_B \ln 1 = 0$ by the Boltzmann formula.
It allows absolute (not just relative) entropies, obtained by integrating $C_p/T$ from 0 K upward using calorimetry.
Some substances (CO, H₂O ice, glasses) retain residual entropy at 0 K because disorder gets frozen in.
The law also forbids reaching absolute zero in a finite number of steps (the unattainability statement).
Key idea: Perfect order has exactly one accessible microstate, so its statistical entropy is exactly zero, giving entropy an absolute zero point.

The core insight

The Third Law is a statement about counting: a flawless crystal at absolute zero occupies one, and only one, quantum microstate, so Boltzmann's $S = k_B \ln W$ forces $S \to 0$ , and this single reference point lets every substance's entropy be measured on an absolute scale via heat-capacity integration from 0 K.

The physical picture

Recall that entropy has two faces you have already met: the statistical definition $S = k_B \ln W$ , where $W$ counts accessible microstates at fixed energy, and the thermodynamic definition built from $\mathrm{d}S = \delta q_{\text{rev}}/T$ . Heating a substance from one temperature to another changes $S$ by an integral of $C_p/T$ over that range — but that only ever gives a difference, $\Delta S$ , unless you know the entropy at the starting point. The Third Law supplies that starting point. Imagine a crystal of, say, solid argon, built from identical atoms locked onto a perfect lattice with no vacancies, no impurities, no orientational freedom. As $T \to 0$ , the system must fall into its single, non-degenerate ground state — every atom "frozen" into one specific arrangement. There is only one way to arrange a perfect crystal in its lowest energy state, so $W = 1$ and $S = k_B \ln 1 = 0$ . Real substances approach this ideal to varying degrees. Metals, ionic crystals, and simple molecular solids with no orientational ambiguity extrapolate smoothly to zero entropy. Others — carbon monoxide, ice, glasses — get kinetically stuck with some disorder frozen in below the temperature where it could anneal out; this leftover entropy is called residual entropy, foreshadowed here and treated in full afterward. The practical payoff is calorimetric: measure $C_p(T)$ from very low $T$ up to 298 K, integrate $C_p/T\,\mathrm{d}T$ , add any phase-transition contributions $\Delta_\text{trs}H/T_\text{trs}$ , and you obtain the absolute Third-Law entropy $S^\circ(298\ \text{K})$ — not just a difference, but a number referenced to true zero.

Figure

Statistical microstates as a perfect crystal is cooled toward 0 K

Figure

Reconstructing absolute entropy from Cp over T integrated from 0 K

The core relations

\lim_{T \to 0} S_{\text{m}}(\text{perfect crystal}) = 0

where: $S_\text{m}$ is molar entropy; "perfect crystal" means a single, non-degenerate ground microstate.

S = k_B \ln W \;\xrightarrow{W \to 1}\; S \to 0

where: $W$ is the number of microstates consistent with the macroscopic state; $k_B$ is Boltzmann's constant.

S^\circ(T) = \int_0^T \frac{C_p(T')}{T'}\,\mathrm{d}T' + \sum_i \frac{\Delta_\text{trs} H_i}{T_{\text{trs},i}}

where: the sum adds entropy jumps at any phase transitions between 0 K and $T$ ; $C_p$ is heat capacity at constant pressure.

C_p \approx a T^3 \quad (T \to 0, \text{ non-metallic solids; Debye law})

where: $a$ is a substance-specific constant, making $C_p/T \propto T^2$ finite and integrable at $T=0$ .

\Delta_r S^\circ = \sum_{\text{products}} \nu S^\circ_\text{m} - \sum_{\text{reactants}} \nu S^\circ_\text{m}

where: $\nu$ are stoichiometric coefficients; this combination only makes sense because each $S^\circ_\text{m}$ is absolute.

Derivation AWhy the Cp/T integral needs the Third Law to converge and to anchor absolute S

Goal. Show why entropy from 0 K to some temperature $T$ is finite despite the $1/T$ singularity, and why the Third Law is essential to call the result "absolute."

Step 1. Start from the thermodynamic definition of entropy change. For a reversible process at constant pressure with no phase change,

\mathrm{d}S = \frac{\delta q_\text{rev}}{T} = \frac{C_p\,\mathrm{d}T}{T}.

Integrating between two temperatures $T_1$ and $T_2$ gives only a difference, $S(T_2) - S(T_1)$ , from thermodynamics alone.

Step 2. Ask what happens as the lower limit $T_1 \to 0$ . Naively $\int C_p/T\,\mathrm{d}T$ looks divergent because $1/T \to \infty$ . Whether the integral actually diverges depends on how fast $C_p$ vanishes as $T\to 0$ .

Step 3. Bring in the low-temperature form of $C_p$ . Experiment and the Debye model of lattice vibrations show that for non-metallic crystalline solids,

C_p \to aT^3 \quad \text{as } T \to 0,

so that

\frac{C_p}{T} \to aT^2,

which vanishes at $T = 0$ rather than blowing up. (For metals, an extra electronic term linear in $T$ appears, $C_p \approx \gamma T + aT^3$ , and $C_p/T \to \gamma$ , a finite constant — still integrable.)

Step 4. Integrate from true zero. Because $C_p/T$ stays finite (indeed goes to zero or to a finite constant) as $T\to 0$ , the integral

\int_0^{T} \frac{C_p(T')}{T'}\,\mathrm{d}T'

converges to a finite number. Thermodynamics alone can compute this finite integral, but it only tells you $S(T) - S(0)$ — it cannot tell you what $S(0)$ itself equals.

Step 5. Invoke the Third Law to fix the additive constant. The Third Law asserts $S(0) = 0$ for a perfect crystal. Substituting this boundary condition converts the difference into an absolute value:

S^\circ(T) = S(T) - S(0) = \int_0^{T} \frac{C_p(T')}{T'}\,\mathrm{d}T'.

Any phase transitions crossed on the way (fusion, solid-solid transitions) contribute additional discrete jumps $\Delta_\text{trs}H/T_\text{trs}$ , added at their respective temperatures.

Step 6. Conclude.

\boxed{S^\circ_\text{m}(T) = \int_0^{T} \frac{C_p(T')}{T'}\,\mathrm{d}T' + \sum_i \frac{\Delta_\text{trs} H_i}{T_{\text{trs},i}}}

This is exactly what makes calorimetric "Third-Law entropies" absolute rather than merely relative, and it is why tables can list a single number $S^\circ_\text{m}(298\ \text{K})$ for every substance.

Worked examples

Example 1 — Foundational: entropy of a solid warmed from near 0 K to its melting point. Suppose measured heat capacities (idealized, smooth data) for a hypothetical simple molecular solid give the following contributions, already evaluated by careful graphical or numerical integration of $C_p/T$ against $T$ :

From 0 K to 10 K, using the Debye $T^3$ law with $C_p = aT^3$ and $a = 0.0021\ \text{J K}^{-4}\text{mol}^{-1}$ :

\int_0^{10} \frac{aT^3}{T}\,\mathrm{d}T = \int_0^{10} aT^2\,\mathrm{d}T = \frac{aT^3}{3}\Big|_0^{10} = \frac{(0.0021)(1000)}{3} = 0.70\ \text{J K}^{-1}\text{mol}^{-1}.

From 10 K to 150 K (below melting), graphical integration of experimental $C_p/T$ data gives $28.4\ \text{J K}^{-1}\text{mol}^{-1}$ .
Fusion occurs at $T_\text{fus} = 150\ \text{K}$ with $\Delta_\text{fus}H = 6.00\ \text{kJ mol}^{-1}$ , contributing

\frac{\Delta_\text{fus}H}{T_\text{fus}} = \frac{6000\ \text{J mol}^{-1}}{150\ \text{K}} = 40.0\ \text{J K}^{-1}\text{mol}^{-1}.

From 150 K (liquid) to 298 K, integration of liquid $C_p/T$ data gives $22.6\ \text{J K}^{-1}\text{mol}^{-1}$ .

Summing every contribution, anchored at $S(0) = 0$ by the Third Law:

S^\circ_\text{m}(298\ \text{K}) = 0.70 + 28.4 + 40.0 + 22.6 = 91.7\ \text{J K}^{-1}\text{mol}^{-1}.

Sanity check: this value is comparable in magnitude to real molar entropies of small molecular liquids at 298 K (e.g., $S^\circ_\text{m}$ for liquid benzene is about 173 J K⁻¹mol⁻¹, for a heavier, more complex molecule), so 91.7 J K⁻¹mol⁻¹ is a physically reasonable number for a smaller, simpler substance. Each term carries units of J K⁻¹mol⁻¹ and the running total only increases, as entropy must upon heating and melting.

Example 2 — Applied: detecting residual entropy by comparing calorimetric and spectroscopic entropies. For carbon monoxide, CO, the Third-Law (calorimetric) entropy obtained by the integration procedure above is measured as

S^\circ_\text{calorimetric}(298\ \text{K}) = 193.0\ \text{J K}^{-1}\text{mol}^{-1}.

Independently, statistical mechanics applied to gas-phase spectroscopic data (rotational and vibrational constants, molecular mass) predicts the ideal entropy a perfectly ordered crystal should give:

S^\circ_\text{spectroscopic}(298\ \text{K}) = 197.9\ \text{J K}^{-1}\text{mol}^{-1}.

The discrepancy is

\Delta S = S^\circ_\text{spectroscopic} - S^\circ_\text{calorimetric} = 197.9 - 193.0 = 4.9\ \text{J K}^{-1}\text{mol}^{-1}.

Interpretation: because CO has a very small dipole moment, molecules in the crystal can sit as ...CO CO CO... or accidentally flip to ...OC CO OC... with almost no energy penalty. Freezing happens too fast for the crystal to sort out a single unique orientation for every molecule, so some orientational disorder is locked in at 0 K instead of relaxing to the single ordered ground state. Estimate the residual entropy per mole using Boltzmann's formula, if each CO molecule were to have exactly two equally likely orientations frozen in independently:

S_\text{res} = k_B \ln(2^{N_A}) = N_A k_B \ln 2 = R \ln 2 = (8.314\ \text{J K}^{-1}\text{mol}^{-1})(0.693) = 5.76\ \text{J K}^{-1}\text{mol}^{-1}.

This simple two-orientation model predicts about $5.8\ \text{J K}^{-1}\text{mol}^{-1}$ of residual entropy, reasonably close to the observed $4.9\ \text{J K}^{-1}\text{mol}^{-1}$ gap (the small difference reflects that not every molecule is perfectly 50:50 disordered in the real crystal). This agreement is the classic evidence that CO crystals are not "perfect" in the Third-Law sense — they carry frozen-in residual entropy, foreshadowing the fuller residual-entropy discussion ahead.

Limiting cases & common traps

Situation	What happens to $S(0)$	Why
Perfect crystal, one substance, one isotope	$S(0) = 0$ exactly	Single non-degenerate ground microstate, $W=1$
Crystal with orientational disorder (CO, N₂O, H₂O ice)	$S(0) = S_\text{res} > 0$	Molecules freeze into multiple near-equivalent orientations
Glasses (frozen non-crystalline solids)	$S(0) > 0$ , often large	No long-range order; many frozen configurations
Solid mixtures / isotopic mixtures	$S(0) > 0$	Positional or isotopic disorder persists at 0 K
Metals	$S(0) = 0$ , but $C_p/T \to \gamma$ (finite, not zero)	Electronic heat capacity linear in $T$ ; still integrable

The chemist behind it

Walther Nernst proposed the Third Law's precursor — the Heat Theorem — in 1906, based on his study of chemical affinities and reaction entropies near absolute zero. He observed that $\Delta S$ for reactions between condensed phases appeared to vanish as $T \to 0$ , and generalized this into a postulate about entropy itself. Max Planck later sharpened the statement into the modern form: the entropy of every perfect crystal approaches the same constant (taken as zero) at absolute zero, not merely that entropy changes vanish. This refinement, plus decades of low-temperature calorimetry by Nernst's laboratory and later by researchers such as William Giauque (who used the law to build absolute entropy tables and studied residual entropy in substances like CO and ice), turned the Third Law from a thermochemical curiosity into the backbone of absolute entropy tabulation used throughout chemical thermodynamics today.

Origins · 1 minWalther Nernst and how this was discovered

Walther Nernst

1864–1941 · German physical chemist who formulated the Heat Theorem, the precursor to the Third Law, and won the 1920 Nobel Prize in Chemistry

Ask the tutor: Want to see how residual entropy in water ice is calculated from its hydrogen-bonded proton disorder using the Pauling model?

Is the Third Law really an independent law, or does it follow from the first two

It is genuinely independent. The First and Second Laws fix energy conservation and the direction of spontaneous change but say nothing about an absolute zero point for entropy — they only define entropy differences via $\mathrm{d}S = \delta q_\text{rev}/T$ . The Third Law adds a boundary condition, $S \to 0$ as $T \to 0$ for a perfect crystal, that cannot be derived from the other two. It is confirmed empirically through calorimetry and justified theoretically through statistical mechanics, specifically the Boltzmann counting of a unique ground state.

Does the Third Law mean absolute zero can never be reached

A related statement, sometimes called the unattainability principle, says absolute zero cannot be reached in a finite number of steps by any process, though it can be approached arbitrarily closely. This follows from the Third Law: since entropy differences between states shrink to zero as $T\to 0$ , isothermal and adiabatic steps used in cooling cycles become progressively less effective, requiring infinitely many steps to reach exactly $T=0$ . Laboratories have reached temperatures in the nanokelvin range, but never exactly zero.

How is the Third Law used to build the entropy tables used in Gibbs energy calculations

Calorimetrists measure $C_p(T)$ from as close to 0 K as feasible up through 298 K, using the Debye $T^3$ law to extrapolate the small residual gap to true zero. They numerically integrate $C_p/T$ over this range and add entropy jumps at any phase transitions. Because the Third Law fixes $S(0)=0$ , this integral directly gives the absolute standard molar entropy $S^\circ_\text{m}(298\ \text{K})$ tabulated in reference books, which you then combine via $\Delta_r S^\circ = \sum \nu S^\circ_\text{products} - \sum \nu S^\circ_\text{reactants}$ .

What is residual entropy and does it break the Third Law

Residual entropy is the nonzero entropy that some solids retain at 0 K because they get trapped in a disordered arrangement rather than relaxing into the single truly ordered ground state. Classic examples are carbon monoxide, ice, and nitrous oxide, where near-symmetric molecules freeze into essentially random orientations. It does not break the Third Law, since the law strictly applies to perfect crystals; residual entropy simply signals that the real solid is not perfectly ordered, and it is measured precisely by comparing calorimetric and spectroscopic entropies.

Why does the entropy of a metal not vanish exactly like the Debye T cubed prediction

Metals have mobile conduction electrons that contribute an additional linear-in- $T$ term to the heat capacity, $C_p \approx \gamma T + aT^3$ , arising from the Fermi-Dirac distribution of electronic energy levels. This makes $C_p/T \to \gamma$ , a small but finite constant, rather than vanishing as $T\to 0$ the way the pure lattice ( $aT^3$ ) term does. Crucially, this finite limit is still integrable from $T=0$ , so the Third Law's requirement, $S(0)=0$ for the perfect metallic crystal, remains satisfied — the integral simply picks up an extra, well-behaved electronic contribution near $T=0$ .