Checkpoint 2 — IQ2: Enthalpy & Hess’s Law
Covering Lessons 06–10: bond energy, enthalpy of formation, Hess’s Law, photosynthesis & respiration, and fuel energy comparisons.
What’s Covered
- Bond dissociation energy
- Bonds broken (endothermic)
- Bonds formed (exothermic)
- ΔH = ΣBE(broken) − ΣBE(formed)
- ΔHf° definition
- ΔHf°(elements) = 0
- ΔH° = ΣΔHf°(products) − ΣΔHf°(reactants)
- Worked formation calculations
- ΔH is path-independent
- Reverse equation → flip sign
- Scale equation → scale ΔH
- NESA multi-step cycles
- ΔH = +2803 / −2803 kJ mol⁻¹
- Reverse reactions, equal & opposite
- ATP coupling as Hess’s Law
- Energy cycle diagrams
- ΔHc via Hess’s Law
- Energy per gram: |ΔHc| ÷ M
- Comparing fuels by kJ g⁻¹
- Three-method comparison
Multiple Choice — 10 marks
1. In calculating ΔH from bond energies, energy is released when:
2. Using bond energies: BE(H–H) = 436 kJ mol⁻¹, BE(Cl–Cl) = 242 kJ mol⁻¹, BE(H–Cl) = 432 kJ mol⁻¹. Calculate ΔH for: H₂(g) + Cl₂(g) → 2HCl(g)
3. The standard enthalpy of formation of liquid water is −285.8 kJ mol⁻¹. This means:
4. Use $\Delta H° = \Sigma\Delta H_f°(\text{products}) - \Sigma\Delta H_f°(\text{reactants})$ with: ΔHf°[CO₂(g)] = −393.5, ΔHf°[H₂O(l)] = −285.8, ΔHf°[CH₄(g)] = −74.8 kJ mol⁻¹. What is ΔH° for: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)?
5. Hess’s Law states that:
6. A thermochemical equation has ΔH = −285.8 kJ mol⁻¹. If the equation is reversed and the coefficients are doubled, what is the new ΔH?
7. Cellular respiration: $\text{C}_6\text{H}_{12}\text{O}_6\text{(aq)} + 6\text{O}_2\text{(g)} \to 6\text{CO}_2\text{(g)} + 6\text{H}_2\text{O(l)}$, ΔH = −2803 kJ mol⁻¹. For photosynthesis (the exact reverse reaction), ΔH is:
8. A metabolic reaction has ΔH = +18.0 kJ mol⁻¹. ATP hydrolysis has ΔH = −30.5 kJ mol⁻¹. The minimum number of moles of ATP hydrolysed to make the overall coupled reaction thermodynamically exothermic is:
9. Ethanol (C₂H₅OH, M = 46.07 g mol⁻¹) has ΔHc = −1366 kJ mol⁻¹. Its energy density in kJ g⁻¹ is approximately:
10. A student wants to calculate ΔHc for propan-1-ol. They have standard enthalpies of formation for all species. Which method is most direct?
Short Answer — 20 marks
Q11 (5 marks)
Using bond energies, calculate ΔH for the combustion of methane:
$\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \to \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}$
Bond energies (kJ mol⁻¹): C–H = 413; O=O = 498; C=O = 805; O–H = 463
Bonds broken (endothermic):
4 × C–H = 4 × 413 = 1652 kJ mol⁻¹
2 × O=O = 2 × 498 = 996 kJ mol⁻¹
Total bonds broken = 1652 + 996 = 2648 kJ mol⁻¹
Bonds formed (exothermic):
2 × C=O (in CO₂) = 2 × 805 = 1610 kJ mol⁻¹
4 × O–H (in 2H₂O) = 4 × 463 = 1852 kJ mol⁻¹
Total bonds formed = 1610 + 1852 = 3462 kJ mol⁻¹
ΔH = bonds broken − bonds formed = 2648 − 3462 = −814 kJ mol⁻¹
Note: This is an approximation — bond energies are average values. The experimental ΔHc[CH₄] = −890 kJ mol⁻¹ (with liquid water products).
Q12 (5 marks)
Use Hess’s Law to calculate ΔH° for: $\text{C(s)} + \frac{1}{2}\text{O}_2\text{(g)} \to \text{CO(g)}$
Given: (1) C(s) + O₂(g) → CO₂(g), ΔH₁ = −393.5 kJ mol⁻¹
(2) CO(g) + ½O₂(g) → CO₂(g), ΔH₂ = −283.0 kJ mol⁻¹
Target: C(s) + ½O₂(g) → CO(g)
Use equation (1) as is: C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ mol⁻¹
Reverse equation (2) to put CO on the products side: CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ mol⁻¹
Add: C(s) + O₂(g) + CO₂(g) → CO₂(g) + CO(g) + ½O₂(g)
Cancel CO₂ and ½O₂: C(s) + ½O₂(g) → CO(g) ✓
ΔH° = −393.5 + 283.0 = −110.5 kJ mol⁻¹
This is the standard enthalpy of formation of CO(g) — it cannot be measured directly by burning carbon in limited oxygen because a mixture of CO and CO₂ would always form.
Q13 (4 marks)
A student claims: "Plants use photosynthesis to store energy and animals use respiration to release it — these are completely different chemical processes." Evaluate this claim using Hess’s Law. In your answer, write both equations with ΔH values and explain their thermodynamic relationship.
The student's claim that they are "completely different" is partially correct chemically (different organisms, different mechanisms) but thermodynamically misleading.
Photosynthesis: 6CO₂(g) + 6H₂O(l) → C₆H₁₂O₆(aq) + 6O₂(g), ΔH = +2803 kJ mol⁻¹ (endothermic)
Respiration: C₆H₁₂O₆(aq) + 6O₂(g) → 6CO₂(g) + 6H₂O(l), ΔH = −2803 kJ mol⁻¹ (exothermic)
Hess’s Law relationship: The two equations are exact reverses of each other — same reactants and products, same magnitudes, opposite signs. By Hess’s Law, since ΔH is path-independent and depends only on the initial and final states, the ΔH values must be equal and opposite. Plants do not "create" energy — they convert light energy into chemical potential energy stored in glucose (ΔH = +2803 kJ mol⁻¹). Animals then release exactly that stored energy through respiration (ΔH = −2803 kJ mol⁻¹). The net result across the biosphere is zero enthalpy change — energy is conserved.
Q14 (6 marks)
Compare hydrogen gas (H₂) and octane (C₈H₁₈) as fuels on the basis of energy per mole and energy per gram.
Given: ΔHc[H₂(g)] = −286 kJ mol⁻¹ (M = 2.02 g mol⁻¹); ΔHc[C₈H₁₈(l)] = −5471 kJ mol⁻¹ (M = 114.23 g mol⁻¹).
(a) Calculate the energy per gram for each fuel. (b) Which fuel is more efficient per gram? (c) Explain why octane has a much larger ΔHc per mole despite being less efficient per gram.
(a) Energy per gram:
H₂: |ΔHc| ÷ M = 286 ÷ 2.02 = 141.6 kJ g⁻¹
C₈H₁₈: |ΔHc| ÷ M = 5471 ÷ 114.23 = 47.9 kJ g⁻¹
(b) Hydrogen is significantly more efficient per gram — it releases ~141.6 kJ g⁻¹ compared to ~47.9 kJ g⁻¹ for octane. Hydrogen provides approximately 3× more energy per gram of fuel.
(c) Octane (C₈H₁₈) has a much larger ΔHc per mole because it is a large molecule with 8 carbon atoms and 18 hydrogen atoms — it forms 8 mol CO₂ and 9 mol H₂O per mole of fuel, releasing energy from 8×(C=O bonds) plus 18×(O–H bonds). The total number of bonds formed in the products is far greater than for H₂. However, one mole of octane also has a much greater mass (114 g) than one mole of H₂ (2 g). When normalised by mass, hydrogen's exceptionally low molar mass means each gram of H₂ contains many more molecules — and therefore contributes more energy per unit mass despite forming only one H₂O molecule per molecule of fuel.
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