Hess's law: the route does not matter

7 minAS Level5.2.1

AI-drafted, independently checked · not yet teacher-reviewed

By the end of this lesson

By the end you can state Hess's law, explain why it is true, and use it to find an enthalpy change by adding up other reactions instead of measuring the one you want.

A problem Hess's law solves

You can measure the enthalpy change of combustion of carbon easily: burn a known mass of carbon in oxygen and measure the heat released.

Now try to measure the enthalpy change of formation of carbon monoxide. You would need carbon and oxygen to combine to give only CO — but some always burns on to CO₂, so you can never get a clean measurement.

Hess's law lets you find it anyway, from reactions you can measure. That is its whole point: it turns impossible measurements into arithmetic.

Definition

Hess's law

The enthalpy change of a reaction is independent of the route taken, provided the initial and final conditions are the same.

Equivalently: the total enthalpy change is the same whether the reaction happens in one step or in several.

Why a route cannot matter

Enthalpy is a state function: its value depends only on the substance, its amount, its state and the conditions — not on how the substance got there.

Think of altitude. Walking from sea level to a hut at 1000 m costs the same height gain whether you take the steep path or the long gentle one. The path changes the effort and the distance, not the height difference.

Enthalpy is the "altitude" here: the enthalpy of the products minus the enthalpy of the reactants is fixed by the start and end points.

Where the analogy stops: you cannot choose to end at a different "altitude" without changing the products. Hess's law only works because the products are identical in every route.

The two moves you need

Once you accept that routes may be swapped, two rules do all the work.

Adding reactions adds their enthalpy changes. If step 1 has $\Delta H_1$ and step 2 has $\Delta H_2$ , the two-step route has $\Delta H_1 + \Delta H_2$ .
Reversing a reaction flips the sign. If $\text{A} \rightarrow \text{B}$ has $\Delta H = -50\ \text{kJ mol}^{-1}$ , then $\text{B} \rightarrow \text{A}$ has $\Delta H = +50\ \text{kJ mol}^{-1}$ .

Also, if you multiply the equation by a number, multiply $\Delta H$ by the same number — because enthalpy change is quoted per mole of the reaction as written.

Worked example: Worked example: finding an impossible measurement

Carbon burns to carbon dioxide, and carbon monoxide burns to carbon dioxide:

$\text{C(s)} + \text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \qquad \Delta H = -393.5\ \text{kJ mol}^{-1}$

$\text{CO(g)} + \tfrac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO}_2\text{(g)} \qquad \Delta H = -283.0\ \text{kJ mol}^{-1}$

Use these to find the enthalpy change of formation of carbon monoxide, $\text{C(s)} + \tfrac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CO(g)}$ .

Have a go first?

Quick check

Multiple choiceQuestion 1 · Recall

CalculateQuestion 2 · Recall

Answer the 3 questions above to continue.