Hess's law and enthalpy cycles

One start, one finish

The enthalpy change from one set of substances to another is the same whatever route is taken between them. Carbon shows it. Graphite burned in plenty of oxygen goes straight to carbon dioxide, releasing 394 kJ for each mole. In a limited supply some stops at carbon monoxide, which burns on to carbon dioxide later with its blue flame. The two steps release 111 kJ and 283 kJ, and their sum is 394 kJ.

C(s) + ½O₂(g) → CO(g) ΔH = −111 kJ mol⁻¹

CO(g) + ½O₂(g) → CO₂(g) ΔH = −283 kJ mol⁻¹

C(s) + O₂(g) → CO₂(g) ΔH = −394 kJ mol⁻¹The sum of the first two equations, and the sum of their ΔH values.

The agreement is not luck. Enthalpy depends only on the state a substance is in, not on how it got there. A mole of carbon and a mole of oxygen have one enthalpy; a mole of CO₂ under the same conditions has another; the difference between them is fixed. That is Hess's law.

If two routes between the same start and finish could differ, a cycle could run forwards by one route and back by the other and release energy from nothing on every turn. Hess's law is the law of conservation of energy applied to chemical reactions.

The first of the three equations shows why the law is useful. The enthalpy of formation of carbon monoxide cannot be measured directly: carbon burned in limited oxygen gives a mixture of CO, CO₂ and unburned carbon, never pure CO. The two measurable combustions give it by subtraction: −394 − (−283) = −111 kJ mol⁻¹.

Hess's law

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

Enthalpy cycle

A diagram showing two or more routes between the same reactants and products, with the enthalpy change for each step, used to calculate an unknown enthalpy change.

Cycles from enthalpies of formation

Every compound can be made, on paper, from its elements. So for any reaction there is a route from the reactants back down to their elements and up again to the products, and the enthalpies of formation of everything involved give the enthalpy change for that route.

Draw the reaction across the top, reactants on the left and products on the right. Write the elements, balanced, in a box underneath. Both arrows from the elements point up: they are the formation of the reactants and the formation of the products. Going the indirect way, from the reactants down to the elements runs against an arrow, so its sign changes; from the elements up to the products runs with one.

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

Describe the drawing

A Hess cycle. Across the top, iron(III) oxide and carbon monoxide on the left, iron and carbon dioxide on the right, joined by an arrow labelled ΔrH. Below, a box holds the elements, two iron, three carbon and three oxygen, with an arrow up to each side: minus 1157 to the reactants and minus 1182 to the products. A dashed route runs down from the reactants and up to the products, giving ΔrH = −25 kJ mol⁻¹.

Each ΔfH° is multiplied by the number of moles of that substance in the equation. Any element in its standard state contributes zero, so O₂, Fe and H₂ never need a value. A missing coefficient costs the method mark for that sum: three moles of CO₂ contribute 3 × (−394), not −394.

Enthalpy of formation of an element

Zero for an element in its standard state, because forming the element from itself involves no change.

The answer is small compared with the values that went into it, and that is common. The reactants lie 1157 kJ below the elements and the products 1182 kJ below them; ΔrH° is the small difference between those two depths. Checking that an answer is the right size for the difference between two sums catches most arithmetic slips.

An enthalpy of formation can be positive, and the formula handles it without change. For the hydrogenation of ethene, C₂H₄(g) + H₂(g) → C₂H₆(g), with ΔfH° of ethene +52 and ethane −84 kJ mol⁻¹, ΔrH° = −84 − (+52) = −136 kJ mol⁻¹. Combined with bond enthalpies, that value gives the strength of the C=C bond in ethene.

Cycles from enthalpies of combustion

Organic compounds are rarely made from their elements, but almost all of them burn, and enthalpies of combustion are easy to measure. So a second kind of cycle uses the combustion products as the common point instead of the elements.

Draw the reaction across the top again. Underneath, write the combustion products, CO₂(g) and H₂O(l), balanced. Now both arrows point down, because each is a combustion: the reactants burning and the products burning. Going the indirect way runs with the left arrow and against the right one, so the formula is the other way round.

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

Describe the drawing

A Hess cycle in the same layout as the formation cycle. Across the top, carbon and hydrogen on the left and methane on the right, joined by an arrow labelled the enthalpy of formation of methane. Both side arrows point down, with two oxygen added, into a box of carbon dioxide and liquid water: minus 966 from the elements and minus 890 from methane, giving ΔfH = −76 kJ mol⁻¹.

Which cycle to draw depends on which data a question gives: enthalpies of formation go in a cycle with the elements underneath, enthalpies of combustion go in a cycle with the combustion products underneath.

Cycles can also be done without a drawing, by adding equations. Write each equation whose ΔH is known, reverse any that run the wrong way and change the sign of its ΔH, multiply any that are needed more than once, then add them so that everything except the target equation cancels. For methane, the combustions of carbon and of two moles of hydrogen are added as written, and the combustion of methane is reversed. Adding those three leaves C(s) + 2H₂(g) → CH₄(g), with ΔH = −394 − 572 + 890 = −76 kJ mol⁻¹, as before.

Either method gives the same number, and a mark scheme accepts either. The drawing makes the direction of each step visible, which makes a dropped sign easier to spot.

Hydrating copper(II) sulfate

White anhydrous copper(II) sulfate turns blue when water is dripped onto it, and warms noticeably as it does. That is the test for water. The reaction makes the hydrated salt:

CuSO₄(s) + 5H₂O(l) → CuSO₄·5H₂O(s)

Its enthalpy change cannot be measured directly. Water added to the solid does not stop at five moles per mole: some of the salt dissolves, some stays anhydrous inside lumps, and the reaction never finishes cleanly. What can be measured is the enthalpy change when each solid dissolves in a large excess of water, because both end as the same solution.

Water of crystallisation

Water molecules held in fixed proportion within the crystal lattice of a hydrated salt.

Anhydrous

Describes a salt with no water of crystallisation, such as white CuSO₄(s).

Enthalpy of solution

The enthalpy change when one mole of a solute dissolves in enough solvent that further dilution causes no further heat change.

Describe the drawing

A Hess cycle for hydrating copper sulfate. Across the top, white anhydrous copper sulfate with five water on the left and the blue hydrate on the right, joined by a dashed arrow labelled ΔH hydration equals question mark. From each side an arrow points down into the same pale blue copper sulfate solution, each beside a small polystyrene cup: minus 65.9 measured on the left, plus 11.7 measured on the right, giving −77.6 kJ mol⁻¹.

The route from anhydrous salt to hydrated salt via the solution goes with the left arrow and against the right one. So the hydration enthalpy is the first enthalpy of solution minus the second. The same pattern, two measured reactions that end in the same place, is used for other changes that cannot be measured directly. The decomposition of potassium hydrogencarbonate is found from the reactions of potassium hydrogencarbonate and of potassium carbonate with hydrochloric acid.

This is not the enthalpy of hydration of an ion, which starts from gaseous ions and is the quantity used in Born–Haber and solution cycles. Here the word means adding water of crystallisation to a solid.

The data-book value is −78 kJ mol⁻¹. A school result is rarely that close, and the cycle shows why. Heat lost while the anhydrous salt dissolves makes ΔH₁ less negative, and so makes the hydration less exothermic. While the hydrated salt dissolves the solution is colder than the room, so heat flows in, the fall is smaller, and ΔH₂ is less positive, which pushes the answer the other way. The two errors partly cancel.

Two other errors do not cancel. Anhydrous copper(II) sulfate takes up water from the air, so a sample left open before weighing is partly hydrated already and releases less heat than its mass suggests. And a fall of 1.4 °C read on a thermometer to ±0.1 °C carries an uncertainty of ±0.2 °C, which is 14% of ΔT: the smaller measured step limits the precision of the whole result.

Decomposing potassium hydrogencarbonate: two reactions, one cycle

Potassium hydrogencarbonate breaks down on strong heating to potassium carbonate, carbon dioxide and steam. A solid being heated strongly cannot sit in a polystyrene cup, and the decomposition is rarely complete, so its enthalpy change is not measured directly. Both solids fizz quickly in dilute hydrochloric acid, though, and both reactions end in the same place: potassium chloride solution, carbon dioxide and water. Edexcel sets this pair as core practical 8, and any board can set it as a Hess calculation with results given.

2KHCO₃(s) → K₂CO₃(s) + CO₂(g) + H₂O(l)The target. Two moles of hydrogencarbonate give one of carbonate.

KHCO₃(s) + HCl(aq) → KCl(aq) + CO₂(g) + H₂O(l) ΔH₁Measured: the acid cools.

K₂CO₃(s) + 2HCl(aq) → 2KCl(aq) + CO₂(g) + H₂O(l) ΔH₂Measured: the acid warms.

In the cycle the target runs across the top. Below it sits 2KCl(aq) + 2CO₂(g) + 2H₂O(l), what two moles of hydrogencarbonate become in acid. The left arrow is 2ΔH₁, because the target starts with two moles of KHCO₃ and ΔH₁ is per mole. The right arrow is ΔH₂: the carbonate reacts with the acid, while the CO₂ and water already on that side pass through unchanged. The indirect route runs with the left arrow and against the right one, so ΔH = 2ΔH₁ − ΔH₂.

Thermal decomposition

The breaking down of a compound into simpler substances by heating it.

Indirect determination

Finding an enthalpy change that cannot be measured directly by combining, in a Hess cycle, enthalpy changes that can be measured.

Enthalpies of formation give the value to compare it with. With ΔfH° / kJ mol⁻¹ of KHCO₃(s) −963, K₂CO₃(s) −1151, CO₂(g) −394 and H₂O(l) −286, the products total −1151 − 394 − 286 = −1831 and the reactants 2 × (−963) = −1926, so ΔH = −1831 − (−1926) = +95 kJ mol⁻¹.

The measured +89 kJ mol⁻¹ is about 6% short of +95, and both measured steps err in the direction that explains the shortfall. In reaction 1 the acid is colder than the room, so heat flows in, the fall is too small, and ΔH₁ is too small; doubled in the cycle, that error counts twice. In reaction 2 heat is lost, the rise is too small, and ΔH₂ is not negative enough. Both make 2ΔH₁ − ΔH₂ smaller than it should be.

Other boards: the practicals built on these cycles

OCR assesses enthalpy determination in PAG 3, where either pair, the potassium salts or the copper(II) sulfate salts, can be used to find an enthalpy change that cannot be measured directly. Cambridge can set the same indirect determination in its practical papers, with results to process and a cycle to draw.