Hess’s Law
Add, scale and reverse step enthalpies to get the ΔH of a reaction you cannot measure directly.
The sum of every step once each is scaled and, where needed, reversed.
−393.5 + 283 = -110.5 kJ
Hess’s law says the enthalpy change of a reaction is the same however you get there, so ΔH values of known steps can simply be added. Burning carbon gives −393.5 kJ and burning CO gives −283.0 kJ; reversing the second and adding gives −110.5 kJ for forming CO.
Why the route does not matter
Enthalpy is a state function: it depends only on where you start and where you finish, never on the path between. Walking up a hill by a winding track or a direct scramble leaves you at the same altitude, and the enthalpy difference between reactants and products behaves the same way.
That is more useful than it sounds, because many reactions cannot be measured directly. Carbon burning to carbon monoxide always produces some CO₂ as well, so the ΔH of that specific reaction cannot be isolated in a calorimeter. But carbon burning to CO₂ can be measured, and CO burning to CO₂ can be measured, and those two known steps reach the same destination by a different route.
Two operations, two rules
Only two things may be done to a step, and each has a consequence for its ΔH. Reversing a reaction flips the sign, because releasing heat one way means absorbing it the other. Multiplying a reaction by a coefficient multiplies ΔH by the same factor, because enthalpy is extensive — twice the substance releases twice the heat. Get those two right and the rest is addition.
reversing a step negates its ΔH; multiplying a step by n multiplies its ΔH by n
- 1 Write the target reaction. Here it is C(s) + ½O₂(g) → CO(g), which cannot be measured cleanly on its own.
- 2 Line up the known steps. C(s) + O₂ → CO₂ has ΔH = −393.5 kJ, and CO + ½O₂ → CO₂ has ΔH = −283.0 kJ.
- 3 Reverse any step pointing the wrong way. CO has to end up as a product, so flip the second step to CO₂ → CO + ½O₂ and its ΔH becomes +283.0 kJ.
- 4 Scale any step whose coefficients do not match. Both steps already use one carbon here, so neither needs multiplying.
- 5 Add the adjusted values. −393.5 + 283.0 = −110.5 kJ, the enthalpy of formation of carbon monoxide.
What each operation does to ΔH
Everything else about the equation stays as written.
| Operation | Effect on ΔH | Example |
|---|---|---|
| Reverse the reaction | Change the sign | −283.0 kJ becomes +283.0 kJ |
| Multiply by 2 | Multiply ΔH by 2 | −393.5 kJ becomes −787.0 kJ |
| Multiply by ½ | Halve ΔH | −283.0 kJ becomes −141.5 kJ |
| Add two steps | Add their ΔH values | −393.5 + 283.0 = −110.5 kJ |
| Leave a step unchanged | ΔH unchanged | −393.5 kJ stays as it is |
The shortcut through formation enthalpies
There is a standard alternative that avoids arranging steps by hand. Tabulated standard enthalpies of formation, ΔH°f, are the enthalpy of making one mole of a compound from its elements in their standard states. With those in hand, ΔH°rxn = ΣΔH°f(products) − ΣΔH°f(reactants), each multiplied by its coefficient in the balanced equation. It is Hess’s law with the steps pre-arranged for you.
Two details matter there. An element in its standard state has ΔH°f of exactly zero by definition — O₂ gas, graphite, solid iron — so those terms drop out. And the subtraction runs products minus reactants; reversing it is the commonest error and flips the sign of the whole answer.