Calculation of enthalpy changes of reaction using bond energies
2 marks · HardA student calculates the enthalpy change for the following reaction using average bond energies:
Bond energies (kJ/mol):
| Bond | Bond energy / kJ mol^{-1} |
|---|---|
| Cl–Cl | 243 |
| H–S | 363 |
| H–Cl | 432 |
State the total energy required to break all the bonds in the reactants, and the total energy released when new bonds form in the products. Show your calculations.
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.
Show answer
Bonds broken: 1 Cl–Cl (243) + 2 H–S (2×363 = 726) = 969 kJ Bonds formed: 2 H–Cl (2×432 = 864)
Also accept: 969 kJ (broken), 864 kJ (formed)
Common mistakes: Mixing up bonds broken and formed; Incorrect multiplication
Show mark scheme and worked solution
Marking points
M1 · 1 mark · Correctly adds bond energies for all bonds broken in the reactants: 243 + (2×363) = 969 kJ.
M2 · 1 mark · Correctly multiplies and sums energies for bonds formed: 2×432 = 864 kJ.
Worked solution
Bonds broken: – 1 Cl–Cl: 243 kJ – 2 H–S: 2 × 363 = 726 kJ Total broken = 243 + 726 = 969 kJ
Bonds formed: – 2 H–Cl: 2 × 432 = 864 kJ
Total energy to break bonds in reactants: 969 kJ Total energy released forming bonds in products: 864 kJ