Calculation of enthalpy changes of reaction using bond energies
2 marks · HardThe following reaction occurs:
Bond energies (in kJ/mol):
| Bond | Bond energy / kJ mol^{-1} |
|---|---|
| H–H | 436 |
| F–F | 158 |
| H–F | 565 |
Calculate the enthalpy change (ΔH) for the reaction using the bond energies given.
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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Also accept: ΔH = –542 kJ mol^{-1}; –542 kJ/mol
Common mistakes: Incorrect subtraction order; Missing negative sign
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Marking points
M1 · 1 mark · Correctly calculates energy for bonds broken and made (shows working): Bonds broken: 436 + 158; bonds made: 2 x 565.
A1 · 1 mark · Arrives at correct answer with sign and units: ΔH = –542 kJ/mol.
Worked solution
Bonds broken: 1 H–H and 1 F–F
Bonds made: 2 H–F
Enthalpy change: ΔH = total bonds broken – total bonds made
Correction: 594 – 1130 = –536 kJ/mol (Not –542; check sums properly if needed).
Final answer (recalculating):