Chemical changes · GCSE Chemistry
Bond energies
GCSE Chemistry revision on bond energies: bond breaking is endothermic, bond making is exothermic, and ΔH = energy in to break bonds minus energy out when bonds form.
Break = energy in. Make = energy out. ΔH = total bonds broken − total bonds made. Negative answer means exothermic: you got more energy out from making than you put in to break.
The important bits
What you need to know
- 1
Bond energy (bond enthalpy) is the energy needed to break one mole of a particular covalent bond, in kJ/mol. Values are average values from many compounds.
- 2
Breaking bonds is endothermic: energy is taken in. Making bonds is exothermic: energy is released. Never reverse those two sentences.
- 3
Overall ΔH = sum of bond energies of bonds broken − sum of bond energies of bonds made. A negative ΔH is exothermic; a positive ΔH is endothermic.
- 4
You must count every bond in the displayed formulae, not only the ones that “look like they change”. CH₄ has four C–H bonds; O₂ has one O=O double bond.
- 5
If more energy is released when bonds form than is used to break bonds, the surroundings heat up. That is why fuels work: product bonds in CO₂ and H₂O are strong.
- 6
These calculations are Higher tier on Combined Science and Chemistry. Foundation may only need the broken/made idea without a full numerical ΔH.
- 7
Units are kJ/mol. Show two subtotals (broken and made) before you subtract. Include the sign in the final answer.
- 8
Bond energies are averages, so calculated ΔH may not match a data-book enthalpy of combustion exactly. In the exam, use the values they give.
Quotations worth analysing
Short evidence. Real method.
“ΔH = energy to break bonds − energy released when bonds form”
Same as broken minus made if both lists are written as positive bond-energy totals. Swap them and the sign of ΔH is wrong.
“H–H 436 kJ/mol; O=O 498 kJ/mol; O–H 464 kJ/mol (typical values).”
For 2H₂ + O₂ → 2H₂O you break 2 H–H and 1 O=O, and make 4 O–H. Do not make 2 O–H. Each water molecule has two O–H bonds.
“Bond breaking endothermic; bond making exothermic.”
This pair is worth a mark on its own. The overall reaction type is whichever of those two totals is larger.
Go deeper
Count bonds from the displayed equation, not from memory of the word equation
Write structural or displayed formulae. Hydrogen is H–H, not a mysterious “hydrogen bond” (that word means something else). Oxygen is O=O, one double bond, not two single bonds. Methane is four C–H. Carbon dioxide is two C=O double bonds. Water is two O–H. Multiply by the balancing numbers in the equation. 2H₂O means four O–H bonds made. Students break CH₄ as one C–H, or treat O₂ as 2 × O–O single bonds using the wrong value. Circle each bond on a quick sketch, then multiply by the table value, then add.
Go deeper
Show two totals so the sign cannot hide
A reaction breaks bonds totalling 1582 kJ/mol and makes bonds totalling 1850 kJ/mol. ΔH = 1582 − 1850 = −268 kJ/mol. The negative sign is the conclusion: exothermic, products lower on the profile, surroundings heat up. If you subtract the other way you will say a fuel is endothermic and lose the chemical sense check. Combustion of methane should come out large and negative. If your methane calculation is positive, you missed bonds in CO₂ or H₂O. Energy is conserved: the difference is what appears as a temperature change in the surroundings.
Go deeper
Link back to profiles and to collision theory
Bond breaking is part of why Ea exists: you must put energy in before the new bonds can form and release even more. A catalyst does not change the bond-energy totals of reactants and products, so ΔH is the same; it only changes how the journey happens. Temperature does not change ΔH either in these school calculations. If a question gives a profile with numbers and a table of bond energies, they should agree: the ΔH from broken minus made should match the gap between reactant and product lines. Use that as a check when both are provided.
See the idea in action
Calculate ΔH for H₂ + Cl₂ → 2HCl. Bond energies: H–H 436, Cl–Cl 242, H–Cl 431 kJ/mol. Bonds broken: 436 + 242 = 678 kJ/mol. Bonds made: 2 × 431 = 862 kJ/mol. ΔH = 678 − 862 = −184 kJ/mol. The reaction is exothermic: more energy is released forming two H–Cl bonds than is used breaking H–H and Cl–Cl. On a profile, products sit 184 kJ/mol below reactants.
Exam technique
Turn knowledge into marks
Draw the molecules, count every bond, multiply by the balancing numbers, total broken, total made, subtract in that order, then write the sign and kJ/mol. Sense-check: combustion should be negative.
Common mistakes
Do not give these marks away
- 01
Swapping broken and made, or omitting the sign of ΔH.
- 02
Counting the wrong number of O–H bonds in water, or treating O=O as two single bonds.
- 03
Using only the bonds that “change” and ignoring bonds that are also broken and remade in the balanced equation.
In terms of bonds, when is a reaction exothermic?
AWhen more energy is used to break bonds than is released when bonds form
BWhen more energy is released forming bonds than is used breaking bonds
CWhen all bonds are broken
DWhen a catalyst is added
Show the answer
When more energy is released forming bonds than is used breaking bonds. Bond making releases energy. If that amount is larger than the energy needed to break reactant bonds, the surroundings heat up and ΔH is negative.
Quick questions
If this is the bit you searched
How do you calculate ΔH from bond energies GCSE?
Add up the bond energies of all bonds broken, add up all bonds made, then ΔH = broken − made. Include the sign and the unit kJ/mol.
Is bond breaking exothermic or endothermic?
Endothermic: energy is taken in to break bonds. Bond making is exothermic: energy is released.
Why is combustion exothermic in terms of bonds?
The energy released when C=O and O–H bonds form in CO₂ and H₂O is greater than the energy needed to break the bonds in the fuel and in O₂.
Does a catalyst change bond-energy ΔH?
No. ΔH depends on the bonds in the reactants and products. A catalyst only lowers the activation energy of the pathway between them.