Calculation of enthalpy changes of reaction using bond energies
4 marks · NormalConsider the following reaction:
Given the following bond energies (kJ/mol):
| Bond | Bond energy |
|---|---|
| C–H | 413 |
| F–F | 158 |
| C–F | 485 |
| H–F | 565 |
List all the bonds to be broken in the reactants.
Calculate the total energy required to break these bonds.
Calculate the total energy released when new bonds are formed in the products.
Calculate the overall enthalpy change (ΔH) for the reaction.
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
(a) 4 C–H bonds, 2 F–F bonds
(b) 4×413 + 2×158 = 1652 + 316 = 1968 kJ
(c) 4 C–F: 4×485 = 1940; 2 H–F: 2×565 = 1130; Total released = 1940 + 1130 = 3070 kJ
(d) ΔH = 1968 – 3070 = –1102 kJ
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Correct list: 4 C–H and 2 F–F bonds broken.
M1 (b) · 1 mark · Correct calculation for total bond breaking energy.
M2 (c) · 1 mark · Correct calculation for bond making energy released. (follow-through)
A1 (d) · 1 mark · Correct ΔH value (–1102 kJ, correct subtraction and sign), accepts ecf. (follow-through)
Worked solution
a) In CH4: 4 C–H bonds; in 2 F2: 2 F–F bonds. Total bonds broken: 4 C–H and 2 F–F.
b) Energy to break = (4×413) + (2×158) = 1652 + 316 = 1968 kJ.
c) In CF4: 4 C–F bonds (4×485 = 1940); in 2 HF: 2 H–F bonds (2×565 = 1130); total = 1940 + 1130 = 3070 kJ released.
d) ΔH = Energy in – Energy out = 1968 – 3070 = –1102 kJ.