SME
General Chemistry I · Problem type
ΔH of methane combustion from bond enthalpies
Use this for “Estimate ΔH for CH₄ + 2O₂ → CO₂ + 2H₂O(g) from the average bond enthalpies C–H 413, O=O 495, C=O 799 and O–H 463 kJ/mol.”
Diagram
Bonds broken: CH₄ + 2O₂ → C + 4H + 4O, +2,642 kJ/mol
Bonds formed: C + 4H + 4O → CO₂ + 2H₂O, −3,450 kJ/mol
ΔH = Bonds broken + Bonds formed = 2,642 + (−3,450) = −808 kJ/mol
Down the ladder: heat is given off (exothermic).
This is an example. Type your own values to start a new problem.
ΔH (kJ/mol)
=
4
×
D(C–H) (kJ/mol)
+
2
×
D(O=O) (kJ/mol)
−
(2
×
D(C=O) (kJ/mol)
+
4
×
D(O–H) (kJ/mol)
)
B
Energy taken in to break the bonds
calculated · kJ/mol
F
Energy given off forming the bonds
calculated · kJ/mol
Formulas
B
=
4
×
D(C–H)
+
2
×
D(O=O)
2,642 = 4 × 413 + 2 × 495
F
=
2
×
D(C=O)
+
4
×
D(O–H)
3,450 = 2 × 799 + 4 × 463
ΔH
=
B
−
F
−808 = 2,642 − 3,450
Assumptions
•
Bond enthalpies are averages over many molecules in the gas phase, so the answer is an estimate, often a few percent off the measured ΔH.
•
Every substance is a gas: the water is steam. Liquid water would also give off its heat of condensation.
•
Breaking a bond takes in its enthalpy and forming one gives the same amount back. CH₄ has 4 C–H bonds, O₂ one O=O, CO₂ two C=O and H₂O two O–H.
Step-by-step
WE KNOW
D(C–H) = 413 kJ/mol, D(O=O) = 495 kJ/mol, D(C=O) = 799 kJ/mol, D(O–H) = 463 kJ/mol
FIND
energy taken in to break the bonds (B), energy given off forming the bonds (F), enthalpy change of the reaction (ΔH)
Step 1 · Find energy taken in to break the bonds (B)
Use B = 4 × D(C–H) + 2 × D(O=O)
Break every bond in the reactants: four C–H in CH₄ and one O=O in each of the two O₂.
B
=
4
×
413
+
2
×
495
B
=
1,652
+
990
B
=
2,642
kJ/mol
Step 2 · Find energy given off forming the bonds (F)
Use F = 2 × D(C=O) + 4 × D(O–H)
Form every bond in the products: two C=O in CO₂ and two O–H in each of the two H₂O.
F
=
2
×
799
+
4
×
463
F
=
1,598
+
1,852
F
=
3,450
kJ/mol
Step 3 · Find enthalpy change of the reaction (ΔH)
Use ΔH = B − F
Energy in to break bonds less energy out from forming them. More out than in means ΔH < 0.
ΔH = 2,642 − 3,450
ΔH = −808 kJ/mol
Check
Put the numbers back in. Both sides should match.
2,642 = 4 × 413 + 2 × 495 ✓
3,450 = 2 × 799 + 4 × 463 ✓
−808 = 2,642 − 3,450 ✓
Topic check
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