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Physical chemistry · Lesson 3 of 7

Hess's law

Enthalpy is a state function: the change depends only on where you start and finish, not the route. So if a reaction can be written as the sum of other reactions, its ΔH is the sum of theirs. That's Hess's law, and it lets you find heats of reactions that are hard to measure.

Carbon burning only partly, to carbon monoxide, is hard to measure because some always goes on to CO₂. But C + O₂ → CO₂ has ΔH = −393.5 kJ, and CO + ½O₂ → CO₂ has −283.0 kJ. Reverse the second (which flips its sign to +283.0) and add: C + ½O₂ → CO, ΔH = −393.5 + 283.0 = −110.5 kJ. If you double a reaction, double its ΔH.

Tables list each compound's standard enthalpy of formation, ΔHf°: the ΔH for making one mole of it from its elements in their normal forms at 25 °C and 1 bar. Elements themselves (O₂, N₂, solid carbon) are zero. Then ΔH of any reaction = sum of ΔHf° of products − sum of ΔHf° of reactants, each times its coefficient.

For methane burning: ΔHf° of CO₂ is −393.5, liquid water −285.8 and methane −74.8 kJ/mol. ΔH = [−393.5 + 2 × (−285.8)] − [−74.8 + 0] = −890.3 kJ/mol. If the water stays as steam (ΔHf° −241.8), it's −802.3: the difference is the heat of condensing two moles of water, which is why boilers that condense their exhaust are more efficient.

Without a table you can estimate from average bond energies: ΔH ≈ energy of bonds broken − energy of bonds made. It's rougher, because a bond's strength varies from molecule to molecule, but it shows where the energy comes from.

Practice

Reaction A has ΔH = −100 kJ. What is ΔH for the same reaction run backwards?

More info

Quick reference

Hess's lawΔH of a sum = sum of the ΔH's.
Reverse a reactionFlip the sign of ΔH.
ΔHf°ΔH to make 1 mol from elements; elements = 0.
ΔH reactionΣ ΔHf°(products) − Σ ΔHf°(reactants)
Bond energiesΔH ≈ broken − made (estimate)

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