AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/9/9-2)
Unit 9 · Topic 9.2
9.2 Absolute Entropy and Entropy Change
Data tables list each substance's standard molar entropy, S°, in J/(mol·K). To find the standard entropy change for a reaction, add up S° for the products and subtract the total for the reactants, each multiplied by its coefficient.
Key terms
- standard molar entropy (S°)
- ΔS° of reaction
- products minus reactants
- J/(mol·K)
Absolute entropies
Unlike enthalpy, entropy has a true zero point: a perfect crystal at absolute zero (0 K). So every substance has a positive absolute entropy above 0 K. The standard molar entropy, S°, is the entropy of one mole of a substance in its standard state, usually at 25 °C. Its units are J/(mol·K).
Elements in their standard states do not have S° = 0. That's a big difference from ΔH°f, where elements are zero. O₂(g), for example, has S° = 205.2 J/(mol·K).
The idea that a perfect crystal at absolute zero has zero entropy is called the third law of thermodynamics. It's the reason tables can list absolute S° values rather than just changes.
Patterns in S° values
- For the same substance: S°(gas) ≫ S°(liquid) > S°(solid). Water vapor is 188.8 J/(mol·K) but liquid water is 69.9 J/(mol·K).
- Among similar substances in the same phase, larger, heavier, more complex molecules usually have higher S°, because they have more ways to move and store energy.
- Hard, tightly bonded solids such as diamond (2.4 J/(mol·K)) have very low S°.
Entropy changes for phase changes
You can use S° values for a physical change too. For H₂O(l) → H₂O(g) at 25 °C, ΔS° = 188.8 − 69.9 = +118.9 J/(mol·K). That's a large positive change, because the molecules go from touching each other to moving freely through a much larger volume.
Compare that with typical reactions: a reaction that changes the number of moles of gas by one or two often has |ΔS°| on the order of 100 to 200 J/K, while one with no change in moles of gas often has a ΔS° of only a few tens of J/K.
Calculating ΔS° for a reaction
ΔS°rxn = ΣnS°(products) − ΣnS°(reactants), where each n is the coefficient from the balanced equation. It's the same products-minus-reactants pattern you used for ΔH°f, but now no species is automatically zero.
The result is in J/K (or J/(mol·K), per mole of reaction as written). Keep it in joules for now; you'll need to watch units when you combine it with ΔH° in kJ (topic 9.3).
Always check your answer against a prediction from topic 9.1. If a reaction makes fewer moles of gas, your calculated ΔS° should be negative.
| Substance | S° (J/(mol·K)) |
|---|---|
| N₂(g) | 191.6 |
| H₂(g) | 130.7 |
| NH₃(g) | 192.5 |
| O₂(g) | 205.2 |
| CO₂(g) | 213.8 |
| CH₄(g) | 186.3 |
| H₂O(l) | 69.9 |
| CaCO₃(s) | 92.9 |
| CaO(s) | 39.8 |
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
ΔS° for the Haber process
Use the table to calculate ΔS° for N₂(g) + 3H₂(g) → 2NH₃(g).
Show the solutionHide the solution
- Step 1: Products: 2 × 192.5 = 385.0 J/K.
- Step 2: Reactants: 191.6 + 3 × 130.7 = 191.6 + 392.1 = 583.7 J/K. N₂ and H₂ are elements, but their S° values are not zero; leaving them out is the classic trap.
- Step 3: ΔS° = 385.0 − 583.7 = −198.7 J/K.
- Step 4: Check: 4 mol of gas → 2 mol of gas, so a negative ΔS° makes sense.
Answer: ΔS° = −198.7 J/K
- Example 2Calculator allowed
ΔS° for decomposing limestone
Calculate ΔS° for CaCO₃(s) → CaO(s) + CO₂(g).
Show the solutionHide the solution
- Step 1: Products: 39.8 + 213.8 = 253.6 J/K.
- Step 2: Reactants: 92.9 J/K.
- Step 3: ΔS° = 253.6 − 92.9 = +160.7 J/K.
- Step 4: Check: a gas is produced from a solid, so a large positive value makes sense.
Answer: ΔS° = +160.7 J/K
- Example 3Calculator allowed
Trap: coefficients and phases
Calculate ΔS° for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l).
Show the solutionHide the solution
- Step 1: Use the liquid-water value, 69.9 J/(mol·K), because the equation shows H₂O(l). Using the gas value (188.8) would be a big error.
- Step 2: Products: 213.8 + 2(69.9) = 213.8 + 139.8 = 353.6 J/K.
- Step 3: Reactants: 186.3 + 2(205.2) = 186.3 + 410.4 = 596.7 J/K.
- Step 4: ΔS° = 353.6 − 596.7 = −243.1 J/K.
- Step 5: Check: 3 mol of gas → 1 mol of gas (the water is liquid), so ΔS° should be clearly negative. It is.
Answer: ΔS° = −243.1 J/K
Common mistakes
- Setting S° = 0 for elements, as you would for ΔH°f.
- Subtracting in the wrong order. It's products minus reactants.
- Forgetting to multiply by coefficients, or using the wrong phase's value.
- Mixing J and kJ when ΔS° is later combined with ΔH°.
On the exam
- Free-response questions often ask you to calculate ΔS° from a table and then use it in ΔG° = ΔH° − TΔS°. Keep track of units at each step.
- A quick sign check against the moles of gas catches most arithmetic errors.
Connected topics
Videos
Check yourself
4 questions on 9.2 Absolute Entropy and Entropy Change. Pick an answer to see if you got it, and why.
For N₂(g) + O₂(g) → 2NO(g), ΔS° = +24.8 J/(mol·K). S° for N₂(g) is 191.6 J/(mol·K) and S° for O₂(g) is 205.0 J/(mol·K). What is S° for NO(g)?
| Substance | S° (J/(mol·K)) |
|---|---|
| C₂H₅OH(l) | 160.7 |
| O₂(g) | 205.0 |
| CO₂(g) | 213.6 |
| H₂O(l) | 69.9 |
Standard molar entropies at 25 °C. Reaction: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l)
What is ΔS° for the reaction as written?
Which of the following best accounts for the sign of ΔS° for this reaction?
If the water were produced as a gas instead, using S° for H₂O(g) = 188.8 J/(mol·K), what would ΔS° be?
0 of 4 answered