JEE Mains Chemistry · Chemical Thermodynamics

Bond Enthalpy and Atomisation Cycles

Estimating a reaction enthalpy as bonds broken minus bonds formed, and reaching bond enthalpies through gaseous atoms and ions.

All Chemical Thermodynamics notes JEE Mains Chemistry strategy

Why this matters

Ten PYQs, seven of them numerical, and two from 2026. Five count the bonds broken and formed, one of them turning a bond energy into the wavelength of light that breaks it. Five go through gaseous atoms or ions: average bond enthalpies, formation from atoms, and ion cycles.

Concept 1 of 2: Reaction enthalpy from bond enthalpies

Breaking a bond always costs energy; forming one always releases it. The enthalpy of a gas-phase reaction is roughly what you pay to break the reactants' bonds minus what you get back from forming the products' bonds. Bonds that survive unchanged cancel, so count only the ones that change.

Definition

$\Delta_r H = \sum BE(\text{bonds broken}) - \sum BE(\text{bonds formed})$ , all species gaseous.
Count every bond: $\mathrm{C_2H_6}$ has 6 C–H and 1 C–C; $\mathrm{C_2H_4}$ has 4 C–H and 1 C=C; $\mathrm{C_2H_2}$ has 2 C–H and 1 C≡C.
Bond enthalpies are averages, so the answer is an estimate.
Atomising methane breaks 4 C–H; atomising ethane breaks 6 C–H and 1 C–C.
One bond: $E = \frac{BE \times 10^3}{N_A}$ J (BE in kJ mol⁻¹). The longest wavelength that breaks it is $\lambda = \frac{hc}{E} = \frac{N_A hc}{BE \times 10^3}$ .

Reaction enthalpy from bond enthalpies

\Delta_r H = \sum BE_{\mathrm{broken}} - \sum BE_{\mathrm{formed}}

Worked example

Estimate

\Delta H

for

\mathrm{C_2H_2(g) + 2H_2(g) \to C_2H_6(g)}

. Bond enthalpies (kJ mol⁻¹): C≡C 839, C–C 348, C–H 413, H–H 436.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 4 April 2024 · Q51Moderate

Example 1 · Chemical Thermodynamics · Bond Enthalpy and Atomisation Cycles

The enthalpy of formation of ethane

\left( C_{2}H_{6} \right)

from ethylene by addition of hydrogen where the bond energies of

C - H, C - C, C = C, H - H

are

414\text{ }kJ, 347\text{ }kJ, 615\text{ }kJ

and

435\text{ }kJ

respectively is ______

kJ

Formed minus broken

The order is broken minus formed. Reversing it gives the right size with the wrong sign, and that value is offered beside the right one.

Bond enthalpies are for gases

The broken-minus-formed rule applies only when every species is a gas. A liquid or solid needs its vaporisation or sublimation enthalpy added first.

Drill 4 more on reaction enthalpy from bond enthalpies

Concept 2 of 2: Average bond enthalpy and atomisation cycles

When bond data are not given, build the change through gaseous atoms. Turning each reactant into atoms costs its atomisation enthalpy; assembling the product from atoms gives back its bond enthalpies. Ions work the same way: split the molecule, add electrons, then hydrate.

Definition

Atomisation enthalpy of a molecule = the sum of all its bond enthalpies. Average bond enthalpy = atomisation enthalpy ÷ number of bonds.
From formation enthalpies: $\Delta_a H(\mathrm{AB}_n) = \Delta_f H(\mathrm{A},g) + n\,\Delta_f H(\mathrm{B},g) - \Delta_f H(\mathrm{AB}_n,g)$ .
$\Delta_f H$ of a gaseous atom such as H(g) is half the dissociation enthalpy of $\mathrm{H_2}$ .
Formation of $\mathrm{CH_4}$ through atoms: $\Delta_f H = \Delta_{\mathrm{sub}}H(\mathrm{C}) + 2\,BE(\mathrm{H{-}H}) - 4\,BE(\mathrm{C{-}H})$ .
Ion in water: $\tfrac{1}{2}\mathrm{X_2(g) \to X(g) \to X^-(g) \to X^-(aq)}$ , so $\Delta H = \tfrac{1}{2}\Delta_{\mathrm{dis}}H + \Delta_{\mathrm{eg}}H + \Delta_{\mathrm{hyd}}H$ .
Heat of solution of a salt = lattice dissociation enthalpy (positive) + the hydration enthalpies of both ions.

Average bond enthalpy from formation enthalpies

\overline{BE} = \frac{\Delta_f H(\mathrm{A},g) + n\,\Delta_f H(\mathrm{B},g) - \Delta_f H(\mathrm{AB}_n,g)}{n}

Worked example

\Delta_f H

(kJ mol⁻¹):

\mathrm{NH_3}(g)

−46, N(g) +473, H(g) +218. Find the average N–H bond enthalpy.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 26 · Q53Moderate

Example 2 · Chemical Thermodynamics · Bond Enthalpy and Atomisation Cycles

The average

S - F

bond energy in

kJmol^{- 1}

SF_{6}

is (Rounded off to the nearest integer) [Given: The values of standard enthalpy of formation of

SF_{6}(\text{ }g),S(g)

and

F(g)

are - 1100, 275 and

80\text{ }kJ{\text{ }mol}^{- 1}

respectively.]

The sign on a lattice enthalpy

If lattice enthalpy is quoted as forming the lattice (−z), breaking it costs +z. With hydration enthalpies −x and −y, the heat of solution is

z - (x + y)

Per mole of compound

\Delta_f H

is per mole of the compound. For

\mathrm{A_2 + B_2 \to 2AB}

, the equation's

\Delta H

is twice

\Delta_f H

(AB); set up the bond balance with that doubled value.

Drill 4 more on average bond enthalpy and atomisation cycles

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Reaction enthalpy from bond enthalpies
Reaction enthalpy from bond enthalpies
$\Delta_r H = \sum BE_{\mathrm{broken}} - \sum BE_{\mathrm{formed}}$
Average bond enthalpy and atomisation cycles
Average bond enthalpy from formation enthalpies
$\overline{BE} = \frac{\Delta_f H(\mathrm{A},g) + n\,\Delta_f H(\mathrm{B},g) - \Delta_f H(\mathrm{AB}_n,g)}{n}$

Watch out for (4)

Formed minus broken
→ Reaction enthalpy from bond enthalpies
Bond enthalpies are for gases
→ Reaction enthalpy from bond enthalpies
The sign on a lattice enthalpy
→ Average bond enthalpy and atomisation cycles
Per mole of compound
→ Average bond enthalpy and atomisation cycles

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Related notes

Hess's law — formation and combustion enthalpies →