7.3 Residual entropy

Residual Entropy and the Third Law

Residual Entropy and the Third Law

The Third Law of Thermodynamics states that the entropy of a perfect crystal approaches zero as temperature approaches absolute zero (0 K). This assumes the crystal has a single, unique ground state with no remaining disorder.

Some systems, however, retain measurable disorder even at 0 K. This leftover disorder is called residual entropy. It occurs in systems where multiple distinct configurations share the same lowest energy, so the system never "settles" into one perfectly ordered arrangement. Residual entropy is best understood as an exception to the idealized conditions of the Third Law, not a contradiction of it. The Third Law still holds for perfect crystals; residual entropy simply tells you that not all real substances form perfect crystals.

Degenerate Ground States

The reason residual entropy exists comes down to degenerate ground states: multiple distinct molecular configurations that all share the exact same lowest energy.

• Because these configurations have equal energy, they're equally probable. The system has no energetic reason to prefer one over another.
• Even at 0 K, the system can be found in any of these states, so some randomness persists.
• The more degenerate ground states a system has, the larger its residual entropy. A system with only one ground state has zero residual entropy (a perfect crystal), while a system with many degenerate states retains significant disorder.

Calculating Residual Entropy

Residual entropy is calculated using the Boltzmann equation:

$S = k_B \ln \Omega$

• $S$ = entropy
• $k_B$ = Boltzmann constant ($1.381 \times 10^{-23} \text{ J/K}$)
• $\Omega$ = number of microstates (distinct configurations the system can adopt)

For residual entropy specifically, $\Omega$ is the number of degenerate ground states. Here's how to apply it:

1. Identify how many equivalent configurations each molecule (or unit) can adopt in the ground state.
2. For $N$ molecules each with $q$ orientations, the total number of microstates is $\Omega = q^N$.
3. Plug into the Boltzmann equation: $S_0 = k_B \ln(q^N) = N k_B \ln q$.
4. To get a molar value, replace $N k_B$ with the gas constant $R$: $S_0 = R \ln q$.

Quick example: If each molecule has two equally likely orientations ($q = 2$), the molar residual entropy is:

$S_0 = R \ln 2 \approx 8.314 \times 0.693 \approx 5.76 \text{ J/(mol·K)}$

This gives you the theoretical maximum for a two-orientation system. Real values are often lower because the orientations aren't always perfectly independent (neighboring molecules can constrain each other).

Examples of Non-Zero Residual Entropy

Ice is the classic example. The oxygen atoms sit on a well-ordered crystal lattice, but the hydrogen atoms don't have fixed positions. Each oxygen is surrounded by four neighbors, and the two hydrogens can sit in multiple arrangements that all satisfy the ice rules (two H atoms close, two far from each oxygen). This creates an enormous number of degenerate ground states across the whole crystal. Linus Pauling estimated the residual entropy of ice at approximately $3.4 \text{ J/(mol·K)}$, which matches experimental measurements closely.

Carbon monoxide (CO) provides a simpler case. In solid CO, each molecule can orient as either CO or OC in the lattice, since the two ends are nearly the same size and have similar interactions. With two orientations per molecule, you'd predict $S_0 = R \ln 2 \approx 5.76 \text{ J/(mol·K)}$. The measured value is approximately $4.6 \text{ J/(mol·K)}$, somewhat lower than the ideal prediction because neighboring molecules aren't completely independent in their orientations.

Both examples confirm that residual entropy is experimentally measurable. It shows up as a discrepancy between the entropy calculated from heat capacity measurements (Third Law method, integrating $C_p/T$ from 0 K) and the entropy calculated from spectroscopic data. That gap is the residual entropy.