Entropy (S) and Microstates
Entropy (\(S\)): A thermodynamic property that quantifies the distribution of available energy among the particles in a system (the number of energetically equivalent microstates \(W\), related via Boltzmann's equation \(S = k_{\text{B}} \ln W\)). A system with more dispersed energy has higher entropy.
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
The entropy of a perfectly crystalline substance at absolute zero (0 K) is exactly zero (\(S = 0\text{ J K}^{-1}\text{mol}^{-1}\)). As temperature rises above 0 K, thermal motion increases microstates and all substances possess positive absolute entropy values (\(S^\circ > 0\)).
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:
Five Principles for Predicting the Sign of \(\Delta S\)
- Phase changes: \(S(\text{solid}) < S(\text{liquid}) \ll S(\text{gas})\). Vaporisation causes a massive increase in entropy.
- Change in gas moles: More moles of gas produced \(\Rightarrow \Delta S > 0\); fewer moles of gas \(\Rightarrow \Delta S < 0\).
- Dissolving an ionic solid: Crystal lattice breaks apart into freely moving aqueous ions \(\Rightarrow \Delta S > 0\).
- Mixing substances: Pure substances mixing increases positional disorder \(\Rightarrow \Delta S > 0\).
- Temperature increase: Higher thermal kinetic energy expands accessible energy levels \(\Rightarrow S\) increases.
Predicting \(\Delta S\): Thermal Decomposition of \( ext{CaCO}_3\)
Reaction: \(\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})\)
Prediction: \(\Delta S > 0\) (positive). A solid reactant decomposes to produce a gas (\(0 \rightarrow 1\text{ mol gas}\)), vastly increasing positional disorder and microstates.
Predicting \(\Delta S\): Ammonia Synthesis (Haber Process)
Reaction: \(\text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightarrow 2\text{NH}_3(\text{g})\)
Prediction: \(\Delta S < 0\) (negative). \(4\text{ moles of gas}\) react to form only \(2\text{ moles of gas}\), reducing the number of independent translational microstates.
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.
Calculating Standard Entropy Change (\(\Delta S^\circ\))
\[ \Delta S^\circ = \sum nS^\circ(\text{products}) - \sum nS^\circ(\text{reactants}) \]
Where \(n\) represents the stoichiometric molar coefficients in the balanced equation.
Where \( n \) is the stoichiometric coefficient of each substance in the balanced equation.
Units of Standard Entropy: J K⁻¹ mol⁻¹
Standard entropy (\(S^\circ\)) is measured in \(\text{J K}^{-1}\text{mol}^{-1}\) (Joules, not kiloJoules!). Always multiply data booklet values by stoichiometric coefficients before summing. When later combining with \(\Delta H\) in the Gibbs equation, remember to convert units.
Worked Example: Combustion of Methane
Calculating Standard Entropy Change of Combustion
Calculate \(\Delta S^\circ\) for the combustion of methane:
\(\text{CH}_4(\text{g}) + 2\text{O}_2(\text{g}) \rightarrow \text{CO}_2(\text{g}) + 2\text{H}_2\text{O}(\text{l})\)
Standard entropy data (\(\text{J K}^{-1}\text{mol}^{-1}\)):
\(S^\circ[\text{CH}_4\text{(g)}] = 186.3\) | \(S^\circ[\text{O}_2\text{(g)}] = 205.2\) | \(S^\circ[\text{CO}_2\text{(g)}] = 213.8\) | \(S^\circ[\text{H}_2\text{O(l)}] = 70.0\)
Step 1: Sum the entropy of products
\(\sum S^\circ_{\text{products}} = (1 \times 213.8) + (2 \times 70.0) = 353.8\text{ J K}^{-1}\text{mol}^{-1}\)
Step 2: Sum the entropy of 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 \(\Delta S^\circ\)
\(\Delta S^\circ = 353.8 - 596.7 = \mathbf{-242.9\text{ J K}^{-1}\text{mol}^{-1}}\)
Interpretation: The negative result confirms the qualitative prediction. Three moles of reactant gas are consumed while forming only one mole of product gas (alongside liquid water), decreasing the dispersal of energy.
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