Cambridge AS Level Chemistry 9701

Chemical energetics

An enthalpy change is the heat taken in or given out at constant pressure. It is negative when a reaction releases heat (exothermic) and positive when it absorbs heat (endothermic) — and almost every mark lost in this topic is a lost minus sign.

Cambridge AS Chemistry 9701 topic 5 is 47 of the 920 Paper 1 questions Quanta has mapped. They come in three kinds: state a standard definition precisely, work an enthalpy change out from a Hess cycle, or work one out from bond energies. This page covers all three, and the definitions matter as much as the arithmetic — examiners mark them word by word.

Updated 22 September 2026

Exothermic and endothermic

	Exothermic	Endothermic
Heat	released to the surroundings	absorbed from the surroundings
Sign of ΔH	negative	positive
Products	lower in energy than reactants	higher in energy than reactants
Surroundings	get warmer	get colder
Examples	combustion, neutralisation	thermal decomposition, photosynthesis

The energy is accounted for by bonds: breaking bonds absorbs energy, making bonds releases it. A reaction is exothermic when the bonds formed are stronger overall than the bonds broken. That one sentence is the explanation the question wants whenever it asks why.

The standard definitions

Standard conditions are 298 K and 100 kPa, with every substance in its standard state — the state it is in under those conditions. The symbol is $Δ H^{⊖}$ .

Each definition below has phrases examiners look for. The killer is almost always one mole, and for formation it is in their standard states.

Formation, ΔHf

The enthalpy change when ONE MOLE of a compound is formed from its constituent elements in their standard states, under standard conditions.

It follows that the enthalpy of formation of any element in its standard state is zero — which is what makes Hess cycles work.

Combustion, ΔHc

The enthalpy change when ONE MOLE of a substance is completely burned in oxygen, under standard conditions.

“Completely” matters: incomplete combustion is a different reaction with a different value. Always negative.

Reaction, ΔHr

The enthalpy change when the molar quantities in the equation as written react, under standard conditions.

It is tied to the equation you were given, so doubling the equation doubles ΔH.

Neutralisation, ΔHn

The enthalpy change when an acid and a base react to form ONE MOLE of water, under standard conditions.

Per mole of water, not per mole of acid — a diprotic acid catches people out.

Atomisation, ΔHat

The enthalpy change when ONE MOLE of gaseous atoms is formed from the element in its standard state.

Always endothermic: you are pulling an element apart.

Hess's law

The total enthalpy change for a reaction is the same whatever route is taken, provided the start and end states are the same. Enthalpy is a state function — it depends on where you are, not how you got there — so an enthalpy change you cannot measure directly can be found by going round a cycle of ones you can.

Two shortcuts follow, and they are worth memorising as they stand:

Δ H_{r} = \sum Δ H_{f} (products) - \sum Δ H_{f} (reactants)

Δ H_{r} = \sum Δ H_{c} (reactants) - \sum Δ H_{c} (products)

Note that they run opposite ways round. With formation data the arrows point up from the elements to both sides, so it is products minus reactants. With combustion data the arrows point down to the combustion products from both sides, so it is reactants minus products. Mixing them up is the single most common error in the topic, and the wrong answer it produces is always on the option list.

And multiply each value by its coefficient in the balanced equation before adding — two moles of water contributes $2 \times Δ H_{f}$ .

Worked example

Use ΔHf values to find the enthalpy change for CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l). ΔHf: CH₄ −74.8, CO₂ −393.5, H₂O(l) −285.8 kJ mol⁻¹.

Spot the element. $O_{2} (g)$ is an element in its standard state, so its enthalpy of formation is zero. It contributes nothing, whatever its coefficient.

Apply products minus reactants, each multiplied by its coefficient:

Δ H_{r} = [(- 393.5) + 2 (- 285.8)] - [(- 74.8) + 2 (0)]

= (- 393.5 - 571.6) - (- 74.8) = - 965.1 + 74.8 = - 890.3 k J m o l^{- 1}

Negative, so exothermic — which is what combustion had to be, and the quickest check that the subtraction went the right way round. Had the terms been reversed the answer would have been $+ 890.3$ , which claims methane absorbs heat when it burns.

Bond energies

A bond energy is the energy needed to break one mole of a bond in the gaseous state. Because a C–H bond in methane is not identical to one in ethanol, tabulated values are averages — and that word is what a question means when it asks why a bond-energy answer differs from the true value.

Δ H = \sum (bonds broken) - \sum (bonds formed)

Breaking is endothermic and comes first with a plus; forming is exothermic and is subtracted. Taking methane again, with C–H 410, O=O 496, C=O 805 and O–H 465 kJ mol⁻¹:

broken = 4 (410) + 2 (496) = 1640 + 992 = 2632

formed = 2 (805) + 4 (465) = 1610 + 1860 = 3470

Δ H = 2632 - 3470 = - 838 k J m o l^{- 1}

Compare that with the $- 890.3$ from the Hess cycle above. The gap is real and examinable: bond energies are averages, and this calculation also assumes every substance is gaseous — while the water in the real reaction condenses to a liquid, releasing more energy still.

Measuring it: q = mcΔT

q = m c Δ T

$m$ is the mass of the solution or water being heated in grams — not the mass of the fuel or the solid dissolved. $c$ is the specific heat capacity, 4.18 J g⁻¹ K⁻¹ for dilute aqueous solutions, and $Δ T$ the temperature change in K (the same number as in °C, since only the difference matters).

Then, to turn joules of heat into an enthalpy change per mole:

Δ H = - \frac{q}{n} {(in kJ mol}^{- 1}, so divide q by 1000)

The minus sign is the step people skip. If the temperature rose, the reaction gave out heat, so $Δ H$ is negative even though $Δ T$ was positive. And $n$ is the moles of the limiting reagent — see moles and stoichiometry.

Why the experimental value is always too small

Heat lost to the surroundings, heat absorbed by the apparatus, incomplete combustion, and evaporation of the fuel. Every one of those makes the measured enthalpy change less exothermic than the true value, and a question asking for an improvement wants insulation, a lid, or a bomb calorimeter.

Common mistakes

1.Formation and combustion cycles run the same way
Formation: products − reactants. Combustion: reactants − products. Draw the arrows before substituting and the direction is obvious.
2.Omitting 'one mole' from a definition
Every standard enthalpy definition is per mole of something specific — one mole of the compound, one mole of water, one mole of gaseous atoms. Leaving it out loses the mark on its own.
3.Giving an element a non-zero ΔHf
An element in its standard state has $Δ H_{f} = 0$ by definition. O₂, N₂, solid carbon — all zero, whatever coefficient they carry.
4.Bond energies subtracted the wrong way
Broken minus formed. Breaking costs energy, so it is the positive term. Reversing it flips the sign of every answer.
5.Using the mass of the fuel in q = mcΔT
$m$ is the mass of the water or solution absorbing the heat. The fuel is what you divide by afterwards, in moles.
6.Forgetting the minus when converting q to ΔH
A temperature rise means an exothermic reaction, so $Δ H$ is negative. Check the sign against the physical situation before writing the final line.

Common questions

What is the standard enthalpy change of formation?

The enthalpy change when one mole of a compound is formed from its constituent elements in their standard states, under standard conditions (298 K, 100 kPa). It follows that an element in its standard state has a formation enthalpy of zero.

What is Hess's law?

That the total enthalpy change of a reaction is independent of the route taken, as long as the initial and final states are the same. It lets you calculate enthalpy changes that cannot be measured directly, by going round a cycle of ones that can.

Why do bond-energy calculations give a different answer?

Because tabulated bond energies are averages across many different compounds, and the calculation assumes every species is gaseous. Where the real reaction produces a liquid, more energy is released than the calculation predicts — which is why methane’s combustion comes out at about −838 by bond energies and −890 from a Hess cycle.

Is ΔH negative or positive for an exothermic reaction?

Negative. The system loses energy to the surroundings, which get warmer. Endothermic reactions have a positive $Δ H$ and cool their surroundings.

Practise chemical energetics against real mark schemes

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