Nitrogen gas makes up 78% of the atmosphere — yet most living things cannot use it directly. The bond holding two nitrogen atoms together requires 945 kJ mol⁻¹ to break, one of the highest bond energies known. Every chemical reaction is fundamentally about breaking old bonds and forming new ones — and bond energies let you calculate ΔH from first principles, without running a single experiment.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
Nitrogen gas (N₂) makes up 78% of the air you breathe. It is right there, in every breath — but plants can't absorb it, most bacteria can't use it, and it took until 1909 for chemists to figure out how to convert it into something biologically useful. The culprit is the N≡N triple bond, which requires 945 kJ mol⁻¹ to break.
Now consider: when nitrogen does react to form ammonia (N₂ + 3H₂ → 2NH₃), new N–H bonds form as the N≡N and H–H bonds are broken. The N–H bond energy is 391 kJ mol⁻¹ and H–H is 436 kJ mol⁻¹.
Before looking at any formula: do you predict the reaction N₂ + 3H₂ → 2NH₃ to be exothermic or endothermic? Think about whether more energy is released forming bonds than is absorbed breaking them.
Type your initial prediction below — you will revisit this at the end of the lesson.
Write your initial prediction in your book. You will revisit it at the end of the lesson.
📚 Core Content
The key insight: ΔH equals the energy absorbed breaking bonds in the reactants minus the energy released forming bonds in the products. Always draw structural formulas first — you must count individual bonds, not just atoms.
Step-by-step method:
The N≡N triple bond (945 kJ mol⁻¹) is the reason 78% of the atmosphere is chemically inert to most life. Plants need nitrogen as ammonium (NH₄⁺) or nitrate (NO₃⁻) — not as N₂. Lightning provides just enough energy to break N≡N and form NO, which eventually reaches soil. The Haber process replicates this industrially: N₂ + 3H₂ → 2NH₃. Without it, roughly half of all nitrogen in human bodies today would not exist — the world couldn't grow enough food to feed its population. The high bond energy of N≡N explains why the process requires 400–500°C, enormous pressure, and a catalyst just to proceed at a useful rate. This anchor reappears in Short Answer Q8.
Bond energy calculations give useful estimates of ΔH, but three specific reasons mean they never match experimental values exactly.
| Reason | Explanation | Effect on answer |
|---|---|---|
| 1. Average bond enthalpies | Bond energy values in tables are averages across many molecules. A C–H bond in CH₄ differs slightly from C–H in C₂H₅OH. Using the average introduces error. | Answer deviates from true ΔH by a few per cent |
| 2. Gaseous state assumption | Bond energies are defined for species in the gaseous state. If water is produced as a liquid (not gas), the latent heat of condensation is not accounted for. This is the most significant error for combustion reactions. | ΔH calculated will be less negative than the true ΔHc° (underestimates energy released) |
| 3. Temperature effects | Bond energy values are typically tabulated at 298 K. At higher temperatures, enthalpies shift slightly. | Minor effect for most HSC calculations |
🔬 Worked Examples
Use bond energy data to calculate the enthalpy change for the combustion of methane:
CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
Bond energies: C–H = 414 kJ mol⁻¹; O=O = 498 kJ mol⁻¹; C=O = 743 kJ mol⁻¹; O–H = 460 kJ mol⁻¹.
Calculate ΔH for the hydrogenation of ethene: C₂H₄(g) + H₂(g) → C₂H₆(g).
Bond energies: C=C = 614; H–H = 436; C–C = 347; C–H = 414 kJ mol⁻¹.
🧪 Activities
a Draw the structural formula of propane (C₃H₈) and count all bonds that must be broken. Calculate ΣB(reactants).
b Count all bonds formed in the products and calculate ΣB(products).
c Calculate ΔH. Is the combustion of propane exothermic or endothermic? Compare your answer to the accepted value of −2220 kJ mol⁻¹ and explain the discrepancy.
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a Calculate ΣB(reactants). Identify which bond accounts for the majority of energy absorbed and explain why this makes N₂ so difficult to react.
b Calculate ΣB(products) and then ΔH. Was your Think First prediction correct?
c If the reaction is exothermic, why does the Haber process require such high temperatures (400–500°C) to operate? Use the concepts from this lesson and Lesson 5.
Type your working and responses below:
Answer in your workbook.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
Wrong: An exothermic reaction has ΔH > 0 because it releases heat.
Right: Exothermic reactions have ΔH < 0 (negative) because the system loses energy to the surroundings. Endothermic reactions have ΔH > 0 (positive) because the system gains energy. The sign convention refers to the system, not the surroundings.
5 random questions from a replayable lesson bank — feedback shown immediately
✍️ Short Answer
6. Use bond energy data to calculate ΔH for the following reaction:
CH₃OH(g) + O₂(g) → CO₂(g) + 2H₂O(g)
Bond energies: C–H = 414; O–H = 460; C–O = 360; O=O = 498; C=O = 743 kJ mol⁻¹.
Note: methanol (CH₃OH) has the structure H₃C–O–H. 4 MARKS
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7. A student calculates ΔH for the combustion of ethene C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(g) and gets −1052 kJ mol⁻¹. The accepted ΔHc° (with H₂O liquid) is −1411 kJ mol⁻¹.
(a) Identify two reasons why the bond energy result (−1052 kJ mol⁻¹) differs from the accepted value (−1411 kJ mol⁻¹). (2 marks)
(b) Explain which reason you identified in (a) is likely responsible for the larger portion of the discrepancy. (2 marks)
4 MARKS
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Answer in your workbook.
8. The Haber process is one of the most important industrial reactions in history: N₂(g) + 3H₂(g) → 2NH₃(g). Approximately half of the nitrogen atoms in every human body entered the biosphere through this reaction.
(a) Using bond energies (N≡N = 945; H–H = 436; N–H = 391 kJ mol⁻¹), calculate ΔH for the Haber process reaction. (3 marks)
(b) Explain why the very high bond energy of N≡N means that N₂ is effectively unreactive at room temperature, even though the reaction to form NH₃ is exothermic. In your answer, distinguish between the thermodynamic and kinetic factors involved. (3 marks)
6 MARKS
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Go back to your Think First prediction for N₂ + 3H₂ → 2NH₃. Now you can verify it precisely:
Type your reflection below:
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(a) C₃H₈: 2 × C–C + 8 × C–H; 5O₂: 5 × O=O
ΣB(reactants) = 2(347) + 8(414) + 5(498) = 694 + 3312 + 2490 = 6496 kJ mol⁻¹
(b) 3CO₂: 6 × C=O; 4H₂O: 8 × O–H
ΣB(products) = 6(743) + 8(460) = 4458 + 3680 = 8138 kJ mol⁻¹
(c) ΔH = 6496 − 8138 = −1642 kJ mol⁻¹ (exothermic). Discrepancy from −2220 kJ mol⁻¹: (1) H₂O was treated as gaseous — latent heat of condensation of 4 mol H₂O(l) not included (~44 × 4 = ~176 kJ additional energy); (2) average bond enthalpies introduce cumulative error across 29 bond values.
(a) ΣB(reactants) = 1(945) + 3(436) = 945 + 1308 = 2253 kJ mol⁻¹. N≡N accounts for 945/2253 = 42% of total energy absorbed. Its extremely high bond energy means the activation energy is very large, making N₂ kinetically inert at room temperature despite the reaction being thermodynamically exothermic.
(b) ΣB(products) = 6(391) = 2346 kJ mol⁻¹. ΔH = 2253 − 2346 = −93 kJ mol⁻¹ — exothermic. The reaction is thermodynamically favourable but only by a small margin (93 kJ mol⁻¹ net); this is why equilibrium conditions must be carefully managed in the Haber process.
(c) Even though ΔH < 0 (thermodynamically favourable), the activation energy is very high because the N≡N triple bond must be overcome. At room temperature, virtually no molecules have sufficient kinetic energy to reach the transition state. High temperature (400–500°C) increases the fraction of molecules with E ≥ Ea; the Fe catalyst lowers Ea by providing an alternative surface mechanism.
1. B — ΔH = ΣB(reactants) − ΣB(products). Reactants first. Option A is the most common error (reversed formula).
2. B — ΣB(reactants) = 436 + 243 = 679; ΣB(products) = 2(432) = 864; ΔH = 679 − 864 = −185 kJ mol⁻¹. Exothermic: more energy released forming 2 H–Cl bonds than absorbed breaking 1 H–H + 1 Cl–Cl.
3. C — Two reasons: (1) average bond enthalpies (same bond type has slightly different energy in different molecules); (2) gaseous state assumption (doesn't account for energy changes when liquid or solid products form). Option B is wrong — the method assumes gaseous state, not liquid.
4. A — The gaseous state assumption means condensation energy of H₂O(g) → H₂O(l) is not counted. The accepted value uses H₂O(l) — this extra energy (latent heat of condensation) makes the true ΔHc° more negative than the bond energy calculation. Option B is the opposite error.
5. D — The N≡N bond energy of 945 kJ mol⁻¹ means the activation energy is enormous — the energy cost to break (or even weaken) this bond at the transition state is very high. At room temperature, too few molecules have sufficient kinetic energy to achieve this. Option C is wrong: N≡N is one of the strongest bonds, not weak.
Q6 (4 marks): Structural formula of CH₃OH: 3 × C–H, 1 × C–O, 1 × O–H.
ΣB(reactants): CH₃OH = 3(414) + 360 + 460 = 1242 + 360 + 460 = 2062 kJ; O₂ = 1(498) = 498 kJ; Total = 2560 kJ mol⁻¹ [1]
ΣB(products): CO₂ = 2(743) = 1486 kJ; 2H₂O = 4(460) = 1840 kJ; Total = 3326 kJ mol⁻¹ [1]
ΔH = 2560 − 3326 = −766 kJ mol⁻¹ [1] → exothermic [1].
(Accepted ΔHc° for methanol with H₂O(l) is −726 kJ mol⁻¹; the difference reflects the gaseous water assumption.)
Q7 (4 marks):
(a) Reason 1: Bond energy values are averages across different molecular environments — the actual bond energies in C₂H₄, O₂, CO₂, and H₂O differ slightly from the tabulated averages, introducing cumulative error [1]. Reason 2: The bond energy method assumed H₂O(g) as a product, but the accepted value uses H₂O(l) — the energy released when water vapour condenses to liquid (latent heat of condensation) is not included in the bond energy calculation, making the calculated ΔH less negative than the true value [1].
(b) The gaseous state assumption (Reason 2) is likely responsible for the larger portion of the discrepancy [1]. For the combustion of ethene producing 2 mol H₂O, condensation releases approximately 2 × 44 kJ = 88 kJ of additional energy. The total discrepancy is 1411 − 1052 = 359 kJ mol⁻¹, suggesting that both reasons contribute, but the accumulation of error across many bond values (average bond enthalpy reason) also contributes substantially [1].
Q8 (6 marks):
(a) ΣB(reactants) = 945 + 3(436) = 945 + 1308 = 2253 kJ mol⁻¹ [1]; ΣB(products) = 6(391) = 2346 kJ mol⁻¹ [1]; ΔH = 2253 − 2346 = −93 kJ mol⁻¹ [1].
(b) Thermodynamic factor: ΔH = −93 kJ mol⁻¹ → the reaction is exothermic and thermodynamically favourable (products are lower in energy than reactants). The law of conservation of energy says this reaction should release energy — it is "downhill" energetically [1]. Kinetic factor: Ea is very high because the N≡N triple bond (945 kJ mol⁻¹) must be overcome at the transition state — even though bonds are also forming as the reaction proceeds, the initial energy barrier is enormous [1]. At room temperature, the proportion of N₂ and H₂ molecules with kinetic energy ≥ Ea is negligibly small, so the reaction rate is essentially zero. The distinction is: ΔH describes the energy difference between start and finish (thermodynamics — the direction of reaction); Ea describes the energy cost to reach the transition state (kinetics — the rate). A reaction can be thermodynamically favourable (ΔH < 0) yet kinetically inert (high Ea) — exactly as for N₂ + 3H₂ → 2NH₃ at ambient conditions [1].
Climb platforms, hit checkpoints, and answer questions on Bond Energy & Enthalpy Change. Quick recall from lessons 1–6.
Tick when you've finished all activities and checked your answers.