Probability & Boltzmann Statistics — Maths for Thermodynamics & Kinetics

Counting the ways energy spreads

Probability and combinatorics become thermodynamics

Probability is about counting outcomes. For a single event with $N$ equally likely possibilities, the probability of any one is $1/ N$ . For combinations — choosing k objects from n without regard to order — the count is $(k n) = \frac{n !}{k ! ( n - k )!}$ . In chemistry, we use this to count microstates: the number of distinct ways to distribute energy quanta over molecules.

For a system of N identical particles with occupation numbers $n_{0}, n_{1}, n_{2}, \dots$ (where $\sum n_{i} = N$ ), the number of microstates giving that particular distribution is:

W = \frac{N !}{n _{0} ! n _{1} ! n _{2} ! \dots} = \frac{N !}{\prod _{i} n _{i} !}

The most probable distribution — the one with the most microstates — is the Boltzmann distribution. It says the ratio of the population of level j to the ground state is:

\frac{n _{j}}{n _{0}} = \frac{g _{j}}{g _{0}} e^{- Δ ε_{j} / k T}

where $g_{j}$ is the degeneracy (number of states at that energy), $k = 1.381 \times 1 0^{- 23} J K^{- 1}$ is the Boltzmann constant, and T is the absolute temperature. A large energy gap or low temperature strongly favours the ground state; heat gives molecules the energy to climb to excited states, which is why reaction rates and spectral line intensities all depend on T.

Boltzmann also gave us the statistical definition of entropy, inscribed on his grave in Vienna: $S = k ln W$ . A larger W (more microstates, more disorder) means higher entropy — and the Second Law of Thermodynamics is simply the statement that systems evolve toward the most probable macrostate.

Combinations

(k n) = \frac{n !}{k ! ( n - k )!}

choose k from n, order irrelevant

Microstates W

W = N! / \prod_{i} n_{i}!

ways to arrange N objects with given nᵢ

Boltzmann factor

e^{- Δ ε / k T}

large gap or low T → near zero

Population ratio

n_{j} / n_{0} = (g_{j} / g_{0}) e^{- Δ ε / k T}

degeneracy g can multiply population

Boltzmann entropy

S = k ln W

k = 1.381 × 10⁻²³ J K⁻¹

Thermal energy scale

k T = 0.0257 eV

at 298 K
or RT = 2.48 kJ mol⁻¹

The multiplicity W is the combinatorial heart of statistical mechanics. For a small system it can be counted directly; for large N Stirling's approximation ( $ln N! \approx N ln N - N$ ) makes the counting tractable. Maximising W subject to the constraints of fixed total energy and fixed N yields the Boltzmann distribution.

When energy levels are degenerate (multiple quantum states at the same energy), the Boltzmann population ratio is scaled by the degeneracy ratio:

\frac{n _{j}}{n _{0}} = \frac{g _{j}}{g _{0}} e^{- Δ ε_{j} / k T}

A 3-fold degenerate excited state ( $g_{j} = 3$ ) is populated three times as much as a non-degenerate state at the same energy. Forgetting degeneracy is the single most common error in NMR and UV-Vis population calculations.

The partition function $q = \sum_{i} g_{i} e^{- ε_{i} / k T}$ is the normalising denominator: $n_{j} / N = g_{j} e^{- ε_{j} / k T} / q$ . The total internal energy is $U = N k T^{2} (\partial ln q / \partial T)_{V}$ — every macroscopic thermodynamic quantity ultimately derives from q.

Common pitfalls

Forgetting degeneracy g. The Boltzmann ratio without g gives the wrong population for any degenerate level. Always write $(g_{j} / g_{0}) e^{- Δ ε / k T}$ .
Using ΔE in kJ/mol but k in J/K. The Boltzmann constant k works with energies in joules per molecule. Use either $Δ ε / k T$ (SI per molecule) or $Δ E / RT$ (per mole with R = 8.314 J mol⁻¹ K⁻¹), never mix them.
Confusing W with ln W. Entropy is $S = k ln W$ , not $S = kW$ . W grows exponentially with N; ln W grows linearly — only ln W is extensive.
Assuming the maximum-multiplicity distribution is the only distribution. At small N, many distributions contribute. The Boltzmann distribution is only overwhelmingly dominant in the thermodynamic limit (N → ∞).

Going deeper

For a two-level system with states at energies 0 and ε, the partition function is $q = 1 + e^{- ε / k T}$ . The fraction in the upper state is $f = e^{- ε / k T} / q$ , which rises from 0 at T = 0 to 0.5 at T → ∞ (both states equally populated). For the two-level spin system of NMR, this plateaus near 0.5 even at room temperature — confirming the near-equal populations and the inherent insensitivity of NMR.

Statistical thermodynamics

Level populations set the intensity of spectroscopic lines, the fraction of molecules with enough energy to react (Arrhenius), and bulk entropy. It all starts with probability and the Boltzmann factor. Open the Boltzmann lab to see populations shift live.

Boltzmann lab

Adjust the energy-level spacing and temperature — watch populations redistribute between three levels.

How molecules spread over three energy levels

61%

ε00 kJ

27%

ε12 kJ

12%

ε24 kJ

\frac{n _{i}}{n _{0}} = e^{- ε_{i} / k T}

level spacing Δε2.0 kJ/mol

temperature T298 K

population ε060.8%

ratio n₁/n₀0.446

thermal energy RT2.48 kJ/mol

When the gap is large compared with RT, almost everything sits in the ground state. Heat the system (or shrink the gap) and the upper levels fill — the reason populations, spectra and reaction rates all depend on T.

Worked example 1Population of the first excited vibrational state of HCl

The fundamental vibrational frequency of HCl is $\tilde{ν} = 2886 c m^{- 1}$ . The energy gap between $v = 0$ and $v = 1$ is $Δ ε = h c \tilde{ν}$ . Calculate the ratio $n_{1} / n_{0}$ at 298 K and at 1000 K. ( $h = 6.626 \times 1 0^{- 34} J s$ , $c = 3.00 \times 1 0^{10} cm s^{- 1}$ , $k = 1.381 \times 1 0^{- 23} J K^{- 1}$ .)

Worked example 2Spin population in a proton NMR experiment

In a 500 MHz proton NMR spectrometer, the energy gap between spin-up and spin-down states is $Δ ε = h ν = 6.626 \times 1 0^{- 34} \times 500 \times 1 0^{6} = 3.31 \times 1 0^{- 25} J$ . With degeneracy $g_{↑} = g_{↓} = 1$ , calculate the population excess of the lower state at 298 K and comment on NMR sensitivity.

Worked examples

Microstates, entropy, and Boltzmann distributions — try each before revealing the solution.

Worked example 3Counting microstates for two energy quanta in three oscillators

Three distinguishable harmonic oscillators share exactly 2 quanta of energy. How many microstates W exist, and what is the entropy of this ensemble?

Worked example 4Entropy of mixing two ideal gases

One mole of N₂ and one mole of Ar are mixed at constant T and P. Using $S = k ln W$ in the macroscopic (molar) form, calculate the molar entropy of mixing $Δ_{mix} S = - n R (x_{1} ln x_{1} + x_{2} ln x_{2})$ .

Worked example 5Boltzmann factor and the Arrhenius activation threshold

The activation energy for a reaction is $E_{a} = 50 kJ mo l^{- 1}$ . What fraction of molecules at 298 K have enough energy to react? How does this change at 400 K?

Worked example 6Degeneracy and population of a rotational energy level

The J = 1 rotational level of CO has energy $ε_{1} = 2 h c B$ above J = 0, where the rotational constant $B = 1.9225 c m^{- 1}$ . The degeneracy of level J is $g_{J} = 2 J + 1$ . Calculate the population ratio $n_{1} / n_{0}$ at 300 K. ( $h c = 1.9864 \times 1 0^{- 23} J cm$ , $k = 1.381 \times 1 0^{- 23} J K^{- 1}$ .)

Worked example 7Entropy of mixing from W — a microstate derivation

Two moles of an ideal gas A and one mole of ideal gas B are mixed at constant T and P. Derive the molar entropy of mixing using $Δ S_{mix} = - R \sum_{i} x_{i} ln x_{i}$ and verify it using $S = k ln W$ for large N (Stirling level).

ChallengeChallenge — two-level population fraction as a function of temperature

An electronic excited state of NO lies $Δ ε = 120 c m^{- 1}$ above the ground state ( $1 c m^{- 1} = 1.9864 \times 1 0^{- 23} J$ ). Both levels are non-degenerate ( $g_{0} = g_{1} = 1$ ).

(i) Calculate the fraction in the excited state at T = 100, 298, 500 and 1000 K.
(ii) At what temperature does the excited state hold 30 % of the population?

Check yourself

Four quick questions on the Boltzmann distribution, microstates, and entropy.

Question 1 of 4 · Score 0

The Boltzmann distribution gives the ratio of populations of two energy levels as:

\frac{n _{1}}{n _{0}} = e^{- Δ ε / k T}

Choose an answer.