Bond energy calculations, and why the answer is only an estimate

8 minAS Level5.1.55.1.6

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can calculate $\Delta H_\text{r}$ from bond energies and explain why your answer may differ from the value in a data book.

The method in one line

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

Energy in on the left, energy out on the right. If the second total is bigger, $\Delta H$ comes out negative — exothermic.

The skill is not the subtraction. It is counting the bonds correctly, especially when a bond appears several times, and when a double or triple bond is involved.

Counting bonds from a structure

Work through the equation one species at a time and ask "which bonds are actually in this molecule?"

$\text{CH}_4$ : four C–H bonds.
$\text{O}_2$ : one O=O double bond. A double bond counts as one bond, but with the double-bond energy (498), not two singles.
$\text{CO}_2$ : two C=O double bonds (its structure is O=C=O).
$\text{H}_2\text{O}$ : two O–H bonds.
$\text{N}_2$ : one N≡N triple bond (945).

Multiply by the coefficient in the equation. In $\text{CH}_4 + 2\text{O}_2$ , there are two moles of O=O, so $2 \times 498$ .

Worked example: Worked example: methane combustion

Use bond energies to calculate $\Delta H$ for

$\text{CH}_4\text{(g)} + 2\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} + 2\text{H}_2\text{O(g)}$

Data: C–H $413$ , O=O $498$ , C=O $805$ , O–H $464$ (all $\text{kJ mol}^{-1}$ ).

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