Thermodynamics extends Energetics (3.1.4). Energetics measured enthalpy changes and used Hess’s law; thermodynamics uses the same cycle-building to reach quantities that cannot be measured directly — above all lattice enthalpy — and then adds a second idea, entropy, to answer a question enthalpy alone cannot: will a reaction actually happen? The whole topic is A-level only and rewards careful bookkeeping of signs and units.
Lattice enthalpy & the key definitions
An ionic lattice is held together by the electrostatic attraction between oppositely charged ions. The energy of that attraction is measured by the lattice enthalpy, which AQA defines in two equal-and-opposite ways. It is the number that puts a figure on the stability of an ionic compound: the more exothermic the lattice enthalpy of formation, the more strongly the ions are held and the more stable the solid.
Lattice enthalpy of formation is the enthalpy change when one mole of a solid ionic compound is formed from its gaseous ions.
Lattice enthalpy of dissociation is the enthalpy change when one mole of a solid ionic compound is separated into its gaseous ions.
Forming a lattice releases energy, so lattice enthalpy of formation is exothermic (negative); pulling the lattice apart requires energy, so lattice enthalpy of dissociation is endothermic (positive). The two values are identical in size and opposite in sign. Lattice enthalpy cannot be measured directly — it is found using a Born–Haber cycle, which needs the five enthalpy terms below.
| Term | Definition (enthalpy change when…) | Sign |
|---|---|---|
| Enthalpy of formation, ΔfH | one mole of a compound forms from its elements in their standard states | usually − |
| Enthalpy of atomisation, ΔatH | one mole of gaseous atoms forms from an element in its standard state | + |
| First ionisation energy | one mole of electrons is removed from one mole of gaseous atoms (forming 1+ ions) | + |
| Bond dissociation enthalpy | one mole of a specified covalent bond is broken in gaseous molecules | + |
| First electron affinity | one mole of gaseous atoms gains one mole of electrons (forming 1− ions) | − |
- Every definition specifies one mole and the correct states (gaseous ions for lattice enthalpy and electron affinity). Omitting either loses the mark.
- The first electron affinity is exothermic, but the second is endothermic: adding an electron to an already negative ion means bringing a negative charge towards a repelling negative ion.
- Lattice enthalpy is more exothermic for ions that are smaller and more highly charged (greater charge density → stronger attraction).
Recap the enthalpy-change sign convention and Hess’s law — Born–Haber cycles are Hess’s law applied to ion formation.
Exam questions
Define the term enthalpy of lattice dissociation.
Show answer
The enthalpy change when one mole of a solid ionic compound… 1 mark
…is separated (dissociates fully) into its gaseous ions. 1 mark
A suitable equation with state symbols for the ions is acceptable. Not “one mole of gaseous ions”.
Explain why the second electron affinity of oxygen has a positive value.
Show answer
The electron is being added to an O− ion, which repels the incoming (negative) electron, so energy must be supplied. 1 mark
Explain why the second ionisation energy of calcium is greater than its first ionisation energy.
Show answer
The second electron is removed from a positive ion (Ca+): the same number of protons attract fewer electrons, so the remaining electrons are held more strongly (and the ion is smaller). 1 mark
State the meaning of the term enthalpy change.
Show answer
A heat (energy) change measured at constant pressure. 1 mark
“Energy change” on its own is ignored — the constant-pressure condition is the mark. Conditions quoted alongside it are ignored even if wrong.
The enthalpies of lattice formation of MgCl2, MgO and BaCl2 are −2018, −2493 and −3889 kJ mol−1, in some order. Select the correct match. Tick (✓) one box.
Explain why the enthalpy of lattice dissociation of sodium oxide is greater than the enthalpy of lattice dissociation of sodium chloride.
Show answer
The oxide ion has a higher (negative) charge than the chloride ion, and is smaller — so it has a higher charge density (higher charge : size ratio). 1 mark
There is therefore a stronger attraction between the oppositely charged ions (O2− and Na+), so more energy is needed to separate them. 1 mark
Ignore references to electronegativity.
Source: AQA A-Level Chemistry past papers.
Born–Haber cycles
A Born–Haber cycle is Hess’s law applied to the formation of an ionic solid. There are two routes from the elements in their standard states to the solid lattice: the direct route (enthalpy of formation), and the indirect route that goes up through gaseous atoms and gaseous ions and then down into the lattice. Because both routes have the same overall enthalpy change, you can find any missing step.
ΔfH = (atomisation + ionisation + electron affinity terms) + ΔLEH(formation)
Rearrange to make the unknown the subject. Two things cause almost every lost mark: forgetting to multiply a step by the number of atoms or ions (two chlorides need 2 × atomisation and 2 × electron affinity), and mishandling a sign. Colour-code each step: energy in is positive, energy out is negative.
Step 1 — write the cycle as a Hess relationship:
ΔfH = ΔatH(Na) + IE1(Na) + ΔatH(Cl) + EA(Cl) + ΔLEH
Step 2 — rearrange for the lattice enthalpy:
ΔLEH = ΔfH − [ΔatH(Na) + IE1(Na) + ΔatH(Cl) + EA(Cl)]
Step 3 — substitute (−411, +109, +494, +121, −364):
ΔLEH = −411 − [(+109) + (+494) + (+121) + (−364)] = −771 kJ mol−1
Interactive — build the Born–Haber cycle
Exam questions
Use the data below to calculate the enthalpy of lattice formation of magnesium oxide.
| Enthalpy of formation of MgO | −602 |
| Enthalpy of atomisation of magnesium | +150 |
| First ionisation energy of magnesium | +736 |
| Second ionisation energy of magnesium | +1450 |
| Bond dissociation enthalpy of oxygen | +496 |
| First electron affinity of oxygen | −142 |
| Second electron affinity of oxygen | +844 |
All values in kJ mol−1.
Show answer
ΔfH = ΔatH(Mg) + ½ BDE(O2) + IE1 + IE2 + EA1(O) + EA2(O) + ΔLEH 1 mark
−602 = 150 + (½ × 496) + 736 + 1450 + (−142) + 844 + ΔLEH 1 mark
ΔLEH(MgO) = −602 − 3286 = −3888 kJ mol−1 1 mark
Use only ½ of the O=O bond enthalpy (one O atom per MgO). +3888 scores 1 mark; missing the ½ gives −4136.
A Born–Haber cycle for caesium iodide uses these data (all kJ mol−1): enthalpy of formation of CsI −337; enthalpy of atomisation of caesium +79; first ionisation energy of caesium +376; electron affinity of iodine −314; enthalpy of lattice formation of CsI −585. Calculate the standard enthalpy of atomisation of iodine.
Show answer
−337 = 79 + ΔatH(I) + 376 + (−314) + (−585) 1 mark
ΔatH(I) = −337 − (−444) = +107 kJ mol−1 1 mark
−107 scores 1 mark. Answer to 2 sig figs or more.
A Born–Haber cycle for caesium iodide is built in this order: the elements in their standard states are atomised, caesium is then ionised, an electron is transferred to the gaseous iodine atom, and finally the gaseous ions come together to form the lattice. Write the formulae, including state symbols, of all the species present (a) on the lowest level of the cycle, the elements in their standard states, and (b) on the highest level of the cycle — reached after atomisation and ionisation, but before the electron affinity step.
Show answer
(a) Cs(s) + ½I2(s) 1 mark
(b) Cs+(g) + e− + I(g) 1 mark
State symbols are required on every species. Iodine is a solid in its standard state, and only ½ mol of I2 is needed for one mole of CsI. The free electron must be shown on the upper level.
A Born–Haber cycle for magnesium oxide starts from Mg(s) + ½O2(g) and finishes at MgO(s). Write the formulae, including state symbols, of all the species present on each of the three levels between them: (a) after both elements have been atomised; (b) after the first and second ionisation energies of magnesium have been supplied; (c) after the first and second electron affinities of oxygen, immediately before the lattice-formation step.
Show answer
(a) Mg(g) + O(g) 1 mark
(b) Mg2+(g) + 2e− + O(g) 1 mark
(c) Mg2+(g) + O2−(g) 1 mark
One mark for each level, and only with correct state symbols. The two electrons released by magnesium must be shown on level (b), because they have not yet been transferred to oxygen.
An incomplete Born–Haber cycle for the formation of calcium chloride has three blank levels: (a) the lowest level, the elements in their standard states; (b) the level reached once both ionisation energies of calcium have been supplied, before the chlorine has been atomised; (c) the level of gaseous ions immediately before the lattice-formation step down to CaCl2(s). Write the formulae, including state symbols, of all the species on each level.
Show answer
(a) Ca(s) + Cl2(g) 1 mark
(b) Ca2+(g) + 2e− + Cl2(g) 1 mark
(c) Ca2+(g) + 2Cl−(g) 1 mark
For (b) the alternative Ca+(g) + e− + 2Cl(g) is also allowed. Note that two chloride ions are needed on level (c).
A different Born–Haber cycle for calcium chloride atomises the chlorine before the electron transfer. Two levels are blank: (i) the level reached once calcium has been fully ionised and the chlorine fully atomised, before the electron affinity step; (ii) the level of gaseous ions immediately before the lattice-formation step. Write the formulae of the missing species on each of these two levels.
Show answer
(i) Ca2+(g) + 2e− + 2Cl(g) 1 mark
(ii) Ca2+(g) + 2Cl−(g) 1 mark
Every ×2 must be shown: two chlorine atoms, then two chloride ions, and two electrons released by the calcium.
Use the cycle in part (a) and the data below to calculate the second ionisation energy of calcium.
| Enthalpy of atomisation of calcium | +193 |
| First ionisation energy of calcium | +590 |
| Enthalpy of atomisation of chlorine | +121 |
| Electron affinity of chlorine | −364 |
| Enthalpy of formation of calcium chloride | −795 |
| Enthalpy of lattice formation of calcium chloride | −2237 |
All values in kJ mol−1.
Show answer
−795 = 193 + 590 + IE2 + (2 × 121) + (2 × −364) + (−2237) 1 mark
The known terms sum to −1940, so −795 = −1940 + IE2
IE2(Ca) = −795 + 1940 = +1145 kJ mol−1 1 mark
Setting out the first line as −795 = −1940 + IE2 also earns the first mark. Both ×2 factors are needed — one on atomisation of chlorine and one on the electron affinity.
Use the data below, from a Born–Haber cycle for strontium chloride (SrCl2), to calculate a value for the electron affinity of chlorine.
| Enthalpy of atomisation of strontium | +164 |
| First ionisation energy of strontium | +548 |
| Second ionisation energy of strontium | +1060 |
| Enthalpy of atomisation of chlorine | +121 |
| Enthalpy of formation of strontium chloride | −828 |
| Enthalpy of lattice formation of strontium chloride | −2112 |
All values in kJ mol−1.
Show answer
−828 = 164 + (2 × 121) + 548 + 1060 + (2 × EA) + (−2112) 1 mark
The known terms sum to −98, so 2 × EA = −828 + 98 = −730 1 mark
EA(Cl) = −730 ÷ 2 = −365 kJ mol−1 1 mark
Two chlorides mean a ×2 on atomisation and on the electron affinity — and then the final answer must be halved back to a per-ion value. Answers of +365, −304.5 or −730 score 2 of the 3 marks; +304.5, +730 or −609 score 1; +609 scores 0.
Use the data below to calculate the enthalpy of lattice dissociation of sodium oxide, Na2O.
| Enthalpy of atomisation of oxygen | +248 |
| Enthalpy of atomisation of sodium | +109 |
| Enthalpy of formation of sodium oxide | −416 |
| First ionisation energy of sodium | +494 |
| First electron affinity of oxygen | −142 |
| Second electron affinity of oxygen | +844 |
All values in kJ mol−1.
Show answer
−416 + ΔLEH(dissociation) = 248 + (2 × 109) + (2 × 494) + (−142) + 844 = 2156 1 mark
ΔLEH(dissociation) = 2156 + 416 = +2572 kJ mol−1 1 mark
−2572 scores 1 mark. Two ×2 factors are needed here (two sodium atoms are atomised and two are ionised), and both electron affinities of oxygen are used. The enthalpy of atomisation of oxygen is given directly, so there is no ½ factor — that only appears when the bond dissociation enthalpy of O2 is supplied instead.
Use the data below to calculate a value for the enthalpy of lattice dissociation of calcium chloride.
| Enthalpy of formation of calcium chloride | −795 |
| Enthalpy of atomisation of calcium | +193 |
| First ionisation energy of calcium | +590 |
| Second ionisation energy of calcium | +1150 |
| Enthalpy of atomisation of chlorine | +121 |
| Electron affinity of chlorine | −364 |
All values in kJ mol−1.
Show answer
−795 + ΔLEH(dissociation) = 193 + 590 + 1150 + (2 × 121) + (2 × −364) = 1447 1 mark
ΔLEH(dissociation) = 1447 + 795 = +2242 kJ mol−1 1 mark
−2242 scores 1 mark. If one or both of the ×2 factors is missing the answer comes out as +2485, +2121 or +2606 — each worth 1 mark.
Source: AQA A-Level Chemistry past papers.
The perfect ionic model & covalent character
Lattice enthalpy can also be calculated theoretically from a model of the crystal. That model makes a simplifying assumption.
The perfect ionic model treats the ions as point charges (perfect, undistorted spheres) with no covalent character.
Comparing the theoretical value with the experimental value from a Born–Haber cycle tests how well that assumption holds:
- If the two values are close, the compound is close to purely ionic — the perfect ionic model works well.
- If the experimental value is significantly more exothermic than the theoretical value, there is additional covalent character: the cation polarises (distorts) the anion’s electron cloud, so the electrons are partly shared and the bonding is stronger than the ionic model predicts.
Covalent character is greatest when the cation is small and highly charged (high polarising power) and the anion is large (easily polarised).
Interactive — one number, three consequences
Charge density here means simply charge ÷ ionic radius, with the radius in pm and the magnitude of the charge used for negative ions. It is a relative comparison between ions, not an absolute physical quantity with units.
Real ions — pick one to set the sliders
The dashed outline is 220 pm, drawn for scale — the largest ion the slider reaches.
Exam questions
The experimental lattice enthalpy of caesium iodide is −585 kJ mol−1; the value from the perfect ionic model is −582 kJ mol−1. Deduce what these values indicate about the bonding in caesium iodide.
Show answer
The values are very close, so caesium iodide is almost purely (perfectly) ionic, with little or no covalent character. 1 mark
Penalise references to atoms or molecules; the species are ions.
For silver chloride the theoretical lattice enthalpy is −770 kJ mol−1 and the experimental value is −905 kJ mol−1. State why there is a difference between these values.
Show answer
Silver chloride has covalent character (partial covalent bonding): the Ag+ ion polarises / distorts the chloride ion. 1 mark
Theoretical values for enthalpies of lattice dissociation can be calculated using a perfect ionic model. State the meaning of the term perfect ionic model.
Show answer
The ions are treated as point charges (perfect spheres) with no covalent character — there is no polarisation of the ions. 1 mark
Do not use the words “atoms” or “molecules” anywhere in this answer — the species are ions.
Source: AQA A-Level Chemistry past papers.
Enthalpies of solution & hydration
When an ionic solid dissolves, the lattice must break apart and the freed ions become surrounded by water molecules. A Hess cycle links the two.
Enthalpy of hydration is the enthalpy change when one mole of gaseous ions is dissolved in water to form one mole of aqueous ions.
Enthalpy of solution is the enthalpy change when one mole of an ionic solid dissolves completely in water to form a very dilute solution.
The dissolving cycle breaks the solid into gaseous ions (lattice dissociation, endothermic) and then hydrates them (exothermic):
ΔsolH = ΔLEH(dissociation) + ΣΔhydH
Lattice dissociation of NaCl = +771; hydration of Na+ = −406; hydration of Cl− = −364 (kJ mol−1).
ΔsolH = (+771) + (−406) + (−364) = +1 kJ mol−1
Hydration is more exothermic for ions with a higher charge density — smaller ions and more highly charged ions attract the polar water molecules (the Oδ− or Hδ+) more strongly.
Interactive — the solution / hydration enthalpy cycle
Exam questions
Use the data to calculate the molar enthalpy change when calcium chloride dissolves in water, then deduce how the temperature changes. Hydration of Ca2+ = −1650; hydration of Cl− = −364; lattice formation of CaCl2 = −2237 (kJ mol−1).
Show answer
ΔsolH = ΔLEH(dissociation) + ΔhydH(Ca2+) + 2ΔhydH(Cl−) = (+2237) + (−1650) + (2 × −364) 1 mark
ΔsolH = −141 kJ mol−1 1 mark
The value is negative (exothermic), so the temperature rises. 1 mark
Lattice dissociation = −(lattice formation) = +2237. Remember the ×2 for two chloride ions.
Calcium bromide dissolves in water. Using enthalpy of solution of CaBr2 = −110, lattice formation of CaBr2 = −2176 and hydration of Ca2+ = −1650 (kJ mol−1), calculate the enthalpy of hydration of bromide ions.
Show answer
−110 = (+2176) + (−1650) + 2ΔhydH(Br−) 1 mark
2ΔhydH(Br−) = −636 1 mark
ΔhydH(Br−) = −318 kJ mol−1 1 mark
Explain why the enthalpy of hydration of fluoride ions is more exothermic than that of chloride ions.
Show answer
The fluoride ion is smaller, so it has a higher charge density. 1 mark
This gives a stronger attraction between the ion and the Hδ+ of the water molecules. 1 mark
Do not accept “fluorine atoms are smaller” or references to ionic bonds.
Magnesium chloride dissolves in water. Give an equation, including state symbols, to represent the process that occurs when the enthalpy of solution of magnesium chloride is measured.
Show answer
MgCl2(s) → Mg2+(aq) + 2Cl−(aq) 1 mark
⇌ is allowed in place of →, and “+ aq” may be written on the left-hand side. The ×2 on the chloride ion is required.
The enthalpy of solution of ammonium nitrate is the enthalpy change for NH4NO3(s) + aq → NH4+(aq) + NO3−(aq), ΔH = +26 kJ mol−1. The enthalpies of hydration are −307 kJ mol−1 for NH4+(g) and −314 kJ mol−1 for NO3−(g). Write the Hess relationship that links these three quantities, naming the species at each corner of the cycle, and use it to calculate the enthalpy of lattice dissociation of ammonium nitrate.
Show answer
The cycle has three corners — NH4NO3(s), the gaseous ions NH4+(g) + NO3−(g), and the aqueous ions NH4+(aq) + NO3−(aq) — so ΔsolH = ΔLEH(dissociation) + ΔhydH(NH4+) + ΔhydH(NO3−) 1 mark
26 = ΔLEH(dissociation) + (−307) + (−314), so ΔLEH(dissociation) = 26 + 307 + 314 1 mark
ΔLEH(dissociation) = +647 kJ mol−1 1 mark
In the original question the first mark was for drawing the labelled cycle itself — three corners with formulae, state symbols and arrows. An equivalent Born–Haber-style energy cycle is accepted. Answers of −647 or ±595 score 2 of the 3 marks.
A student measures 25.0 cm3 of distilled water into a beaker, records its temperature, adds 4.00 g of solid NH4NO3, stirs, and records the lowest temperature reached. The initial temperature is 20.2 °C and the lowest temperature is 12.2 °C. Calculate the enthalpy of solution, in kJ mol−1, for ammonium nitrate in this experiment. Assume the specific heat capacity of the solution c = 4.18 J K−1 g−1 and the density of the solution = 1.00 g cm−3. Mr(NH4NO3) = 80.0.
Show answer
q = mcΔT = 25.0 × 4.18 × (20.2 − 12.2) = 25.0 × 4.18 × 8.0 = 836 J 1 mark
n(NH4NO3) = 4.00 ÷ 80.0 = 0.0500 mol 1 mark
ΔsolH = 836 ÷ 0.0500 = 16 720 J mol−1 = +16.7 kJ mol−1 1 mark
Use the mass of water, 25.0 g — using m = 29 g is not allowed and gives +19.4 (2 of 3 marks). The temperature fell, so the change is endothermic and the answer is positive; −16.7 scores 2 of 3. Two significant figures or more.
The uncertainty in each of the temperature readings from the thermometer used in this experiment is ±0.1 °C. Calculate the percentage uncertainty in the temperature change.
Show answer
(2 × 0.1 ÷ 8.0) × 100 = 2.5% 1 mark
Two readings are taken to find one temperature change, so the ±0.1 °C uncertainty is counted twice.
Suggest a change to the student’s method, using the same apparatus, that would reduce the percentage uncertainty in the temperature change. Give a reason for your answer.
Show answer
Change: use a larger mass of NH4NO3 (or a smaller volume of water). 1 mark
Reason: the temperature change is then greater (the final temperature is lower), so the same ±0.1 °C is a smaller percentage of it. 1 mark
The two marking points are independent. “Larger volume of water” and “final temperature is higher” are not accepted; higher concentration, and any change of apparatus such as adding insulation, are ignored.
Another student obtained a value of +15 kJ mol−1 using the same method. Suggest the main reason for the difference between this experimental value and the correct value of +26 kJ mol−1.
Show answer
Heat was gained from the surroundings (or the solid did not dissolve completely). 1 mark
“Heat loss” is not accepted. The change is endothermic, so the beaker is colder than its surroundings and thermal energy flows in, making the measured temperature drop too small. References to mistakes in the method are ignored.
The enthalpies of hydration of Li+(g), Na+(g) and K+(g) are −519, −406 and −322 kJ mol−1 respectively. Explain why the enthalpy of hydration becomes less exothermic from Li+ to K+.
Show answer
From Li+ to K+ the size of the ion increases (more shells, larger ionic radius), so the charge density decreases. 1 mark
The electrostatic attraction between the metal ion and the Oδ− of the water molecules is therefore weaker. 1 mark
Do not write “atomic radius” or refer to molecules — these are ions. The second mark does not depend on the first, and converse arguments are accepted. Note that a cation is attracted to the Oδ− end of water, whereas an anion is attracted to the Hδ+ end.
Give a reason why data books do not contain a value for the enthalpy of solution of sodium oxide.
Show answer
Sodium oxide reacts with water (to form a solution of NaOH) rather than simply dissolving. 1 mark
“It dissolves in water” is not accepted — the answer must say that a reaction occurs.
Source: AQA A-Level Chemistry past papers.
Entropy, ΔS
Enthalpy alone cannot explain why some endothermic reactions still happen. The missing idea is entropy, S — a measure of how disordered a system is, or how many ways its particles and energy can be arranged. Nature tends towards greater disorder.
Entropy rises as particles gain freedom of movement, so solid < liquid < gas. It also increases when a solid dissolves and when a reaction produces more moles of gas. A perfect crystal at 0 K has zero entropy.
Calculate an entropy change as products minus reactants, using absolute entropy values (in J K−1 mol−1):
ΔS = ΣS(products) − ΣS(reactants)
Interactive — ΔS from standard entropies
Pick a reaction to load its species, coefficients and standard entropies — then edit any value and calculate again.
Same substance, three states: entropy rises solid → liquid → gas, and the jump on boiling is far bigger than the jump on melting — which is why the moles of gas dominate the sign of ΔS.
Exam questions
Calculate the entropy change for the reaction 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g). Standard entropies (J K−1 mol−1): NH3 193, O2 205, NO 211, H2O 189.
Show answer
ΔS = [(4 × 211) + (6 × 189)] − [(4 × 193) + (5 × 205)] = 1978 − 1797 1 mark
ΔS = +181 J K−1 mol−1 1 mark
(a) Explain why the standard entropy of carbon dioxide is greater than that of carbon. (b) State the temperature at which the standard entropy of solid aluminium is 0 J K−1 mol−1.
Show answer
(a) Carbon dioxide is a gas, so it is more disordered than solid carbon. 1 mark
(b) 0 K (absolute zero / −273 °C). 1 mark
Source: AQA A-Level Chemistry past papers.
Gibbs free energy & feasibility
Enthalpy and entropy are combined in the Gibbs free-energy change, which decides whether a reaction is thermodynamically feasible (able to happen):
ΔG = ΔH − TΔS
T is the temperature in kelvin. A reaction is feasible when ΔG is zero or negative. A units mismatch is easy to make here, because ΔH is quoted in kJ mol−1 but ΔS in J K−1 mol−1.
- Convert ΔS to kJ (divide by 1000) before substituting, or the TΔS term is a thousand times too big. Entropy is almost always given in joules.
- Consider a reaction with ΔH = +30 kJ mol−1 and ΔS = +50 J K−1 mol−1 at 298 K. Correctly, ΔG = 30 − 298 × 0.050 = +15 kJ mol−1 (not feasible). Forget to convert and you get 30 − 298 × 50 = −14 870, wrongly “feasible”.
- Give the answer to 3 significant figures and keep the minus sign on exothermic values.
Because ΔG = ΔH − TΔS, temperature matters. The reaction is on the boundary of feasibility when ΔG = 0, which rearranges to give the temperature at which it just becomes feasible:
at ΔG = 0: T = ΔH / ΔS
Plotting ΔG against T gives a straight line: the intercept is ΔH and the gradient is −ΔS.
Interactive — is it feasible? ΔG = ΔH − TΔS
Enter ΔH and ΔS exactly as a question gives them — ΔH in kilojoules, ΔS in joules — and pick K or °C for the temperature. Every conversion is shown line by line. Leave the temperature box empty to see only the crossover temperature.
Exam questions
For CO2(g) + 3H2(g) ⇌ CH3OH(g) + H2O(g), calculate ΔG at 890 K. Data: ΔfH / kJ mol−1: CO2 −394, H2 0, CH3OH −201, H2O −242. S / J K−1 mol−1: CO2 214, H2 131, CH3OH 238, H2O 189.
Show answer
ΔH = [(−201) + (−242)] − [(−394) + 0] = −49 kJ mol−1 1 mark
ΔS = [238 + 189] − [214 + (3 × 131)] = 427 − 607 = −180 J K−1 mol−1 1 mark
ΔG = ΔH − TΔS 1 mark
Convert ΔS: −180 J = −0.180 kJ K−1 mol−1 1 mark
ΔG = −49 − (890 × −0.180) = −49 + 160 1 mark
ΔG = +111 kJ mol−1 1 mark
Leaving ΔS in J gives +160 151 — the classic unit error.
For a gas-phase reaction, a straight-line graph of ΔG against T has a y-intercept of +145 kJ mol−1 and a gradient of −0.170 kJ K−1 mol−1. Deduce ΔH and ΔS, and state the temperature above which the reaction is feasible.
Show answer
ΔH = intercept = +145 kJ mol−1. 1 mark
Gradient = −ΔS, so ΔS = +0.170 kJ K−1 mol−1 = +170 J K−1 mol−1. 1 mark
Feasible when ΔG ≤ 0: T = ΔH / ΔS = 145 / 0.170 ≈ 853 K, so feasible above about 850 K. 1 mark
For 2Al2O3(s) + 3C(s) → 4Al(s) + 3CO2(g), calculate the minimum temperature at which the reaction becomes feasible. Data: ΔfH / kJ mol−1: Al2O3 −1669, Al 0, C 0, CO2 −394. S / J K−1 mol−1: Al2O3 51, Al 28, C 6, CO2 214.
Show answer
ΔH = [3 × (−394)] − [2 × (−1669)] = −1182 + 3338 = +2156 kJ mol−1 1 mark
ΔS = [(4 × 28) + (3 × 214)] − [(2 × 51) + (3 × 6)] = 754 − 120 = +634 J K−1 mol−1 1 mark
At the boundary ΔG = 0, so T = ΔH / ΔS 1 mark
Convert ΔS: +634 J = +0.634 kJ K−1 mol−1 1 mark
T = 2156 / 0.634 = 3400 K (to 3 s.f.) 1 mark
The reaction is endothermic with a positive ΔS, so it becomes feasible only above this temperature.
For 4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g), ΔH = −905 kJ mol−1 and ΔS = +181 J K−1 mol−1. Calculate a value for the Gibbs free-energy change (ΔG), in kJ mol−1, at 600 °C.
Show answer
ΔG = ΔH − TΔS = −905 − [(600 + 273) × 181 × 10−3] 1 mark
ΔG = −905 − 158 = −1063 kJ mol−1 1 mark
Allow −1060. T must be converted to kelvin: 600 °C = 873 K. Using 600 K instead loses the mark.
The reaction between ammonia and oxygen was then carried out at a higher temperature. Explain how this change affects the value of ΔG for the reaction.
Show answer
ΔG becomes more negative (less positive). 1 mark
ΔS is positive, so raising T makes TΔS bigger, and −TΔS more negative. 1 mark
“ΔG increases / decreases / gets larger / gets smaller” on its own is ignored — say more negative or less positive.
The enthalpy of formation of caesium iodide is −337 kJ mol−1. Use the entropy data below to show that the reaction CsI(s) → Cs(s) + ½I2(s) is not feasible at 298 K.
| CsI(s) | Cs(s) | I2(s) | |
|---|---|---|---|
| S / J K−1 mol−1 | 130 | 82.8 | 117 |
Show answer
ΔS = [82.8 + (½ × 117)] − 130 = 141.3 − 130 = +11.3 J K−1 mol−1 1 mark
ΔG = ΔH − TΔS, and this reaction is the reverse of the formation of CsI, so ΔH = +337 kJ mol−1 1 mark
ΔG = 337 − (298 × 11.3 × 10−3) = 337 − 3.37 1 mark
ΔG = +334 kJ mol−1 — positive, so the reaction is not feasible at 298 K. 1 mark
Any negative answer loses the final mark. The third mark is specifically for the unit conversion — ΔS into kJ K−1 mol−1 or ΔH into J mol−1. Note the ½ on the entropy of I2(s), matching the equation.
Ammonium nitrate dissolves as NH4NO3(s) + aq → NH4+(aq) + NO3−(aq), ΔH = +26 kJ mol−1. Using the entropy data below, calculate ΔG at 298 K and explain what your value indicates about the feasibility of this reaction. Assume that for the solvent, water, the entropy change ΔS = 0.
| NH4NO3(s) | 151 |
| NH4+(aq) | 113 |
| NO3−(aq) | 146 |
All entropies in J K−1 mol−1.
Show answer
ΔS = (113 + 146) − 151 = +108 J K−1 mol−1 1 mark
ΔG = ΔH − TΔS = 26 − (298 × 108 × 10−3) = 26 − 32.2 1 mark
ΔG = −6.18 kJ mol−1 1 mark
ΔG is negative, so the reaction is feasible (spontaneous) at 298 K. 1 mark
Allow −6.184 or −6.2. The final mark stands alone. This is the case worth remembering: the change is endothermic (ΔH positive), yet it is still feasible because the positive entropy change makes TΔS larger than ΔH.
Calculate the temperature, in °C, at which the reaction NaCl(s) → Na(s) + ½Cl2(g) becomes feasible. ΔH = +411 kJ mol−1 and ΔS = +90.1 J K−1 mol−1.
Show answer
At the boundary ΔG = 0, so T = ΔH ÷ ΔS 1 mark
T = 411 ÷ (90.1 × 10−3) = 4562 K 1 mark
Temperature = 4562 − 273 = +4289 °C 1 mark
Allow 4290 °C. The question asks for degrees Celsius, so the −273 conversion carries the final mark on its own — stopping at 4562 K scores 2 of 3.
Source: AQA A-Level Chemistry past papers.
The thermodynamics checklist
On this topic the chemistry is rarely the hard part — the signs, units and × factors are. Working every calculation the same way keeps those under control.
- Convert units first — entropy from J to kJ if enthalpy is in kJ.
- Substitute carefully, keeping every sign and every × factor.
- Evaluate the sign of the answer — and sanity-check it (exothermic should be negative).
- Interpret: is ΔG ≤ 0 (feasible)? Then give the answer to 3 significant figures with units.
- Examiner reports identify three preventable losses on this topic: mixing entropy in joules with enthalpy in kilojoules, dropping the minus sign on an exothermic value, and not giving three significant figures.
- In Born–Haber work the same care applies: every ×2 (for two ions) and every sign must be right, and lattice enthalpies of formation and dissociation differ only by sign.
Interactive — build the mark-scheme answer
“Ammonium nitrate dissolves in water: NH4NO3(s) + aq → NH4+(aq) + NO3−(aq), ΔH = +26 kJ mol−1. Standard entropies / J K−1 mol−1: NH4NO3(s) 151, NH4+(aq) 113, NO3−(aq) 146. Calculate ΔG at 298 K and explain what it shows about feasibility.” [4 marks] — select every step that earns a mark, and nothing that doesn’t. Order doesn’t matter: AQA credits each point on its own.
the working, assembled
Capstone question
The table gives lattice enthalpies of dissociation (kJ mol−1) from the perfect ionic model and from Born–Haber cycles. Discuss the values: compare the three perfect-ionic values, then compare each perfect-ionic value with its Born–Haber value. Marked by levels of response.
| Perfect ionic model | Born–Haber cycle | |
|---|---|---|
| Calcium chloride | 2223 | 2237 |
| Potassium chloride | 690 | 701 |
| Silver chloride | 770 | 905 |
Show answer
Comparing the perfect-ionic values: CaCl2 is much larger than KCl and AgCl because Ca2+ has a higher charge (and is a small ion), giving stronger attraction between the oppositely charged ions. Stage 1
Where the two methods agree: CaCl2 and KCl have very similar perfect-ionic and Born–Haber values, so their bonding is close to perfectly ionic (the ions behave as spherical point charges). Stage 2
Where they differ: AgCl shows a large difference (770 vs 905), so it has significant covalent character — the Ag+ ion is polarising and distorts the chloride ion’s electron cloud. Stage 3
Source: AQA A-Level Chemistry past papers.
The same ΔG ≤ 0 feasibility test reappears later in the A2 course for electrochemical cells (3.1.11), where feasibility is judged from electrode potentials.
- Lattice enthalpy: formation (gaseous ions → solid) is exothermic; dissociation (solid → gaseous ions) is endothermic; equal and opposite. More exothermic for smaller, higher-charge ions.
- Born–Haber cycle = Hess’s law for ion formation: ΔfH = (atomisation + ionisation + bond + electron affinity terms) + lattice enthalpy of formation. Watch × factors and signs.
- Perfect ionic model treats ions as point charges / perfect spheres. A large gap between the theoretical and Born–Haber (experimental) value indicates covalent character (the cation polarises the anion).
- Enthalpy of solution = ΔLE(dissociation) + ΣΔhydH. Enthalpy of hydration is more exothermic for smaller, higher-charge ions.
- Entropy ΔS = ΣS(products) − ΣS(reactants), in J K−1 mol−1. Disorder rises solid < liquid < gas.
- Feasibility: ΔG = ΔH − TΔS; a reaction is feasible when ΔG ≤ 0. At the boundary ΔG = 0, so T = ΔH / ΔS. Convert ΔS to kJ before substituting.