Factorising the Molecular Partition Function

Introduction

Each mode of molecular motion now has its own partition function. The obvious next question is how to combine them. The answer is remarkably simple: when the energy of a molecule is a sum of independent parts, its partition function is a product of the separate partition functions. This single step turns an impossible sum over every quantum state of a molecule into four manageable factors, and it is why thermodynamic properties such as entropy and energy can be split into translational, rotational, vibrational and electronic contributions.

Core explanation

From sums to products. Suppose every state of a molecule is labelled by independent quantum numbers for each mode, and its energy is

ε = ε T(n) + ε R(J) + ε V(v) + ε E(e)

The partition function sums over all combinations:

q = Σ n Σ J Σ v Σ e e^(−β[ε T + ε R + ε V + ε E])

Because the exponential of a sum is a product of exponentials, and each sum involves only its own index, the multiple sum splits:

q = (Σ n e^(−βε T))(Σ J g J e^(−βε R))(Σ v e^(−βε V))(Σ e g e e^(−βε E)) = q T q R q V q E

Degeneracies travel with their own mode, and the nuclear spin factor (2I + 1)² can be appended as a further constant factor that cancels in chemistry.

Logarithms become sums. Thermodynamic functions depend on ln q, so

ln q = ln q T + ln q R + ln q V + ln q E

and quantities such as U and S become sums of independent mode contributions. This is the statistical basis for statements such as "the rotational contribution to the heat capacity of N₂ is R".

The approximations. Factorisation is exact only if the modes are truly independent. The key assumptions are:

- Born–Oppenheimer separation of electrons from nuclei, excellent because nuclei are thousands of times heavier than electrons. - Translation independent of internal motion , exact for an ideal gas since the centre of mass moves freely. - Rigid rotor and harmonic oscillator , the weakest link: a vibrating bond changes the moment of inertia, and a spinning molecule stretches (centrifugal distortion). These couplings are small at ordinary temperatures.

Sizes of the factors. For a typical small molecule at room temperature, q T is astronomically large (about 10³⁰ for a molar volume), q R is tens to thousands, q V is close to 1, and q E is a small integer. Translation dominates the number of accessible states, which is why it dominates the entropy of gases.

Formulae

q = q T q R q V q E. q T = V/Λ³ with Λ = h/(2πmkT)^½. q R = kT/(σhcB̃) (linear, high T). q V = 1/(1 − e^(−θ V/T)). q E = g₀ (typically). ln q = Σ ln q mode.

Step-by-step reasoning

To build the molecular partition function of a diatomic gas:

1. Calculate Λ from the molar mass and temperature, then q T = V/Λ³. 2. Calculate q R from B̃ and σ. 3. Calculate q V from ν̃. 4. Read q E from the ground term symbol. 5. Multiply the four factors, and note which factor dominates.

Visual explanation

Imagine a four-dimensional grid of molecular states, with one axis for each mode. Each axis carries its own set of Boltzmann weights. Because the weights multiply, the total count of accessible states is the product of the counts along each axis — like the volume of a box being length × width × height × depth.

Real-world analogy

A restaurant offers 5 starters, 8 mains, 4 desserts and 2 drinks, and any choice can be combined with any other. The number of possible meals is 5 × 8 × 4 × 2 = 320. Independent choices multiply; that is exactly why independent energy modes give a product of partition functions.

Real-world example

Standard thermodynamic tables (such as the JANAF tables used in rocket and combustion design) are largely computed, not measured, for small gas molecules. Spectroscopists supply B̃, ν̃ and electronic levels; the factorised partition function then gives entropies, heat capacities and enthalpies over thousands of kelvin, often more accurately than calorimetry can.

Why?

Why does a sum of energies produce a product of partition functions? Because e^(−β(a+b)) = e^(−βa) × e^(−βb). Summing over all combinations of independent indices then gives the product of the individual sums, exactly as expanding (x₁ + x₂)(y₁ + y₂) produces every cross term x iy j.

Common misconception

"Adding the partition functions of each mode gives the total." Adding would count states from different modes as if they were alternatives. In reality a molecule has a translational state and a rotational state and a vibrational state simultaneously, so the counts multiply, and only the logarithms add.

Worked example

Question: Estimate q for N₂ at 298 K in a volume of 0.02479 m³ (the molar volume at 1 bar). Data: m = 4.65 × 10⁻²⁶ kg, B̃ = 1.998 cm⁻¹, ν̃ = 2358 cm⁻¹, ground state ¹Σg⁺.

Reasoning: Λ = h/(2πmkT)^½ = 6.626 × 10⁻³⁴/(2π × 4.65 × 10⁻²⁶ × 1.381 × 10⁻²³ × 298)^½ = 1.91 × 10⁻¹¹ m, so Λ³ = 6.98 × 10⁻³³ m³ and q T = 0.02479/6.98 × 10⁻³³ = 3.55 × 10³⁰. q R = 207.2/(2 × 1.998) = 51.9. q V = 1/(1 − e^(−11.4)) ≈ 1.00001. q E = 1.

Answer: q ≈ 3.55 × 10³⁰ × 51.9 ≈ 1.8 × 10³², dominated by translation.

Quick check

1. If q T, q R, q V and q E are 10³⁰, 100, 1.2 and 3, what is ln q and which mode contributes most? Answer: ln q = 69.08 + 4.61 + 0.18 + 1.10 ≈ 74.97, with translation contributing by far the most.

Exam focus

Be able to prove q = q T q R q V q E from ε = Σ ε mode and state the assumptions. Numerical questions often ask for one factor at a time; keep careful track of σ, units of B̃ and ν̃, and the energy zero used for q V.

Advanced insight

Factorisation breaks down where modes couple strongly. Large-amplitude internal rotations (for example the methyl torsion in ethane) mix vibration and rotation, and are treated as hindered rotors. At high temperatures, anharmonicity and rotation–vibration interaction add correction terms, often written as small multiplicative factors to q. In liquids and solids, molecules interact and the single-molecule q no longer describes the system; the canonical partition function Q must be used instead.

Summary

When a molecule's energy is a sum of independent contributions, its partition function is the product q = q T q R q V q E, and ln q is a sum. This lets thermodynamic properties be split into mode contributions. The result rests on the Born–Oppenheimer approximation and the rigid rotor–harmonic oscillator model. For gases at room temperature, q T dominates, q R is moderate, and q V and q E are close to small integers.

Practice questions

1. Why is the factorisation of q T from the internal partition function exact for an ideal gas? Answer: Centre-of-mass motion of a free molecule does not interact with its internal coordinates, so ε T is strictly additive to the internal energy. 2. Which assumption behind factorisation is least accurate for a hot diatomic gas? Answer: The rigid rotor–harmonic oscillator assumption, since anharmonicity and rotation–vibration coupling grow at high temperature. 3. Explain why ln q, not q itself, is split into mode contributions for U and S. Answer: Thermodynamic functions depend on ln q, and the logarithm of a product is a sum, so each mode adds its own term. 4. By what factor does q for N₂ change if the volume is doubled at constant temperature? Answer: It doubles, because only q T depends on volume and q T is proportional to V.