A thermodynamic quantity that measures the dispersal of matter and energy among the available microstates of a system. The greater the number of possible arrangements, the higher the entropy. Units: J K⁻¹ mol⁻¹
Entropy is often described as "disorder," but the IB syllabus defines it more precisely as the number of ways energy can be distributed among particles. Every substance has a measurable absolute entropy value \( S^\circ \), and crucially, entropy values are always positive (unlike enthalpy, which can be positive or negative).
Third Law of Thermodynamics
A perfect crystal at absolute zero (0 K) has an entropy of exactly zero. This provides the baseline from which all absolute entropy values are measured. This is why standard entropy values \( S^\circ \) are always positive.
Predicting Entropy Changes
Entropy Increases With...
Phase Changes and Entropy
Any process that transitions matter from a more ordered state to a more dispersed state results in a positive entropy change (\( \Delta S > 0 \)).
| Process | Direction | ΔS |
|---|---|---|
| Melting | Solid → Liquid | Positive (+) |
| Boiling / Evaporation | Liquid → Gas | Positive (+) |
| Sublimation | Solid → Gas | Positive (+) |
| Condensation | Gas → Liquid | Negative (−) |
| Freezing | Liquid → Solid | Negative (−) |
| Deposition | Gas → Solid | Negative (−) |
Predicting ΔS in Chemical Reactions
Beyond phase changes, predicting entropy changes in reactions requires examining the balanced equation. The key indicators, in order of importance:
Rules for Predicting ΔS
- Change in gaseous moles (most important): Gas particles have far higher molar entropy than liquids or solids. An increase in moles of gas (\( \Delta n_\text{gas} > 0 \)) means \( \Delta S > 0 \). A decrease means \( \Delta S < 0 \).
- Dissolution of a solid: Dissolving an ionic solid (e.g. NaCl) into water breaks the ordered lattice and disperses ions throughout the solution. This generally gives \( \Delta S > 0 \).
- Molecular complexity: If gas moles stay constant, breaking complex molecules into simpler fragments increases entropy. Polymerisation (simple → complex) decreases it.
Think About It
CaCO₃(s) → CaO(s) + CO₂(g). Predict the sign of ΔS.
ΔS > 0 (positive): a gas is produced from a solid. The number of moles of gas increases from 0 to 1, greatly increasing the dispersal of matter.
Think About It
N₂(g) + 3H₂(g) → 2NH₃(g). Predict the sign of ΔS.
ΔS < 0 (negative): four moles of gas are compressed into two moles of gas, severely restricting the overall dispersal of matter.
Calculating Standard Entropy Changes (ΔS°)
The standard entropy change for a reaction is calculated using absolute standard entropy values \( S^\circ \), which are found in Section 12 of the IB Data Booklet.
Where \( n \) is the stoichiometric coefficient of each substance in the balanced equation.
Worked Example: Combustion of Methane
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Data from the IB Data Booklet:
- \( S^\circ[\text{CH}_4\text{(g)}] = 186.3 \text{ J K}^{-1}\text{mol}^{-1} \)
- \( S^\circ[\text{O}_2\text{(g)}] = 205.2 \text{ J K}^{-1}\text{mol}^{-1} \)
- \( S^\circ[\text{CO}_2\text{(g)}] = 213.8 \text{ J K}^{-1}\text{mol}^{-1} \)
- \( S^\circ[\text{H}_2\text{O(l)}] = 70.0 \text{ J K}^{-1}\text{mol}^{-1} \)
Step 1: Sum the entropy of the products (multiply by coefficients):
\( \sum S^\circ_\text{products} = (1 \times 213.8) + (2 \times 70.0) = 353.8 \text{ J K}^{-1}\text{mol}^{-1} \)
Step 2: Sum the entropy of the reactants:
\( \sum S^\circ_\text{reactants} = (1 \times 186.3) + (2 \times 205.2) = 596.7 \text{ J K}^{-1}\text{mol}^{-1} \)
Step 3: Calculate ΔS°:
\( \Delta S^\circ = 353.8 - 596.7 = -242.9 \text{ J K}^{-1}\text{mol}^{-1} \)
Interpretation: The negative result confirms the qualitative prediction. Three moles of reactant gas are consumed but only one mole of product gas is formed (alongside a liquid). This net reduction in gaseous moles decreases the dispersal of matter, lowering entropy.