9701 Chemistry · Topic 5 · AS + A Level
Energetics and Thermodynamics Cheat Sheet — A Level Chemistry 9701
Energetics rewards precision more than almost any other 9701 topic: the definitions have to be quoted exactly, the signs have to be right, and the cycle has to be drawn before the arithmetic starts. These two sheets take you from q = mcΔT and Hess’s law at AS through to Born–Haber cycles, entropy and ΔG = ΔH − TΔS at A2, with the full working shown for every example.
✅ 19 worked sections
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What’s on this cheat sheet
Energetics I — Enthalpy
01 · Enthalpy basics
Exothermic — heat released to the surroundings, ΔH negative, products lower in energy, temperature of surroundings rises. Endothermic — heat absorbed, ΔH positive.
Bond breaking is endothermic, bond making exothermic. A reaction is exothermic when the bonds formed are stronger than the bonds broken.
Standard conditions (°): 100 kPa, 298 K, solutions 1 mol dm⁻³, all substances in their standard states. Always give state symbols and units of kJ mol⁻¹.
02 · Definitions to quote
ΔHf° — enthalpy change when one mole of a compound forms from its elements in their standard states. Zero for an element.
ΔHc° — enthalpy change on complete combustion of one mole of a substance in excess oxygen.
ΔHneut° — enthalpy change when one mole of water forms from acid and base. About −57 kJ mol⁻¹ for strong acid + strong base.
Bond enthalpy — energy to break one mole of a bond in the gaseous state; the value quoted is a mean over many compounds.
03 · Calorimetry
m = mass of water or solution in g (never the fuel or the solid), c = 4.18 J g⁻¹ K⁻¹, ΔT in K or °C (same size). q comes out in J — divide by 1000 for kJ.
n = moles of the limiting reactant. The minus sign converts “heat gained by water” into “enthalpy change of the reaction”. Assume the solution has the density and specific heat capacity of water.
04 · Worked example — calorimetry
q = 150.0 × 4.18 × 22.5 = 14 108 J = 14.1 kJ
n = 0.850 ÷ 60.1 = 0.01414 mol
ΔHc = −14.1 ÷ 0.01414 = −999 kJ mol⁻¹
The data‑booklet value is −2021 kJ mol⁻¹; the shortfall is heat lost to the air and apparatus and incomplete combustion.
05 · Hess’s law
The enthalpy change of a reaction is independent of the route taken, because enthalpy is a state function. Add the steps of any complete path.
Reverse a step and the sign flips; multiply a step and the value multiplies. Arrows pointing the same way around a cycle add; one pointing against the cycle is subtracted.
06 · Worked example — Hess cycle
ΔHc[C(s)] = −394 and ΔHc[CO(g)] = −283 kJ mol⁻¹.
Both routes end at CO₂(g), so using the cycle above:
ΔH₁ = ΔH₂ + ΔH₃ → −394 = ΔH₂ + (−283)
ΔH₂ = −394 + 283 = −111 kJ mol⁻¹
07 · Exam‑style question — draw the cycle
Q. Use the enthalpies of formation below to find ΔH for
CaCO₃(s) → CaO(s) + CO₂(g).
ΔHf: CaCO₃ −1207, CaO −635, CO₂ −394 kJ mol⁻¹.
ΔH = ΣΔHf(products) − ΣΔHf(reactants)
ΔH = (−635 − 394) − (−1207) = +178 kJ mol⁻¹
Endothermic — which is why limestone needs a kiln.
08 · The two standard routes
ΔH = ΣΔHf(products) − ΣΔHf(reactants)
From bond enthalpies:
ΔH = Σ(bonds broken) − Σ(bonds formed)
Multiply every term by its coefficient. Bond enthalpies only apply to gases, so a reaction involving liquids or solids gives an approximate answer — say so if asked to compare.
09 · Worked example — bond enthalpies
C–H 414, O=O 498, C=O 804, O–H 463 kJ mol⁻¹
Broken: 4(414) + 2(498) = 1656 + 996 = 2652
Formed: 2(804) + 4(463) = 1608 + 1852 = 3460
ΔH = 2652 − 3460 = −808 kJ mol⁻¹
Note H₂O is taken as a gas — using H₂O(l) would need the enthalpy of vaporisation as well.
Marks lost here
— Using the mass of the fuel or solid in q = mcΔT instead of the mass of water or solution.
— Forgetting the minus sign, or leaving the answer in J when the question asks for kJ mol⁻¹.
— Dividing by the total moles rather than the moles of the limiting reactant.
— Reversing a Hess step without flipping the sign.
Energetics II — Born–Haber, Entropy, Gibbs
10 · Born–Haber terms
| Step | Sign |
|---|---|
| Atomisation / sublimation | + |
| Bond dissociation (½ per Cl) | + |
| Ionisation energy | + |
| Electron affinity | − (1st) |
| Lattice enthalpy (formation) | − |
| Enthalpy of formation | usually − |
Lattice enthalpy of formation — enthalpy change when one mole of an ionic solid forms from its gaseous ions. Quoted as dissociation in the data booklet, so watch which direction the question wants: the two differ only in sign.
11 · Born–Haber cycle — NaCl
Vertical axis is enthalpy; the diagram is schematic, not to scale.
12 · Worked example — Born–Haber
ΔHf = ΔHat + IE₁ + ½D + Eea + ΔHlat
−411 = 107 + 496 + 121 + (−349) + ΔHlat
−411 = 375 + ΔHlat
ΔHlat = −411 − 375 = −786 kJ mol⁻¹
For MgCl₂ double the ½D term, add IE₁ and IE₂, and take 2 × Eea.
13 · Lattice enthalpy trends
Magnitude ↑ with larger ionic charge and smaller ionic radius: MgO ≫ NaCl ≫ KI. Charge matters more than radius.
The theoretical value assumes a perfectly ionic lattice of point charges. Where the experimental (Born–Haber) value is much larger, the bonding has covalent character from polarisation of the anion — e.g. AgI.
14 · Enthalpy of solution cycle
ΔHsol = +786 + (−424 − 359) = +3 kJ mol⁻¹
Very small, so dissolving NaCl barely changes the temperature; it is driven by entropy.
15 · Entropy
Units J K⁻¹ mol⁻¹ — note the J, not kJ. Entropy measures the dispersal of energy among available states; S° of a perfect crystal at 0 K is zero, and S° values are positive for everything, elements included.
ΔS positive when: solid → liquid → gas, moles of gas increase, a solid dissolves, or one particle becomes many. Negative when gas moles fall, as in 2NO + O₂ → 2NO₂.
16 · Gibbs free energy
T in kelvin; divide ΔS by 1000 to convert J to kJ before subtracting. ΔG negative → spontaneous.
| ΔH | ΔS | Spontaneous? |
|---|---|---|
| − | + | always |
| − | − | low T only |
| + | + | high T only |
| + | − | never |
At the crossover ΔG = 0, so T = ΔH ÷ ΔS — this is the melting or boiling point when the change is a phase change.
17 · Worked example — Gibbs
ΔH° = +178 kJ mol⁻¹, ΔS° = +161 J K⁻¹ mol⁻¹
At 298 K: ΔG = 178 − (298 × 0.161) = 178 − 48.0 = +130 kJ mol⁻¹ → not spontaneous.
Crossover: T = 178 ÷ 0.161 = 1106 K. Above this temperature ΔG turns negative and the carbonate decomposes.
Data booklet & marks lost
Look up: ΔHf, ΔHc and S° values, bond enthalpies, ionisation energies, electron affinities, lattice and hydration enthalpies, c = 4.18 J g⁻¹ K⁻¹.
— Mixing J and kJ in ΔG = ΔH − TΔS. This is the single most common slip.
— Using the data booklet’s lattice dissociation value without flipping the sign.
— Forgetting ½ for the bond dissociation of Cl₂, or omitting the second ionisation energy for a 2+ ion.
— Calling a spontaneous reaction fast: ΔG says nothing about rate.
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Energetics and Thermodynamics — Frequently Asked Questions
Which mass goes into q = mcΔT?
The mass of the water or solution being heated — never the mass of the fuel or the solid that reacts. This single confusion accounts for more lost marks in calorimetry questions than anything else.
Why is an experimental enthalpy of combustion always less exothermic than the data book value?
Heat is lost to the surroundings and the apparatus, combustion is usually incomplete, and some of the fuel evaporates without burning. All three make the measured temperature rise smaller than it should be.
How do you tell whether a reaction is feasible?
A reaction is feasible when ΔG is negative, where ΔG = ΔH − TΔS. Remember to convert ΔS from J K⁻¹ mol⁻¹ to kJ K⁻¹ mol⁻¹ before subtracting, and that T is in kelvin.
What is the difference between lattice enthalpy of formation and of dissociation?
They describe the same lattice in opposite directions and differ only in sign. Formation is the enthalpy change when one mole of the ionic solid forms from its gaseous ions and is negative; dissociation is the reverse and is positive. The 9701 data booklet quotes dissociation values.
