Entropy and energetics

Why does sodium chloride exist, why is it CaCl₂ and not CaCl, and why do some reactions that take in heat still happen by themselves? This topic answers with two tools: Born–Haber and solution cycles, which measure the energy holding ionic solids together, and entropy, which counts the ways energy can be shared out and decides whether a change is feasible at all.

The enthalpy changes behind an ionic lattice

The lattice energy of an ionic compound measures how strongly its ions hold each other. It cannot be measured directly, because you cannot collect a mole of free gaseous ions and let them crash together. Instead it is calculated with Hess's law from enthalpy changes that can be measured. You need to know each of them precisely. All are standard values: measured at 100 kPa and a stated temperature, usually 298 K, with substances in their standard states.

• Why is the first electron affinity negative? The incoming electron is attracted by the nucleus of the neutral atom, so energy is released. Going down group 7 from chlorine to iodine it becomes less exothermic: the atoms are bigger, with more shielding, so the added electron is held less strongly.
• Why is the second electron affinity positive? The electron is being added to an ion that is already negative. Energy has to be put in to overcome the repulsion between them. For oxygen it is +798 kJ mol⁻¹, much larger than the −141 kJ mol⁻¹ released by the first.
• Atomisation of a diatomic element is per mole of atoms formed, so Δ_atH(Cl) is half the Cl–Cl bond enthalpy: about +122 kJ mol⁻¹ from a bond enthalpy of about +243 kJ mol⁻¹.

Born–Haber cycles

A Born–Haber cycle is a Hess's law cycle that links the formation of an ionic solid directly from its elements with an indirect route through gaseous atoms and gaseous ions. Both routes start and finish in the same place, so their total enthalpy changes are equal:

• Group 2 halides such as MgCl₂: include both the first and second ionisation energies of the metal, and twice the atomisation enthalpy and electron affinity of chlorine, because there are two chloride ions. (2 × Δ_atH(Cl) equals the Cl–Cl bond enthalpy.)
• Oxides such as MgO: include the first and second electron affinities of oxygen. The second is endothermic, so its arrow points up in the cycle.
• Any one value in the cycle can be found if all the others are known: exam questions often ask for an electron affinity or an atomisation enthalpy instead of the lattice energy.

Why calcium chloride is CaCl₂

Born–Haber cycles can test compounds that don't exist. A theoretical lattice energy can be estimated for CaCl and CaCl₃ from the sizes and charges of the ions, and the cycle then predicts their enthalpies of formation.

The enthalpy of formation is largely a contest between the energy needed to make the cation and the energy released when the lattice forms. Going from Ca⁺ to Ca²⁺ costs an extra 1145 kJ mol⁻¹, but the doubly charged ion attracts two chloride ions far more strongly, and the lattice energy grows by over 1500 kJ mol⁻¹: CaCl₂ is much more exothermic than CaCl. Removing a third electron would mean breaking into the full inner shell of calcium, which costs 4912 kJ mol⁻¹. The extra lattice energy cannot repay that, so CaCl₃ would be strongly endothermic. Even CaCl, though exothermic, would not survive: 2CaCl(s) → Ca(s) + CaCl₂(s) releases about 440 kJ mol⁻¹.

What makes a lattice energy large?

Lattice energy comes from the electrostatic attraction between oppositely charged ions. That attraction is stronger when the charges are larger and when the ions can get closer together. So lattice energy becomes more exothermic when:

• the ions are smaller: their centres are closer, so they attract more strongly. Lattice energies become less negative down a group.
• the ions have larger charges: the attraction depends on the product of the charges, so going from 1+ and 1− ions to 2+ and 2− ions roughly quadruples it.

Theoretical and experimental lattice energies

A lattice energy can also be calculated purely from theory, using the perfect ionic model: the ions are treated as perfect spheres with their charge spread evenly, held together only by electrostatic attraction. Comparing this theoretical value with the experimental value from a Born–Haber cycle tests whether the model is right.

• If the two values agree closely, the compound is almost purely ionic.
• If the Born–Haber value is more exothermic than the theoretical one, the bonding has some covalent character. The cation pulls the anion's electron cloud towards itself (polarisation), concentrating electron density between the ions; this extra attraction makes the real lattice stronger than the model predicts.
• Polarising power of a cation is greatest when it is small and highly charged (high charge density): Mg²⁺ polarises more than Na⁺, and Al³⁺ more still.
• Polarisability of an anion is greatest when it is large (and highly charged): its outer electrons are far from the nucleus and easily pulled. I⁻ is far more polarisable than F⁻.

Enthalpies of solution and hydration

Water molecules are polar. The δ− oxygen atoms cluster round positive ions and the δ+ hydrogen atoms point at negative ions. These ion–dipole attractions release energy, so hydration enthalpies are always negative.

Hydration enthalpy is more exothermic for ions with a higher charge density: a smaller radius or a larger charge lets the water molecules get closer and attract more strongly.

The solution cycle

Dissolving can be imagined in two stages: break the lattice into gaseous ions (the reverse of lattice formation, so +|LE|), then hydrate the ions. Hess's law gives:

Lattice energy and total hydration enthalpy are usually large and similar in size, so Δ_solH, the small difference between them, can come out either side of zero. A very endothermic Δ_solH, where the lattice is far stronger than the hydration can repay, usually means the compound is insoluble. But enthalpy is not the whole story: ammonium nitrate dissolves readily even though Δ_solH is +26 kJ mol⁻¹ and the solution goes cold. To see why, we need entropy.

Entropy

A spontaneous (feasible) change is one that happens on its own once started, without anything continuing to drive it: a gas spreading through a room, ice melting on a warm day, an acid neutralising an alkali. Most feasible reactions are exothermic, but not all. Ice melting and ammonium nitrate dissolving both take in heat. The quantity that decides the direction of every change is entropy, S.

Energy comes in small packets (quanta) that particles can swap among themselves. With more quanta, or more particles to share them, the number of possible arrangements grows enormously. A system left to itself drifts into the most probable state: the one with the most arrangements, which is why energy spreads out and never concentrates by itself.

Entropy of substances

• Gases have far higher entropies than liquids, and liquids higher than solids, because their particles are free to move and can be arranged in vastly more ways.
• Entropy increases with temperature: more energy quanta means more ways to arrange them, and particles vibrate and move more.
• There is a sudden jump at each change of state. The jump at boiling is much bigger than at melting, because a gas is so much more disordered than a liquid.
• More complex molecules, with more atoms that can move in more ways, have higher entropies than simple ones; mixtures have higher entropies than pure substances.
• Only a perfect crystal at 0 K has zero entropy. Every substance at 298 K, elements included, has a positive standard entropy, S^⦵, measured in J K⁻¹ mol⁻¹ (note: joules, not kilojoules).

Predicting the sign of ΔS_system

The system is the reacting chemicals; everything else (the flask, the water, the air) is the surroundings. You can usually predict the sign of the system's entropy change from the equation. ΔS_system is positive when a solid or liquid turns into a gas, and when the number of moles of gas increases; it is negative when the reverse happens.

• NH₄Cl(s) → NH₃(g) + HCl(g): a solid becomes two gases, so ΔS_system is large and positive.
• 2Mg(s) + O₂(g) → 2MgO(s): a gas is used up to make a solid, so ΔS_system is negative.
• N₂(g) + 3H₂(g) → 2NH₃(g): four moles of gas become two, so ΔS_system is negative.
• NaCl(s) + aq → Na⁺(aq) + Cl⁻(aq): an ordered lattice breaks into freely moving ions, so ΔS_system is usually positive.

Calculating ΔS_system

Entropy of the surroundings and total entropy

An exothermic reaction releases energy into the surroundings. Those extra quanta can be shared among the surrounding particles in many more ways, so the entropy of the surroundings increases. An endothermic reaction takes energy from the surroundings and lowers their entropy. The effect is bigger when the surroundings are cold: adding a given amount of energy to something cold makes a larger relative difference to the number of ways it can be shared than adding it to something already hot.

Changes of state

At its melting or boiling temperature, a substance is in equilibrium between the two states, and ΔS_total = 0. That makes T = ΔH/ΔS_system a way to calculate the temperature of a change of state. For water boiling, ΔH = +40.7 kJ mol⁻¹ and ΔS_system = +109 J K⁻¹ mol⁻¹, so T = 40 700 / 109 = 373 K. Below 373 K, ΔS_surroundings is more negative than ΔS_system is positive, and liquid water is the stable state; above it, the reverse.

How temperature affects feasibility

Since ΔS_total = ΔS_system − ΔH/T, raising the temperature shrinks the −ΔH/T term. Whether that helps depends on the signs:

If ΔH is close to zero, −ΔH/T is small at any temperature, and feasibility depends almost entirely on ΔS_system.

Gibbs energy

Chemists usually combine ΔH and ΔS_system into one quantity that refers only to the system: the Gibbs energy change, ΔG.

Gibbs energy and equilibrium constants

ΔG is linked to the equilibrium constant K (K_c or K_p) of the reaction by

A negative ΔG means ln K is positive, so K > 1 and products predominate at equilibrium. The more negative ΔG, the larger K. A reaction with ΔG of about −60 kJ mol⁻¹ at 298 K has K of about 10¹⁰: it goes essentially to completion. A positive ΔG gives K < 1, but not zero, so a little product still forms.

Feasible does not mean fast

ΔG and ΔS_total say whether a reaction can happen, not how quickly. A mixture of hydrogen and oxygen at room temperature has a hugely negative ΔG for forming water, yet nothing happens until a spark supplies the activation energy. Diamond should turn into graphite at room temperature (ΔG is negative), but the rate is immeasurably slow. Such systems are thermodynamically unstable but kinetically stable (kinetically inert): a high activation energy stops the feasible reaction from happening at a noticeable rate. Remember too that the calculations use standard data and assume ΔH and ΔS stay the same at other temperatures, which is only approximately true.

Entropy and solubility

Now return to dissolving. When an ionic solid dissolves, the ions leave an ordered lattice and spread through the solution, so ΔS_system is usually positive.

• If Δ_solH is negative, ΔS_surroundings is positive too, so ΔS_total is positive and the solid dissolves at any temperature.
• If Δ_solH is positive, ΔS_surroundings is negative. The solid still dissolves if the positive ΔS_system is bigger. Ammonium nitrate: Δ_solH = +25.7 kJ mol⁻¹, ΔS_system = +108.7 J K⁻¹ mol⁻¹. At 298 K, ΔS_surroundings = −25 700/298 = −86.2, so ΔS_total = +22.5 J K⁻¹ mol⁻¹: it dissolves, and cold packs use exactly this.
• Raising the temperature makes the negative −Δ_solH/T term smaller in size, so salts with an endothermic Δ_solH generally become more soluble when heated.
• If Δ_solH is very endothermic, because the lattice energy is far larger than the hydration enthalpies, ΔS_surroundings is too negative to overcome and the solid is insoluble.

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