Bond Energy Calculations

The bond energy calculation is the most predictable 3-marker on Paper 4: a balanced equation, a table of bond energies, “calculate the enthalpy change”. The method never varies, yet examiner reports show a third of candidates reversing the final subtraction and reporting +184 instead of −184. The fix is a fixed layout, used every single time.

The principle: breaking costs, making pays

During a reaction, the bonds in the reactants break and new bonds form in the products.

• Bond breaking is endothermic: energy must be supplied to separate bonded atoms.
• Bond making is exothermic: energy is released when new bonds form.

The overall reaction is exothermic if making the new bonds releases more energy than breaking the old ones cost; endothermic if breaking costs more than making returns. This is the particle-level reason behind the temperature changes described in Exothermic and Endothermic Reactions, and it earns 2 marks whenever a question says “explain, in terms of bond energies, why the reaction is exothermic”.

The calculation method

Bond energy is the energy needed to break one mole of a particular bond, in kJ/mol. The recipe:

1. Write or copy the balanced equation, ideally as displayed formulae so every bond is visible.
2. Count and total the energy of all bonds broken (reactant side).
3. Count and total the energy of all bonds made (product side).
4. ΔH = energy in (bonds broken) − energy out (bonds made).

Worked example (hydrogen burning in chlorine): H2 + Cl2 → 2HCl, with bond energies H–H = 436, Cl–Cl = 242, H–Cl = 431 kJ/mol.

Bonds broken (in)	Bonds made (out)
1 × H–H = 436	2 × H–Cl = 862
1 × Cl–Cl = 242
Total in = 678 kJ	Total out = 862 kJ

ΔH = 678 − 862 = −184 kJ/mol. Negative, so exothermic: more energy was released forming the two H–Cl bonds than was absorbed breaking the H–H and Cl–Cl bonds.

The two-column table is the layout to copy. It makes coefficients impossible to miss and it shows the examiner your method, which protects partial marks if one bond count slips.

Counting bonds without dropping any

The coefficient multiplies every bond in that molecule. In CH4 + 2O2 → CO2 + 2H2O: break 4 C–H and 2 O=O; make 2 C=O and 4 O–H (each water has two O–H bonds, and there are two waters). Double bonds count once at their own (larger) energy value: O=O is one bond at 498 kJ/mol, not two at 249. Drawing the displayed formulae before counting removes most counting errors, and displayed formulae themselves are revised under covalent bonding in Atoms, Elements and Compounds.

Connecting to the diagram

On a reaction pathway diagram, the climb to the peak corresponds to bond breaking and the descent to bond making; ΔH is the net difference between the two lines. A question can present the same chemistry either way (numbers in a table or arrows on a diagram), and the Energy Level Diagrams page covers the drawing half. The whole topic in sequence is on the Chemical Energetics pillar.

Worked exam question

Hydrogen reacts with bromine: H2 + Br2 → 2HBr. Bond energies in kJ/mol: H–H 436, Br–Br 193, H–Br 366. (a) Calculate the energy needed to break the bonds in the reactants. [1] (b) Calculate the energy released when the bonds in the products form. [1] (c) Calculate the enthalpy change, ΔH, and state whether the reaction is exothermic or endothermic. [2]

Model answer: (a) 436 + 193 = 629 kJ (1). (b) 2 × 366 = 732 kJ (1). (c) ΔH = 629 − 732 = −103 kJ/mol (1); negative, so the reaction is exothermic (1).

Mark-by-mark: (a) and (b) are pure counting. The only trap is forgetting the 2 in 2HBr. In (c) the first mark needs the correct value with the minus sign; “103” unsigned typically loses it. The final mark links sign to classification, and error carried forward applies: a wrong (b) used correctly in (c) still earns the (c) method marks.

The mistakes that cost marks

1. Reversing the subtraction. ΔH = broken − made, in that order. Made − broken flips the sign and the classification with it.
2. Ignoring coefficients. 2HBr means two H–Br bonds; 2H2O means four O–H bonds. Multiply everything.
3. Dropping the sign or units. The answer is −103 kJ/mol, not 103.
4. Saying “breaking bonds releases energy”. It is the single most penalised reversal in this topic: breaking absorbs, making releases.

How examiners want it phrased

Student wording	Mark-scheme wording
”It gives out energy because of the bonds"	"More energy is released making bonds in the products than is taken in breaking bonds in the reactants"
"Breaking bonds is hard"	"Bond breaking is endothermic; energy must be taken in"
"The answer is 103"	"ΔH = −103 kJ/mol, so the reaction is exothermic"
"New bonds give energy back"	"Bond making is exothermic; energy is released when bonds form”

The sentence pair “bond breaking is endothermic / bond making is exothermic” plus the comparison “more energy released than taken in” covers nearly every explain-mark this subtopic offers. If the arithmetic is fine but the signs keep flipping, that is a method-layout problem, exactly the kind of thing a free 1-hour trial lesson with a Chemistry specialist fixes by rebuilding the two-column habit on past-paper questions.

Test yourself

Answer all three on paper before clicking. The answers stay hidden until you do.

Q1 (3 marks). Ammonia is made from nitrogen and hydrogen: N2 + 3H2 → 2NH3. Bond energies in kJ/mol: N≡N 945, H–H 436, N–H 391. Calculate the enthalpy change, ΔH, and state whether the reaction is exothermic or endothermic.

Show answer

• energy in = 945 + (3 × 436) = 2253 kJ [1] • energy out = 6 × 391 = 2346 kJ [1] • ΔH = 2253 − 2346 = −93 kJ/mol, so the reaction is exothermic [1]. The 2NH3 means six N–H bonds, not three.

Q2 (2 marks). Explain, in terms of bond breaking and bond making, why the complete combustion of methane is exothermic.

Show answer

• bond breaking is endothermic and bond making is exothermic [1] • more energy is released making the bonds in the products than is taken in breaking the bonds in the reactants [1]

Q3 (3 marks). Ethene reacts with hydrogen: C2H4 + H2 → C2H6. Bond energies in kJ/mol: C=C 612, C–C 347, C–H 413, H–H 436. Calculate ΔH for this reaction.

Show answer

• bonds broken = 1 × C=C + 4 × C–H + 1 × H–H = 612 + 1652 + 436 = 2700 kJ [1] • bonds made = 1 × C–C + 6 × C–H = 347 + 2478 = 2825 kJ [1] • ΔH = 2700 − 2825 = −125 kJ/mol [1]. Counting only the bonds that change (C=C + H–H in; C–C + 2 C–H out) gives the same −125.