Calculation of enthalpy changes of reaction using bond energies
3 marks · EasyBond energies are given below:
| Bond | Bond energy / kJ mol^{-1} |
|---|---|
| N≡N | 946 |
| H–H | 436 |
| N–H | 388 |
Use the bond energies to answer the following about the reaction:
Calculate the total energy needed to break all bonds in the reactants.
Calculate the total energy released by forming all bonds in the products.
Determine the enthalpy change (ΔH) for the reaction.
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(a) 946 + (3 × 436) = 946 + 1308 = 2254 kJ
(b) 2 × (3 × 388) = 6 × 388 = 2328 kJ
(c) –74 kJ/mol
Common mistakes: Wrong sign on final answer; Missing units or steps
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Marking points
M1 (a) · 1 mark · Correctly calculates total bond breaking energy (946 + 3 × 436 = 2254 kJ).
M2 (b) · 1 mark · Correctly calculates total bond forming energy (6 × 388 = 2328 kJ).
A1 (c) · 1 mark · Correctly calculates ΔH = 2254 – 2328 = –74 kJ/mol with correct sign.
Worked solution