HomeLearning HubA Level ChemistryAS 5: Chemical energetics
AS 5

Chemical energetics

AS Level · Physical chemistry · Papers 1, 2 and 3 · extended in A2 23

🎯What you need to be able to do

  • Recognise exothermic (ΔH negative) and endothermic (ΔH positive) reactions, and draw and interpret reaction pathway diagrams showing ΔH and the activation energy.
  • Define standard conditions and the standard enthalpy changes of reaction, formation, combustion and neutralisation.
  • Explain enthalpy changes as the balance between breaking and making bonds, and calculate ΔHr from bond energies, knowing which values are exact and which are averages.
  • Calculate enthalpy changes from experimental results using q = mcΔT and ΔH = −mcΔT/n.
  • Apply Hess’s law to construct energy cycles and calculate enthalpy changes that cannot be measured directly, including from bond energy data.

📚The chemistry

5.1 Enthalpy change, ΔH

Enthalpy change, ΔH, is the heat energy transferred in a reaction at constant pressure. The sign is written from the point of view of the chemicals:

  • Exothermic — energy is transferred to the surroundings; the products have less enthalpy than the reactants; ΔH is negative; the surroundings warm up. Combustion, neutralisation, respiration.
  • Endothermic — energy is taken from the surroundings; the products have more enthalpy; ΔH is positive; the surroundings cool down. Thermal decomposition, photosynthesis, dissolving ammonium nitrate.

Reaction pathway diagrams

A reaction pathway diagram plots enthalpy against the progress of the reaction. Reactants and products are drawn as horizontal levels; the curve between them rises to a peak. Label two things:

  • ΔH — an arrow from the reactant level to the product level (pointing down for exothermic, up for endothermic);
  • the activation energy, Ea — an arrow from the reactant level up to the top of the peak. It is the minimum energy colliding particles must have to react.

Ea is always measured from the reactants, and is always positive, even for an exothermic reaction. A catalyst lowers the peak but leaves both levels, and so ΔH, unchanged.

Two enthalpy against progress-of-reaction diagrams. In the exothermic one the products are below the reactants, with a downward arrow for delta H and an upward arrow from the reactants to the peak for the activation energy; a dashed catalysed curve has a lower peak between the same two levels. In the endothermic one the products are above the reactants and delta H points upwards.
Exothermic and endothermic pathways. Ea is measured from the reactants; a catalyst lowers the peak but not ΔH.

Standard enthalpy changes

Enthalpy changes depend on conditions, so they are quoted under standard conditions, shown by the symbol ⦵: this syllabus takes those to be 298 K and 101 kPa, with every substance in its standard state (its normal physical state under those conditions) and solutions at 1 mol dm−3.

  • Standard enthalpy change of reaction, ΔHr — the enthalpy change when the amounts shown in the equation react under standard conditions.
  • Standard enthalpy change of formation, ΔHf — the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions. For water: H2(g) + ½O2(g) → H2O(l). The ΔHf of an element in its standard state is zero by definition.
  • Standard enthalpy change of combustion, ΔHc — the enthalpy change when one mole of a substance is completely burned in oxygen under standard conditions. Always negative.
  • Standard enthalpy change of neutralisation, ΔHneut — the enthalpy change when an acid and an alkali react to form one mole of water under standard conditions. For any strong acid with any strong alkali it is about −57 kJ mol−1, because the reaction is always H+ + OH → H2O.

Notice what each definition fixes as “one mole”: the compound formed, the substance burned, the water formed. Fractions in equations such as ½O2 are therefore normal and correct.

Bond energies

Energy is transferred in reactions because bonds are broken and made. Breaking a bond needs energy (endothermic, ΔH positive); making a bond releases it (exothermic). If the bonds made are stronger than the bonds broken, the reaction is exothermic overall.

\[ \Delta H_\mathrm{r} = \sum(\text{energy of bonds broken}) - \sum(\text{energy of bonds made}) \]

Some bond energies are exact and some are averages. The bond in H2 or HCl exists in only one molecule, so its energy has one value. A C–H bond, though, has a slightly different energy in methane, ethane and ethanol, depending on what else is attached; the data section gives an average over many compounds. The data section itself makes the distinction: diatomic molecules have exact values, polyatomic ones average values. So a ΔHr calculated from average bond energies is only approximate. It also assumes every substance is a gas, because bond energies are defined for the gaseous state.

Measuring enthalpy changes: calorimetry

Carry out the reaction in, or under, a known mass of water and measure its temperature change. The energy transferred to the water is

\[ q = mc\Delta T \]

where m is the mass of water or solution being heated (in g, taking 1 cm3 of a dilute solution as 1 g), c = 4.18 J g−1 K−1 and ΔT is the temperature rise. Divide by the moles that reacted, and add the sign:

\[ \Delta H = -\frac{mc\Delta T}{n} \]

The minus sign is there because a temperature rise means the reaction released energy. Two standard set-ups:

  • Reactions in solution (neutralisation, displacement, dissolving): mix in an insulated polystyrene cup with a lid. m is the total volume of the solutions.
  • Combustion: burn a fuel in a spirit burner under a metal can of water; weigh the burner before and after. Heat losses to the air are large, so the value is always far less exothermic than the true one.

5.2 Hess’s law

Hess’s law: the total enthalpy change for a reaction is independent of the route taken, provided the starting and finishing conditions are the same. It follows from conservation of energy. It lets you calculate enthalpy changes that cannot be measured directly — because the reaction is too slow, produces a mixture, or cannot be made to happen at all — by going round an energy cycle.

Two cycles cover most questions:

  • From enthalpies of formation. Both reactants and products can be formed from the same elements, so \( \Delta H_\mathrm{r} = \sum \Delta H_\mathrm{f}(\text{products}) - \sum \Delta H_\mathrm{f}(\text{reactants}) \).
  • From enthalpies of combustion. Both reactants and products burn to the same combustion products, so \( \Delta H_\mathrm{r} = \sum \Delta H_\mathrm{c}(\text{reactants}) - \sum \Delta H_\mathrm{c}(\text{products}) \) — the other way round.

Rather than memorise which way round, draw the cycle every time and follow the arrows: go with an arrow, add its ΔH; go against it, subtract. Remember to multiply each value by the number of moles in the equation.

A standard example: ΔHf of propane cannot be measured, because carbon and hydrogen do not simply combine to give propane. But all three burn. With ΔHc: C −394, H2 −286, C3H8 −2220 kJ mol−1:

\[ \Delta H_\mathrm{f}(\mathrm{C_3H_8}) = 3(-394) + 4(-286) - (-2220) = -106\ \mathrm{kJ\ mol^{-1}} \]
A Hess's law triangle. Three carbon plus four hydrogen at top left and propane at top right, joined by an arrow labelled delta H f equals unknown. Both have arrows down to three carbon dioxide plus four water at the bottom: from the elements, 3 times minus 394 plus 4 times minus 286, which is minus 2326; from propane, minus 2220 kilojoules per mole. The result, minus 106, is written underneath.
The propane cycle drawn out: −2326 − (−2220) = −106 kJ mol−1.

✏️Worked example

A spirit burner containing ethanol, C2H5OH, is used to heat 200 g of water in a copper can. The mass of the burner falls by 0.920 g and the temperature of the water rises by 26.0 K. (a) Calculate the enthalpy change of combustion of ethanol from these results. (b) Use the bond energies in the data section to estimate the enthalpy change of combustion, assuming all species are gases. (c) The data-book value is −1367 kJ mol−1. Explain why both your answers differ from it. [c = 4.18 J g−1 K−1; bond energies / kJ mol−1: C–C 350, C–H 410, C–O 360, O–H 460, O=O 496, C=O in CO2 805]

(a) Energy transferred to the water:

\[ q = mc\Delta T = 200 \times 4.18 \times 26.0 = 21\,736\ \mathrm{J} = 21.74\ \mathrm{kJ} \]

Mr(C2H5OH) = 2(12.0) + 6(1.0) + 16.0 = 46.0, so n = 0.920 / 46.0 = 0.0200 mol.

\[ \Delta H_\mathrm{c} = -\frac{21.74}{0.0200} = -1087\ \mathrm{kJ\ mol^{-1}} \approx -1090\ \mathrm{kJ\ mol^{-1}} \]

(b) C2H5OH + 3O2 → 2CO2 + 3H2O. Ethanol contains 1 C–C, 5 C–H, 1 C–O and 1 O–H bond.

Broken: 350 + 5(410) + 360 + 460 + 3(496) = 4708 kJ
Made: 4(805) + 6(460) = 5980 kJ
\[ \Delta H_\mathrm{c} = 4708 - 5980 = -1272\ \mathrm{kJ\ mol^{-1}} \]

(c) The experimental value is much less exothermic because of heat losses: much of the energy from the flame heats the air and the can rather than the water, and some ethanol may burn incompletely (soot on the can) or evaporate from the wick. The bond-energy estimate differs because it uses average bond energies, not the values for the bonds in these particular molecules, and because it assumes ethanol and water are gases. Under standard conditions both are liquids, and condensing them releases extra energy — which is why the true value is more exothermic than the estimate.

Check it. Signs first: combustion is exothermic, so both answers must be negative, and the experiment warmed the water, which agrees. Magnitudes second: combustion enthalpies of small organic molecules are around −1000 to −2000 kJ mol−1, so an answer of −1.09 or −1 087 000 means a J/kJ slip.
Using the mass of the fuel in q = mcΔT. m is the mass of the water being heated — 200 g here, not 0.920 g. The mass of fuel belongs in n. The same mistake in solution calorimetry is using the mass of a solid added instead of the volume of the solution. And do not forget the minus sign: ΔH is the enthalpy change of the reaction, which lost the energy the water gained.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Define the standard enthalpy change of formation, and write the equation it refers to for ethanol, C2H5OH(l).
The standard enthalpy change of formation is the enthalpy change when one mole of a compound is formed from its elements in their standard states, under standard conditions (298 K, 101 kPa). For ethanol: 2C(s) + 3H2(g) + ½O2(g) → C2H5OH(l). Carbon is graphite, the standard state; the ½O2 is needed to make exactly one mole of product; and the state symbols are part of the answer.
2. 50.0 cm3 of 1.00 mol dm−3 HCl is mixed with 50.0 cm3 of 1.00 mol dm−3 NaOH in a polystyrene cup. The temperature rises by 6.8 K. Calculate the enthalpy change of neutralisation.
Total volume 100 cm3, so m = 100 g. q = 100 × 4.18 × 6.8 = 2842 J = 2.842 kJ. n(H2O) = n(HCl) = 1.00 × 0.0500 = 0.0500 mol. ΔHneut = −2.842 / 0.0500 = −57 kJ mol−1 (to 2 s.f., limited by the 6.8 K). The mass is the total volume of solution, 100 g, not 50 g: both solutions warm up.
3. Use bond energies from the data section to estimate the enthalpy change for CH4(g) + 2O2(g) → CO2(g) + 2H2O(g). [C–H 410, O=O 496, C=O in CO2 805, O–H 460 kJ mol−1]
Broken: 4 C–H + 2 O=O = 4(410) + 2(496) = 1640 + 992 = 2632 kJ. Made: 2 C=O + 4 O–H = 2(805) + 4(460) = 1610 + 1840 = 3450 kJ. ΔH = 2632 − 3450 = −818 kJ mol−1. The value measured with gaseous water is about −802 kJ mol−1; the difference is because C–H and O–H are average bond energies.
4. Sketch and label a reaction pathway diagram for an endothermic reaction, and show how a catalyst changes it.
Axes: enthalpy (y) against progress of reaction (x). Draw a reactant level on the left and a higher product level on the right. Join them with a curve rising to a peak above both. Label ΔH with an arrow pointing up from the reactant level to the product level (positive). Label Ea with an arrow from the reactant level up to the peak. For the catalysed reaction, draw a second curve between the same two levels with a lower peak: Ea is smaller, ΔH is unchanged.
5. Use these standard enthalpy changes of formation to calculate ΔHr for Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g). [ΔHf / kJ mol−1: Fe2O3(s) −824, CO(g) −111, CO2(g) −394]
ΔHr = ΣΔHf(products) − ΣΔHf(reactants). Products: 2(0) + 3(−394) = −1182. Reactants: (−824) + 3(−111) = −1157. ΔHr = −1182 − (−1157) = −25 kJ mol−1. Iron is an element in its standard state, so its ΔHf is zero — it still appears in the cycle, contributing nothing.
6. The enthalpy change for MgSO4(s) + 7H2O(l) → MgSO4·7H2O(s) cannot be measured directly. Explain why not, and how Hess’s law allows it to be found.
Adding exactly seven moles of water to the anhydrous salt does not cleanly give the solid hydrate: some of the salt dissolves, the reaction is slow and incomplete, and the temperature change cannot be tied to one process. Instead, measure two enthalpy changes that can be measured: dissolve the anhydrous salt in a large excess of water (ΔH1), and separately dissolve the hydrated salt in the same amount of water (ΔH2). Both give the same solution, so by Hess’s law the route anhydrous → hydrate → solution has the same total as anhydrous → solution. Therefore ΔH(hydration) = ΔH1 − ΔH2.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • Chemguide (Jim Clark) — the energetics pages, with Hess’s law cycles drawn out step by step
  • Royal Society of Chemistry — the practical guides for measuring enthalpy changes of neutralisation and combustion
  • PhET — Energy Forms and Changes, for the idea of energy flowing between a system and its surroundings