Physical

The Boltzmann distribution: populations at temperature

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Take a molecule with two energy levels separated by ΔE. Put it at temperature T. The probability of finding the molecule in the higher level, relative to the lower, is exactly:

PhighPlow=e−ΔE/kBT\frac{P_{high}}{P_{low}} = e^{-\Delta E / k_B T}

This is the Boltzmann distribution. It is the single most important result of statistical mechanics, and it appears in every corner of physical chemistry: IR band intensities, NMR sensitivity, kinetic isotope effects, semiconductor carrier populations, and the very meaning of temperature. This guide derives it, applies it to four canonical problems, and clarifies why populations decide which measurements are sensitive at which temperature.

Prerequisites: what entropy actually is — Boltzmann’s S = k ln W is the entropy expression on which the distribution rests. If Gibbs free energy reads well, the exponent RT here is the same RT that appears in ΔG.

Where the exponential comes from

Consider N distinguishable molecules distributed among energy levels ε_i. The number of ways to arrange n_i molecules in level i (with total ∑n_i = N) is the multinomial:

W=N!∏ini!W = \frac{N!}{\prod_i n_i!}

Statistical mechanics postulates that the most probable macrostate (largest W) is the one you observe. Maximise ln W subject to two constraints: fixed particle number ∑n_i = N and fixed energy ∑n_i ε_i = U. Lagrange multipliers give the result:

ni∝e−βεin_i \propto e^{-\beta \varepsilon_i}

where β is the Lagrange multiplier for the energy constraint. Comparison with thermodynamics (equating dU with T dS) identifies β = 1/k_B T. So:

ninj=e−(εi−εj)/kBT\frac{n_i}{n_j} = e^{-(\varepsilon_i - \varepsilon_j)/k_B T}

For per-mole quantities, replace k_B with R = N_A k_B = 8.314 462 618 J K⁻¹ mol⁻¹ (CODATA 2018) and use ΔE in J/mol instead of J.

Degeneracies — when levels are not “one state each”

Sometimes one energy value corresponds to multiple microstates (say, three rotational orientations at the same energy). Then the population ratio includes the degeneracy g:

ninj=gigje−(εi−εj)/kBT\frac{n_i}{n_j} = \frac{g_i}{g_j} e^{-(\varepsilon_i - \varepsilon_j)/k_B T}

Rotational levels of a linear molecule have g_J = 2J+1 (Atkins 11e §16); an increase in J both raises the energy (working against the exponential) and boosts the degeneracy (working for it). The peak of the rotational-state population lies at J_max ≈ √(k_B T / (2Bhc)) − 1/2 (McQuarrie §18.6), and not at J = 0.

Worked example — IR band intensity from ground-state population

Fraction of CO in the v=0 vibrational state at 298 K
  1. Formula
  2. Substitute
  3. Note

    At 298 K, almost every CO molecule is in v=0. The IR absorption from v=0 → v=1 (at 2143 cm⁻¹) is intense because n₀ is enormous; the 'hot band' from v=1 → v=2 is 30,000× less populated at the initial state and is essentially invisible. For a lower-frequency vibration (say a Si-Si stretch at 400 cm⁻¹), the ratio is e^(−1.94) = 0.14 — 12% of molecules are in v=1 at room temperature, and hot bands become measurable.

  4. Result

The temperature-frequency trade-off

The Boltzmann exponent ΔE/k_B T is the single dimensionless number that controls everything. Some useful benchmarks at 298 K (k_B T = 4.11 × 10⁻²¹ J = 207 cm⁻¹ = 2.48 kJ/mol = 0.0257 eV):

  • ΔE < k_B T — the two levels are nearly equally populated. Rotational states of a small molecule.
  • ΔE ~ k_B T — measurable population in both. Low-frequency vibrations, some spin states in a magnetic field.
  • ΔE >> k_B T — nearly all population in the ground state. High-frequency vibrations, electronic states, most bond energies.

For rotational spectroscopy in the microwave (ΔE ~ 1 cm⁻¹), room temperature is “hot” and many levels are populated. For UV-vis electronic spectroscopy (ΔE ~ 30,000 cm⁻¹), room temperature is “cold” and only the ground electronic state matters. This is why a UV-vis spectrum shows a single sharp absorption band from the ground state, while an IR spectrum can show hot bands and combination bands at higher temperature.

NMR — the least sensitive spectroscopy, explained

NMR asks about the population difference between two spin states in a magnetic field. For ¹H at B₀ = 9.4 T (400 MHz), the energy gap is:

ΔE=hν=(6.626×10−34)(4×108)=2.65×10−25 J\Delta E = h\nu = (6.626 \times 10^{-34})(4 \times 10^8) = 2.65 \times 10^{-25} \text{ J}

At 298 K, k_B T = 4.11 × 10⁻²¹ J. The ratio:

n+n−=e−6.4×10−5≈1−6.4×10−5\frac{n_+}{n_-} = e^{-6.4 \times 10^{-5}} \approx 1 - 6.4 \times 10^{-5}

The population difference is 64 parts per million. That’s the fundamental sensitivity limit of NMR: only 64 out of every million protons contribute to the signal. Cranking up B₀ (higher-field magnets) improves the population difference linearly; cooling the sample (probe cryogenics, DNP) also helps; but the underlying “why NMR needs so much sample” is Boltzmann. Compare with UV-vis, where nearly 100% of the population is in the ground state at every moment: UV-vis needs 10⁶× less analyte.

The Arrhenius equation is Boltzmann in disguise

The activation energy in the Arrhenius equation is exactly the ΔE that gates the reactive population:

k=Ae−Ea/RTk = A e^{-E_a / RT}

Fraction of molecules with kinetic energy ≥ E_a in a Maxwell-Boltzmann distribution equals e^(−E_a/RT). That fraction, multiplied by a collision frequency A, gives the rate constant. See rate laws from mechanism for the full derivation and transition state theory for the ΔG‡ refinement.

Every “raise temperature to speed up reaction” fact you know is Boltzmann-driven: a 10 K rise near room temperature roughly doubles k for a reaction with E_a ≈ 50 kJ/mol, because the exponential is fantastically sensitive to T.

Kinetic isotope effects — one exponent, two vibrations

The heavier isotope has lower zero-point energy (ZPE) than the lighter isotope for the same bond. When a bond breaks at the transition state, the ZPE difference maps onto an activation-energy difference:

kHkD=e−(Ea,H−Ea,D)/RT\frac{k_H}{k_D} = e^{-(E_{a,H} - E_{a,D})/RT}

For a C-H versus C-D bond breaking, ΔZPE ≈ 4.5 kJ/mol (Atkins 11e §22), which at 298 K gives k_H/k_D ≈ 6-7 — a “primary” KIE that is the diagnostic for C-H bond cleavage in the rate-limiting step. All of that comes out of Boltzmann on the appropriate vibrational-state population difference between H and D.

What T actually means

Temperature is defined thermodynamically as (∂U/∂S)_V, but Boltzmann gives it a molecular reading:

1T=kBd(ln⁡W)dU\frac{1}{T} = k_B \frac{d(\ln W)}{dU}

Higher T means adding energy increases the number of accessible microstates more slowly, so β = 1/k_B T is smaller and the exponent is milder. In the extreme, T → ∞ gives β = 0, and all levels are equally populated regardless of energy — a “population inversion” is impossible at finite positive T, which is why laser action requires negative-temperature (non-Boltzmann) inversions.

Negative temperature is not “colder than absolute zero”; it is “hotter than infinite temperature”, meaning the population is inverted — more molecules in high-energy than low-energy levels. It shows up in laser gain media and nuclear-spin systems but not in translational or vibrational manifolds (which have infinitely many levels and can’t invert).

Common pitfalls

Using ΔE in the wrong units. ΔE and k_B T must have matching units. If ΔE is in J/mol, divide by RT (not k_B T). If ΔE is in kJ/mol at 298 K, RT = 2.48 kJ/mol.

Forgetting degeneracies. Rotational, spin, and electronic degeneracies all enter as ratios g_i / g_j. The bare exp(−ΔE/k_B T) is only correct when both levels are non-degenerate or the ratio is being taken between single microstates, not energy levels.

Applying to non-equilibrium systems. The Boltzmann distribution is derived assuming thermal equilibrium. A shocked gas, a laser cavity, or a molecular beam does not follow it — a “vibrational temperature” and a “rotational temperature” can differ by hundreds of kelvin in a supersonic expansion.

Mixing up n_i (population) with P_i (probability). For N total molecules, n_i = N × P_i. The ratios are the same, but if you want an absolute count, multiply by N.

Practice

Problem 1 Easy A vibration has ν̃ = 500 cm⁻¹. What fraction of molecules is in v=1 at 298 K?

Answer: ΔE = hcν̃ / (k_B T) = (500)(1.986 × 10⁻²³) / (4.11 × 10⁻²¹) = 2.42. Population ratio n₁/n₀ = e⁻² ᐧ ⁴² = 0.089. Fraction in v=1: 0.089 / (1 + 0.089) = 8.2%. Low-frequency vibrations show substantial v=1 population at room temperature; this is what enables hot-band structure in IR spectra of small molecules.

Problem 2 Intermediate Two conformers of a molecule have energies 0 (conformer A) and 8 kJ/mol (conformer B). What is the equilibrium ratio [A]:[B] at 298 K and at 373 K?
Conformer populations vs temperature
  1. Formula
  2. Substitute
  3. Note

    Heating shifts the equilibrium toward the higher-energy conformer. This is why NMR line broadening from conformer interconversion becomes visible at elevated temperature — populations of both conformers grow enough that both contribute to the spectrum. Cooling to 173 K would give [A]:[B] ≈ 260:1, essentially locking the molecule in A.

  4. Result
Problem 3 Hard Rotational constant B of HCl is 10.44 cm⁻¹. Compute the most populated rotational level J at 298 K. Compare with the ground state J=0.

Answer: For a linear rotor with g_J = 2J+1 and ε_J = J(J+1)hcB, differentiating (2J+1) e^(−J(J+1)hcB/k_B T) with respect to J and setting to zero gives J_max = √(k_B T / (2hcB)) − 1/2 = √(207/(2×10.44)) − 0.5 = √9.91 − 0.5 = 2.65. J_max = 3. The population at J=3 is exp(−3×4×10.44/207)/exp(0) × (2·3+1)/(2·0+1) = e⁻⁰·⁶⁰⁵ × 7 = 3.82× the J=0 population. Rotational states peak well above the ground state because the degeneracy 2J+1 initially outweighs the Boltzmann suppression.

Where this leads

The Boltzmann distribution is the ground floor of statistical mechanics:

  • Partition functions — Q = ∑g_i e^(−ε_i/k_B T) collects the whole distribution into one sum; every thermodynamic property comes from Q.
  • Maxwell-Boltzmann distribution of molecular speeds — the same exponent applied to translational kinetic energy.
  • Quantum statistics — Fermi-Dirac (electrons) and Bose-Einstein (photons, ⁴He) modify Boltzmann when quantum-degenerate; the modification is only relevant at low T or high density.
  • Chemical equilibrium — K_eq = e^(−ΔG°/RT) is Boltzmann applied to the free energy gap between products and reactants.

Every one of them starts by writing e^(−ΔE / k_B T) and reading off populations.

Sources

  • Atkins & de Paula, Physical Chemistry, 11th ed. (2018), Ch. 15-16
  • McQuarrie & Simon, Physical Chemistry: A Molecular Approach (1997), Ch. 17-18
  • Callen, Thermodynamics and an Introduction to Thermostatistics, 2nd ed. (1985), Ch. 15
  • CODATA 2018 — k_B = 1.380 649 × 10⁻²³ J/K (exact)
  • L. Boltzmann (1877) — original derivation of S = k ln W