Question
Calculate ΔH for the combustion of methane:
CH4(g)+2O2(g)→CO2(g)+2H2O(l)
given the following enthalpies of formation (in kJ/mol):
- ΔHf∘(CH4)=−74.8
- ΔHf∘(CO2)=−393.5
- ΔHf∘(H2O,l)=−285.8
- ΔHf∘(O2)=0 (element in standard state)
Solution — Step by Step
The reaction enthalpy is products minus reactants:
ΔHrxn=∑nΔHf∘(products)−∑nΔHf∘(reactants)
Products side:
1×(−393.5)+2×(−285.8)=−393.5−571.6=−965.1kJ
Reactants side:
1×(−74.8)+2×0=−74.8kJ
ΔHrxn=−965.1−(−74.8)=−965.1+74.8=−890.3kJ/mol
Final: ΔH=−890.3kJ/mol.
Why This Works
Hess’s law lets us treat enthalpy as a state function — the total ΔH depends only on initial and final states, not the path. By imagining a hypothetical path that goes “products → elements → reactants” and adding the reverse, we get a clean accounting using formation enthalpies.
The negative sign confirms an exothermic reaction. Methane combustion releases 890 kJ per mole, matching real-world fuel data.
Alternative Method
Use bond enthalpies (slightly less accurate but quicker if formation values are not given):
ΔH≈∑(bonds broken)−∑(bonds formed)
For methane combustion, bond enthalpies give about −820 kJ/mol — close but not exact, because bond enthalpies are averages.
Common Mistake
Students forget to multiply by stoichiometric coefficients. H2O has a coefficient of 2, so its formation enthalpy contributes 2×(−285.8), not −285.8. Missing this single multiplication is the most common -2 marks in NEET thermochemistry.