IB Chemistry HLTopic 4 — Energy CyclesPaper 1 & 2Core skill~11 min read
Bond Enthalpy Calculations
Every chemical reaction is really just two things happening at once: old bonds coming apart and new bonds snapping together. One of those costs energy, the other pays energy back. Whichever wins decides whether your reaction heats the room up or cools it down — and once you see it that way, the calculation is simple arithmetic.
📘 What you need to know
Breaking a bond always needs energy in. Bond breaking is endothermic, so bond enthalpies are always positive.
Making a bond always gives energy out. Bond making is exothermic, so those values are negative.
Breaking and making the same bond involve the same size of energy, just opposite signs.
Average bond enthalpy is the energy needed to break one mole of a bond in gaseous molecules, averaged over a range of similar compounds.
The calculation: ΔH = (sum of bonds broken) − (sum of bonds made), using data booklet values.
Answers from bond enthalpies are estimates, because the values are averages and they only apply to gases.
Bonds that appear unchanged on both sides cancel out — you can leave them out and save time.
Why breaking costs and making pays
A covalent bond is a shared pair of electrons sitting between two positive nuclei. Both nuclei are pulling on that pair, and that attraction is what holds the atoms together. To pull the atoms apart you have to fight that attraction, and fighting an attraction costs energy. So bond breaking always takes energy in from the surroundings — it is endothermic.
Run the film backwards and the logic flips. When two atoms fall together into a bond, they are moving with the attraction, not against it. Energy is released to the surroundings. So bond making is always exothermic.
The data booklet gives you one number per bond. Whether you write it as plus or minus depends entirely on which direction you are going.
Which side wins?
A reaction does both jobs. So compare the two totals:
If more energy comes out of bond making than went into bond breaking, there is energy left over. It escapes to the surroundings and the reaction is exothermic — ΔH is negative. The products sit lower in energy, so they are more stable than the reactants.
If bond breaking costs more than bond making pays back, the shortfall has to be borrowed from the surroundings. The reaction is endothermic — ΔH is positive, and the products are less stable.
Notice the trick this diagram uses: pretend the reaction pulls every atom completely apart, then rebuilds. Enthalpy does not care about the route, so the answer is still right.
The two profiles look different but the sums are identical. Do not memorise two formulas — there is only one, and the sign of the answer tells you which picture you were in.
Why the word “average” matters
Methane has four C—H bonds and they all look identical on paper. They are not. Pulling the first hydrogen off CH4 is easier than pulling the second off CH3, because once one hydrogen is gone the carbon holds the remaining three a little more tightly. Each successive bond costs a different amount.
You cannot measure them individually, so chemists measure the total needed to shatter the whole molecule into atoms and divide by four. Then they repeat the exercise across many different C—H containing compounds and average again. That final number is the average bond enthalpy in your data booklet.
Average bond enthalpy
The energy needed to break one mole of a particular bond in gaseous molecules, averaged over similar compounds
This is why bond enthalpy answers are only estimates. The C—H in your molecule is probably not exactly 414 kJ mol−1, it is just close. Expect your answer to be within a few percent of the true value, not identical to it.
The calculation, step by step
🧩 The method that never fails
Balance the equation and check the state symbols. Bond enthalpies only work for gases.
Draw out the displayed formulas of everything. Every single bond, including the ones inside water and carbon dioxide. This is where marks are lost.
Count the bonds broken (left-hand side), multiply each by its data booklet value and by the equation coefficient. Total these as a positive number.
Count the bonds made (right-hand side) the same way. Total these as a negative number.
Add the two totals. The sign of the answer tells you exothermic or endothermic.
The only equation you need
ΔH = Σ(bonds broken) − Σ(bonds made)
Some people prefer to write it as ΔH = Σ(bonds broken) + Σ(bonds made) and make the “made” total negative themselves. Both give the same answer. Pick one and stick to it, because mixing the two mid-question is how sign errors happen.
WORKED EXAMPLE
Complete combustion of methane
Use average bond enthalpies to estimate ΔH for the complete combustion of methane, producing gaseous water:
CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)
Bond
C—H
O=O
C=O
O—H
Average bond enthalpy / kJ mol−1
414
498
804
463
Step 1: count what you breakCH₄ has 4 C—H bonds. 2O₂ has 2 O=O bonds.4 × 414 = 16562 × 498 = 996bonds broken = +2652 kJ mol−¹Step 2: count what you makeCO₂ is O=C=O, so 2 C=O. Each H₂O has 2 O—H, so 2 waters give 4.2 × 804 = 16084 × 463 = 1852bonds made = −3460 kJ mol−¹Step 3: add themΔH = 2652 − 3460ΔH = −808 kJ mol−¹negative, so exothermic — exactly what you expect from burning a fuel
The classic slip here is writing CO2 as having one C=O. It has two. Always sketch O=C=O before you count.
WORKED EXAMPLE
Hydrogenation of ethene — and a shortcut worth knowing
The long way: count everythingBroken: 1 C=C + 4 C—H + 1 H—H. Made: 1 C—C + 6 C—H.broken = 614 + (4 × 414) + 436 = 2706made = 346 + (6 × 414) = 2830ΔH = 2706 − 2830ΔH = −124 kJ mol−¹The short way: ignore bonds that surviveFour C—H bonds are on both sides, so they cancel. Only the C=C, the H—H and the two new bonds actually change.broken = 614 + 436 = 1050made = 346 + (2 × 414) = 1174ΔH = 1050 − 1174ΔH = −124 kJ mol−¹same answer, half the arithmetic — but only cancel if you are certain the bond is unchanged
Use the shortcut with care. It is a gift on Paper 1 where time is tight, but on Paper 2 a full bond-by-bond table earns method marks even if you fumble the final number. If in doubt, count everything.
WORKED EXAMPLE
Working backwards to find a bond enthalpy
For H2(g) + Cl2(g) → 2HCl(g), ΔH = −184 kJ mol−1. Given H—H = 436 and Cl—Cl = 242 kJ mol−1, find the H—Cl bond enthalpy.
Step 1: write the usual equation, leaving the unknown as xBroken: 1 H—H + 1 Cl—Cl. Made: 2 H—Cl, so 2x.−184 = (436 + 242) − 2xStep 2: rearrange−184 = 678 − 2x2x = 678 + 184 = 862H—Cl = +431 kJ mol−¹bond enthalpies are always positive — if your x comes out negative, you flipped a sign somewhere
Why your answer will not match the data book value
Look up the enthalpy of combustion of methane and you will find roughly −891 kJ mol−1, not −808. That is not a mistake in your working. There are two honest reasons:
The values are averages. They are borrowed from other molecules and will never be a perfect fit for yours.
Bond enthalpies only describe gases. Our calculation produced water as a gas. The real enthalpy of combustion is quoted with liquid water, and condensing water releases a further chunk of energy. That difference alone accounts for most of the gap.
If a question asks why a bond-enthalpy answer differs from an experimental one, those two sentences are the whole mark scheme: averaged values, and gaseous species only. Say both.
💡 Exam tip
Draw the displayed formula every time. It takes fifteen seconds and it is the difference between counting 4 C—H and counting 3.
Multiply by the coefficients. The “2” in front of H2O doubles four O—H bonds to eight electrons worth of bonding — that is 4 O—H total, and students forget it constantly.
Set your work out as a balance sheet: broken total, made total, then subtract. Examiners can follow it and award method marks.
These questions often carry 3 marks, and two of them are for the working. Show every line even if you are unsure of the final number.
Check the sign is sensible. Combustion, neutralisation and most bond-forming reactions are exothermic. A positive answer for burning a fuel means you subtracted the wrong way round.
If the question says “estimate”, it is signposting bond enthalpies. If it gives you formation or combustion data instead, use a Hess cycle.
⚠ Common mix-up
Giving bond breaking a negative sign. Breaking always costs energy, so it is always positive. No exceptions.
Subtracting the wrong way. It is broken minus made, not made minus broken. Get this backwards and every answer has the wrong sign.
Missing the second C=O in carbon dioxide, or only counting one O—H per water molecule.
Using bond enthalpies for liquids or solids. If the equation shows H2O(l) or a solid reactant, bond enthalpies are the wrong tool — you need Hess’s Law.
Confusing bond enthalpy with bond strength direction. A bigger bond enthalpy means a stronger bond that is harder to break, not one that releases more when broken.
Saying “the products have stronger bonds so the reaction is endothermic”. Stronger product bonds means more energy released, so exothermic. Read your own sentence back before you write it.
Up next: Hess’s Law — what to do when the reaction you want cannot be measured in a lab at all.
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