Worked examples — Bond enthalpies — estimating ΔH_rxn from bond energies

The scenario matrix

Before we solve anything, here is the full space of cases this topic can throw at you. Each later example is tagged with the cell it fills.

Bond-enthalpy table used throughout (kJ/mol):

Case A — clearly exothermic

Forecast: we break one H–H (436) and one Cl–Cl (242) — both fairly weak — and make two H–Cl bonds (431 each = 862). More strong bonds out than weak bonds in ⟹ guess exothermic (negative).

1. Bonds broken: $1\times$ H–H $+1\times$ Cl–Cl $=436+242=678$. Why this step? The whole method is "atomise reactants first" — list every reactant bond that disappears.
2. Bonds formed: $2\times$ H–Cl $=2\times 431=862$. Why this step? The coefficient 2 means two HCl molecules, each with one H–Cl bond ⟹ 2 bonds.
3. Apply master equation: $\Delta H=678-862=-184$ kJ/mol. Why this step? Broken − Formed. Formed carries the "energy released" so it subtracts.

Verify: sign is negative ✅ matching the forecast. Sanity: experimental $\Delta H_f^{\circ }(\text{HCl})=-92$ kJ/mol per mole, and we made 2 mol ⟹ $2\times (-92)=-184$. Exact match. ✅

Case B — clearly endothermic

Forecast: we must break a strong C–C single (348) and two C–H (826) to convert ethane's 6 C–H + 1 C–C into ethene's 4 C–H + 1 C=C plus a new H–H. We spend on breaking, and the C=C we make (614) is weaker than the C–C + 2 C–H we destroyed. Guess endothermic (positive).

1. Count bonds in ethane (reactant): 1 C–C, 6 C–H. Why this step? Always inventory both structures from their Lewis Structures before subtracting.
2. Count bonds in products: ethene = 1 C=C + 4 C–H; hydrogen = 1 H–H.
3. Bonds broken (all reactant bonds): $348+6(413)=348+2478=2826$.
4. Bonds formed (all product bonds): $614+4(413)+436=614+1652+436=2702$.
5. Master equation: $\Delta H=2826-2702=+124$ kJ/mol. Why this step? Broken − Formed.

Verify: $+124$ is exactly $-1\times (-124)$, the negative of the parent's hydrogenation answer. Reversing a reaction flips the sign — a state-function check (Hess's Law). ✅ Positive ⟹ endothermic, matches forecast.

Case C — the stoichiometric-coefficient trap

Forecast: we break one very strong N≡N (945!) and three H–H (1308), then form ammonia's N–H bonds. Because N≡N is huge, this one is a close call — let's compute.

1. Bonds broken: $1\times$ N≡N $+3\times$ H–H $=945+3(436)=945+1308=2253$. Why this step? The coefficient 3 on H₂ means three H–H bonds break, not one. This is the classic trap.
2. Count N–H bonds formed: each NH₃ has 3 N–H bonds, and we make 2 molecules ⟹ $2\times 3=6$ N–H bonds. Why this step? Bond count = (bonds per molecule) × (molecule count). Look at the figure: 6 green bonds.
3. Bonds formed: $6\times$ N–H $=6\times 391=2346$.
4. Master equation: $\Delta H=2253-2346=-93$ kJ/mol.

Verify: experimental $\Delta H_{rxn}\approx -92$ kJ/mol — our estimate of $-93$ is within 1 kJ. ✅ If you'd wrongly used 1 H–H (436) and 3 N–H (1173) you'd get $945+436-1173=+208$: badly wrong sign. That single mistake is the whole reason this cell exists.

Case D — degenerate / near-zero ΔH

Forecast: if the count and type of every bond is the same on both sides, the sums cancel. Guess $\Delta H\approx 0$.

1. Bonds broken: $1\times$ C–H $+1\times$ C–C $=413+348=761$. Why this step? Same routine, no shortcut — list them.
2. Bonds formed: $1\times$ C–H $+1\times$ C–C $=413+348=761$. Why this step? Same types reappear, just in new positions.
3. Master equation: $\Delta H=761-761=0$ kJ/mol.

Verify: exactly zero by the bond-enthalpy method. But — real isomers differ by small amounts (strain, environment) that average bond values cannot see. This is a live demonstration of the parent's "only an estimate" warning: the true answer is a few kJ, not exactly zero. The degenerate case is precisely where the method's blindness is loudest. ✅

Case E — work backwards for a missing bond energy

Forecast: we're inverting Ex 3. Since we already know N≡N ≈ 945, our answer should land there — this checks that the algebra is reversible.

1. Write the master equation symbolically: $\Delta H=\underset{\text{broken}}{\underbrace{(x+3\times 436)}}-\underset{\text{formed}}{\underbrace{(6\times 391)}}$ Why this step? Treat the unknown bond energy as a variable; everything else is a number.
2. Substitute knowns: $-93=(x+1308)-2346=x-1038$. Why this step? Collect the constants so the unknown is alone.
3. Solve: $x=-93+1038=945$ kJ/mol. Why this step? Add 1038 to both sides — pure rearrangement.

Verify: $x=945$ matches the table's N≡N value exactly. ✅ Any exam "find the missing bond energy" is this same one-variable solve.

Case F — real-world word problem

Forecast: combustion of a hydrocarbon is strongly exothermic (that's why it's a fuel) — guess a large negative number, maybe $\approx -2000$ kJ/mol, and 44 g is exactly 1 mole so the mass part is a clean read-off.

1. Inventory propane bonds: C₃H₈ is H₃C–CH₂–CH₃ ⟹ 2 C–C bonds and 8 C–H bonds. Why this step? Draw the Lewis Structures chain: two C–C links between three carbons, eight H's.
2. Bonds broken: propane + 5 O₂ $=2(348)+8(413)+5(498)=696+3304+2490=6490$. Why this step? Coefficient 5 ⟹ five O=O bonds break.
3. Bonds formed: in 3 CO₂ each C=O counts, 2 per molecule ⟹ 6 C=O; in 4 H₂O, 2 O–H each ⟹ 8 O–H. $6(799)+8(463)=4794+3704=8498$.
4. Master equation: $\Delta H=6490-8498=-2008$ kJ/mol. (a) answer.
5. Scale by amount: $44\text{g}÷44\text{g/mol}=1$ mol, so heat released $=1\times 2008=2008$ kJ. (b) answer. Why this step? Translate the word "44 g" into moles, then multiply by per-mole energy — the classic word-problem bridge.

Verify: experimental $\Delta H_c^{\circ }$(propane, gas products) $\approx -2043$ kJ/mol; our $-2008$ is within ~2% — good for an average-bond estimate. Units: (kJ/mol)×(mol) = kJ ✅. Sign negative ⟹ exothermic, as any fuel must be. ✅

Case G — exam twist: estimate vs reality (resonance)

Forecast: three C=C's are hydrogenated, like tripling Ex 1 of the parent note (each addition ≈ −124 kJ/mol), so guess ≈ −360 to −370 kJ/mol from bonds alone.

1. Bonds broken: benzene's 3 C=C + the 3 H–H added. (The 3 C–C and 6 C–H survive, so ignore them — they're on both sides.) $3(614)+3(436)=1842+1308=3150$. Why this step? Only bonds that change matter; unchanged bonds cancel in Broken − Formed.
2. Bonds formed: the 3 former C=C become C–C singles, and 6 new C–H appear. $3(348)+6(413)=1044+2478=3522$. Why this step? Each addition of H₂ across a double bond: 1 C=C→C–C plus 2 new C–H.
3. Master equation: $\Delta H_{\text{estimate}}=3150-3522=-372$ kJ/mol.

Verify (the twist): the measured enthalpy of hydrogenation of real benzene is only about $-208$ kJ/mol. Our estimate is $\approx 164$ kJ too exothermic. Why? Real benzene is more stable than "cyclohexatriene" by its Resonance Energy (~150 kJ/mol) — the delocalised ring doesn't have three isolated double bonds. Bond-enthalpy tables can't see delocalisation, so they overshoot. This is the parent note's "ignores resonance" limit made numeric. ✅

$\Delta H_{\text{resonance}}\approx \Delta H_{\text{estimate}}-\Delta H_{\text{measured}}=-372-(-208)=-164\text{ kJ/mol}$

Active Recall

Connections

• 2.5.09 Bond enthalpies — estimating ΔH_rxn from bond energies (Hinglish) — the parent, Hinglish version.
• Hess's Law — why reversing a reaction flips the sign (Ex 2).
• Standard Enthalpy of Formation — the accurate cross-check for Ex 3, Ex 6.
• Resonance Energy — the missing term in Ex 7.
• Exothermic vs Endothermic Reactions — the sign axis of the whole matrix.
• Lewis Structures — how you count bonds correctly (Ex 3, Ex 6).
• Enthalpy and the First Law of Thermodynamics — the state-function backbone.

Case Map