The big idea: Instead of measuring every reaction in a calorimeter, chemists store the enthalpy of each substance as a standard value and add them up. Two of these matter here:
- Standard enthalpy of formation, ΔHf⊖ — the enthalpy change when one mole of a compound forms from its elements in their standard states. - Standard enthalpy of combustion, ΔHc⊖ — the enthalpy change when one mole of a substance is completely burned in oxygen.
Because enthalpy is a state function, you can take a 'paper' route through these values to find a reaction's ΔH⊖ without doing the experiment.
What 'standard' (⊖) means: Standard conditions are 100 kPa pressure and a stated temperature (usually 298 K), with all substances in their standard states.
- ΔHf⊖ of an element in its standard state is zero (e.g. O2(g), C(graphite)) — there is nothing to form. - ΔHc⊖ values are always negative — combustion is exothermic.
The formation route: form the products from elements, minus form the reactants from elements. The lower box is the common starting point — the elements in their standard states.
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To find a reaction's enthalpy change from formation data, take products minus reactants, each multiplied by its coefficient in the balanced equation.
- standard enthalpy change of the reaction (kJ mol⁻¹)
- standard enthalpy of formation of a substance (kJ mol⁻¹)
- sum over all species, each multiplied by its coefficient in the equation
Worked example — combustion of methane from ΔHf⊖
Calculate ΔH⊖ for CH4(g) + 2O2(g) → CO2(g) + 2H2O(l). Use ΔHf⊖: CH4(g) = −74.6, CO2(g) = −393.5, H2O(l) = −285.8 kJ mol⁻¹.
Solution
- Formula first — products minus reactants:
- ΔHf⊖ of O2(g) is zero (an element), so substitute the products (note 2 H2O) and the reactants:
- Add the products:
- Subtract the reactants — work it out:
Final answer
ΔH⊖ = −890.5 kJ mol⁻¹ (exothermic, as expected for combustion).
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When you are given combustion data instead, the cycle flips: take reactants minus products (the opposite of the formation rule), because everything burns down to the same products.
- standard enthalpy change of the reaction (kJ mol⁻¹)
- standard enthalpy of combustion of a substance (kJ mol⁻¹)
- sum over all species, each multiplied by its coefficient in the equation
Worked example — making ethanol from ΔHc⊖
Calculate ΔH⊖ for C2H4(g) + H2O(l) → C2H5OH(l). Use ΔHc⊖: C2H4(g) = −1411, C2H5OH(l) = −1367 kJ mol⁻¹. (Water does not combust.)
Solution
- Formula first — reactants minus products for combustion data:
- Only C2H4 and C2H5OH burn (H2O has no combustion enthalpy), so substitute:
- Work it out — keep the sign and unit:
Final answer
ΔH⊖ = −44 kJ mol⁻¹ (slightly exothermic).
Don't mix the two rules up: The sign is the only difference, and it is easy to swap:
- Formation data → ΔH⊖ = products − reactants. - Combustion data → ΔH⊖ = reactants − products.
A quick check: combustion of a fuel should come out negative. If your sign looks wrong, you have probably used the wrong rule.
How this is tested: Energy-cycle calculations are a staple of Paper 2, with the formation/combustion data supplied in a table.
- Paper 2: a part-marks 'calculate ΔH⊖' question — set up the sum, then substitute and solve. - Paper 1A (MCQ): identify the correct expression (which terms are products, which are reactants) or the sign of ΔH⊖.
Most marks are lost on stoichiometric coefficients and the element-is-zero rule — not on the arithmetic.
Three easy marks: (1) Write the correct sum before the numbers (products − reactants for ΔHf⊖). (2) Multiply each value by its coefficient in the balanced equation. (3) Set ΔHf⊖ = 0 for any element in its standard state.
IB-style question — combustion of propane
Calculate the standard enthalpy change for the complete combustion of propane: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l). Use ΔHf⊖: C3H8(g) = −104.7, CO2(g) = −393.5, H2O(l) = −285.8 kJ mol⁻¹.
Solution
- Formula first — products minus reactants:
- Use the coefficients 3 and 4 for the products; O2(g) is an element so its ΔHf⊖ = 0:
- Add the products:
- Subtract the reactants — work it out:
Final answer
ΔH⊖ = −2219 kJ mol⁻¹ (3 s.f.) — strongly exothermic.