Chemistry · Statistical Thermodynamics · Energy States

Boltzmann Distribution and Partition Function

The Boltzmann distribution gives equilibrium populations of energy states; the partition function normalizes their statistical weights.

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Formula

p_{i} = \frac{g _{i} e ^{- E_{i} / (RT)}}{Q}, Q = j \sum g_{j} e^{- E_{j} / (RT)}

E_i in J/mol · R in J·mol⁻¹·K⁻¹ · T in K · Q, p_i and g_i dimensionless

Variables & units

Variables & units
Symbol	Meaning	Unit
$p_{\mathrm{i}}$	Probability or fraction in energy level i	dimensionless
$g_{\mathrm{i}}$	Degeneracy, number of states at the same energy	dimensionless
$E_{\mathrm{i}}$	Molar energy of state i relative to the chosen zero	J/mol
$R$	Universal gas constant	J·mol⁻¹·K⁻¹
$T$	Absolute temperature	K
$Q$	Canonical partition function over all states	dimensionless

LaTeX

p_i=\frac{g_i e^{-E_i/(RT)}}{Q},\quad Q=\sum_j g_j e^{-E_j/(RT)}

QUANTA STUDYquanta-study.de

Two energy levels E₀ and E₁ = E₀ + 5 kJ/mol; the upper level has degeneracy three. Adjacent bars show normalized occupations p₀ = 0.715 and p₁ = 0.285 at 298 K; this is an idealized two-state model. — Illustrative canonical two-state model at 298 K: degeneracy g₁ = 3 raises occupation of the upper level despite ΔE = 5 kJ mol⁻¹. Bars show normalized probabilities, not measurements for a specific molecule.

Derivation & background

The canonical Boltzmann distribution applies at thermal equilibrium and fixed temperature. Each level contributes its degeneracy multiplied by an exponential energy weight; the partition function Q sums these weights and makes the probabilities add to one. Energies in this entry are molar, so R must be used; if energies are per molecule, replace R with kB. The equation describes populations of molecular states, not the equilibrium between chemical species by itself.

Exam blueprint

Validity range

Canonical equilibrium at fixed temperature; discrete molecular states, common energy zero and included degeneracies. Molar energies require R; per-particle energies require kB.

Derivation steps

Each state receives a weight from degeneracy and the Boltzmann factor; Q normalizes all weights into probabilities.

1Calculate each level weight wi=gi e^(−Ei/RT).
2Sum the weights to obtain Q=Σwi.
3Divide each wi by Q and check that Σpi=1.

Rearrangements

Energy from population

E_i=-RT\ln\left(\frac{p_iQ}{g_i}\right)

Keep the reference energy, degeneracy and molar units consistent.

Task variant

Levels: E0=0, g0=1; E1=5.00 kJ/mol, g1=3; T=298 K. Find p1.

Q=1+3e^(−5000/(8.314·298))≈1.399; p1≈0.285.
The remainder p0≈0.715, total ≈1.

Common mistakes

Use per-molecule E with R or omit degeneracy when counting states.

Put E and the constant on the same particle or molar scale, include gi, and normalize by Q.

Exam context

Two-level systems, spectroscopy and statistical interpretation of temperature-dependent molecular populations.

Formula cluster

Energetics and statistical thermodynamics

The partition function collects microstates and yields normalized populations.

μ_i = μ_i° + RT ln a_i

Chemical Potential of a Component

ΔG = ΔH − TΔS

Gibbs Energy (Free Enthalpy)

ln(K₂/K₁) = −ΔH°/R·(1/T₂ − 1/T₁)

Van 't Hoff Equation (Reaction Isobar)

Worked example

For two levels with E0=0, g0=1 and E1=5.00 kJ/mol, g1=3 at 298 K, Q=1+3e^(−5000/(8.314·298))≈1.399 and p1≈0.285. Check: p0≈0.715 and p0+p1≈1.000.

Applications

Explain temperature-dependent molecular populations and connect microscopic states to thermal properties.

How do you calculate with Boltzmann Distribution and Partition Function?

Here is how to work through a typical Boltzmann Distribution and Partition Function (pᵢ = gᵢe⁻ᴱⁱ⁄ᴿᵀ/Q; Q = Σgⱼe⁻ᴱʲ⁄ᴿᵀ) task step by step:

1
Task
Levels: E0=0, g0=1; E1=5.00 kJ/mol, g1=3; T=298 K. Find p1.
Solution path
1. Q=1+3e^(−5000/(8.314·298))≈1.399; p1≈0.285.
2. The remainder p0≈0.715, total ≈1.

Frequently asked questions about Boltzmann Distribution and Partition Function

What is the partition function Q?+

Q is the sum of statistical weights of all energy states, each degeneracy multiplied by a Boltzmann factor. It normalizes populations: pi=gi e^(−Ei/RT)/Q, so fractions sum to one. For two levels at 0 and 5.00 kJ/mol with degeneracies 1 and 3 at 298 K, Q≈1.399 and the excited level has population about 0.285. Q encodes which states are thermally accessible at a temperature. It is not a sum of chemical-species concentrations. Include only states accessible in the model; a common energy zero cancels consistently.

How do energy and temperature change populations?+

A higher-energy state receives a smaller weight at equal degeneracy because its exponent is more negative. Higher temperature weakens that difference: thermal energy makes excited states more probable. In the example, the level lies 5 kJ/mol above the ground state and has about 28.5 percent population at 298 K despite threefold degeneracy. Its fraction rises as T increases. The ratio also depends on gi, so a highly degenerate level can be more populated than energy spacing alone suggests. Compare populations only at the same temperature and under the same state definition.

Why is degeneracy gi included?+

Degeneracy counts how many distinct microstates share one energy. A level with three equal-energy states contributes three statistical weights, not one. That is why gi multiplies the exponential term. Omitting degeneracy can substantially underestimate a population, especially when states have multiple orientations or quantum numbers. The example with g1=3 shows how degeneracy partly offsets an excited energy level. Check whether your table lists individual states or energy levels whose degeneracies are already included. The degeneracy must match the energy-level table used in the calculation.

When should I use R versus kB?+

The exponent must be dimensionless. If Ei is molar energy in J/mol, use R in J/(mol·K). If you use the energy of one molecule in joules, use kB in J/K. Do not mix these scales: using molar energy with kB makes the exponent wrong by roughly Avogadro’s number. The forms are equivalent because R=N_AkB. Record energy units and reference basis before calculating, then check that E/(RT) is dimensionless. Convert energies before taking the exponential. The exponent must be dimensionless. The exponent is unitless.

Does the distribution apply to chemical equilibria between species?+

The displayed equation distributes molecules of a system among internal energy states at canonical equilibrium. It is not by itself the equilibrium equation for a reaction among distinct chemical species; chemical potentials, standard states and conservation laws also matter. The partition function can provide thermodynamic quantities such as energy and entropy and enter statistical-mechanical derivations of equilibria. Use this page for state populations, spectroscopy and statistical-thermodynamics foundations. For reaction direction, also use ΔrG, Q and K. Reaction equilibrium also requires activities and chemical potentials.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

More formulas: Gases and energetics

p \cdot V = n \cdot R \cdot T

Ideal Gas Law

\Delta H_R^0 = \sum \Delta H_f^0(\text{Produkte}) - \sum \Delta H_f^0(\text{Edukte})

Reaction Enthalpy (Hess's Law)

\Delta H_{\text{Gitter}} = \Delta H_f^0 - \Delta H_{\text{Sub}} - I - \tfrac{1}{2} D - E_{\text{EA}}

Lattice Energy (Born-Haber Cycle)

C_g = k_H \cdot p_g

Henry's Law (Concentration Convention)

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Boltzmann Distribution and Partition Function

The Boltzmann distribution gives equilibrium populations of energy states; the partition function normalizes their statistical weights.

UniversityUpper secondaryUniversity/collegeBachelor foundationsSubject specialization

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Variables & units

Symbol

Meaning

Unit

p_{\mathrm{i}}

Probability or fraction in energy level i

dimensionless

g_{\mathrm{i}}

Degeneracy, number of states at the same energy

dimensionless

E_{\mathrm{i}}

Molar energy of state i relative to the chosen zero

J/mol

R

Universal gas constant

J·mol⁻¹·K⁻¹

T

Absolute temperature

Q

Canonical partition function over all states

dimensionless

Derivation & background

Exam blueprint

Validity range

Canonical equilibrium at fixed temperature; discrete molecular states, common energy zero and included degeneracies. Molar energies require R; per-particle energies require kB.

Derivation steps

Each state receives a weight from degeneracy and the Boltzmann factor; Q normalizes all weights into probabilities.

1Calculate each level weight wi=gi e^(−Ei/RT).
2Sum the weights to obtain Q=Σwi.
3Divide each wi by Q and check that Σpi=1.

How do you calculate with Boltzmann Distribution and Partition Function?

Here is how to work through a typical Boltzmann Distribution and Partition Function (pᵢ = gᵢe⁻ᴱⁱ⁄ᴿᵀ/Q; Q = Σgⱼe⁻ᴱʲ⁄ᴿᵀ) task step by step:

Task

Levels: E0=0, g0=1; E1=5.00 kJ/mol, g1=3; T=298 K. Find p1.

Solution path

Q=1+3e^(−5000/(8.314·298))≈1.399; p1≈0.285.
The remainder p0≈0.715, total ≈1.

Frequently asked questions about Boltzmann Distribution and Partition Function

What is the partition function Q?+

How do energy and temperature change populations?+

Why is degeneracy gi included?+

When should I use R versus kB?+

Does the distribution apply to chemical equilibria between species?+

Retain Boltzmann Distribution and Partition Function for exams

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential