The enthalpy of atomisation for the reaction CH4(g) → C(g) + 4H(g) is 1665 kJ mol^-1. What is the bond energy

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Q. The enthalpy of atomisation for the reaction CH4(g) \to C(g) + 4H(g) is 1665 kJ mol^-1. What is the bond energy of C-H bond?

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Concept understanding — Bond Enthalpy

Bond Enthalpy: The Energy Cost of Breaking a Bond

Think of a chemical bond as a spring holding two atoms together. To pull the atoms apart, you have to do work — you have to put energy in. That energy, per mole of bonds broken, is the bond enthalpy. The stronger the bond, the more energy you need to supply, and the larger the bond enthalpy.

The intuition is simple: breaking bonds costs energy; forming bonds releases energy. A reaction is exothermic if the energy released in forming new bonds is greater than the energy consumed in breaking old ones.

The Precise Definition

Bond enthalpy (symbol: $\Delta H_{\text{bond}}$ or $B.E.$ ) is defined under a specific set of conditions:

The average enthalpy change when one mole of a particular covalent bond is broken, with all species in the gaseous state.

Three key points are packed into that sentence:

One mole of bonds — we measure the energy for Avogadro's number of bonds, not just one.
Gaseous state — the atoms or molecules must be in the gas phase. This removes complications from intermolecular forces (like hydrogen bonding in liquid water) and lets us isolate the bond energy itself.
Average — in a molecule like water ( $H_2O$ ), the two O–H bonds are not identical in energy. The first O–H bond in $H_2O$ requires about 502 kJ/mol to break, but the second (in the remaining OH radical) requires about 427 kJ/mol. So we report the average bond enthalpy for O–H: roughly 464 kJ/mol.

Important

Bond enthalpy is always positive — it is the energy absorbed to break a bond. Bond formation releases the same amount of energy (negative enthalpy change).

How It's Used: Estimating Reaction Enthalpies

You can estimate the enthalpy change of a reaction ( $\Delta H_{\text{rxn}}$ ) using bond enthalpies:

\Delta H_{\text{rxn}} = \sum \text{(bond enthalpies of bonds broken)} - \sum \text{(bond enthalpies of bonds formed)}

The logic: you put energy in to break bonds (positive), and you get energy out when bonds form (negative). So:

\Delta H_{\text{rxn}} = \text{Energy in} - \text{Energy out}

Watch out

This method gives an estimate, not an exact value. Bond enthalpies are averages taken from many different molecules, so they don't account for the exact molecular environment. For precise work, use standard enthalpies of formation.

Example: Combustion of Methane

Consider: $CH_4(g) + 2O_2(g) \rightarrow CO_2(g) + 2H_2O(g)$

Bonds broken (energy in):

4 C–H bonds: $4 \times 413 = 1652$ kJ/mol
2 O=O bonds: $2 \times 498 = 996$ kJ/mol
Total in: $1652 + 996 = 2648$ kJ/mol

Bonds formed (energy out):

2 C=O bonds: $2 \times 799 = 1598$ kJ/mol
4 O–H bonds: $4 \times 464 = 1856$ kJ/mol
Total out: $1598 + 1856 = 3454$ kJ/mol

\Delta H_{\text{rxn}} = 2648 - 3454 = -806 \text{ kJ/mol}

The negative sign tells you the reaction is exothermic — more energy is released in forming bonds than was consumed in breaking them. …

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The enthalpy of atomisation of methane represents the total energy required to break all four C-H bonds. Dividing this total energy by the number of bonds gives the average C-H bond energy, which is $\boxed{416.25 \text{ kJ mol}^{-1}}$ .

When we talk about the enthalpy of atomisation of a molecule, we are referring to the energy required to break all the chemical bonds within one mole of that substance to produce individual gaseous atoms. This process is always endothermic, meaning it requires energy input.

For a molecule like methane ( $\text{CH}_4$ ), which consists of one carbon atom bonded to four hydrogen atoms, the enthalpy of atomisation is the sum of the energies required to break each of these four C-H bonds. Since all the bonds are of the same type (C-H), we can consider the enthalpy of atomisation as four times the average C-H bond energy.

Important

The enthalpy of atomisation ( $\Delta_a H^\circ$ ) of a polyatomic molecule is the sum of the bond dissociation energies of all the bonds present in one mole of the molecule.

For a molecule with $n$ identical bonds, the average bond energy is $\frac{\Delta_a H^\circ}{n}$ .

Let's break down the calculation:

Understand the given reaction:

The reaction provided is:

$\text{CH}_4(\text{g}) \rightarrow \text{C}(\text{g}) + 4\text{H}(\text{g})$

This reaction explicitly shows one mole of gaseous methane breaking down into one mole of gaseous carbon atoms and four moles of gaseous hydrogen atoms. This is precisely the definition of the enthalpy of atomisation of methane.

The enthalpy change for this reaction is given as $1665 \text{ kJ mol}^{-1}$ .
Identify the bonds being broken:

In one molecule of $\text{CH}_4$ , there are four C-H single bonds. To atomise one mole of $\text{CH}_4$ , all four moles of C-H bonds must be broken.
Relate enthalpy of atomisation to bond energy:

The total energy required to break all four C-H bonds in one mole of methane is $1665 \text{ kJ mol}^{-1}$ .

If we assume that all C-H bonds in methane have the same average energy, then the enthalpy of atomisation is simply four times the average C-H bond energy. …

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Q.The enthalpy of atomisation for the reaction CH4(g) → C(g) + 4H(g) is 1665 kJ mol^-1. What is the bond energy of C-H bond?

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