Enthalpy of Reaction from Bond Energy | chemca

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Enthalpy of Reaction from Bond Energy

Calculating heat of reaction by analyzing bond breaking and bond formation.

By chemca Team • Updated Jan 2026

A chemical reaction essentially involves the breaking of old bonds in reactants and the formation of new bonds in products. By using Bond Bond Dissociation Energies (B.E.), we can estimate the standard enthalpy of a reaction ($\Delta H_{rxn}^\circ$).

1. The Basic Concept

• Bond Breaking: Requires energy input. This is an Endothermic process ($+$ve).
• Bond Formation: Releases energy. This is an Exothermic process ($-$ve).

Therefore, the net enthalpy change is the energy required to break all bonds in the reactants minus the energy released when bonds form in the products.

The Formula:

$$ \Delta H_{rxn} = \sum (B.E.)_{reactants} - \sum (B.E.)_{products} $$

Note: This is "Reactants minus Products", unlike most other thermodynamic formulas (like $\Delta H_f^\circ$) which are "Products minus Reactants".

2. Conditions and Limitations

Crucial Requirement: This method is strictly valid only when all reactants and products are in the Gaseous State.

If liquids or solids are involved, you must also include the enthalpies of vaporization, fusion, or sublimation to convert them into the gaseous state before applying bond energies.

Also, tabulated bond energies are often Average Bond Enthalpies. For example, the C-H bond energy in methane is slightly different from that in ethane, so calculations are approximations.

3. Solved Example

Problem: Calculate the enthalpy change for the formation of HCl.

$$ H_{2(g)} + Cl_{2(g)} \rightarrow 2HCl_{(g)} $$

Given Bond Energies:

• $H-H = 436 \, kJ/mol$
• $Cl-Cl = 242 \, kJ/mol$
• $H-Cl = 431 \, kJ/mol$

Solution:

1. Bonds Broken (Reactants):
1 mole of H-H + 1 mole of Cl-Cl
Energy = $436 + 242 = 678 \, kJ$

2. Bonds Formed (Products):
2 moles of H-Cl
Energy = $2 \times 431 = 862 \, kJ$

3. Calculate $\Delta H$:

$$ \Delta H = \Sigma B.E._R - \Sigma B.E._P $$ $$ \Delta H = 678 - 862 = -184 \, kJ $$

Since $\Delta H$ is negative, the reaction is Exothermic.

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