Bond Energies and Enthalpy Changes

Every reaction breaks some bonds and makes others, and the balance between the two decides whether energy is released or absorbed. If you know how much energy each bond is worth, you can estimate the enthalpy change of almost any gas-phase reaction without doing an experiment. This note explains why bond breaking is endothermic and bond making exothermic, the difference between exact and average bond energies, how to calculate $\Delta H_r$ from bond energies, why the answer is only approximate, and how to use bond energies inside Hess cycles to find unknown values. Expect at least one bond-energy calculation on Paper 2.

Bond energy

Bond energies are always positive. Breaking a bond means pulling two nuclei apart from the shared pair they are both attracted to; that always needs energy. The reverse process, forming a bond, releases exactly the same amount of energy.

Exact and average bond energies

For a diatomic molecule there is only one bond and nothing else in the molecule, so the bond energy can be measured exactly:

$$\ce{H2(g) -> 2H(g)} \qquad \Delta H = +436\ \text{kJ mol}^{-1}$$

For a bond that occurs in many different molecules, such as C–H, the energy needed to break it depends slightly on its environment: the C–H bond in methane is not quite the same as the C–H bond in ethanol or in chloroform. Even within methane, breaking the first, second, third and fourth C–H bonds needs different amounts of energy. So the data booklet gives an average bond energy, averaged over many compounds.

Data booklet values

bond	energy / kJ mol⁻¹	bond	energy / kJ mol⁻¹
H–H	436	C–H	410
Cl–Cl	242	C–C	350
Br–Br	193	C=C	610
H–Cl	431	C–O	360
H–Br	366	C=O	740
O=O	496	C=O in $\ce{CO2}$	805
N≡N	944	O–H	460
N–N	160	N–H	390
O–O	150	C–Cl	340

These are the values used throughout this note. In an exam, always use the values the question or the data booklet gives you.

Calculating ΔH from bond energies

Imagine the reaction happening in two steps: first, every bond in the reactants is broken to give separate gaseous atoms; then those atoms join to form every bond in the products. By Hess's law, the overall enthalpy change is the sum of the two steps.

If more energy is released making bonds than is used breaking them, $\Delta H_r$ is negative: exothermic.

Why bond-energy answers are approximate

Calculated values usually differ from accurate experimental values (found by calorimetry and Hess's law) by a few per cent. There are two main reasons:

1. Average bond energies are used for most bonds. The actual energy of a particular bond in a particular molecule may be higher or lower than the average.
2. Bond energies apply to gases. If any substance is a liquid or solid under the conditions of the real reaction (for example water in a combustion, or liquid ethanol), extra energy is involved in changing state (overcoming intermolecular forces), which the bond energy calculation ignores.

The second reason is why bond-energy values for combustion are less exothermic than $\Delta H_c^{\ominus}$: the standard value forms liquid water, releasing extra energy as the steam condenses.

Bond energies inside Hess cycles

Bond energies can also be one side of a Hess cycle. This lets you find an unknown bond energy from an enthalpy change of formation, or an enthalpy change of a reaction involving a liquid by adding the energy for a change of state. You may also need the enthalpy change of atomisation, $\Delta H_{at}$: the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state. For carbon (graphite), $\Delta H_{at} = +717\ \text{kJ mol}^{-1}$; for a diatomic gas, it is half the bond energy (for hydrogen, $\tfrac{1}{2} \times 436 = 218\ \text{kJ mol}^{-1}$).

Worked examples

Practice

Answers

1. The energy required to break one mole of a particular covalent bond in the gaseous state. The C–H bond occurs in many different compounds, and its strength varies slightly with its environment; the quoted value is the mean over many compounds.
2. Broken: $436 + 193 = 629$. Formed: $2 \times 366 = 732$. $\Delta H = 629 - 732 = -103\ \text{kJ mol}^{-1}$.
3. Broken: C–C $350$, $6 \times$ C–H $2460$, $3.5 \times$ O=O $1736$; total $4546$. Formed: $4 \times$ C=O (in $\ce{CO2}$) $3220$, $6 \times$ O–H $2760$; total $5980$. $\Delta H = 4546 - 5980 = -1434\ \text{kJ mol}^{-1}$.
4. Broken: C–H $410$, Cl–Cl $242$; total $652$. Formed: C–Cl $340$, H–Cl $431$; total $771$. $\Delta H = 652 - 771 = -119\ \text{kJ mol}^{-1}$.
5. Broken: C=C $610$, one O–H $460$; total $1070$. Formed: C–C $350$, C–H $410$, C–O $360$; total $1120$. (The other O–H and four C–H bonds are unchanged.) $\Delta H = 1070 - 1120 = -50\ \text{kJ mol}^{-1}$.
6. Broken: N–N $x$, $4 \times$ N–H $1560$, O=O $496$; total $x + 2056$. Formed: N≡N $944$, $4 \times$ O–H $1840$; total $2784$. $-568 = x + 2056 - 2784$, so $x = 160\ \text{kJ mol}^{-1}$.
7. The C–H and O–H bond energies are averages, not the exact values in these molecules. Under standard conditions water is a liquid; the bond energy calculation assumes gaseous water and so omits the energy released when steam condenses.
8. Broken: $2 \times (2\ \text{O–H} + 1\ \text{O–O}) = 2 \times (920 + 150) = 2140$. Formed: $4 \times$ O–H $1840$ and O=O $496$; total $2336$. $\Delta H = 2140 - 2336 = -196\ \text{kJ mol}^{-1}$.
9. $\ce{1/2N2(g) + 3/2H2(g) -> NH3(g)}$. Atomising the elements: $\tfrac{1}{2}(944) + \tfrac{3}{2}(436) = 472 + 654 = 1126\ \text{kJ}$. Hess: $1126 - 3E = -46$, so $3E = 1172$ and $E = 391\ \text{kJ mol}^{-1}$.
10. Gas-phase reaction $\ce{CH3OH(g) + 3/2O2(g) -> CO2(g) + 2H2O(g)}$: broken $3(410) + 360 + 460 + 1.5(496) = 2794$; formed $2(805) + 4(460) = 3450$; $\Delta H = -656\ \text{kJ mol}^{-1}$. For liquid methanol, first vaporise it ($+38$): $-656 + 38 = -618$. For liquid water, condense two moles ($2 \times -44 = -88$): $-618 - 88 = -706\ \text{kJ mol}^{-1}$. This is close to $-726$; the remaining difference is because the bond energies used are average values, not the exact values for the C–H, C–O and O–H bonds in methanol and water.