Skip to content
Dabour HubLEARN · PRACTICE · EXPLORESign inStudent

4.2.2Unit 4. Rates, Equilibria and Further Organic Chemistry

Entropy Changes — IAL Edexcel Linear (Chemistry) Revision Notes & Topical Questions

5 min read 6 parts Lesson 20 of 56 Unit quiz

📚 IAL Edexcel Linear (Chemistry) Hub / Topical Revision
🩺
100% Curriculum Specification Match Lesson Doctor Verified
The only platform with 100% curriculum specification match and verified interactive question banks.
0%read

Part 1 of 6Entropy Change of the System

The entropy change of the system is found in the same way as an enthalpy change from enthalpies of formation.

Formula
  • Formula ΔS°system = ΣS°(products) − ΣS°(reactants), with each value multiplied by its coefficient in the balanced equation.
Units
  • Units The answer is in J K⁻¹ mol⁻¹, the same unit as the individual standard entropy values.
Sign
  • Sign A positive value means the system becomes more disordered, and a negative value means it becomes more ordered.
Coefficients
  • Coefficients Always multiply S° by the number of moles in the equation, or the result will be wrong.

Part 2 of 6Predicting the Sign

Predicting the Sign
Gas moles increase
  • Gas moles increase ΔS is large and positive, for example in thermal decomposition producing a gas.
Gas moles decrease
  • Gas moles decrease ΔS is large and negative, for example when ammonia is formed from nitrogen and hydrogen.
Solid to liquid or gas
  • Solid to liquid or gas ΔS is positive because the particles become freer to move.
No change in gas moles
  • No change in gas moles ΔS is small and its sign is hard to predict without data.

Part 3 of 6Calculation Steps

1
Write the balanced equation and include state symbols, as the state changes the entropy.
2
List the S° of every reactant and product from the data supplied.
3
Multiply each S° by its coefficient and sum the products, and sum the reactants separately.
4
Subtract the total for the reactants from the total for the products.
5
State the answer with its sign and unit, and check the sign against your prediction.

Part 4 of 6Examples

Ammonia synthesis
  • Ammonia synthesis N₂(g) + 3H₂(g) → 2NH₃(g) reduces 4 mol of gas to 2 mol, so ΔS is negative.
Decomposition
  • Decomposition CaCO₃(s) → CaO(s) + CO₂(g) produces a gas from a solid, so ΔS is positive.
Combustion
  • Combustion Burning a hydrocarbon to CO₂ and H₂O usually increases the number of gas molecules when the water is a vapour.
Dissolving
  • Dissolving NH₄NO₃(s) dissolving in water gives positive ΔS as the lattice breaks into freely moving ions.

Part 5 of 6Worked Example

Question Calculate ΔS°system for N₂(g) + 3H₂(g) → 2NH₃(g), given S° values of 191.6 (N₂), 130.6 (H₂) and 192.3 (NH₃) J K⁻¹ mol⁻¹.

  1. Products2 × 192.3 = 384.6 J K⁻¹ mol⁻¹.
  2. Reactants191.6 + 3 × 130.6 = 191.6 + 391.8 = 583.4 J K⁻¹ mol⁻¹.
  3. ΔS = 384.6 − 583.4 = −198.8 J K⁻¹ mol⁻¹.

Answer ΔS°system = −198.8 J K⁻¹ mol⁻¹

Part 6 of 6Exam Checklist & Tips

4.2.2 – Key Points for the Exam

  • ΔS°system The standard entropy change of the reacting chemicals, in J K⁻¹ mol⁻¹.
  • Standard entropy The entropy of one mole of a substance under standard conditions.
  • Coefficient The number in front of a formula in a balanced equation.
  • Disorder A measure of the randomness in the arrangement of particles.
  • State symbol The label (s), (l), (g) or (aq) showing the physical state.

Common Mistakes

  • Forgetting to multiply S° values by the coefficients in the equation.
  • Subtracting products from reactants instead of reactants from products.
  • Giving the answer in kJ K⁻¹ mol⁻¹ without noticing the unit change.
  • Ignoring the state symbols when predicting the sign.

Last updated:

\n