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IGCSE · Practice questions · Bond breaking, bond making and bond energies

Count every bond.

Six original Cambridge-style questions on bond energies: the energy of breaking and making bonds and step-by-step ΔH calculations. Extended content.

Original questions All questions on this page are original work, written in the Cambridge IGCSE style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Broken minus made.

01
Explain
[3 marks]

(Extended) In terms of bond breaking and bond making, explain why some reactions are exothermic and others endothermic.

  • Bond breaking is endothermic and bond making is exothermic. ✓
  • If more energy is released making new bonds than is taken in breaking old bonds, the reaction is exothermic. ✓
  • If more energy is taken in breaking bonds than is released making bonds, the reaction is endothermic. ✓
02
Calculation
[3 marks]

(Extended) Calculate ΔH for H2 + Cl2 → 2HCl. (Bond energies in kJ/mol: H–H 436, Cl–Cl 243, H–Cl 432)

  • Bonds broken: 436 + 243 = 679 kJ. ✓
  • Bonds made: 2 × 432 = 864 kJ. ✓
  • ΔH = 679 − 864 = −185 kJ/mol. ✓
03
Calculation
[4 marks]

(Extended) Calculate ΔH for the complete combustion of methane, CH4 + 2O2 → CO2 + 2H2O. (Bond energies in kJ/mol: C–H 410, O=O 496, C=O 805, O–H 460)

  • Bonds broken: 4 × C–H + 2 × O=O = 1640 + 992 = 2632 kJ. ✓
  • Bonds made: 2 × C=O + 4 × O–H = 1610 + 1840 = 3450 kJ. ✓
  • ΔH = 2632 − 3450 = −818 kJ/mol. ✓
  • The value is negative, so the reaction is exothermic. ✓
04
Calculation
[3 marks]

(Extended) Ethene reacts with bromine: C2H4 + Br2 → C2H4Br2. Calculate ΔH. (Bond energies in kJ/mol: C=C 610, Br–Br 193, C–C 350, C–Br 290; the C–H bonds are unchanged)

  • Bonds broken: C=C + Br–Br = 610 + 193 = 803 kJ. ✓
  • Bonds made: C–C + 2 × C–Br = 350 + 580 = 930 kJ. ✓
  • ΔH = 803 − 930 = −127 kJ/mol. ✓
05
Calculation
[3 marks]

(Extended) Calculate ΔH for the decomposition 2NH3 → N2 + 3H2, and state whether it is exothermic or endothermic. (Bond energies in kJ/mol: N≡N 945, H–H 436, N–H 390)

  • Bonds broken: 6 × N–H = 2340 kJ. ✓
  • Bonds made: N≡N + 3 × H–H = 945 + 1308 = 2253 kJ. ✓
  • ΔH = 2340 − 2253 = +87 kJ/mol: endothermic. ✓
06
Explain
[2 marks]

(Extended) A student calculates ΔH for a reaction as −2632 kJ/mol by adding up only the energies of the bonds in the reactants. Explain the mistake.

  • Breaking the bonds in the reactants takes in energy, so it should be counted as positive. ✓
  • The student must also subtract the energy released when the bonds in the products are made: ΔH = bonds broken − bonds made. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.