Bond enthalpies — estimating ΔH from bonds broken and formed — Chemistry

Before this

This leads to

Calorimetry
Collision theory
The Born–Haber cycle · coming soon

A molecule nobody has measured

Hess's law lets you find ΔH for almost any reaction, but only if someone has already measured the enthalpies of formation or combustion of every substance in it. Now suppose a chemist designs a new fuel molecule that has never been made. There is no table entry for it. Is there any way to estimate how much energy it would release?

There is, and it needs only about twenty numbers. Every molecule, however new, is built from a small number of kinds of bond: C–H, C–C, O–H, C=O and a few others. If we know roughly how much energy each kind of bond holds, we can estimate the energy change of any reaction between gases by counting bonds.

The last lecture promised another route between the same start and finish, by breaking and making bonds. This lecture builds that route, shows how well it works, and shows exactly where its small errors come from.

Why breaking a bond always takes energy

A covalent bond is a shared pair of electrons, attracted to the nuclei of both atoms. To pull the two atoms apart, you must work against that attraction. So breaking a bond always takes in energy from the surroundings. It is endothermic.

Run the film backwards. When two separate atoms come together and form a bond, the attraction pulls them in, and energy is released to the surroundings. Forming a bond is exothermic. By conservation of energy, forming a bond releases exactly as much energy as breaking the same bond takes in.

Stop and think

A student writes: "Burning methane releases energy because the bonds in methane break." What is wrong with this?

Show the answer

Breaking bonds never releases energy. It always takes energy in. Burning methane releases energy because of the bonds that form afterwards, in carbon dioxide and water. Those new bonds release more energy than was needed to break the old ones in methane and oxygen. The overall reaction is exothermic only because the "making" part is bigger than the "breaking" part.

Mean bond enthalpy

The energy needed to break one particular bond in one particular molecule is its bond dissociation enthalpy. For the first C–H bond in methane, $CH_{4} (g) \to CH_{3} (g) + H (g)$ , it is about 439 kJ/mol.

But the four C–H bonds in methane do not break equally easily, one after another. Removing each hydrogen in turn takes about 439, 460, 423 and 339 kJ/mol. Their average is about 415 kJ/mol. And a C–H bond in ethane or in ethanol has a slightly different value again, because its neighbouring atoms are different.

So data tables give an average over many molecules. This is the mean bond enthalpy: the average energy needed to break one mole of a particular type of bond, with everything in the gas state. Some books call it the average bond enthalpy, or simply the bond energy. Because it is always energy taken in, it is always positive.

Water shows the same thing. Pulling the first hydrogen off a water molecule, $H_{2} O (g) \to OH (g) + H (g)$ , takes about 493 kJ/mol. Pulling the second one off the OH that is left takes only about 424 kJ/mol. The two bonds looked identical in the molecule, but once one has gone, the other is in a different situation. Their average is about 459 kJ/mol. The table's O–H value, 467, is higher, because it also averages O–H bonds in alcohols and other molecules. No single number can be right for every molecule. That is the whole reason the method is an estimate.

Bond	Mean bond enthalpy (kJ/mol)	Bond	Mean bond enthalpy (kJ/mol)
H–H	436	O=O	498
C–H	416	C=O in CO₂	803
C–C	346	C=O in other molecules	695
C=C	598	N≡N	946
O–H	467	N–H	391
H–Cl	431	Cl–Cl	242

These values come from one table (LibreTexts, from Cotton, Wilkinson and Gaus, 1995). Other books differ by a few kJ/mol, so in an examination always use the values the question gives. Notice that the C=O bond in carbon dioxide has its own value. It is much stronger there than in most other molecules, so a single average would be poor for it.

The route through gaseous atoms

Now the promise from the Hess's law lecture. Take any reaction between gases. Instead of going straight from reactants to products, imagine a two-step route. First, break every bond in the reactants, so that you have a cloud of separate gaseous atoms. Second, let those atoms join up into the products, forming every bond in them.

The atoms are the same at the top of the route, whichever way you go, because a reaction only rearranges atoms. So Hess's law applies: the direct ΔH equals the sum of the two steps. Step one costs the bond enthalpies of every bond broken. Step two releases the bond enthalpies of every bond formed. Walking along the route, we keep the sign of step one, and step two goes down, so its sign is negative:

Δ H \approx \sum (bonds broken) - \sum (bonds formed)

The "≈" matters. The formula is exact for the route through atoms. The approximation comes only from using mean values instead of the true bond enthalpies in these particular molecules.

Burning methane by way of gaseous atoms. Breaking every bond takes in about 2660 kJ/mol. Forming every bond in the products releases about 3474 kJ/mol. The overall change is about 2660 − 3474 = −814 kJ/mol. The products end up lower because more energy is released than is taken in.

Estimating ΔH from bond enthalpies

For a reaction in which every substance is a gas, $Δ H \approx \sum (bonds broken) - \sum (bonds formed)$ , using mean bond enthalpies.

If the bonds formed release more energy than the bonds broken take in, ΔH is negative, and the reaction is exothermic.

That is it. That is the whole idea. A reaction's energy change is the difference between what it costs to take the old bonds apart and what the new bonds give back.

Count every bond: draw the molecules

Almost every mistake in these calculations is a counting mistake. The safest method is to draw the full structural formula of every molecule, showing every bond, and to multiply by the numbers in the balanced equation.

The full structural formulae for burning methane,

CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (g)

. Broken: 4 C–H and 2 O=O. Formed: 2 C=O and 4 O–H (two in each water molecule).

Worked Example 1 — Hydrogen chloride

Estimate ΔH for $H_{2} (g) + Cl_{2} (g) \to 2 HCl (g)$ .

Broken: 1 H–H and 1 Cl–Cl, so $436 + 242 = 678$ kJ.

Formed: 2 H–Cl, so $2 \times 431 = 862$ kJ.

$Δ H \approx 678 - 862 = - 184$ kJ/mol. Exothermic.

Check against the data book: ΔH_f^⦵[HCl(g)] = −92.3 kJ/mol, so for 2 mol of HCl the value is −184.6 kJ/mol. The estimate is almost exact, because H–H, Cl–Cl and H–Cl exist in only one molecule each, so their "mean" values are the true values.

Worked Example 2 — Methane, and what to compare it with

Estimate ΔH for $CH_{4} (g) + 2 O_{2} (g) \to CO_{2} (g) + 2 H_{2} O (g)$ .

Broken: $4 \times 416 + 2 \times 498 = 1664 + 996 = 2660$ kJ.

Formed: $2 \times 803 + 4 \times 467 = 1606 + 1868 = 3474$ kJ.

$Δ H \approx 2660 - 3474 = - 814$ kJ/mol.

Is that good? The predict box below asks you to decide before you read on.

Predict, then reveal — how good is −814?

The bond method gives ΔH ≈ −814 kJ/mol for burning methane. A data book gives the standard enthalpy of combustion of methane as −890.3 kJ/mol. How good is the bond estimate? Choose before you read on.

The bond estimate for burning methane is −814 kJ/mol, and the data book gives −890.3 kJ/mol. The data-book value is for liquid water. With water as a gas, the true value is −890.3 + 2 × 44.0 = −802.3 kJ/mol, so the bond estimate is only about 1.5% away.

The comparison must be fair. The data-book value ΔH_c^⦵ = −890.3 kJ/mol is for liquid water, because water is a liquid in its standard state. The bond method can only describe gases, so it gives the value for water vapour. The Hess's law lecture found that turning 1 mol of liquid water into vapour takes +44.0 kJ/mol. So, with two moles of water as gas, the true value is $- 890.3 + 2 \times 44.0 = - 802.3$ kJ/mol. The bond estimate, −814, is only about 1.5% away.

Worked Example 3 — Ethane, with a step-by-step count

Estimate ΔH for $C_{2} H_{6} (g) + \frac{7}{2} O_{2} (g) \to 2 CO_{2} (g) + 3 H_{2} O (g)$ . Step through the count below.

Step through — counting every bond in burning ethane

Press Next to add one step. Read the reason before you go on.

Each step gives one line of the count for C₂H₆(g) + 7/2 O₂(g) → 2CO₂(g) + 3H₂O(g), and the reason for it.

Draw ethane: H₃C–CH₃. Two carbons joined by one C–C bond, and three H on each carbon. So ethane has 1 C–C and 6 C–H bonds.
Bonds broken in ethane: 346 + 6 × 416 = 346 + 2496 = 2842 kJ.
Bonds broken in oxygen: 7/2 × 498 = 3.5 × 498 = 1743 kJ. Half a mole of O=O bonds is fine, because the numbers are moles.
Total broken: 2842 + 1743 = 4585 kJ.
Bonds formed in 2CO₂: each O=C=O has 2 C=O, so 4 C=O: 4 × 803 = 3212 kJ (the value for CO₂).
Bonds formed in 3H₂O: each has 2 O–H, so 6 O–H: 6 × 467 = 2802 kJ.
Total formed: 3212 + 2802 = 6014 kJ.
ΔH ≈ 4585 − 6014 = −1429 kJ/mol (broken minus formed).

The data-book combustion value for ethane is −1561 kJ/mol, again with liquid water. With three moles of water as vapour, the true value is $- 1561 + 3 \times 44.0 = - 1429$ kJ/mol. The estimate matches it to the nearest kJ/mol. That is partly luck: small errors in different bonds happened to cancel.

Worked Example 4 — Where the method is weaker

Estimate ΔH for adding hydrogen to ethene: $C_{2} H_{4} (g) + H_{2} (g) \to C_{2} H_{6} (g)$ .

Look at what changes. Four C–H bonds in ethene survive into ethane, so leave them out of both sides. Broken: 1 C=C and 1 H–H, so $598 + 436 = 1034$ kJ. Formed: 1 C–C and 2 new C–H, so $346 + 2 \times 416 = 1178$ kJ.

$Δ H \approx 1034 - 1178 = - 144$ kJ/mol.

The Hess's law lecture found −135.8 kJ/mol from combustion data. The estimate is about 6% too exothermic. The reason is the averages: the real C=C bond in ethene, and the real C–H bonds in ethane, are not exactly the table's mean values. When only a few bonds change, a small error in one of them is a large part of the answer.

Which route should you use?

You now have two routes between the same start and finish. Hess's law with enthalpies of formation or combustion uses measured values for the actual substances, so it is accurate, and it works for liquids and solids. The bond route uses a short list of averages, so it is only an estimate, and only for gases. But it works for molecules that have never been measured, and it shows why a reaction gives out energy: which bonds cost the most, and which give back the most. When data for the real substances are available, use them. When they are not, count bonds.

Three classic mistakes

Trap 1 — Formed minus broken

The formula is broken minus formed. If you reverse it, every sign flips, and burning methane comes out as +814, an endothermic fire. Picture the route: up to the atoms (broken, positive), then down to the products (formed, negative).

Trap 2 — Counting one molecule instead of the equation

Multiply by the numbers in the balanced equation. In $2 H_{2} O$ there are four O–H bonds, not two. In $\frac{7}{2} O_{2}$ there are three and a half moles of O=O bonds, which is fine, because the numbers are moles.

Trap 3 — Using bond enthalpies for liquids, or the wrong C=O

Mean bond enthalpies are for gases only. A bond estimate for a reaction that makes liquid water must be compared with the value for water vapour, or corrected by 44.0 kJ/mol for each mole of water. And the C=O in carbon dioxide uses 803, not the 695 of other molecules.

Try it yourself

Use the table of mean bond enthalpies above, with H–F = 566 and F–F = 158 kJ/mol. Work each one on paper first. Open a hint only when you are stuck.

Estimate ΔH for $H_{2} (g) + F_{2} (g) \to 2 HF (g)$ . The data-book value of ΔH_f^⦵[HF(g)] is −273.3 kJ/mol. How close is your estimate?

Hint 1
Broken: one H–H and one F–F. Formed: two H–F.

Hint 2
The data value is per mole of HF; the equation makes 2 mol.

Full answer
Broken: $436 + 158 = 594$ kJ. Formed: $2 \times 566 = 1132$ kJ. $Δ H \approx 594 - 1132 = - 538$ kJ/mol. The data value for 2 mol is $2 \times (- 273.3) = - 546.6$ kJ/mol. The estimate is about 1.6% less exothermic. The weak F–F bond (only 158) is one reason the reaction is so violently exothermic.
Estimate ΔH for $N_{2} (g) + 3 H_{2} (g) \to 2 NH_{3} (g)$ . Compare it with the data-book value, ΔH_f^⦵[NH₃(g)] = −45.9 kJ/mol.

Hint 1
Each NH₃ has three N–H bonds, and the equation makes two NH₃.

Hint 2
Broken: 1 N≡N and 3 H–H. Formed: 6 N–H.

Full answer
Broken: $946 + 3 \times 436 = 946 + 1308 = 2254$ kJ. Formed: $6 \times 391 = 2346$ kJ. $Δ H \approx 2254 - 2346 = - 92$ kJ/mol. The data value for 2 mol is $2 \times (- 45.9) = - 91.8$ kJ/mol, so the estimate is very close. Notice how small the answer is compared with the 2254 kJ needed to break the bonds first. That huge first step, mostly the N≡N bond, is why this reaction is so slow without a catalyst.
For $H_{2} (g) + Br_{2} (g) \to 2 HBr (g)$ , ΔH = −103 kJ/mol. The H–H bond enthalpy is 436 and the Br–Br bond enthalpy is 193 kJ/mol. Find the H–Br bond enthalpy.

Hint 1
Write the formula with the unknown: −103 = (436 + 193) − 2x.

Hint 2
Rearrange: 2x = 436 + 193 + 103.

Full answer
$- 103 = 629 - 2 x$ , so $2 x = 629 + 103 = 732$ , and $x = 366$ kJ/mol. Check: $436 + 193 - 2 \times 366 = 629 - 732 = - 103$ . ✓ This is how bond enthalpies are found in the first place: from measured enthalpy changes, worked backwards.
Compare hydrogen and methane as fuels, with water as a gas in both. (a) Estimate ΔH for $H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (g)$ . (b) Using −814 kJ/mol for methane, find the energy released per gram of each fuel (M: H₂ = 2, CH₄ = 16 g/mol). (c) Which fuel gives more energy per gram, and by roughly how many times? (d) The data values with water as a gas are −241.8 kJ/mol for hydrogen and −802.3 kJ/mol for methane. Does your conclusion survive?

Hint 1
For (a): broken 1 H–H and ½ O=O; formed 2 O–H.

Hint 2
Energy per gram = energy per mole ÷ molar mass.

Full answer
(a) Broken: $436 + 0.5 \times 498 = 685$ kJ. Formed: $2 \times 467 = 934$ kJ. $Δ H \approx 685 - 934 = - 249$ kJ/mol. (b) Hydrogen: $249 \div 2 \approx 124.5$ kJ/g. Methane: $814 \div 16 \approx 50.9$ kJ/g. (c) Hydrogen, by about $124.5 \div 50.9 \approx 2.4$ times. Hydrogen molecules are very light, and every bond they form is a strong O–H bond. (d) Yes: $241.8 \div 2 = 120.9$ kJ/g and $802.3 \div 16 \approx 50.1$ kJ/g, also about 2.4 times. A rough method was enough to answer a real design question correctly.

The bigger idea

Bond enthalpies connect the energy of a whole reaction to the strength of single bonds, and they explain behaviour that ΔH alone cannot. Making ammonia is exothermic, yet nitrogen gas barely reacts at room temperature, because the N≡N bond (946 kJ/mol) must start to break before any new bonds can form. That first cost is the start of the idea of activation energy, in the lectures on reaction rates. Bond strengths also hold surprises. A C≡C triple bond (813) is weaker than three C–C single bonds (3 × 346 = 1038). But N≡N (946) is far stronger than three N–N single bonds (3 × 160 = 480), because the N–N single bond is unusually weak. Numbers like these are clues that chemists use to understand how bonds really work.

Key concepts in this lesson

Breaking a covalent bond always takes in energy, and forming the same bond releases the same amount, so bond enthalpies are always positive.
The mean bond enthalpy is the average energy to break one mole of a type of bond in the gas state, so no single molecule matches it exactly.
For reactions between gases, $Δ H \approx \sum (bonds broken) - \sum (bonds formed)$ , which is Hess's law applied to a route through gaseous atoms.
The estimate is only approximate because mean values are used, and it must be compared with data for the same states, such as water as a gas.
Draw every structural formula and multiply by the numbers in the balanced equation, because most errors are counting errors.