Short on time? This page is the recap. For explanations, the data table and worked examples, use the full study guide.
Recap
- Absolute entropy is measured from a true zero: a perfect crystal at 0 K. So every substance at 298 K has a positive entropy.
- Standard molar entropy, S°, is the absolute entropy of 1 mol of a substance at 1 bar, usually listed at 298 K.
- Gases have much larger S° values than liquids and solids. For one substance, S°(g) > S°(l) > S°(s).
- The entropy change of any process, chemical or physical, comes from the S° values of the species before and after it.
Key relationships
| Relationship | Symbols and units | Use it to |
|---|---|---|
| ΔS° = ΣnS°(products) − ΣnS°(reactants) | S° in J K⁻¹ mol⁻¹; n = coefficients | calculate ΔS° for a reaction or phase change |
| ΔS° (kJ K⁻¹ mol⁻¹) = ΔS° (J K⁻¹ mol⁻¹) ÷ 1000 | — | prepare ΔS° for combining with ΔH° (Topic 9.3) |
| ΔS for a sample = moles of reaction × ΔS° | result in J K⁻¹ | scale to a real amount |
| Reverse the equation → ΔS° changes sign; double it → ΔS° doubles | — | adapt a given value |
Assumptions behind the numbers
- Values are for standard conditions: 1 bar and the listed temperature (usually 298 K).
- Each species is in the state shown in the equation. Changing a state symbol changes the S° value.
- ΔS° is per mole of reaction for the equation exactly as written.
Mistakes to avoid
- Setting S° = 0 for elements such as O₂(g) or H₂(g). Only ΔH°f is zero for elements.
- Forgetting coefficients. Multiply every S° by its coefficient before adding.
- Reactants minus products. It is always products minus reactants.
- Wrong state symbol. H₂O(l) is 70.0; H₂O(g) is 188.8 J K⁻¹ mol⁻¹.
- J vs kJ. S° is in joules; divide by 1000 before mixing with ΔH° in kJ.
- Skipping the sign check. If the sign disagrees with the change in moles of gas, recheck.
- Accepting a negative S° when working backwards. That means a sign slip.
Quick self-check
Use S°: CH₃OH(l) 126.8, O₂(g) 205.2, CO₂(g) 213.8, H₂O(l) 70.0 J K⁻¹ mol⁻¹.
- ΔS° for 2 CH₃OH(l) + 3 O₂(g) → 2 CO₂(g) + 4 H₂O(l)? (−161.6 J K⁻¹ mol⁻¹)
- The same value in kJ K⁻¹ mol⁻¹? (−0.1616 kJ K⁻¹ mol⁻¹)
- ΔS° for CO(g) + 2 H₂(g) → CH₃OH(l) is −332.3 J K⁻¹ mol⁻¹. What is ΔS° for 2 CH₃OH(l) → 2 CO(g) + 4 H₂(g)? (+664.6 J K⁻¹ mol⁻¹: reversed and doubled)
Next: practice questions.