Chemical energetics

Cambridge International AS & A Level Chemistry 9701 · A Level

Every objective in this topic is printed under "A Level subject content" in the 9701 syllabus and carries the tier "A Level". None of it is AS, and it is a separate topic from the AS topic 5 of the same name, which it builds on. It is examined on Paper 4. Paper 1 is the AS multiple-choice paper and the whole 9701 bank on this site comes from it, so no practice here is tagged to this topic.

The definitions, stated exactly

Marks in this topic are lost on definitions more often than on calculations, and the reason is that every one of them specifies a state and a quantity. Get those two right and the rest follows.

• Enthalpy change of atomisation, ΔH_at: the enthalpy change when one mole of gaseous atoms is formed from the element in its standard state. Always endothermic, because bonds must be broken. Note "one mole of gaseous atoms", so for chlorine it is ½Cl₂(g) → Cl(g), with the half, and not the full bond enthalpy.
• First ionisation energy: the energy to remove one electron from each of one mole of gaseous atoms to form one mole of gaseous 1+ ions. Endothermic.
• First electron affinity, EA: the enthalpy change when one electron is added to each of one mole of gaseous atoms to form one mole of gaseous 1- ions. Usually exothermic, because the added electron is attracted by the nucleus.
• Lattice energy, ΔH_latt: the enthalpy change when one mole of an ionic solid is formed from its gaseous ions. Always exothermic, and strongly so, because oppositely charged ions attract as they come together. The syllabus defines it in this direction, from gaseous ions to solid, so the value is always negative.
• Enthalpy change of hydration, ΔH_hyd: the enthalpy change when one mole of gaseous ions is dissolved in water to give an infinitely dilute solution. Always exothermic, because ion-dipole attractions form between the ion and water molecules.
• Enthalpy change of solution, ΔH_sol: the enthalpy change when one mole of a solute dissolves in a solvent to give an infinitely dilute solution. May be exothermic or endothermic, and it is usually small, because it is the difference between two large numbers.

Second electron affinities

The first electron affinity of oxygen is exothermic, but the second is strongly endothermic:

O⁻(g) + e⁻ → O²⁻(g)

The reason is worth understanding rather than memorising: the electron is being added to a species that is already negative, so it must be forced in against electrostatic repulsion. That is why the O²⁻ term in a Born-Haber cycle for an oxide points upwards.

Trends in electron affinity

Electron affinity becomes less exothermic down a group, because the added electron enters a shell further from the nucleus, with more shielding, so the attraction is weaker.

Across the relevant groups, two anomalies matter:

• In Group 17, chlorine has a more exothermic first electron affinity than fluorine, even though fluorine is smaller. Fluorine's 2p subshell is so compact that the electrons already there repel the incoming one strongly, which more than offsets the shorter distance to the nucleus.
• In Group 16, oxygen and sulfur show the same anomaly for the same reason: sulfur's first electron affinity is more exothermic than oxygen's.

Group 16 first electron affinities are less exothermic than the corresponding Group 17 values, because a Group 16 atom has one less proton and a lower nuclear charge.

Born-Haber cycles

A Born-Haber cycle is Hess's law applied to the formation of an ionic compound. It exists because lattice energy cannot be measured directly, so it is calculated from quantities that can be.

▶

The cycle relates the enthalpy change of formation of the solid to the sum of the steps that go via the gaseous ions:

ΔH_f = ΔH_at(metal) + IE(metal) + ΔH_at(non-metal) + EA(non-metal) + ΔH_latt

Rearranged for the unknown:

ΔH_latt = ΔH_f - ΔH_at(metal) - IE(metal) - ΔH_at(non-metal) - EA(non-metal)

Worked example, sodium chloride. Given ΔH_f = -411, ΔH_at(Na) = +107, first ionisation energy of Na = +496, ΔH_at(Cl) = +122, first electron affinity of Cl = -349, all in kJ mol⁻¹.

Sum of the steps to reach the gaseous ions:

107 + 496 + 122 - 349 = 376

Then:

-411 - 376 = -787

so the lattice energy of sodium chloride is -787 kJ mol⁻¹.

Worked example with a 2+ cation and a 2- anion. For magnesium oxide, both ionisation energies of magnesium are needed, and both electron affinities of oxygen, and the atomisation of oxygen is ½O₂(g) → O(g).

Given ΔH_f = -602, ΔH_at(Mg) = +148, IE₁ = +738, IE₂ = +1451, ΔH_at(O) = +249, EA₁ = -141, EA₂ = +798.

Sum of the steps:

148 + 738 + 1451 + 249 - 141 + 798 = 3243

ΔH_latt = -602 - 3243 = -3845

so about -3845 kJ mol⁻¹, nearly five times the value for sodium chloride. The reason is the next section.

Two habits prevent most errors here. Count the charges first, so you know how many ionisation energies and how many electron affinities you need. And check every sign as you write the cycle: atomisation and ionisation are positive, first electron affinity is negative, second electron affinity is positive, lattice energy and usually formation are negative.

What makes a lattice energy large

Lattice energy depends on the strength of the electrostatic attraction between the ions, and therefore on two things:

• Ionic charge. Larger charges attract more strongly, so the lattice energy is more exothermic. This is the dominant factor. MgO, with 2+ and 2-, has a lattice energy several times that of NaCl, with 1+ and 1-.
• Ionic radius. Smaller ions can approach more closely, so the attraction is stronger and the lattice energy is more exothermic. Down Group 17 the lattice energies of the sodium halides become less exothermic as the halide ion grows.

Combine them as charge density: a small, highly charged ion gives a large lattice energy. That single idea also explains the melting points of ionic solids, which is why questions often ask for both together.

Enthalpy of solution, and its cycle

Dissolving an ionic solid takes two steps, and the energy cycle simply sets one route against the other:

• Break the lattice into gaseous ions. This is the reverse of the lattice energy, so it is endothermic and numerically equal to it.
• Hydrate the gaseous ions. This is exothermic.

ΔH_sol = -ΔH_latt + ΣΔH_hyd

or equivalently, ΔH_sol = ΔH_hyd(cation) + ΔH_hyd(anion) - ΔH_latt.

Worked example. For sodium chloride, ΔH_latt = -787, ΔH_hyd(Na⁺) = -406 and ΔH_hyd(Cl⁻) = -378 kJ mol⁻¹.

Sum of the hydration enthalpies:

-406 - 378 = -784

ΔH_sol = 787 - 784 = 3

so +3 kJ mol⁻¹, very slightly endothermic, which is why a beaker of water gets marginally cooler when salt dissolves in it.

That result illustrates the general point. ΔH_sol is the small difference between two large numbers, so a compound dissolves endothermically or exothermically depending on which of the two just wins, and a small error in either input changes the sign of the answer.

What makes hydration enthalpy large. The same two factors as lattice energy, for the same electrostatic reason:

• Greater ionic charge means stronger ion-dipole attraction to water, so more exothermic.
• Smaller ionic radius means the water molecules approach closer, so more exothermic.

So Mg²⁺ is hydrated far more exothermically than Na⁺, being both smaller and doubly charged, and hydration enthalpies become less exothermic down any group.

Entropy

Entropy, S, is a measure of the number of possible arrangements of the particles and of their energy in a system. A system with more ways of arranging itself has a higher entropy, and systems tend towards states with more arrangements simply because there are more of them.

Standard entropies are quoted in J K⁻¹ mol⁻¹, note the joules, and, unlike enthalpies, S is never negative for a substance: a perfect crystal at 0 K has S = 0 and everything else is above it.

Predicting the sign of ΔS

• Changes of state. Solid to liquid to gas increases the freedom of the particles enormously, so melting, boiling and sublimation all have positive ΔS, and the gas step is much the largest. Freezing and condensing are negative.
• Dissolving. Usually positive, because an ordered lattice is broken up and the ions are dispersed through the solvent. Note the qualification: for a small, highly charged ion the water molecules become strongly ordered around it, which can make ΔS small or even negative.
• Temperature rise. Positive. The particles have a wider range of energies available, so there are more ways of distributing the energy. A sharp rise occurs at each change of state.
• Change in the number of gaseous molecules. This dominates any reaction involving gases. More moles of gas on the right means positive ΔS, fewer means negative. Count only the gases; solids and liquids barely matter by comparison.

Worked example of the reasoning. For CaCO₃(s) → CaO(s) + CO₂(g), a gas is produced from a solid, so ΔS is clearly positive. For N₂(g) + 3H₂(g) → 2NH₃(g), four moles of gas become two, so ΔS is negative.

Calculating ΔS

ΔS = ΣS(products) - ΣS(reactants)

Worked example. For N₂(g) + 3H₂(g) → 2NH₃(g), with S values of 192, 131 and 193 J K⁻¹ mol⁻¹ respectively.

Products:

2 × 193 = 386

Reactants:

192 + 3 × 131 = 585

ΔS = 386 - 585 = -199

so -199 J K⁻¹ mol⁻¹, negative as predicted.

Remember to multiply each entropy by its stoichiometric coefficient, which is the commonest arithmetic slip here.

Gibbs free energy and feasibility

ΔG = ΔH - TΔS

with T in kelvin. The units trap is unavoidable and is worth stating plainly: ΔH is in kJ mol⁻¹ and ΔS is in J K⁻¹ mol⁻¹, so ΔS must be divided by 1000 before it is used, or ΔH multiplied by 1000. Almost every wrong answer in this section comes from that one step.

The rule of feasibility:

• ΔG negative: the reaction is feasible, meaning thermodynamically spontaneous.
• ΔG positive: not feasible.
• ΔG = 0: the system is at equilibrium, and the temperature at which this happens is the point at which feasibility changes.

Worked example. For the decomposition of calcium carbonate, ΔH = +178 kJ mol⁻¹ and ΔS = +161 J K⁻¹ mol⁻¹. Is it feasible at 298 K?

Convert the entropy term:

298 × 161 / 1000 = 47.978

ΔG = 178 - 47.978 = 130.022

so ΔG is about +130 kJ mol⁻¹, positive, and the reaction is not feasible at room temperature. Limestone does not decompose in a field.

Finding the temperature at which it becomes feasible. Set ΔG = 0, so ΔH = TΔS:

T = 178 × 1000 / 161 = 1105.6

so about 1106 K, or 833 °C, above which decomposition becomes feasible. That is why a lime kiln has to be hot.

The four cases

Working this table out from the equation is faster and safer than memorising it. The TΔS term grows with temperature, so temperature decides the outcome only when ΔH and TΔS have the same sign.

Feasible is not the same as fast

A reaction with a negative ΔG will go, but thermodynamics says nothing about how quickly. The conversion of diamond to graphite has a negative ΔG at room temperature and yet does not happen in any observable time, because the activation energy is enormous. Feasibility is about the energetics, rate is about the kinetics, and confusing the two is a favourite trap in a longer answer.

Common mistakes

• Defining lattice energy in the wrong direction, so the sign comes out positive. The syllabus defines it as gaseous ions to solid, which is exothermic.
• Using a bond enthalpy where an enthalpy of atomisation is wanted. Atomisation makes one mole of gaseous atoms, so it is half the bond enthalpy for a diatomic molecule.
• Making the second electron affinity negative. It is endothermic, because the electron is added to a negative ion.
• Forgetting the second ionisation energy in a Born-Haber cycle for a 2+ cation.
• Saying charge and radius affect lattice energy without saying which way, or saying "the lattice energy is bigger" when the value is negative and what is meant is more exothermic.
• Adding the lattice energy to the hydration enthalpies instead of subtracting it in a solution cycle.
• Assuming ΔS is positive for every dissolving process.
• Forgetting to multiply an entropy by its coefficient in the balanced equation.
• Mixing kJ and J in ΔG = ΔH - TΔS.
• Using degrees Celsius for T.
• Saying a reaction with negative ΔG must be fast.

Not the topic you were looking for? Describe what you are stuck on in your own words and we will take you to the notes that answer it.