Coaching Centre
Hess’s law
Hess's law as a consequence of conservation of energy, the stepped reaction of carbon to carbon dioxide via carbon monoxide, rules for manipulating thermochemical equations, solving Hess's law problems, and the heat of formation of magnesium oxide investigation.
Try measuring the enthalpy change when carbon burns to carbon monoxide. You can't: burn carbon in limited oxygen and you get a mixture of CO, CO₂ and leftover carbon, with no way of telling how much heat came from which. Yet the data book lists the value to one decimal place: −110.5 kJ mol⁻¹. It was never measured directly. It was calculated, from two reactions that can be measured, using a law published by Germain Hess in 1840.
Hess's law
Hess's law: the enthalpy change for a reaction is the same whether it happens in one step or in several steps. It depends only on the initial and final states, not on the route.
This isn't a separate law of nature, it's the law of conservation of energy in disguise. If two routes between the same reactants and products had different ΔH values, you could go out by one route and back by the other and create energy from nothing. Enthalpy behaves like height on a mountain: the height you gain from the car park to the summit is the same whether you take the steep track or the zigzag.
A stepped reaction: carbon to carbon dioxide
Carbon can burn straight to CO₂, or in two steps via CO:
$$\begin{aligned} &\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) && \Delta H = -393.5 \text{ kJ} \\ &\text{C}(s) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g) && \Delta H_1 = \;? \\ &\text{CO}(g) + \tfrac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) && \Delta H_2 = -283.0 \text{ kJ} \end{aligned}$$
The first and third reactions can be measured in a bomb calorimeter, because both go to completion in excess oxygen. Hess's law says the two-step route must give the same total: $\Delta H_1 + \Delta H_2 = -393.5$, so $\Delta H_1 = -393.5 - (-283.0) = -110.5$ kJ mol⁻¹.
You can check the same stepped reaction with bond energies (4E). CO contains a triple bond (1072 kJ mol⁻¹), so for CO + ½O₂ → CO₂: broken 1072 + ½(498) = 1321 kJ, formed 2(805) = 1610 kJ, giving −289 kJ, close to the measured −283.0 kJ. The steps involving solid carbon can't be done this way, because bond energies apply only to gases.
Manipulating thermochemical equations
The strategy for any problem:
- Write the target equation at the top.
- For each substance in the target, find a given equation that contains it. Reverse that equation if the substance is on the wrong side; multiply it to get the right coefficient.
- Add up the adjusted equations, cancelling anything that appears on both sides. Check you get exactly the target.
- Add up the adjusted ΔH values.
Calculate ΔH for C(graphite) → C(diamond), given:
(1) C(graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ
(2) C(diamond) + O₂(g) → CO₂(g), ΔH = −395.4 kJ
Find ΔH for 2C(s) + H₂(g) → C₂H₂(g), given:
(1) C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ
(2) H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ
(3) C₂H₂(g) + 5/2 O₂(g) → 2CO₂(g) + H₂O(l), ΔH = −1299.6 kJ
- Puzzle CO from C: press Reverse (2). Does the sum match the target? What is ΔH?
- Puzzle Methane: you need ×2 on one equation and a reversal on another. The answer is −74.8 kJ mol⁻¹, the enthalpy of formation of methane.
- Puzzle NO₂ from N₂: here nothing needs reversing. Why does NO cancel?
- Puzzle Ethyne: get a match and check it against the worked example (+226.8 kJ).
Testing Hess's law in the lab
A second classic check uses sodium hydroxide: dissolve solid NaOH in water and then neutralise the solution with HCl (two steps), or add solid NaOH directly to the acid (one step). Within experimental error the two routes give the same total ΔH, about −102 kJ mol⁻¹. (Neutralisation itself is studied in detail in Year 12 Module 6.)
Test yourself
Given S(s) + O₂(g) → SO₂(g), ΔH = −296.8 kJ, and 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = −197.8 kJ, what is ΔH for S(s) + 3/2 O₂(g) → SO₃(g)?
Keep the first (−296.8) and halve the second (−98.9): −296.8 − 98.9 = −395.7 kJ.
Hess's law is a consequence of
If two routes had different ΔH you could create energy by going one way and returning the other.
A + B → C, ΔH = −80 kJ; C → D, ΔH = +30 kJ. What is ΔH for D → A + B?
A + B → D has ΔH = −80 + 30 = −50 kJ; reversing gives +50 kJ.
Why must ΔfH of carbon monoxide be found indirectly?
You can't stop the reaction cleanly at CO, so the measured heat would come from an unknown mixture of products.
Label the enthalpy changes in the carbon cycle.
Put the steps for solving a Hess's law problem in order.
Hess's law states that the enthalpy change of a reaction is independent of the taken. Reversing an equation changes the of ΔH, and multiplying an equation by 2 makes ΔH as large. ΔH for C(s) + ½O₂(g) → CO(g) is kJ mol⁻¹.
Use the following data to calculate ΔH for the formation of ethanol from its elements, 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l). Show all working.
(1) C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ
(2) H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ
(3) C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH = −1367 kJ
2 × (1): 2C(s) + 2O₂(g) → 2CO₂(g), ΔH = −787.0 kJ
3 × (2): 3H₂(g) + 3/2 O₂(g) → 3H₂O(l), ΔH = −857.4 kJ
Reverse (3): 2CO₂(g) + 3H₂O(l) → C₂H₅OH(l) + 3O₂(g), ΔH = +1367 kJ
Adding: 2C + 3H₂ + 7/2 O₂ + 2CO₂ + 3H₂O → 2CO₂ + 3H₂O + C₂H₅OH + 3O₂. Cancelling 2CO₂, 3H₂O and 3O₂ leaves 2C(s) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l), the target.
ΔH = −787.0 − 857.4 + 1367 = −277 kJ mol⁻¹ (data value −277.6 kJ mol⁻¹).