What is resonance energy and how does it relate to stability?

Resonance energy is the extra stabilization a molecule gains from electron delocalization beyond what any single Kekule structure predicts. It equals the difference between experimental and bond-energy-calculated enthalpies of formation. Larger resonance energy means greater stability - benzene has about -152 kJ per mole.

Hess’s Law and Bond Energy

Q: Why can't the enthalpy of formation of CO be measured directly?

Burning carbon in oxygen almost always gives complete combustion to CO2, not CO. Instead we measure combustion of C to CO2 and combustion of CO to CO2, then subtract using Hess's law to obtain the enthalpy of formation of CO.

Q: What is the difference between bond energy and bond dissociation energy?

Bond dissociation energy is for a specific bond in a specific molecule. Bond energy is the average across many molecules. They coincide for diatomics but differ for polyatomics like water because breaking successive bonds changes the environment.

Q: Why must all species be in the gas phase for the bond energy method?

Bond energies are defined for gas-phase species. If reactants or products are condensed phases, you must add vaporization or sublimation enthalpies for reactants and subtract them for products to use bond energies correctly.

Q: In the Born-Haber cycle, why does electron affinity enter with a negative sign?

Electron affinity is usually tabulated as positive (energy released when a gas atom gains an electron). But the cycle tracks enthalpy added to the system, so the actual enthalpy change of that step is minus EA.

ChemistryChemical ThermodynamicsFor JEE aspirants

Hess's law of constant heat summation is a direct consequence of enthalpy being a state function: the total enthalpy change of a reaction is the same whether the reaction proceeds in one step or several. This lets us calculate enthalpy changes that are hard to measure directly by cleverly combining known reactions. Bond energy (or bond enthalpy) is the average energy required to break one mole of a specific type of bond in the gaseous state. Together, Hess's law and bond energies let us predict enthalpies of formation, reaction, and resonance stabilization. This concept also covers the Born-Haber cycle for the lattice energy of ionic crystals.

Key Formulas - Quick Reference

Hess's law: $Δ H$ of a reaction depends only on initial and final states, not on the path. If a reaction is broken into steps, $Δ H_{overall} = Δ H_{1} + Δ H_{2} + \dots$
Enthalpy of reaction from formation enthalpies: $Δ H_{r x n}^{\circ} = \sum ν_{i} Δ H_{f}^{\circ} (products) - \sum ν_{i} Δ H_{f}^{\circ} (reactants)$
Enthalpy of reaction from bond enthalpies: $Δ H = \sum (BE of bonds broken) - \sum (BE of bonds formed)$
Bond enthalpy for diatomics: $Δ H_{B E} = Δ H_{dissociation}$
Born-Haber cycle for MX: $Δ H_{f}^{\circ} (MX) = Δ H_{s u b} + \frac{1}{2} Δ H_{d i ss} + IE + EA + Δ H_{l a tt i ce}$
Resonance energy: $Δ H_{resonance} = Δ H_{f, expt} - Δ H_{f, calc from bond energies}$

1. Hess's law of constant heat summation

Statement. The heat absorbed or released in a chemical process is the same whether the process takes place in one step or several steps.

If a reaction

A \to B

has enthalpy change

Δ H

, and the same net change can be achieved by intermediate steps

A \to C \to D \to B

with enthalpy changes

Δ H_{1}, Δ H_{2}, Δ H_{3}

, then

Δ H = Δ H_{1} + Δ H_{2} + Δ H_{3}

Why Hess's law is true

Enthalpy is a state function: its value depends only on the state of the system, not on the path. So $Δ H$ for a reaction depends only on initial reactants and final products - never on the route between them. Hess's law is a direct corollary of the first law of thermodynamics.

Using Hess's law: rules for manipulating thermochemical equations

Multiplying/dividing a chemical equation by any factor multiplies/divides its $Δ H$ by the same factor.
Reversing a chemical equation changes the sign of $Δ H$ (but not its magnitude).
Adding chemical equations adds the corresponding $Δ H$ values.

Solved Example 1

Calculate the enthalpy of formation of CO(g) from the following data:
(i)

C(s) + O_{2} (g) \to CO_{2} (g), Δ H_{1} = - 393 kJ mol^{- 1}

(ii)

CO (g) + \frac{1}{2} O_{2} (g) \to CO_{2} (g), Δ H_{2} = - 283 kJ mol^{- 1}

Solution:

Target: $C(s) + \frac{1}{2} O_{2} (g) \to CO (g), Δ H_{f} = ?$

Subtract equation (ii) from equation (i):

$C(s) + O_{2} (g) - [CO (g) + \frac{1}{2} O_{2} (g)] \to CO_{2} (g) - CO_{2} (g)$

which simplifies to $C(s) + \frac{1}{2} O_{2} (g) \to CO (g)$ .

$Δ H_{f} = Δ H_{1} - Δ H_{2} = - 393 - (- 283) = - 110 kJ mol^{- 1}$ .

2. Applications of Hess's law

(a) Enthalpies of formation of hard-to-synthesize compounds

Many compounds cannot be synthesized directly from their elements (like C $_{6}$ H $_{6}$ , C $_{2}$ H $_{6}$ , CO), so their $Δ H_{f}^{\circ}$ cannot be measured directly. Hess's law lets us calculate them from combustion data or other measurable reactions.

(b) Enthalpies of reactions

Once we have $Δ H_{f}^{\circ}$ tabulated for all relevant compounds, any reaction's $Δ H^{\circ}$ is computed as:

Δ H_{r x n}^{\circ} = \sum ν_{i} Δ H_{f}^{\circ} (products) - \sum ν_{i} Δ H_{f}^{\circ} (reactants)

Solved Example 2

Calculate the enthalpy of reaction at

2 5^{\circ} C

F_{2} O (g) + H_{2} O (g) \to O_{2} (g) + 2 HF (g)

. Given

Δ H_{f}^{\circ}

: F

_{2}

(g) = + 5.5 kcal

, H

_{2}

(g) = - 57 kcal

, HF

(g) = - 64 kcal

Solution:

$Δ H^{\circ} = [Δ H_{f}^{\circ} (O_{2}) + 2Δ H_{f}^{\circ} (HF)] - [Δ H_{f}^{\circ} (F_{2} O) + Δ H_{f}^{\circ} (H_{2} O)]$

$= [0 + 2 (- 64)] - [(+ 5.5) + (- 57)]$

$= - 128 - (- 51.5) = - 76.5 kcal$ .

(c) Enthalpies of transitions and phase changes

Enthalpies of transitions between allotropes (graphite $\to$ diamond, monoclinic $\to$ rhombic sulphur) can be calculated by subtracting combustion enthalpies of the two forms.

(d) Bond energies and lattice energies

Hess's law underlies both bond energy calculations (below) and the Born-Haber cycle (Section 5).

3. Bond energy (bond enthalpy)

The bond energy is the enthalpy required to break one mole of a specified type of bond in a molecule in the gas phase, giving the atoms (or radicals) as free gaseous species.

For a diatomic molecule: bond energy = bond dissociation energy. Example:

H_{2} (g) \to 2 H (g), Δ H_{B E} (H-H) = + 436 kJ mol^{- 1}

Bond energy vs bond dissociation enthalpy for polyatomics

In a polyatomic molecule like $H_{2} O$ , the two O-H bonds have slightly different dissociation enthalpies because the "second" O-H is being broken in a different chemical environment (the OH radical, not H $_{2}$ O):

H_{2} O (g) \to H (g) + OH (g), Δ H_{1} = 501.9 kJ mol^{- 1}

OH (g) \to O (g) + H (g), Δ H_{2} = 423.4 kJ mol^{- 1}

The bond energy of O-H is the average:

$Δ H_{B E} (O-H) = \frac{501.9 + 423.4}{2} = 462.6 kJ mol^{- 1}$ .

Trends in bond energy. Higher bond order

\to

higher bond energy:

C-C \approx 348

C=C \approx 615

C \equiv C \approx 812 kJ mol^{- 1}

. Shorter bonds are generally stronger.

4. Enthalpy of reaction from bond energies

Δ H_{r x n} = \sum (BE of bonds broken in reactants) - \sum (BE of bonds formed in products)

This applies only when all species involved are in the gas phase. If reactants or products are liquid/solid, additional atomization/vaporization steps must be included.

Solved Example 3

Calculate the bond energy of H-Cl given: bond energy of H-H =

430 kJ mol^{- 1}

, Cl-Cl =

242 kJ mol^{- 1}

, and

Δ H_{f}^{\circ} (HCl, g) = - 91 kJ mol^{- 1}

Solution:

Formation reaction: $\frac{1}{2} H_{2} (g) + \frac{1}{2} Cl_{2} (g) \to HCl (g), Δ H_{f}^{\circ} = - 91 kJ mol^{- 1}$ .

Using bond energies: $Δ H_{f}^{\circ} = \frac{1}{2} (BE H-H) + \frac{1}{2} (BE Cl-Cl) - (BE H-Cl)$

$- 91 = \frac{1}{2} (430) + \frac{1}{2} (242) - BE(H-Cl)$

$- 91 = 215 + 121 - BE(H-Cl) = 336 - BE(H-Cl)$

$BE(H-Cl) = 336 + 91 = 427 kJ mol^{- 1}$ .

Solved Example 4

Given the bond energies N

\equiv

N =

945

, H-H =

436

, and N-H =

391 kJ mol^{- 1}

, calculate the enthalpy of the reaction:

N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g)

Solution:

Bonds broken (in reactants): 1 N $\equiv$ N + 3 H-H.

Bonds formed (in products): 6 N-H (each NH $_{3}$ has 3 N-H, and there are 2 molecules).

$Δ H = [945 + 3 (436)] - [6 (391)]$

$= [945 + 1308] - [2346] = 2253 - 2346 = - 93 kJ$ .

5. Born-Haber cycle for lattice energy

Lattice energy ( $Δ H_{L}$ ) is the enthalpy change when one mole of an ionic solid is formed from its gaseous ions. It cannot be measured directly, but can be calculated using the Born-Haber cycle - a Hess's law cycle that connects lattice energy to measurable quantities.

Figure: Born-Haber cycle for MX. Each numbered step is a measurable enthalpy change: steps 1-3 are endothermic (orange, upward), steps 4-5 are exothermic (green, downward). By Hess's law they sum to the overall standard enthalpy of formation (purple, dashed):

Δ H_{f}^{\circ} = Δ H_{1} + \frac{1}{2} Δ H_{2} + Δ H_{3} + Δ H_{4} + Δ H_{5}

Steps of the Born-Haber cycle for MX(s)

Sublimation of metal: $M (s) \to M (g), Δ H_{1} = Δ H_{s u b}$ .
Dissociation of $\frac{1}{2}$ mole X $_{2}$ : $\frac{1}{2} X_{2} (g) \to X (g), Δ H_{2} = \frac{1}{2} Δ H_{d i ss}$ .
Ionization of gaseous metal atom: $M (g) \to M^{+} (g) + e^{-}, Δ H_{3} = IE$ .
Electron affinity of X: $X (g) + e^{-} \to X^{-} (g), Δ H_{4} = - EA$ (sign convention: EA reported as positive when energy released).
Lattice formation: $M^{+} (g) + X^{-} (g) \to MX (s), Δ H_{5} = Δ H_{L}$ (negative, energy released).

By Hess's law:

Δ H_{f}^{\circ} (MX) = Δ H_{s u b} + \frac{1}{2} Δ H_{d i ss} + IE - EA + Δ H_{L}

Rearranging gives lattice energy in terms of the other (measurable) quantities.

Solved Example 5

Write the Born-Haber expression for the lattice energy of CaBr

_{2}

Solution:

Note that Ca is a group-2 metal: two ionization energies (IE $_{1}$ , IE $_{2}$ ) needed. Br $_{2}$ is diatomic: dissociation gives 2Br atoms.

Steps:

(1) Ca $(s) \to$ Ca $(g)$ : $Δ H_{s u b} = S$

(2) Br $_{2} (g) \to 2$ Br $(g)$ : $Δ H_{d i ss} = D$

(3) Ca $(g) \to$ Ca $^{+} (g) + e^{-}$ : IE $_{1}$

(4) Ca $^{+} (g) \to$ Ca $^{2 +} (g) + e^{-}$ : IE $_{2}$

(5) $2$ Br $(g) + 2 e^{-} \to 2$ Br $^{-} (g)$ : $- 2 EA$

(6) Ca $^{2 +} (g) + 2$ Br $^{-} (g) \to$ CaBr $_{2} (s)$ : $Δ H_{L} = U$

By Hess's law: $Δ H_{f}^{\circ} (CaBr_{2}) = S + D + IE_{1} + IE_{2} - 2 EA + U$ .

6. Determination of resonance energy

Some molecules like benzene are stabilized by resonance - electron delocalization gives them extra stability beyond what individual bond energies predict. The resonance energy is the difference between the calculated enthalpy of formation (using bond energies for a specific Kekule structure) and the experimentally observed value.

Δ H_{resonance} = Δ H_{f, expt} - Δ H_{f, calc}

(or equivalently,

Δ H_{resonance} = Δ H_{combustion,calc} - Δ H_{combustion,expt}

)

Solved Example 6

Calculate the resonance energy of benzene. Given: standard enthalpy of formation of cyclohexane

(ℓ) = - 156 kJ mol^{- 1}

, standard enthalpy of formation of benzene

(ℓ) = + 49 kJ mol^{- 1}

, standard enthalpy of hydrogenation of cyclohexene

(ℓ) = - 119 kJ mol^{- 1}

Solution:

Enthalpy of formation of 3 C=C bonds (in benzene ring, ignoring resonance):

$Δ H_{3 C=C, calc} = Δ H_{f} (benzene) - Δ H_{f} (cyclohexane) = 49 - (- 156) = + 205 kJ$

So observed energy of forming 3 C=C from 3 C-C is $+ 205 kJ$ .

From the hydrogenation of cyclohexene: adding H $_{2}$ across one C=C releases $119 kJ$ (so formation of one C=C from one C-C+H $_{2}$ requires $+ 119 kJ$ ). Extrapolating: 3 C=C bonds would require $3 \times 119 = + 357 kJ$ if resonance were absent.

$Δ H_{resonance} = Δ H_{expt} - Δ H_{calc} = 205 - 357 = - 152 kJ mol^{- 1}$ .

The negative value means benzene is $152 kJ mol^{- 1}$ more stable than the (hypothetical) Kekule structure.

7. Bond enthalpies and enthalpies of formation

For any organic molecule in the gas phase, its enthalpy of formation from elements can be estimated by:

Atomize the elemental reactants: e.g. C(graphite) $\to$ C(g), $Δ H_{s u b, C}$ ; and $\frac{1}{2}$ H $_{2} (g) \to$ H $(g)$ , $\frac{1}{2} Δ H_{d i ss, H_{2}}$ .
Form the target molecule from these gaseous atoms - releasing energy equal to the sum of all bond energies formed.

Δ H_{f}^{\circ} (compound, g) = \sum (atomization enthalpies of elements) - \sum (bond energies of bonds in compound)

Solved Example 7

Calculate the heat of formation of benzene from the following data, assuming no resonance stabilization. Bond energies: C-C =

83 kcal

, C=C =

140 kcal

, C-H =

99 kcal

. Heat of atomization: C =

170.9 kcal

, H =

52.1 kcal

Solution:

Target: $6 C(s) + 3 H_{2} (g) \to C_{6} H_{6} (g)$ .

Atomization: $6 \times 170.9 + 6 \times 52.1 = 1025.4 + 312.6 = 1338 kcal$ absorbed.

Benzene (Kekule) has 3 C-C + 3 C=C + 6 C-H bonds:

Energy released on bond formation: $3 (83) + 3 (140) + 6 (99) = 249 + 420 + 594 = 1263 kcal$ .

$Δ H_{f}^{\circ} (benzene, calc) = 1338 - 1263 = + 75 kcal$ (endothermic).

Note this ignores resonance; experimental value is much less positive because of resonance stabilization.

8. Common Mistakes to Avoid

Watch out

Applying bond energy method to reactions in solution or condensed phases. Bond energies are defined for gas-phase species only. If reactants/products are liquid or solid, include atomization/vaporization enthalpies.
Confusing bond dissociation energy with bond energy. Bond dissociation energy is for a specific bond in a specific molecule; bond energy is an average across many molecules. For diatomics they coincide.
Sign errors when reversing equations. If you reverse a reaction, flip the sign of $Δ H$ . Do this before adding equations.
Not multiplying $Δ H$ when scaling coefficients. If you double an equation to balance a Hess's law scheme, its $Δ H$ also doubles.
Wrong sign convention for electron affinity in Born-Haber. EA is usually tabulated as positive (energy released when an electron is added). In the cycle, the step $X (g) + e^{-} \to X^{-} (g)$ has $Δ H = - EA$ .
Forgetting the second (or higher) ionization energy for divalent cations. For MgO or CaBr $_{2}$ , you need IE $_{1}$ + IE $_{2}$ , not just IE $_{1}$ .
Assuming all C-H bonds have the same energy in a specific molecule. The tabulated $416$ or $413 kJ mol^{- 1}$ is an average. Individual C-H bond dissociation values differ by tens of kJ.

Frequently Asked Questions

Q1. What is Hess's law and why is it true?

Hess's law states that the total enthalpy change of a reaction is the same whether the reaction proceeds in one step or many. It is true because enthalpy is a state function - it depends only on the initial and final states, not on the path taken. This is a direct consequence of the first law of thermodynamics.

Q2. Why can't the enthalpy of formation of CO be measured directly?

When you burn carbon in oxygen, the product is almost always CO $_{2}$ (complete combustion), not CO. To get pure CO you would need very carefully controlled conditions that are hard to reproduce. Instead we measure combustion of C to CO $_{2}$ and combustion of CO to CO $_{2}$ (both easy) and subtract to get formation of CO - a classic Hess's law application.

Q3. What is the difference between bond energy and bond dissociation energy?

Bond dissociation energy is the exact energy needed to break a specific bond in a specific molecule, giving definite fragments. Bond energy is the average value of dissociation energies for a given type of bond across many molecules. For diatomics like $H_{2}$ or HCl, the two are identical. For polyatomics like H $_{2}$ O, they differ because breaking successive bonds changes the environment.

Q4. What is lattice energy and why can't it be measured directly?

Lattice energy is the enthalpy change when one mole of ionic solid is formed from gaseous ions. It cannot be measured directly because we cannot generate free gaseous ions of common metals and non-metals in a lab and then let them combine into a crystal in a controlled way. The Born-Haber cycle uses Hess's law to obtain lattice energy from measurable quantities like ionization energy, electron affinity, and enthalpy of formation.

Q5. Why is bond energy called an average value?

Because the exact energy to break a bond depends on the surrounding chemical environment. A C-H bond in methane, ethane, benzene, and acetylene all have slightly different dissociation energies. The tabulated bond energy is the mean value from many such molecules and is used for estimation - not exact predictions.

Q6. What is resonance energy and how is it related to stability?

Resonance energy is the extra stabilization a molecule gains from electron delocalization (resonance) beyond what any single Kekule/Lewis structure would predict. It equals the difference between the experimental and calculated (bond-energy-based) enthalpies of formation. Larger resonance energy means greater stability - benzene, for instance, has a resonance energy of about $- 152 kJ mol^{- 1}$ .

Q7. Why must all species be in the gas phase for the bond energy method?

Bond energies are defined as the enthalpy to break bonds in the gas phase, giving free atoms/radicals. If reactants or products are liquid or solid, additional energy is involved in vaporization or solidification. To use bond energies correctly for a condensed-phase reaction, you must add enthalpies of vaporization/sublimation for reactants and subtract them for products.

Q8. In the Born-Haber cycle, why does electron affinity enter with a negative sign?

Electron affinity is usually tabulated as a positive number (the amount of energy released when a gas-phase atom accepts an electron). But in the Born-Haber cycle we track enthalpy added to the system at each step. Since the electron affinity step actually releases energy, the enthalpy change of that step is $-$ EA (a negative number).

Previous year questions on Hess’s Law and Bond Energy

16 questions from past papers, each with a step-by-step solution.

Show all 16 questions

Ready to master Chemical Thermodynamics?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test