Chemistry • Year 11 • Module 4 • Lesson 11
Entropy — Definition, Modelling & Predicting ΔS
Apply the Δn(gas) rule, interpret standard entropy bar-chart data, analyse a real-world Australian case study using entropy reasoning, and predict the sign of ΔS across a range of scenarios.
1. Interpret standard entropy data — S° bar chart
The bar chart below shows the standard molar entropy (S°) at 298 K for six substances commonly encountered in HSC chemistry. Use the chart to answer the questions. 8 marks
Figure 1.1. Standard molar entropy S° at 298 K for selected substances. Data from NIST WebBook.
1.1 Identify the two substances with the highest S° values. State their physical states and use lesson content to explain why gases have much higher entropy than solids at the same temperature. 3 marks
1.2 Diamond and graphite are both pure carbon yet have different S° values (2.4 vs 5.7 J K−1 mol−1). This contradicts the common misconception that S° for all elements equals zero. Explain this, with reference to the Third Law of Thermodynamics. 2 marks
1.3 Predict the sign of ΔS for the reaction C(diamond, s) → C(graphite, s) and justify your prediction using the data in the chart. 3 marks
2. Cause-and-effect chain — CSR sugar dissolving
A spoonful of CSR sugar (sucrose, C₁₂H₂₂O₁₁) is stirred into hot tea. Complete the cause-and-effect chain below by filling in the empty boxes. The first cause is provided. 5 marks
3. Compare and contrast — enthalpy vs entropy
Complete the table to compare enthalpy (H) and entropy (S) as thermodynamic state functions. 7 marks
| Feature | Enthalpy (H) | Entropy (S) |
|---|---|---|
| What it measures | ||
| SI unit | ||
| Reference point (zero) | ||
| Can absolute values be tabulated? | ||
| Value for a pure element in standard state | ||
| Is it a state function? | ||
| Increases when phase changes from solid → liquid → gas? |
4. Case study — dry ice at Australian food festivals
At many Australian food festivals, solid CO&sub2; (dry ice, −78.5 °C) is placed in warm punch bowls or used for theatrical smoke effects. The solid immediately begins to sublime (convert directly from solid to gas) without passing through the liquid phase. The white “smoke” is actually water vapour condensing around the cold CO&sub2; gas cloud. 6 marks
4.1 Write the equation for the sublimation of dry ice and predict the sign of ΔS. Justify your prediction by applying the Δn(gas) rule. 3 marks
4.2 The sublimation of dry ice is endothermic (ΔH > 0), yet it occurs spontaneously at room temperature. Use the Second Law of Thermodynamics to explain how an endothermic process can be spontaneous, with reference to ΔS(system) and ΔS(surroundings). 3 marks
Q1.1 — Highest S° substances
CO&sub2;(g) (213.8 J K−1 mol−1) and N&sub2;(g) (191.6 J K−1 mol−1) are both gases. Gases have much higher entropy than solids or liquids because gas particles have vastly more possible positions and speeds (translational freedom), creating far more microstates. Solid diamond (2.4) has the lowest value because its atoms are locked in a rigid lattice with only vibrational freedom — very few microstates.
Q1.2 — S° of diamond vs graphite (Third Law)
The Third Law states S = 0 only for a perfect crystal at absolute zero (0 K). At 298 K, all substances have S° > 0 because thermal energy introduces disorder. Diamond and graphite have different crystal structures (hence different numbers of microstates at 298 K), giving different S° values. Unlike ΔH°f, S° for elements is never zero at room temperature.
Q1.3 — ΔS for diamond → graphite
ΔS = S°(graphite) − S°(diamond) = 5.7 − 2.4 = +3.3 J K−1 mol−1. ΔS is positive because graphite has more entropy than diamond (graphite has a layered structure with more vibrational modes; diamond is an extremely rigid lattice with fewer microstates). The reaction increases disorder, so ΔS > 0.
Q2 — Cause-and-effect chain (CSR sugar)
Effect 1 / Cause 2: “freely” / “microstates”.
Effect 2 / Cause 3: “positive” / “greater”.
Effect 3 / Cause 4: “positive” / “positive” (small positive for surroundings if slightly exothermic; even if slightly endothermic, ΔS(system) dominates).
Overall outcome: “spontaneous” / “Second”.
Note on marks: Accept any chemically consistent wording that demonstrates understanding of microstates, entropy sign, and Second Law.
Q3 — Compare table (sample answers)
| Feature | Enthalpy (H) | Entropy (S) |
|---|---|---|
| Measures | Heat content at constant pressure; energy stored in bonds | Dispersal of energy across microstates |
| SI unit | kJ mol−1 | J K−1 mol−1 |
| Reference zero | Convention: ΔH°f of elements = 0 (arbitrary) | Third Law: S = 0 for a perfect crystal at 0 K (absolute) |
| Absolute values? | No — only ΔH can be measured | Yes — absolute S° values can be tabulated |
| Pure element value | ΔH°f = 0 by convention | S° ≠ 0 at 298 K (e.g. graphite = 5.7) |
| State function? | Yes | Yes |
| Increases s → l → g? | Yes (heat is absorbed in melting and boiling) | Yes (more microstates in each successive phase) |
Q4.1 — Dry ice sublimation sign of ΔS
Equation: CO&sub2;(s) → CO&sub2;(g). Δn(gas) = 1 − 0 = +1. One mole of gas is produced where none existed in the reactant. Therefore ΔS > 0 (positive). Gas particles have far more possible positions and speeds than solid particles, dramatically increasing the number of microstates.
Q4.2 — Second Law and spontaneous endothermic process
By the Second Law, a process is spontaneous when ΔS(universe) = ΔS(system) + ΔS(surroundings) > 0. For dry ice sublimation, ΔS(system) is very large and positive (solid → gas is an enormous increase in microstates). Even though the process is endothermic (ΔH > 0) which causes a small ΔS(surroundings) < 0 (heat is absorbed from surroundings), the large gain in ΔS(system) far outweighs the loss in ΔS(surroundings). Therefore ΔS(universe) > 0 and the process is spontaneous.