Trouton’s Rule states that, for many simple liquids, the molar entropy of vaporization at the normal boiling point is nearly constant, usually about 85–88 J mol⁻¹ K⁻¹. It means that when one mole of such a liquid changes into vapor at its boiling point, the entropy increase is approximately the same. ^[1]

Formula

During vaporization, a liquid changes into a gas. In the liquid state, molecules are close together, and their movement is restricted by intermolecular forces. When the liquid changes into vapor, the molecules move much farther apart and gain greater freedom of motion. This increase in molecular freedom causes an increase in entropy. ^[1–4]

Trouton’s Rule is based on the relationship between entropy change, enthalpy change, and temperature during vaporization:

ΔS_vap = ΔH_vap/T_b

Here,

ΔS_vap: Molar entropy of vaporization

ΔH_vap: Molar enthalpy of vaporization

T_b: Boiling point of the liquid in Kelvin

The unit of ΔS_vap is usually J mol⁻¹ K⁻¹.

Thermodynamic Basis

At the boiling point, the liquid and vapor phases are in equilibrium. This means that the Gibbs free energy change for vaporization is zero: ^[1]

ΔG = 0

The relationship between Gibbs free energy, enthalpy, temperature, and entropy is:

ΔG = ΔH − TΔS

For vaporization at the normal boiling point, assuming liquid–vapor equilibrium, this becomes:

0 = ΔH_vap − T_b​ΔS_vap

Rearranging the equation gives:

ΔS_vap = ΔH_vap/T_b​​

For many simple liquids, this value is approximately:

ΔS_vap ≈ 85−88 J mol⁻¹ K⁻¹

This approximate value is known as Trouton’s constant.

Trouton’s Rule can also be used to estimate the enthalpy of vaporization of a liquid if its boiling point is known. Rearranging the formula gives:

ΔH_vap ≈ ΔS_vap x T_b

Using Trouton’s constant:

ΔH_vap ≈ 88 × T_b

Here, T_b​ must be in Kelvin, and the answer will be in J mol⁻¹.

Liquids That Follow Trouton’s Rule

Trouton’s Rule works best for simple, non-associated liquids that do not show strong hydrogen bonding or complex molecular interactions. ^[1]

Examples include benzene, carbon tetrachloride, and diethyl ether. Similar non-associated organic liquids usually show values near Trouton’s constant.

Exceptions to Trouton’s Rule

Trouton’s Rule gives poor results for liquids with strong intermolecular forces or molecular association. ^[1]

Common exceptions include water, alcohols, and carboxylic acids. Water and alcohols form strong hydrogen bonds, while carboxylic acids may associate with each other and form dimers. Because of these unusual associations, their entropy of vaporization can differ from the Trouton value.

Therefore, Trouton’s Rule should be used as an approximate guide for simple liquids, not as a universal rule for all substances.

Worked Examples

Problem 1

A liquid has an enthalpy of vaporization of 30.8 kJ mol⁻¹ and boils at 350 K. Calculate its entropy of vaporization. Check if it follows Trouton’s Rule.

Solution

We use the formula:

ΔS_vap = ΔH_vap/T_b​​

Given:

ΔH_vap = 30.8 kJ mol⁻¹

T_b = 350 K

Since entropy is usually expressed in J mol⁻¹ K⁻¹, first convert kilojoules into joules:

30.8 kJ mol⁻¹ = 30,800 J mol⁻¹

Now substitute the values into the formula:

ΔS_vap = 30,800 J mol⁻¹/350 K

=> ΔS_vap = 88 J mol⁻¹ K⁻¹

The entropy of vaporization of the liquid is 88 J mol⁻¹ K⁻¹, which is close to Trouton’s value. Therefore, the liquid approximately follows Trouton’s Rule.

Problem 2

A simple liquid boils at 360 K. Estimate its enthalpy of vaporization using Trouton’s Rule.

Solution

According to Trouton’s Rule, the entropy of vaporization of many simple liquids is approximately:

ΔS_vap ≈ 88 J mol⁻¹ K⁻¹

We use the formula:

ΔH_vap ≈ ΔS_vap x T_b​

Given:

T_b = 360 K

Now substitute the values:

ΔH_vap ≈ 88 J mol⁻¹ K⁻¹ x 360 K

=> ΔH_vap ≈ 31,680 J mol−1

Convert joules into kilojoules:

31,680 J mol⁻¹ = 31.68 kJ mol⁻¹

Therefore, the estimated enthalpy of vaporization is 31.68 kJ mol⁻¹.

This is an approximate value because Trouton’s Rule is an estimation rule, not an exact law.