Chemical energetics

Lattice energy and enthalpy change of atomisation

An ionic solid is held together by the attraction between its ions, so any energy cycle for an ionic compound has to start from separated gaseous ions. Cambridge defines lattice energy as the enthalpy change when one mole of a solid ionic lattice is formed from its gaseous ions, so for sodium chloride the change of interest is Na+(g) + Cl-(g) -> NaCl(s) and the value is negative, about -788 kJ mol-1, because energy is released as the ions come together. Some textbooks and some questions use the opposite, endothermic definition, one mole of solid separated into gaseous ions, which has the same magnitude but a positive sign. Read the direction the question gives you and match the sign to it.

Before any of this can happen the elements have to be turned into separate gaseous atoms. The enthalpy change of atomisation, delta H_at, is the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state, measured under standard conditions. For sodium metal it is the energy needed to separate the atoms from the metallic lattice, +108 kJ mol-1. For a diatomic element you must form one mole of atoms, which means half a mole of molecules, so delta H_at(chlorine) is half of the Cl-Cl bond dissociation enthalpy: the Cl-Cl bond needs +244 kJ mol-1, so the atomisation enthalpy is +122 kJ mol-1.

The size of a lattice energy follows the electrostatic attraction between two charges, which increases with the product of the charges and decreases as the distance between the centres of the ions increases. Doubling both charges therefore multiplies the attraction four times over, which is why magnesium oxide, with 2+ and 2- ions, has a lattice energy of roughly -3844 kJ mol-1 compared with about -788 kJ mol-1 for sodium chloride. Making the ions smaller has a smaller effect, because the radius appears in the expression once, so a radius falling by a fifth increases the attraction by about a quarter rather than four times: lithium chloride (small Li+ ion) is about -846 kJ mol-1, sodium chloride about -788 kJ mol-1 and potassium chloride about -711 kJ mol-1.

Charge beats radius every time in an exam question. If you are asked which of two compounds has the more exothermic lattice energy, compare the charges first, and only compare the radii when the charges match. If you are asked to explain the comparison, you must quote the charges and the radii and say what happens to the attraction between the ions.

• Enthalpy change of atomisation: one mole of gaseous atoms from the element in its standard state, always endothermic for a metal or a diatomic molecule.
• Lattice energy, as used in Born-Haber cycles: one mole of solid lattice from the gaseous ions, exothermic and therefore negative.
• More charge, more attraction: 2+ with 2- is about four times as exothermic as 1+ with 1- at the same distance.
• Smaller ions, stronger attraction: the attraction falls as the distance between the ion centres increases.

Electron affinity: definition, factors and the Group 16 and 17 trends

The first electron affinity, EA, is the enthalpy change when one mole of gaseous atoms each gain one electron to form one mole of gaseous 1- ions. For chlorine the change is Cl(g) + e- -> Cl-(g), which releases about 349 kJ mol-1: the incoming electron is attracted to the nucleus, and the energy it releases when it settles into a vacant orbital makes the process exothermic, so the value is written -349 kJ mol-1. Electron affinity is the mirror image of ionisation energy, which removes an electron instead of adding one.

Three factors set the size of an electron affinity. The nuclear charge pulls the incoming electron in and makes the value more exothermic. The distance of the incoming electron from the nucleus matters, so a larger atom with the outer shell further out has a weaker attraction and a less exothermic value. The shielding by inner shells reduces the effective nuclear charge, again making the value less exothermic. Finally, the electron cannot go into an orbital that is already full, and it is repelled by electron density that is already there, so a very small, electron-dense atom can have a surprisingly small electron affinity.

Going down Group 17 the atoms get larger and the outer shell is better shielded, so the attraction for the incoming electron falls and the first electron affinity becomes less exothermic: about -328 kJ mol-1 for fluorine, -349 kJ mol-1 for chlorine, -325 for bromine and -295 for iodine. Fluorine breaks the smooth pattern, and that is the point examiners test. Fluorine has a much smaller atom than chlorine, so an incoming electron is forced into a region that is already crowded with electron density, and the repulsion between them cancels much of the extra attraction from the higher nuclear charge. Chlorine releases more energy because its larger atom spreads that electron density out.

The Group 16 elements have one proton fewer than the Group 17 element of the same period, so the nuclear attraction for the incoming electron is smaller and the first electron affinity is less exothermic: oxygen's is about -142 kJ mol-1 against fluorine's -328 kJ mol-1. The second electron affinity is a different matter. Adding an electron to an ion that is already negatively charged means forcing two negative charges together, so the change X-(g) + e- -> X2-(g) is endothermic: for oxygen it is about +798 kJ mol-1. Sulphur shows the same pattern, with a positive second electron affinity that is smaller in magnitude than oxygen's, because sulphur's larger ion spreads the charge over a bigger volume. Every Born-Haber cycle for a compound with a 2- anion therefore carries a large positive term.

• First electron affinity is usually exothermic; the value is more exothermic for a higher nuclear charge or a smaller, less shielded atom.
• Down Group 17 the first electron affinity becomes less exothermic from chlorine to iodine, because the outer shell is further from the nucleus and better shielded.
• Fluorine is the exception: its atom is so small and electron-dense that the incoming electron is repelled, so its value is less exothermic than chlorine's.
• The second electron affinity is always endothermic, because a second electron must be forced onto an already negative ion.

Constructing and using Born-Haber cycles

A Born-Haber cycle applies Hess's law to an ionic solid. There are two routes from the elements to the compound. The direct route is the standard enthalpy change of formation, which can be measured. The indirect route goes: atomise the elements, remove electrons from the metal to make its cation (ionisation energies), give electrons to the non-metal (electron affinities), and bring the gaseous ions together into the lattice. Because Hess's law says the total enthalpy change is the same whichever route you take, the measured formation enthalpy fixes whichever value in the indirect route cannot be measured.

The cycle is the reason lattice energy and electron affinity are often found indirectly. Lattice energy cannot be measured directly, because you cannot separate a solid into a gas of ions and measure the heat change; the second electron affinity of oxygen certainly cannot be measured, because oxygen 2- ions are not stable on their own. If one of the values is known, the cycle gives you the other. The layout matters: put the elements and the compound on the same horizontal line, with the gaseous ions below, and add the steps with their correct signs.

For sodium chloride the arithmetic is: delta H_f(NaCl) = delta H_at(Na) + delta H_at(Cl) + IE(Na) + EA(Cl) + lattice energy. Using -411 = +108 + 122 + 496 - 349 + L gives L = -788 kJ mol-1. Notice that every step in the indirect route is written in the direction that takes you from the elements to the solid, which is why the electron affinity is negative here: the electron is being gained, not lost.

Two traps catch most candidates. For a 2+ cation you must add both ionisation energies, and the second is much larger than the first; for a 2- anion you must add both electron affinities, and the second one is positive. For magnesium oxide the cycle runs -602 = (+148) + (+249) + (+738) + (+1451) + (-142) + (+798) + L, so the sum of the endothermic steps is 3384 kJ mol-1, the electron affinities contribute -142 and +798, and L = -602 - 3242 = -3844 kJ mol-1. The huge negative lattice energy is exactly what makes the formation of an oxide of a 2+ metal so favourable.

• Write the direct route and the indirect route, then set them equal: delta H_f = atomisation terms + ionisation energies + electron affinities + lattice energy.
• Keep every step in the same direction, elements to compound, so the signs take care of themselves.
• A 2+ cation needs both ionisation energies, and a 2- anion needs both electron affinities, the second of which is positive.
• Show the substitution as well as the answer; a correct number with no cycle earns fewer marks than the method.

Enthalpy change of solution and enthalpy change of hydration

Dissolving an ionic solid can be split into two imaginary steps. First the lattice must be broken up into gaseous ions, which costs the lattice energy, so this step has the enthalpy change -delta H_latt and is endothermic. Then the gaseous ions are hydrated: water molecules gather round each ion and are attracted to it by ion-dipole forces, releasing energy. The enthalpy change of hydration, delta H_hyd, is the enthalpy change when one mole of gaseous ions is dissolved in water to form one mole of aqueous ions, and it is exothermic. The enthalpy change of solution, delta H_sol, is the enthalpy change when one mole of an ionic solid dissolves in water to give aqueous ions, with the water in large excess so the ions are fully separated and cannot recombine into a lattice.

Link the three through an energy cycle: delta H_sol = (-delta H_latt) + delta H_hyd(cation) + delta H_hyd(anion). For potassium chloride, +711 - 322 - 364 = +25 kJ mol-1: the cycle needs 711 kJ mol-1 to break the lattice, the two ions release 322 and 364 kJ mol-1 when they are hydrated, and the net result is endothermic, so the water cools as the solid dissolves. Run the same cycle for potassium fluoride, whose lattice energy is about -821 kJ mol-1 but whose small fluoride ion releases about 515 kJ mol-1 when it is hydrated: +821 - 322 - 515 = -16 kJ mol-1, so potassium fluoride dissolves exothermically even though its lattice energy is larger in magnitude. Which salt dissolves with a temperature rise depends on whether the hydration step wins.

The magnitude of an enthalpy change of hydration follows the same electrostatic ideas as lattice energy, but here the attraction is between an ion and the partial charges on water molecules. More charge and less distance both make it more exothermic, so hydration enthalpies are roughly -406 kJ mol-1 for Na+, -322 for K+, -515 for F-, -364 for Cl- and a striking -1921 kJ mol-1 for Mg2+. A small, highly charged ion such as Mg2+ has a very concentrated charge and pulls the water molecules in strongly, so it releases far more energy than a singly charged ion of the same size.

Notice that a 2+ ion hydrates with about five times the energy of a 1+ ion of similar size, and also forms a lattice with about four times the lattice energy. When the two large numbers nearly cancel, a 2+ salt can have a small enthalpy of solution either way, and the entropy change, not the enthalpy, decides whether it dissolves.

• delta H_sol = -delta H_latt + delta H_hyd(cation) + delta H_hyd(anion).
• Breaking the lattice is endothermic, hydrating the ions is exothermic, and the sign of the sum decides whether the solution warms or cools.
• More charge and smaller radius make both the lattice energy and the enthalpy of hydration more exothermic.
• A hydrate with an enthalpy of solution very close to zero dissolves with almost no temperature change, and then the entropy change decides.

Entropy: the number of possible arrangements

Entropy, S, is a measure of the number of possible arrangements of the particles and their energy in a given system. A solid in which the ions sit in a fixed, repeating lattice has very few arrangements, so its entropy is low. A gas, whose molecules fly about at a wide spread of speeds and directions, has an enormous number of arrangements, so its entropy is high. This is why entropy increases in the order solid, then liquid, then gas, and why entropy is measured in J K-1 mol-1: an amount of energy per mole for each kelvin, not a total amount of energy.

The sign of an entropy change is usually the point of the question. Whenever the particles gain freedom of movement, the number of arrangements goes up and delta S is positive: melting, boiling, subliming, and dissolving a solid all do this, because the ordered arrangement is replaced by a more disordered one. The reverse changes, such as freezing and condensing, lock the particles back into place and give a negative delta S.

A change of state gives a large entropy change, because the freedom of movement of every particle changes at once. Heating a substance without changing its state gives a smaller positive entropy change, because the particles move faster and their energies spread over more values. A reaction gives a positive entropy change when the number of gaseous molecules increases, and a negative one when it falls: the decomposition CaCO3(s) -> CaO(s) + CO2(g) turns an ordered solid into another solid plus a mole of gas, and its entropy change is about +160 J K-1 mol-1, while the Haber reaction N2(g) + 3H2(g) -> 2NH3(g) uses four moles of gas to make two, so its entropy change is about -198 J K-1 mol-1. A reaction in which a solid or liquid turns into a gas is almost always positive, because a gas is so much more disordered than a condensed phase.

• Gas beats liquid beats solid: more freedom of movement means more arrangements and a higher entropy.
• Melting, boiling and temperature rise: positive. Freezing, condensing and cooling: negative.
• More moles of gas on the product side: positive. Fewer moles of gas: negative.
• Entropy has the unit J K-1 mol-1, so it is never quoted in kJ mol-1 without a conversion factor.

Calculating the entropy change of a reaction

Standard entropies of individual substances are measured and tabulated, and the entropy change of a reaction is the entropy of the products minus the entropy of the reactants: delta S = sum of S(products) - sum of S(reactants). Take the values from the table, multiply each by the number of moles in the balanced equation, and subtract. Standard entropies are always positive for a real substance at 298 K, so unlike an enthalpy of formation you never have to worry about a negative sign in the sum itself; a negative delta S can only come from the balance of the equation.

Worked example. For CaCO3(s) -> CaO(s) + CO2(g), the standard entropies are +92.9, +39.7 and +213.6 J K-1 mol-1 respectively. So delta S = (39.7 + 213.6) - 92.9 = +160.4 J K-1 mol-1. The positive sign is what you should expect without doing any arithmetic, because one mole of gas has appeared on the right-hand side, and the calculation confirms it. Now reverse the reaction, CaO(s) + CO2(g) -> CaCO3(s), and the entropy change is -160.4 J K-1 mol-1: reversing a reaction reverses the sign of delta S.

The units are the trap. Standard entropies are in J K-1 mol-1 but enthalpies are almost always in kJ mol-1, so before you feed both into the Gibbs equation you must convert: either divide the entropy change by 1000 to put it in kJ K-1 mol-1, or multiply the enthalpy change by 1000 to put it in J mol-1. Getting this wrong changes the answer by a factor of a thousand and is the single most common arithmetic error in this topic. Keep the units attached to every line of working, and check that your final delta G comes out in kJ mol-1.

• delta S = sum of S(products) - sum of S(reactants), with the balancing numbers included.
• Standard entropies at 298 K are positive, so a negative delta S must come from the equation, usually from fewer moles of gas on the right.
• Convert between J K-1 mol-1 and kJ K-1 mol-1 before using the Gibbs equation.
• Reversing a reaction reverses the sign of delta S; doubling the equation doubles the entropy change.

Gibbs free energy and feasibility

Enthalpy and entropy pull in different directions: a reaction tends to go if it releases energy, and it also tends to go if it increases disorder. The Gibbs equation combines the two into a single quantity, the Gibbs free energy change: delta G = delta H - T delta S, where T is the temperature in kelvin and delta H and delta S are the standard values for the reaction. A process is feasible when delta G is zero or negative; if delta G is positive, the process cannot take place under those conditions, whatever the rate question. Feasibility is a statement about energy and disorder, nothing else.

For the decomposition of calcium carbonate, delta H = +178 kJ mol-1 and delta S = +160.4 J K-1 mol-1. At 298 K the entropy term is T delta S = 298 x 0.1604 = 47.8 kJ mol-1, so delta G = 178 - 47.8 = +130.2 kJ mol-1. The value is strongly positive, so limestone does not decompose at room temperature. Raise the temperature and the term you subtract grows: the two sides are equal when T = delta H / delta S = 178 / 0.1604 = 1110 K, and above that temperature delta G is negative and the decomposition becomes feasible. That is the thermodynamic reason a lime kiln has to reach well over 1000 K to turn limestone into quicklime.

Every sign combination is possible, and each one has a fixed consequence; the table below is worth learning as four rules rather than deriving it each time. Where delta H and delta S have the same sign there is a boundary temperature, T = delta H / delta S, at which delta G is exactly zero. Below it the reaction is feasible if both signs are negative, and above it the reaction is feasible if both signs are positive. delta H and delta S themselves change only slightly with temperature, which is why using the standard values at all temperatures gives a good approximation.

Always finish a feasibility question by linking the sign of delta G to the temperature. And remember what feasibility does not say: a reaction with a negative delta G may be far too slow to observe. The Haber process is the classic case, because with delta H = -92 kJ mol-1 and delta S = -198 J K-1 mol-1, delta G becomes more negative as the temperature falls, so the thermodynamics favour the lowest possible temperature, yet the industrial plant runs at about 700 K with an iron catalyst because at 298 K the rate would be negligible.

• delta G = delta H - T delta S, with T in kelvin, and the answer in the same units as delta H once the entropy change has been converted.
• delta G equal to zero or negative: feasible. Positive delta G: not feasible under those conditions.
• A reaction with positive delta H and positive delta S becomes feasible only above T = delta H / delta S; with negative delta H and negative delta S, only below it.
• Feasibility says nothing about rate: a feasible reaction can still be so slow that it is never seen.