Entropy Calculations
Thermodynamics Worksheet
Standard Entropy Data
Use the table below to determine the entropy change of the following reactions.
| Substance | Entropy / J mol⁻¹ K⁻¹ | Substance | Entropy / J mol⁻¹ K⁻¹ |
|---|---|---|---|
| C₂H₄ (g) | 219.3 | HCl (g) | 186.9 |
| C₆H₁₂O₆ (s) | 209.2 | Mg (s) | 32.7 |
| C₆H₁₄ (l) | 296.1 | MgCl₂ (s) | 89.6 |
| CH₃COCl (l) | 186.3 | N₂ (g) | 191.5 |
| CH₃COOH (l) | 158.0 | Na (s) | 51.2 |
| Cl₂ (g) | 223.1 | NH₃ (g) | 192.6 |
| CO₂ (g) | 51.1 | NH₄⁺ (aq) | 113.4 |
| H⁺ (aq) | 0.00 | O₂ (g) | 205.2 |
| H₂ (g) | 130.5 | OH⁻ (aq) | -10.8 |
| H₂O (g) | 188.7 | PCl₅ (g) | 364.2 |
| H₂O (l) | 16.8 | POCl₃ (l) | 325.5 |
| H₂O₂ (l) | 55.7 | SO₄²⁻ (aq) | 20.1 |
Calculations
1. Formation of one mole of magnesium chloride.
Equation: Mg (s) + Cl₂ (g) → MgCl₂ (s)
Calculation: ΔS = ΣS(products) – ΣS(reactants)
ΔS = S(MgCl₂) – [S(Mg) + S(Cl₂)]
ΔS = 89.6 – [32.7 + 223.1]
ΔS = 89.6 – 255.8
ΔS = -166.2 J mol⁻¹ K⁻¹
2. Condensation of water.
Equation: H₂O (g) → H₂O (l)
Calculation:
ΔS = S(H₂O(l)) – S(H₂O(g))
ΔS = 16.8 – 188.7
ΔS = -171.9 J mol⁻¹ K⁻¹
3. Combustion of one mole of hexane, as per the enthalpy of combustion.
Equation: C₆H₁₄ (l) + 9.5O₂ (g) → 6CO₂ (g) + 7H₂O (l)
Calculation:
ΔS = [6 × S(CO₂) + 7 × S(H₂O)] – [S(C₆H₁₄) + 9.5 × S(O₂)]
ΔS = [(6 × 51.1) + (7 × 16.8)] – [296.1 + (9.5 × 205.2)]
ΔS = [306.6 + 117.6] – [296.1 + 1949.4]
ΔS = 424.2 – 2245.5
ΔS = -1821.3 J mol⁻¹ K⁻¹
4. The reaction between one mole of ammonia and excess sulfuric acid.
Equation (ionic): NH₃(g) + H⁺(aq) → NH₄⁺(aq)
Calculation:
ΔS = S(NH₄⁺) – [S(NH₃) + S(H⁺)]
ΔS = 113.4 – [192.6 + 0.00]
ΔS = -79.2 J mol⁻¹ K⁻¹
5. Photosynthesis.
Equation: 6CO₂ (g) + 6H₂O (l) → C₆H₁₂O₆ (s) + 6O₂ (g)
Calculation:
ΔS = [S(C₆H₁₂O₆) + 6 × S(O₂)] – [6 × S(CO₂) + 6 × S(H₂O)]
ΔS = [209.2 + (6 × 205.2)] – [(6 × 51.1) + (6 × 16.8)]
ΔS = [209.2 + 1231.2] – [306.6 + 100.8]
ΔS = 1440.4 – 407.4
ΔS = +1033 J mol⁻¹ K⁻¹
6. Decomposition of one mole of hydrogen peroxide.
Equation: H₂O₂ (l) → H₂O (l) + ½O₂ (g)
Calculation:
ΔS = [S(H₂O) + 0.5 × S(O₂)] – S(H₂O₂)
ΔS = [16.8 + (0.5 × 205.2)] – 55.7
ΔS = [16.8 + 102.6] – 55.7
ΔS = 119.4 – 55.7
ΔS = +63.7 J mol⁻¹ K⁻¹
7. The reaction between ethanoic acid and phosphorous pentachloride to form ethanoyl chloride, hydrogen chloride gas and phosphorous oxychloride (POCl₃).
Equation: CH₃COOH(l) + PCl₅ (g) → CH₃COCl (l)+ HCl(g) + POCl₃ (l)
Calculation:
ΔS = [S(CH₃COCl) + S(HCl) + S(POCl₃)] – [S(CH₃COOH) + S(PCl₅)]
ΔS = [186.3 + 186.9 + 325.5] – [158.0 + 364.2]
ΔS = 698.7 – 522.2
ΔS = +176.5 J mol⁻¹ K⁻¹
8. The formation of one mole of ammonia.
Equation: ½ N₂ (g) + 1½ H₂ (g) → NH₃ (g)
Calculation:
ΔS = S(NH₃) – [0.5 × S(N₂) + 1.5 × S(H₂)]
ΔS = 192.6 – [(0.5 × 191.5) + (1.5 × 130.5)]
ΔS = 192.6 – [95.75 + 195.75]
ΔS = 192.6 – 291.5
ΔS = -98.9 J mol⁻¹ K⁻¹
9. What can you deduce, if anything, about the reactions’ enthalpy changes based on the information provided.
- Must be exothermic to be feasible as the entropy is negative.
- Must be exothermic to be feasible as the entropy is negative.
- Must be exothermic to be feasible as the entropy is negative.
- Must be exothermic to be feasible as the entropy is negative (Corrected from original).
- Could be exothermic or endothermic as the entropy is positive.
- Could be exothermic or endothermic as the entropy is positive.
- Could be exothermic or endothermic as the entropy is positive.
- Must be exothermic to be feasible as the entropy is negative.