1.8.5Physical Chemistry
Entropy Change, ∆S
5 min read 6 parts Lesson 44 of 63
Part 1 of 6Why Entropy Is Needed
Why Entropy Is Needed
Limits of ΔH
- Enthalpy alone cannot explain why some endothermic changes, such as dissolving or decomposition, happen spontaneously.
Entropy
- Entropy, S, measures the number of ways that particles and their energy can be arranged, so it measures disorder.
Increase
- A more disordered system has more possible arrangements and higher entropy, and this favours the change taking place.
Units
- Entropy is measured in J K⁻¹ mol⁻¹, a much smaller unit than the kJ mol⁻¹ used for enthalpy.
Part 2 of 6Predicting the Sign of ΔS
State change
- Entropy rises from solid to liquid to gas, because particles gain freedom of movement and spread out.
Gas moles
- A reaction that produces more moles of gas than it consumes has a large positive entropy change.
Temperature
- Raising the temperature increases the number of ways energy is shared among particles, so entropy increases.
Dissolving
- Dissolving a solid normally increases entropy because ions are released and spread through the solvent.
Fewer particles
- Combining many gas molecules into few, such as N₂ + 3H₂ → 2NH₃, gives a negative entropy change.
| Change | ΔS sign |
|---|---|
| Melting or boiling | Positive |
| CaCO₃(s) → CaO(s) + CO₂(g) | Positive |
| 2H₂(g) + O₂(g) → 2H₂O(l) | Negative |
Part 3 of 6Calculating ΔS
1
Look up the standard molar entropy, S⦵, of each substance in J K⁻¹ mol⁻¹ from the data given.
2
Multiply each value by its coefficient in the balanced equation, because entropy depends on the amount present.
3
Add the entropies of all products and the entropies of all reactants separately.
4
Subtract: ΔS = ΣS(products) − ΣS(reactants), keeping the sign.
5
State the answer in J K⁻¹ mol⁻¹ and check that the sign agrees with the change in moles of gas.
Part 4 of 6Features of Standard Entropies
Features of Standard Entropies
Elements
- Elements have non-zero standard entropies, unlike standard enthalpies of formation which are zero for elements.
Gases
- Gases have much higher values than solids of similar size, because their particles are widely spaced and fast.
Complexity
- Larger molecules with more atoms have more ways to store energy and generally higher entropy.
Part 5 of 6Worked Example
Question Calculate ΔS⦵ for CaCO₃(s) → CaO(s) + CO₂(g) using S⦵: CaCO₃ = 92.9, CaO = 39.7, CO₂ = 213.6 J K⁻¹ mol⁻¹.
- ΣS(products) = 39.7 + 213.6 = 253.3.
- ΣS(reactants) = 92.9.
- ΔS = 253.3 − 92.9.
Answer ΔS⦵ = +160.4 J K⁻¹ mol⁻¹ (positive because a gas is formed)
Part 6 of 6Exam Checklist & Tips
1.8.5 – Key Points for the Exam
- Entropy, S A measure of the disorder of a system, linked to the number of arrangements of particles and energy.
- Standard molar entropy The entropy of one mole of a substance under standard conditions, in J K⁻¹ mol⁻¹.
- Spontaneous change
- A change that can take place without a continuous input of energy;
- it may still be slow.
- A change that can take place without a continuous input of energy;
- Disorder The extent to which particles and energy are spread out among available arrangements.
Common Mistakes
- Quoting entropy in kJ rather than J K⁻¹ mol⁻¹, then combining it with ΔH without converting.
- Stating that elements have zero standard entropy.
- Forgetting to multiply by balancing coefficients.
- Saying a positive ΔS alone means a reaction is feasible; ΔH and temperature also matter.