ΔH = Σ(bond enthalpies of broken) - Σ(bond enthalpies of formed). Example: H₂ + Cl₂ → 2HCl.

Calculate ΔH Using Hess's Law — Bond Enthalpy Method

Question

Calculate the enthalpy change (ΔH) for the reaction:

H_2(g) + Cl_2(g) \rightarrow 2HCl(g)

Given bond enthalpies: H–H = 436 kJ/mol, Cl–Cl = 242 kJ/mol, H–Cl = 431 kJ/mol.

Solution — Step by Step

In any reaction, reactant bonds break (energy absorbed) and product bonds form (energy released). For this reaction, we break one H–H bond and one Cl–Cl bond, and form two H–Cl bonds.

This is the entire logic behind the bond enthalpy method — it’s just an energy ledger.

\Delta H_{\text{broken}} = \text{BE}_{H-H} + \text{BE}_{Cl-Cl}

= 436 + 242 = 678 \text{ kJ/mol}

We’re putting energy in to break these bonds, so this is positive.

\Delta H_{\text{formed}} = 2 \times \text{BE}_{H-Cl} = 2 \times 431 = 862 \text{ kJ/mol}

Two moles of HCl are formed, so we multiply by 2. Energy is released when bonds form.

\Delta H = \Sigma \text{BE(broken)} - \Sigma \text{BE(formed)}

\Delta H = 678 - 862 = -184 \text{ kJ/mol}

The negative sign tells us the reaction is exothermic — more energy is released forming H–Cl bonds than is absorbed breaking H–H and Cl–Cl bonds. Makes sense: HCl is a stable molecule.

ΔH = −184 kJ/mol

Why This Works

Every chemical bond stores a certain amount of energy — the bond enthalpy. When a bond breaks, that stored energy goes back into the surroundings (as kinetic energy, heat, etc.), and when a bond forms, energy is released.

The bond enthalpy method is essentially Hess’s Law applied at the molecular level. We imagine the reaction happening in two theoretical steps: first, atomise all reactants (break every bond), then build all products from scratch (form every bond). The net ΔH is the difference.

This method gives approximate values because bond enthalpies are averages. For example, the C–H bond enthalpy in CH₄ is slightly different from C–H in CHCl₃. For exact values, use formation enthalpies (ΔH°f). In JEE, if they say “using bond enthalpies”, this method is what they want.

Alternative Method — Using Hess’s Law with Formation Enthalpies

If bond enthalpies aren’t given but standard enthalpies of formation are, we use:

\Delta H_{\text{rxn}} = \Sigma \Delta H_f^\circ(\text{products}) - \Sigma \Delta H_f^\circ(\text{reactants})

For $H_2 + Cl_2 \rightarrow 2HCl$ :

$\Delta H_f^\circ(HCl) = -92.3$ kJ/mol
$H_2$ and $Cl_2$ are elements in standard state, so $\Delta H_f^\circ = 0$

\Delta H = 2(-92.3) - [0 + 0] = -184.6 \text{ kJ/mol}

Notice this matches our bond enthalpy answer almost exactly. When the question says “bond enthalpy method”, use the first approach. When it gives $\Delta H_f^\circ$ values, use this one.

Common Mistake

The most common error: students subtract in the wrong order and write $\Delta H = \Sigma\text{BE(formed)} - \Sigma\text{BE(broken)}$ , flipping the sign. They get +184 kJ/mol and label it exothermic anyway, which is doubly wrong.

Remember the logic: energy in (breaking) minus energy out (forming). If forming releases more than breaking absorbs, the difference is negative — exothermic. Keep the formula as broken minus formed.

Also watch the stoichiometry: two moles of HCl means 2 × BE(H–Cl). Missing the coefficient “2” here is a classic one-mark drop in CBSE boards.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Hess’s Law with Formation Enthalpies

Common Mistake

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