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Bond enthalpy questions look like quick arithmetic, but they test two ideas students often get backwards: which way energy flows when a bond breaks, and how far an average value can be trusted. This post works three examples, one forwards, one backwards and one that shows the method's limits, using NCERT's table of bond enthalpies.
What does the syllabus ask for?
The JEE (Main) 2026 syllabus lists "enthalpies of bond dissociation" with Hess's law in the thermodynamics unit, and the NEET (UG) 2026 syllabus has the same line. Check your own year's syllabus. If Hess's law cycles and formation enthalpies are new to you, read our Hess's law guide first; it includes a basic bond enthalpy example.
Does breaking a bond release energy?
No. The NCERT Class 11 Thermodynamics chapter states it plainly: "Energy is required to break a bond and energy is released when a bond is formed."
The opposite belief is widespread. Galley surveyed students at the start of an introductory physical chemistry course over several years and found that the vast majority believed bond breaking is exothermic. Over 80% chose bond breaking in the reactants as the reason a simple combustion releases heat. Galley reports that pointing out the misconception early in the course markedly reduced the errors on later exams. So say it to yourself before every problem: breaking costs, forming pays, and ΔH is the balance.
That gives the working formula NCERT uses for gas-phase reactions: ΔrH ≈ Σ(bond enthalpies of reactants) − Σ(bond enthalpies of products). Note the order. It is reactants minus products, the reverse of the formation-enthalpy formula.
What is a mean bond enthalpy?
NCERT shows that the four C–H bonds in methane do not need the same energy to break one after another: 427, 439, 452 and 347 kJ mol⁻¹. Together they add to 1665 kJ mol⁻¹, the enthalpy of atomisation, so the mean C–H bond enthalpy in methane is 1665 ÷ 4 = 416 kJ mol⁻¹. C–H bonds in other compounds differ slightly, and NCERT's table lists an average of 414 kJ mol⁻¹. Every value in the table is an average of this kind.
Worked example 1: hydrogenating ethene
C₂H₄(g) + H₂(g) → C₂H₆(g). NCERT's mean values: C=C 611, H–H 435.8, C–C 347, C–H 414 kJ mol⁻¹.
Draw the structures. Ethene has one C=C and four C–H; ethane has one C–C and six C–H.
Only the changing bonds count. Broken: one C=C and one H–H = 611 + 435.8 = 1046.8 kJ mol⁻¹.
Formed: one C–C and two new C–H = 347 + 2 × 414 = 1175 kJ mol⁻¹.
ΔH ≈ 1046.8 − 1175 ≈ −128 kJ mol⁻¹.
For comparison, the NIST Chemistry WebBook lists the enthalpy of formation of ethene gas as 52.4 kJ mol⁻¹, and of ethane gas as −84.0 kJ mol⁻¹, which gives −136.4 kJ mol⁻¹. The estimate is about 6% low: good enough for most purposes.
Worked example 2: finding an unknown bond enthalpy
NCERT gives N₂(g) + 3H₂(g) → 2NH₃(g) with ΔrH = −92.4 kJ mol⁻¹. Use N≡N 946 and H–H 435.8 kJ mol⁻¹ to find the N–H bond enthalpy.
Broken: one N≡N and three H–H = 946 + 3 × 435.8 = 2253.4 kJ mol⁻¹.
Formed: six N–H = 6x.
−92.4 = 2253.4 − 6x, so x = 2345.8 ÷ 6 = 391 kJ mol⁻¹.
NCERT's table gives 389 kJ mol⁻¹, so the back-calculation agrees within 1%.
Worked example 3: where the estimate fails
Estimate ΔH for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g) with NCERT's values: C–H 414, O=O 498, C=O 741, O–H 464 kJ mol⁻¹.
Broken: 4 × 414 + 2 × 498 = 2652 kJ mol⁻¹.
Formed: 2 × 741 + 4 × 464 = 3338 kJ mol⁻¹.
ΔH ≈ −686 kJ mol⁻¹.
Now use formation enthalpies from NIST: methane −74.6, carbon dioxide −393.5 and water vapour −241.8 kJ mol⁻¹. That gives −393.5 + 2(−241.8) + 74.6 = −802.5 kJ mol⁻¹. The estimate misses by about 116 kJ mol⁻¹.
The main culprit is the C=O value. The table's 741 kJ mol⁻¹ is an average over compounds with a C=O group. You can work out the value for carbon dioxide itself from NIST's enthalpies of formation of gaseous carbon atoms (716.7 kJ mol⁻¹) and oxygen atoms (249.2 kJ mol⁻¹): atomising CO₂ takes 716.7 + 2 × 249.2 + 393.5 = 1608.6 kJ mol⁻¹, or about 804 kJ mol⁻¹ per C=O bond. Use 804 in step 2 and the estimate becomes −812 kJ mol⁻¹, much closer. This is why NCERT calls the relationship approximate.
Which mistakes should you avoid?
Products minus reactants. For bond enthalpies it is reactants minus products.
Counting all bonds instead of only those that change. Both work, but the full count invites arithmetic slips.
Using bond enthalpies for liquids. NCERT restricts the method to gases; H₂O(l) needs its enthalpy of vaporisation as well.
Treating a single bond value as exact. If formation enthalpies are given, use them instead.
Bond strength links back to bond order and structure; our profile of Linus Pauling and the nature of the chemical bond tells that story. If the gap is in bonding itself, the Fermi Concept Library covers every concept in physics, chemistry, biology, mathematics and computing, each with its own lesson.
A practice routine
Day 1: draw full structures for ten small molecules and count each bond type.
Day 2: four forward estimates, listing bonds broken and formed in two columns.
Day 3: three back-calculations of an unknown bond enthalpy.
Day 4: two estimates compared with formation-enthalpy answers; write one line on why they differ.
Today, estimate ΔH for H₂(g) + Cl₂(g) → 2HCl(g) using NCERT's H–H, Cl–Cl and H–Cl values.
