Calculation of enthalpy changes of reaction using bond energies
3 marks · NormalCalculate the enthalpy change () for the reaction:
| Bond | Bond energy / kJ mol^{-1} |
|---|---|
| H–H | 436 |
| O=O | 498 |
| O–H | 464 |
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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ΔH = [2×436 + 1×498] – [4×464] = 1370 – 1856 = –486 kJ (±3)
Also accept: –486 kJ/mol
Common mistakes: Omitting to multiply bonds correctly; Wrong sign for ΔH
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Marking points
M1 · 1 mark · Correct process: lists all bonds broken and formed.
M2 · 1 mark · Correct substitution: [2×436 + 498] and [4×464].
A1 · 1 mark · Correct final answer (–486 kJ), with sign and unit. (follow-through)
Worked solution
Find bonds broken (left): 2 H–H, 1 O=O.
Bonds formed (right): 4 O–H.
ΔH = Energy in – Energy out = 1370 – 1856 = –486 kJ.