Calculation of enthalpy changes of reaction using bond energies
4 marks · NormalThe following data show bond energies for several bonds:
| Bond | Bond Energy / kJ mol^{-1} |
|---|---|
| C–H | 413 |
| O=O | 498 |
| C=O | 743 |
| O–H | 464 |
Consider the equation:
Count and list all the bonds broken in the reactants.
Count and list all the bonds formed in the products.
Calculate the total energy absorbed in breaking the bonds in the reactants.
Calculate the enthalpy change () 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) Bonds broken: 2 C–H, 1 C=O, 2 H–H, 1 O=O
(b) Bonds formed: 2 C=O, 4 O–H
(c) 2×413 + 1×743 + 2×436 + 1×498 = 826 + 743 + 872 + 498 = 2939 kJ
(d) (Energy in) – (Energy out) = 2939 – [2×743 + 4×464] = 2939 – [1486 + 1856] = 2939 – 3342 = –403 kJ
Show mark scheme and worked solution
Marking points
B1 (a) · 1 mark · Correct count of all bonds broken in reactants.
B2 (b) · 1 mark · Correct count of all bonds formed in products.
M1 (c) · 1 mark · Correct calculation of total energy absorbed in breaking bonds. (follow-through)
A1 (d) · 1 mark · Correct calculation of enthalpy change using values from previous parts. (follow-through)
Worked solution
a) CH2O(g): 2 C–H and 1 C=O; H2(g): 2 H–H; O2(g): 1 O=O. Bonds broken: 2 C–H, 1 C=O, 2 H–H, 1 O=O.
b) CO2(g): 2 C=O; 2H2O(g): 4 O–H. Bonds formed: 2 C=O, 4 O–H.
c) Total energy in: 2×413 + 1×743 + 2×436 + 1×498 = 826 + 743 + 872 + 498 = 2939 kJ.
d) Total energy out: 2×743 + 4×464 = 1486 + 1856 = 3342 kJ. ΔH = 2939 – 3342 = –403 kJ.