Bond Energy & Enthalpy Change
In 1909, Fritz Haber at the University of Karlsruhe achieved the first laboratory synthesis of ammonia from nitrogen and hydrogen — but only after calculating that the N≡N bond energy of 945 kJ mol⁻¹ and the H–H bond energy of 436 kJ mol⁻¹ would require enormous energy input on every cycle. His bond energy calculations predicted ΔH ≈ −92 kJ mol⁻¹ for the reaction, confirming it was exothermic overall — yet the activation energy was so high that he required an iron catalyst and 400°C to achieve a practical rate. Bond energies told him whether the reaction was possible; activation energy told him how hard it was to start.
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
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.
Key Facts
- Bond energy = average energy to break 1 mol of a bond in the gaseous state (kJ mol⁻¹)
- Bond breaking is always endothermic; bond forming is always exothermic
- ΔH = ΣB(reactants) − ΣB(products): reactants first
Concepts
- Why bond energy calculations give approximate ΔH values (average bonds, gaseous state assumption)
- How the sign of ΔH follows from comparing energy in vs energy out
- Why N≡N's high bond energy makes nitrogen so unreactive
Skills
- Use structural formulas to count bonds in reactants and products
- Calculate ΔH from bond energy data using ΔH = ΣB(reactants) − ΣB(products)
- Explain two reasons why bond energy ΔH differs from experimental values
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:
- Write the balanced chemical equation
- Draw structural formulas (Lewis structures) for each reactant and product to count individual bonds
- List all bonds broken in reactants, multiply by bond energy × stoichiometric coefficient
- List all bonds formed in products, multiply by bond energy × stoichiometric coefficient
- Apply: ΔH = ΣB(reactants) − ΣB(products)
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.
ΔH (bond energy method) = Σ(bond energies broken in reactants) − Σ(bond energies formed in products). Draw structural formulas first to count individual bonds. Bond breaking requires energy (endothermic); bond forming releases energy (exothermic).
Pause — copy the highlighted definition into your book before moving on.
Quick check: When calculating ΔH using bond energies, which of the following correctly describes the sign rule?
We just saw how to calculate ΔH by summing bond energies broken minus bond energies formed. That raises a question: if the method is so systematic, why doesn’t it match experimental results exactly? This card answers it → because tabulated bond energies are averages, not exact values for any specific molecular environment.
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 |
Bond energy calculations give estimates only because tabulated values are averages across many compounds — not the exact bond energy for a specific molecule. Results differ from experimental values because bond energies vary with molecular environment and phase of reactants/products.
Add the highlighted point to your notes before the check below.
Explain it: A student calculates ΔH for the combustion of methane using bond energies and gets −674 kJ mol⁻¹. The accepted standard enthalpy of combustion is −890 kJ mol⁻¹. Explain in your own words why there is a discrepancy, identifying which reason is likely responsible for the larger portion of the difference.
Worked examples · reveal as you go
Combustion of Methane. 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.
GIVEN: Balanced equation, bond energy table as above
FIND: ΔH using ΔH = ΣB(reactants) − ΣB(products)
CH₄: 4 × C–H bonds
2O₂: 2 × O=O bonds (one per O₂ molecule)
ΣB(reactants) = 4(414) + 2(498)
= 1656 + 996
= 2652 kJ mol⁻¹
CO₂: 2 × C=O bonds (it's O=C=O — two double bonds)
2H₂O: 2 molecules × 2 O–H = 4 × O–H bonds
ΣB(products) = 2(743) + 4(460)
= 1486 + 1840
= 3326 kJ mol⁻¹
ΔH = 2652 − 3326 = −674 kJ mol⁻¹
Negative → exothermic. More energy is released forming bonds in CO₂ and H₂O than is absorbed breaking bonds in CH₄ and O₂.
Comparison: The accepted standard enthalpy of combustion of CH₄ is −890 kJ mol⁻¹ (with H₂O(l) as product). Our calculated value (−674 kJ mol⁻¹) is less negative because: (1) we used H₂O(g) — the extra energy released when steam condenses to liquid is not counted; (2) average bond energies were used.
Hydrogenation of Ethene. 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.
GIVEN: Bond energy data above
FIND: ΔH for C₂H₄(g) + H₂(g) → C₂H₆(g)
C₂H₄ (ethene): H₂C=CH₂ → 1 × C=C bond + 4 × C–H bonds
H₂: 1 × H–H bond
ΣB(reactants) = 1(614) + 4(414) + 1(436)
= 614 + 1656 + 436
= 2706 kJ mol⁻¹
C₂H₆ (ethane): H₃C–CH₃ → 1 × C–C bond + 6 × C–H bonds
ΣB(products) = 1(347) + 6(414)
= 347 + 2484
= 2831 kJ mol⁻¹
ΔH = 2706 − 2831 = −125 kJ mol⁻¹
Exothermic — the hydrogenation of ethene releases energy. This is consistent with what we expect: adding H₂ across a double bond to form a more stable single bond releases energy.
Accepted value ≈ −137 kJ mol⁻¹. Discrepancy of 12 kJ mol⁻¹ is due to the use of average bond enthalpies.
Complete this bond energy calculation for H&sub2; + Cl&sub2; → 2HCl.
Bonds broken: 1× H–H + 1× Cl–Cl = 436 + 243 = kJ mol¹
Bonds formed: 2× H–Cl = 2 × 432 = kJ mol¹
ΔH = broken − formed = 679 − 864 = kJ mol¹
Formula Reference — This Lesson
Common errors · the 3 traps that cost marks
Reversing the formula: products minus reactants
Students write ΔH = ΣB(products) − ΣB(reactants), getting the wrong sign and an exothermic result when it should be endothermic (or vice versa).
Fix: Bond energy is reactants minus products. Memory aid: bond energy = bonds you break (reactants) minus bonds you make (products). Note this is opposite to the enthalpy of formation formula in L07 — knowing both formulas prevents the confusion.
Miscounting bonds in polyatomic molecules
Students write CO₂ has one C=O bond (molecular formula CO₂ looks like 1 carbon and 2 oxygens), or forget that 2H₂O means 4 O–H bonds total.
Fix: Always draw Lewis/structural formulas first. CO₂ is O=C=O — two double bonds. 2H₂O means 2 × 2 = 4 O–H bonds. Propane (C₃H₈) has 2 C–C and 8 C–H bonds. Never count bonds from the molecular formula alone.
Giving only one reason why bond energy results are approximate
Students answer "because average bond enthalpies are used" and stop there, missing a second mark.
Fix: HSC always expects two reasons: (1) average bond enthalpies — same bond type varies in different molecules; (2) gaseous state assumption — doesn't account for latent heat when H₂O(l) or other liquids are produced. Stating both earns both marks.
Quick-fire practice · 5 reps +2 XP per reveal
Combustion of propane (C₃H₈):
C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(g)
Bond energies: C–H = 414; C–C = 347; O=O = 498; C=O = 743; O–H = 460 kJ mol⁻¹.
Draw the structural formula of propane and count all bonds. Then calculate ΣB(reactants).
Bonds in C₃H₈: 2 × C–C bonds + 8 × C–H bonds
Bonds in 5O₂: 5 × O=O bonds
ΣB(reactants) = 2(347) + 8(414) + 5(498)
= 694 + 3312 + 2490
= 6496 kJ mol⁻¹
Continuing the combustion of propane reaction above. Count all bonds formed in the products (3CO₂ + 4H₂O) and calculate ΣB(products).
4H₂O: 4 × 2 O–H = 8 × O–H bonds
ΣB(products) = 6(743) + 8(460)
= 4458 + 3680
= 8138 kJ mol⁻¹
Using the values from drills 1 and 2, calculate ΔH for the combustion of propane. Is it exothermic or endothermic? Compare to the accepted value of −2220 kJ mol⁻¹ and explain the discrepancy in two sentences.
Exothermic — ΔH is negative; more energy is released forming bonds in products than absorbed breaking bonds in reactants.
Comparison to accepted −2220 kJ mol⁻¹: (1) We used H₂O(g) — the accepted value uses H₂O(l) as product; the condensation of 4 mol of steam releases significant additional energy (~44 kJ mol⁻¹ × 4 = ~176 kJ) not counted in our calculation. (2) Average bond enthalpies were used throughout — the actual bond energies in propane, oxygen, and CO₂ differ from the tabulated averages, accumulating error across 29 bond values.
Haber process: N₂(g) + 3H₂(g) → 2NH₃(g)
Bond energies: N≡N = 945; H–H = 436; N–H = 391 kJ mol⁻¹.
Calculate ΔH. Is the reaction exothermic or endothermic? How does this compare to the Think First prediction you made at the start of the lesson?
ΣB(products) = 6(391) = 2346 kJ mol⁻¹
ΔH = 2253 − 2346 = −93 kJ mol⁻¹ → exothermic
N≡N accounts for 945/2253 = 42% of total energy absorbed. Most students predict endothermic because breaking N≡N sounds so expensive — but forming six N–H bonds just barely wins out (by only 93 kJ mol⁻¹). This slim margin explains why NH₃ is relatively unstable and can decompose back to N₂ and H₂.
You've just shown that N₂ + 3H₂ → 2NH₃ is exothermic (ΔH = −93 kJ mol⁻¹). Yet this reaction doesn't occur at room temperature — N₂ is effectively inert at ambient conditions. How can a reaction be both thermodynamically exothermic and kinetically inert? Distinguish between ΔH and Ea in your answer.
The N≡N triple bond (945 kJ mol⁻¹) must be weakened or broken at the transition state — this makes Ea enormous, far exceeding the thermal energy of molecules at 25°C. At room temperature, virtually no N₂ and H₂ molecules have sufficient kinetic energy to overcome this barrier, so the reaction rate is essentially zero.
A reaction can be thermodynamically favourable (ΔH < 0) yet kinetically inert (high Ea). The Haber process solves this with temperature (400–500°C, to increase the fraction of molecules with E ≥ Ea) and an iron catalyst (to lower Ea).
Download the PDF for classwork, homework or revision. Includes key ideas, activities, questions, an extend task and success-criteria proof.
Go back to your Think First prediction for N₂ + 3H₂ → 2NH₃. Now you can verify it precisely — exactly as Fritz Haber did at the University of Karlsruhe in 1909:
- ΣB(reactants) = 1(945) + 3(436) = 945 + 1308 = 2253 kJ mol⁻¹ — breaking the N≡N and three H–H bonds costs enormous energy. Haber knew the N≡N bond (945 kJ mol⁻¹) was one of the strongest bonds in chemistry — this is why atmospheric nitrogen is so inert.
- ΣB(products) = 6(391) = 2346 kJ mol⁻¹ — forming six N–H bonds releases slightly more energy than was absorbed.
- ΔH = 2253 − 2346 = −93 kJ mol⁻¹ → exothermic. Haber's calculation confirmed the reaction was energetically feasible — but only barely. The slim energy margin is why NH₃ can also decompose back to N₂ and H₂, which became the key challenge in scaling up the Haber process.
- Most students predict endothermic because breaking N≡N sounds so expensive. The counterintuitive result — that forming six N–H bonds wins out — is exactly what bond energy calculations quantify where intuition fails.
Pick your answer, then rate your confidence — that tells the system what to drill next.
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.
Wrong: The bond energy formula is ΔH = ΣB(products) − ΣB(reactants).
Right: ΔH = ΣB(reactants) − ΣB(products). Reactants first. This is opposite to the enthalpy of formation formula — knowing both prevents confusion.
Wrong: Bond energy calculations are approximate only because average values are used.
Right: Two reasons are required for HSC: (1) average bond enthalpies; (2) gaseous state assumption. The gaseous state assumption typically causes the larger error in combustion reactions because latent heat of condensation of water is not accounted for.
Q6. 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
Q7. 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
Q8. 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
Show comprehensive answers ▼
Drill Activity — Propane Combustion
(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.
Drill Activity — Haber Process
ΔH calculation: ΣB(reactants) = 1(945) + 3(436) = 945 + 1308 = 2253 kJ mol⁻¹. ΣB(products) = 6(391) = 2346 kJ mol⁻¹. ΔH = 2253 − 2346 = −93 kJ mol⁻¹ — exothermic, but only by a small margin.
Kinetics vs thermodynamics: 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.
Multiple Choice
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.
Short Answer Model Answers
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].
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