Edexcel A-level Chemistry (9CH0) · Topic 8: Energetics I
Mini-Lesson
Energetics I
This mini-lesson covers the whole of Edexcel Topic 8: enthalpy changes and standard conditions, the standard enthalpies of formation, combustion and neutralisation, Hess's law cycles, mean bond enthalpies, and calorimetry — including where the errors come from.
Work through each screen, answer the questions as you go (several are calculations) and collect ⭐ stars. Every calculation is worked through for you first. Press Start when you're ready.
Enthalpy
Enthalpy change and standard conditions
Enthalpy (H) is the heat content of a system at constant pressure. You cannot measure H itself — only the change, ΔH, when a reaction happens.
Because enthalpy changes depend on conditions, we quote standard values, written ΔH⦵, measured under standard conditions:
a pressure of 100 kPa (1 bar);
a stated temperature, normally 298 K (25 °C);
all substances in their standard states — the physical state they exist in under those conditions (e.g. H₂O(l), Br₂(l), C as graphite);
any solutions at a concentration of 1 mol dm⁻³.
Exam habit: the "per mole" in kJ mol⁻¹ means per mole of reaction as written in the equation. If you double the equation, you double ΔH.
Exo and endo
Exothermic and endothermic
Every reaction breaks bonds (which costs energy — always endothermic) and makes bonds (which releases energy — always exothermic). The sign of ΔH just tells you which one won.
Exothermic (ΔH negative) — energy released to the surroundings; the products are lower in enthalpy than the reactants; the surroundings get hotter. Bond making releases more than bond breaking cost.
Endothermic (ΔH positive) — energy taken in from the surroundings; the products are higher in enthalpy; the surroundings get colder.
The vertical axis is enthalpy. ΔH is measured from reactants to products — the hump in between is the activation energy.Quick check
Standard conditions
1Which set of conditions is correct for a standard enthalpy change, ΔH⦵?
Definitions
The three enthalpy changes you must define
Standard enthalpy of formation, ΔfH⦵ — the enthalpy change when 1 mol of a compound is formed from its elements in their standard states, under standard conditions. e.g. 2C(s, graphite) + 3H₂(g) + ½O₂(g) → C₂H₅OH(l)
Standard enthalpy of combustion, ΔcH⦵ — the enthalpy change when 1 mol of a substance is burned completely in excess oxygen, under standard conditions. e.g. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l)
Standard enthalpy of neutralisation, ΔneutH⦵ — the enthalpy change when an acid and an alkali react to form 1 mol of water, under standard conditions. H⁺(aq) + OH⁻(aq) → H₂O(l), ΔH ≈ −57 kJ mol⁻¹ for strong acid + strong alkali
Key consequence: ΔfH⦵ of an element in its standard state is zero — "forming" O₂(g) from O₂(g) involves no change at all.
Why neutralisation is nearly constant for strong acid + strong alkali: both are fully ionised, so the reaction is always the same one — H⁺(aq) + OH⁻(aq) → H₂O(l). With a weak acid, energy is absorbed to ionise it further, so the value is less exothermic.
Quick check
Which equation is a formation equation?
2Which equation represents the standard enthalpy of formation of methane, CH₄(g)?
Calorimetry
Measuring ΔH: q = mcΔT
In the lab you cannot measure enthalpy — you measure a temperature change in a known mass of water (or solution) and convert it into heat energy.
q = m c ΔTq = heat energy in J · m = mass of solution in g · c = 4.18 J g⁻¹ K⁻¹ · ΔT = temperature change in K (= °C)
ΔH = − q ÷ nn = moles of the limiting reactant (or the moles of fuel burned). Divide q by 1000 to get kJ.
The minus sign is doing real work: if the solution warms up it gained heat, so the reaction lost it — ΔH is negative.
Worked example — an exothermic displacement
50.0 cm³ of 0.200 mol dm⁻³ CuSO₄(aq) is reacted with excess zinc powder. The temperature rises by 10.2 °C.
q = 50.0 × 4.18 × 10.2 = 2131.8 J = 2.1318 kJ (mass of solution ≈ 50.0 g, since density ≈ 1 g cm⁻³)
n(CuSO₄) = 0.0500 dm³ × 0.200 mol dm⁻³ = 0.0100 mol (zinc is in excess, so CuSO₄ is limiting)
ΔH = −2.1318 ÷ 0.0100 = −213 kJ mol⁻¹ (3 s.f.)
Calculate
Your turn — heat released
350.0 cm³ of 1.00 mol dm⁻³ HCl is mixed with 50.0 cm³ of 1.00 mol dm⁻³ NaOH. The temperature rises by 6.80 °C. Calculate q, the heat energy released, in J. Take the density of the mixture as 1.00 g cm⁻³ and c = 4.18 J g⁻¹ K⁻¹.
J
Hint: the mass is the total mixture, 50.0 + 50.0 = 100 g. q = 100 × 4.18 × 6.80.
Calculate
Your turn — enthalpy of neutralisation
4Using q = 2842 J from the last screen, calculate ΔneutH in kJ mol⁻¹ to 3 significant figures. Remember to include the sign.
kJ mol⁻¹
Hint: n(HCl) = 0.0500 × 1.00 = 0.0500 mol, and n(NaOH) is the same, so 0.0500 mol of H₂O forms. ΔH = −(2.842 kJ) ÷ 0.0500 mol.
Hess's law
Hess's law — the enthalpy short cut
Hess's law: the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same.
That lets you calculate a ΔH you could never measure directly (like the formation of methane from its elements) by going the long way round through data you do have.
With formation data the arrows point up from the elements — so ΔrH = ΣΔfH(products) − ΣΔfH(reactants).
ΔrH = ΣΔfH(products) − ΣΔfH(reactants)with combustion data the arrows point down, so it flips: ΔrH = ΣΔcH(reactants) − ΣΔcH(products)
Never forget the coefficients. If the equation has 2H₂O(l), you use 2 × ΔfH(H₂O, l).
A mean bond enthalpy is the energy needed to break 1 mol of a given covalent bond in the gaseous state, averaged over a range of different compounds. Bond breaking is always endothermic, so the values are always positive.
ΔrH = Σ(bonds broken) − Σ(bonds made)i.e. energy in − energy out. Everything must be gaseous for this to be valid.
Worked example — combustion of methane (all gaseous)
Why isn't that −890? Because these are mean values (averaged over many compounds, so not exact for this one), and because the equation makes H₂O(g), not H₂O(l) — condensing the water would release more energy still.
Calculate
Your turn — bond enthalpy calculation
7Calculate ΔrH for H₂(g) + Cl₂(g) → 2HCl(g) in kJ mol⁻¹. Bond enthalpies: H–H = 436, Cl–Cl = 242, H–Cl = 431.
8A ΔH calculated from mean bond enthalpies rarely matches the value from a Hess cycle using ΔfH data. The best reason is that…
Practical
Burning a fuel: the calorimetry practical
A spirit burner containing a liquid fuel is weighed, then used to heat a known mass of water in a copper calorimeter. You record the mass of fuel burned and the temperature rise of the water.
Worked example — burning methanol
0.480 g of methanol (CH₃OH, Mr = 32.0) raises the temperature of 100 g of water by 24.0 °C.
q = 100 × 4.18 × 24.0 = 10 032 J = 10.032 kJ
n(CH₃OH) = 0.480 ÷ 32.0 = 0.0150 mol
ΔcH = −10.032 ÷ 0.0150 = −669 kJ mol⁻¹ (3 s.f.)
Data-book value: −726 kJ mol⁻¹. The experimental value is always less exothermic than the true one — see the next screen.
Calculate
Your turn — enthalpy of combustion
90.500 g of ethanol (C₂H₅OH, Mr = 46.0) is burned and heats 150 g of water by 22.0 °C. Calculate ΔcH in kJ mol⁻¹ (3 s.f., include the sign). c = 4.18 J g⁻¹ K⁻¹.
Heat loss to the surroundings — from the flame, the calorimeter walls and the top of the water. The biggest error by far, and it always makes ΔH appear less exothermic.
Heat absorbed by the apparatus — the copper can, thermometer and stirrer warm up too, but q = mcΔT only counts the water.
Incomplete combustion — a yellow, sooty flame means some carbon becomes CO or C rather than CO₂, releasing less energy. Again → less exothermic.
Evaporation of the fuel from the wick between weighings — the mass burned is overstated.
Non-standard conditions — the experiment is not done at 100 kPa and exactly 298 K, and the "water" in a neutralisation is treated as pure water (c = 4.18, density = 1.00).
Improvements: insulate (polystyrene cup + lid), shield the flame from draughts, use a bomb calorimeter, or use the extrapolation method — plot temperature against time and extrapolate the cooling line back to the moment of mixing to find the true maximum ΔT.
Quick check
Which way does the error push?
10A student's experimental ΔcH for ethanol is −1269 kJ mol⁻¹; the data-book value is −1367 kJ mol⁻¹. Which explanation fits?
Quick check
Bond breaking and bond making
11Which statement about bonds and enthalpy is correct?
Quick check
A weak acid
12Neutralising NaOH with ethanoic acid gives ΔneutH = −56.1 kJ mol⁻¹, whereas HCl gives −57.9 kJ mol⁻¹. Why is the weak-acid value less exothermic?
Sort it
Exothermic, endothermic — or zero?
Tap a card, then tap the bin its enthalpy change belongs in.
🔥 Exothermic (ΔH < 0)
❄️ Endothermic (ΔH > 0)
0️⃣ ΔH = 0 by definition
Match it
Match the term to its definition
Tap an item on the left, then its partner on the right.
Term
Definition
Recap
The big ideas to know
ΔH: heat change at constant pressure; negative = exothermic, positive = endothermic; units kJ mol⁻¹
Standard conditions: 100 kPa, stated T (usually 298 K), standard states, solutions at 1 mol dm⁻³
ΔfH: 1 mol of compound from its elements in standard states — and it is zero for an element
ΔcH: 1 mol of substance burned completely in excess O₂ · ΔneutH: 1 mol of H₂O formed