Calculation of enthalpy changes of reaction using bond energies
4 marks · NormalThe reaction between ethane and chlorine is represented by the following equation:
Use the bond energies given below for your calculations.
| Bond | Bond energy / kJ mol^{-1} |
|---|---|
| C–H | 413 |
| C–C | 348 |
| C–Cl | 338 |
| Cl–Cl | 243 |
| H–Cl | 432 |
List all bonds that are broken and all bonds that are formed in this reaction.
Calculate the total energy required to break the bonds in the reactants.
Calculate the total energy released when new bonds are formed in the products.
Calculate the enthalpy change () for the reaction. State whether the reaction is exothermic or endothermic.
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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(a) Bonds broken: 1 C–H, 1 Cl–Cl; Bonds formed: 1 C–Cl, 1 H–Cl
(b) 413 + 243 = 656 kJ
(c) 338 + 432 = 770 kJ (released) kJ
(d) ΔH = 656 – 770 = –114 kJ; exothermic kJ
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Marking points
B1 (a) · 1 mark · Correct identification of bonds broken and formed: 1 C–H, 1 Cl–Cl broken; 1 C–Cl, 1 H–Cl formed.
M1 (b) · 1 mark · Correct calculation of total energy to break bonds in reactants (1×C–H + 1×Cl–Cl).
M2 (c) · 1 mark · Correct calculation of total energy released in products (1×C–Cl + 1×H–Cl).
A1 (d) · 1 mark · Correct enthalpy change (–114 kJ), and correct exothermic identification. (follow-through)
Worked solution
a) Bonds broken: 1 C–H (from ethane) and 1 Cl–Cl. Bonds formed: 1 C–Cl and 1 H–Cl.
b) Total energy to break bonds = (1 × 413) + (1 × 243) = 656 kJ
c) Total energy released in products = (1 × 338) + (1 × 432) = 770 kJ
d) ΔH = Energy in – Energy out = 656 – 770 = –114 kJ. Negative, so the reaction is exothermic.