IB Chemistry SLTopic 4 — Energy CyclesPaper 1 & 2Core skill~12 min read
Bond Enthalpy Calculations
Strip a reaction down and it is only two things happening: old bonds coming apart and new ones snapping together. Pulling atoms apart costs energy, letting them join releases it — and the gap between those two amounts is the enthalpy change.
📚 What you need to know
Breaking bonds is endothermic (energy in, positive). Making bonds is exothermic (energy out, negative). No exceptions.
Bond enthalpy is the energy needed to break one mole of a bond in the gaseous state.
ΔH = Σ(bonds broken) – Σ(bonds formed), in kJ mol–1.
Every value must be multiplied by how many of that bond there are, including the coefficients in the equation.
The data booklet quotes average bond enthalpies, so answers from this method are always approximate.
The method assumes everything is a gas. A liquid reactant or product makes the answer noticeably wrong.
Breaking costs, making pays
A covalent bond is an attraction between two nuclei and a shared pair of electrons. To separate the atoms you have to pull against that attraction, and pulling against an attraction always takes energy. So bond breaking is endothermic.
Run it backwards and the logic reverses. Two atoms falling together into a bond release energy, exactly as a ball releases energy falling towards the ground. So bond making is exothermic.
Bond enthalpy
the energy required to break one mole of a bond, in the gaseous state
Because breaking and making are the same event in opposite directions, they involve the same amount of energy with the opposite sign. Breaking one mole of H–Cl bonds takes in 431 kJ; forming one mole gives out 431 kJ. That symmetry is what makes the whole calculation work.
Data booklet values are all positive, because they are quoted for breaking. You are the one who has to decide where the minus signs go — the booklet will not do it for you.
Where the formula comes from
Imagine taking the reaction by a deliberately silly route. First rip every reactant molecule apart into separate gaseous atoms. Then let those atoms rebuild themselves as the products. You end up in exactly the same place, so the energy change must be the same.
Up the hill to separated atoms, back down to the products. The difference between the climb and the drop is the enthalpy change.
The climb up costs you every bond in the reactants. The drop down pays you back every bond in the products. Subtract one from the other:
Enthalpy change from bond enthalpies
ΔH = Σ(bonds broken) – Σ(bonds formed)
Read the result like a balance sheet. If the products’ bonds give out more than the reactants’ bonds cost, the answer comes out negative and the reaction is exothermic. Stronger bonds in the products means a more stable product mixture and a more negative ΔH.
You may see the same idea written as ΔH = Σ(bonds broken) + Σ(bonds formed), with the “formed” values entered as negatives. Same maths, more chances to lose a minus sign. Pick one version and stick to it.
Why the values are only averages
A C–H bond in methane is not quite the same bond as a C–H bond in ethanol, because the rest of the molecule pulls on the electrons differently. Even inside a single methane molecule the four C–H bonds do not break equally easily: once one hydrogen has gone, the remaining ones are held differently.
Four bonds, four different values. The number in the data booklet is the average of these, further averaged across many similar compounds.
So the data booklet cannot give you the C–H bond enthalpy. It gives an average bond enthalpy: the mean energy needed to break one mole of that bond, averaged over a range of similar compounds. That has two consequences you should be ready to state:
Answers calculated this way are approximate — usually within a few percent, not exact.
The method is only strictly valid when every species is a gas. If water is produced as a liquid, the real reaction releases extra energy as it condenses, which bond enthalpies know nothing about.
If a question gives you an equation with a state symbol of (l) and asks you to comment on your answer, that state symbol is the hint. Bond enthalpies are gas-phase numbers.
Doing the calculation
🧩 The method, every time
Write the balanced equation. For an enthalpy of combustion, scale it so that one mole of the fuel burns, even if that means halves.
Draw the displayed formula of every substance, showing every single bond.
Add up the bonds broken, multiplying each bond enthalpy by how many there are.
Add up the bonds formed the same way.
ΔH = broken – formed. Give the answer in kJ mol–1 with its sign.
The coefficients are part of the count. Two water molecules mean four O–H bonds, and it is astonishingly easy to write two.
WORKED EXAMPLE
Use the bond enthalpies H–H = 436, Cl–Cl = 242 and H–Cl = 431 kJ mol–1 to calculate ΔH for H2(g) + Cl2(g) → 2HCl(g).
Step 1 — bonds broken1 × H–H = 436 1 × Cl–Cl = 242 total = 678Step 2 — bonds formed2 × H–Cl = 2 × 431 = 862Step 3 — subtract678 − 862 = −184ΔH = −184 kJ mol⁻¹The 2 in front of HCl doubles that bond enthalpy. Miss it and you get +247, which is not even the right sign.
WORKED EXAMPLE
Calculate ΔH for the hydrogenation of ethene, C2H4(g) + H2(g) → C2H6(g). Use C–H = 414, C=C = 614, C–C = 346, H–H = 436 kJ mol–1.
Step 1 — bonds broken (ethene + hydrogen)4 × C–H = 1656 1 × C=C = 614 1 × H–H = 436total = 2706Step 2 — bonds formed (ethane)6 × C–H = 2484 1 × C–C = 346 total = 2830Step 3 — subtract2706 − 2830 = −124ΔH = −124 kJ mol⁻¹Shortcut worth spotting: the four C–H bonds appear on both sides, so they cancel. Only C=C + H–H against 2 C–H + C–C actually matters.
WORKED EXAMPLE
Calculate the enthalpy of combustion of methane, CH4(g) + 2O2(g) → CO2(g) + 2H2O(g), using C–H = 414, O=O = 498, C=O = 804, O–H = 463 kJ mol–1. Compare it with the accepted value of –891 kJ mol–1.
Step 1 — bonds broken4 × C–H = 1656 2 × O=O = 996 total = 2652Step 2 — bonds formedCO₂ has two C=O bonds, and each of the two waters has two O–H bonds.2 × C=O = 1608 4 × O–H = 1852 total = 3460Step 3 — subtract2652 − 3460 = −808ΔH = −808 kJ mol⁻¹About 80 kJ less exothermic than the accepted value. Most of that gap is the water: the accepted figure is for liquid water, and condensing 2 mol of steam would release roughly another 88 kJ. The rest is the use of averaged bond enthalpies.
💡 Exam tip
Draw the displayed formulae. It takes twenty seconds and it is the only reliable way to count bonds you cannot see in a formula like C2H6.
Multiply by the coefficients. 2H2O is four O–H bonds; 5O2 is five O=O bonds.
Enthalpy of combustion questions want one mole of fuel, so half-integer coefficients like 2½O2 are fine and often necessary.
Show every line of working. These questions usually carry three marks and two of them are for the tallies, so a slip at the end still scores.
Asked why your answer differs from the data booklet? Say average bond enthalpies are used, and check whether a substance is not a gas.
⚠️ Common mix-up
Getting the subtraction the wrong way round. It is broken minus formed, not formed minus broken.
Counting molecules instead of bonds. One O2 molecule is one bond, but one CH4 molecule is four.
Ignoring bonds that “do not change”. They cancel, which is fine — but only if you counted them on both sides in the first place.
Treating a double bond as two singles. C=O is one bond with its own value, not two C–O bonds.
Expecting an exact answer. Average values cannot give one, and examiners award marks for saying so.
Up next: Hess’s Law — because bond enthalpies are really just one particular route through a reaction, and once you see that, you can build a route through almost anything.
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