Hess’s Law from Combustion Data

Learning Objectives (OCR Spec 3.2.1)

By the end of this lesson you should be able to:

• Construct a Hess cycle using standard enthalpy changes of combustion (ΔcH°)
• Apply the algebraic shortcut ΔrH° = ΣΔcH°(reactants) − ΣΔcH°(products)
• Compare and contrast the combustion-cycle and formation-cycle approaches
• Solve problems involving organic compounds where only combustion data are tabulated

Why Combustion Cycles?

Enthalpies of combustion are easy to measure experimentally with a bomb calorimeter - the fuel is ignited in excess oxygen in a sealed steel vessel surrounded by water, and the temperature rise gives q directly. For this reason, extensive tables of ΔcH° exist for organic compounds, hydrogen, carbon and many other fuels.

If we want the enthalpy change for a reaction involving such substances - for example, the formation of ethanol from its elements - we can construct a Hess cycle in which the common intermediate is the set of combustion products (CO2, H2O, SO2 etc.). This is the second major type of Hess cycle you must master for OCR.

Constructing the Combustion Cycle

Unlike the formation cycle, the arrows now point away from reactants and products down towards the combustion products:

graph TD
  R[Reactants] -->|ΔrH° = ?| P[Products]
  R -->|ΣΔcH° reactants| C[Combustion products: CO2 + H2O ...]
  P -->|ΣΔcH° products| C

The indirect route from reactants to products is: combust the reactants (downward, ΔcH° of reactants) then reverse the combustion of the products (upward, −ΔcH° of products).

By Hess’s law:

ΔrH° = ΣΔcH°(reactants) − ΣΔcH°(products)

Note the reversal compared to the formation cycle. Many students mix these up, so always draw the arrows first.

Memory Aid

• Formation cycle: arrows go UP from elements to compounds → products minus reactants.
• Combustion cycle: arrows go DOWN to combustion products → reactants minus products.

A mnemonic: "F–UP" (formation = up arrows, products − reactants) and "C–DOWN" (combustion = down arrows, reactants − products).

Worked Example 1 – Enthalpy of Formation of Methane

Q: Use the following combustion data to calculate ΔfH°(CH4(g)):

Substance	ΔcH° / kJ mol⁻¹
C(s, graphite)	−393.5
H2(g)	−285.8
CH4(g)	−890.3

A:

Target equation: C(s) + 2H2(g) → CH4(g), ΔrH° = ΔfH°(CH4)

Using ΔrH° = ΣΔcH°(reactants) − ΣΔcH°(products):

ΔfH°(CH4) = [ΔcH°(C) + 2 × ΔcH°(H2)] − [ΔcH°(CH4)] = [−393.5 + 2 × (−285.8)] − [−890.3] = [−393.5 + (−571.6)] − [−890.3] = −965.1 + 890.3 = −74.8 kJ mol⁻¹

Matches the literature value. This is how combustion data yield formation data that cannot be measured directly (you cannot make methane cleanly from graphite and hydrogen in a calorimeter).

Worked Example 2 – Enthalpy of Hydrogenation of Ethene

Q: Use the data to find ΔrH° for C2H4(g) + H2(g) → C2H6(g).

Substance	ΔcH° / kJ mol⁻¹
C2H4(g)	−1411
H2(g)	−286
C2H6(g)	−1560

A:

ΔrH° = ΣΔcH°(reactants) − ΣΔcH°(products) = [−1411 + (−286)] − [−1560] = −1697 + 1560 = −137 kJ mol⁻¹

Exothermic - consistent with the fact that hydrogenation (industrial margarine production) releases heat.

Worked Example 3 – Fermentation of Glucose

Q: Fermentation: C6H12O6(s) → 2C2H5OH(l) + 2CO2(g). Calculate ΔrH° given:

Substance	ΔcH° / kJ mol⁻¹
C6H12O6(s)	−2803
C2H5OH(l)	−1367

(CO2 cannot be "combusted" further, so it does not appear in the cycle.)

A:

ΔrH° = ΣΔcH°(reactants) − ΣΔcH°(products) = [−2803] − [2 × (−1367) + 2 × 0] = −2803 − [−2734] = −2803 + 2734 = −69 kJ mol⁻¹