Cambridge International AS & A Level Chemistry 9701 · A Level topic 23

Chemical energetics

What this chapter covers23.1–23.4

Sodium burns in chlorine with a brilliant yellow flame and leaves a white crust of salt. Magnesium ribbon flares white in air and crumbles to magnesium oxide. Both reactions are strongly exothermic, and both produce ionic solids. Yet every step that turns a metal atom into a positive ion costs energy, and removing the second electron from magnesium costs almost twice as much as removing the first. Something in the formation of an ionic solid must repay those costs with interest. This chapter identifies that something — the lattice energy — and shows how it is measured, even though it can never be measured directly.

The chapter then asks two further questions that ordinary enthalpy changes cannot answer. Why do some salts, such as ammonium nitrate or potassium chloride, dissolve readily even though dissolving them makes the solution colder? And why does calcium carbonate refuse to decompose at room temperature but decompose steadily in a lime kiln, even though the enthalpy change hardly alters between the two temperatures? The answers need a second quantity, entropy, and a way of combining it with enthalpy, the Gibbs free energy change.

Enthalpy change of atomisation23.1.1(a)

The first step in building an ionic solid from its elements is to turn each element into separate gaseous atoms. For a metal this means pulling the atoms out of the metallic lattice; for a non-metal such as chlorine it means breaking the covalent bonds in the diatomic molecules.

Two features of the definition are worth fixing now. First, the quantity is defined per mole of atoms produced, not per mole of element used. For chlorine, one mole of Cl atoms comes from half a mole of Cl₂ molecules, so ΔH_at for chlorine is half the Cl–Cl bond energy: 242 ÷ 2 = 121 kJ mol⁻¹. For oxygen, atomisation is half the O=O bond energy, ½ × 496 = +248 kJ mol⁻¹. Second, atomisation is always endothermic. Whatever holds the atoms together — metallic bonding, covalent bonds, or the intermolecular forces in solid iodine or liquid bromine — has to be overcome, and that always requires energy.

Lattice energy23.1.1(b)

An ionic solid is a regular three-dimensional array of oppositely charged ions. Each ion is attracted by all the oppositely charged neighbours around it and repelled by the similarly charged ions a little further away; the net effect is a large attraction that holds the lattice together. The lattice energy measures how much energy is released when that lattice assembles from free ions.

Lattice energy defined this way is always exothermic: oppositely charged ions attract, and energy is released as they come together. The more negative the value, the more strongly the ions are held.

Look carefully at the equation for MgCl₂. One mole of compound is formed from one mole of Mg²⁺ ions and two moles of Cl⁻ ions. It is therefore wrong to say that lattice energy is "formed from one mole of gaseous ions"; the one mole refers to the compound. Examiners have reported both errors — omitting "one mole" altogether, and attaching it to the ions instead of the compound.

Lattice energy cannot be measured directly, because there is no practical way of assembling a mole of gaseous ions into a crystal and catching the heat released. It is found indirectly, by Hess's law, from quantities that can be measured — which is what a Born–Haber cycle does.

First electron affinity23.1.2(a)

Forming the anions in an ionic solid means adding electrons to gaseous non-metal atoms. The energy change for this is the electron affinity.

The definition mirrors that of ionisation energy, and examiners regularly report candidates confusing the two. The distinction is simply the direction: ionisation energy removes an electron from a gaseous atom and is always endothermic; electron affinity adds an electron to a gaseous atom. Both are defined per mole of gaseous atoms, and both require the gaseous state.

For most non-metals the first electron affinity is exothermic. The incoming electron is attracted by the nucleus of the neutral atom, because the atom's own electrons do not completely shield the nuclear charge from an electron arriving at the outer shell. Energy is released as it is captured.

Second electron affinity

Oxide and sulfide ions carry a 2− charge, so forming them needs a second electron to be added:

The second electron affinity is endothermic, and large. The electron is now being added to a species that is already negatively charged, so it is repelled; energy must be supplied to force it on. This is why, in a Born–Haber cycle for an oxide, the total energy needed to make O²⁻(g) from O(g) is strongly positive (−141 + 798 = +657 kJ mol⁻¹). The oxide ion exists in solids only because the very large lattice energy of a 2+/2− lattice more than repays that cost.

What affects electron affinity: Groups 16 and 1723.1.2(b), (c)

The size of a first electron affinity depends on how strongly the nucleus attracts an electron arriving at the outer shell. The same three factors that govern ionisation energy are at work, and they act in the same directions:

• nuclear charge — more protons attract the incoming electron more strongly;
• distance — the further the outer shell is from the nucleus, the weaker the attraction;
• shielding — more complete inner shells between nucleus and outer shell reduce the attraction the incoming electron feels.

Down Group 17

From chlorine to bromine to iodine, the first electron affinity becomes less exothermic (−349, then about −325, then about −295 kJ mol⁻¹). Nuclear charge rises, but each step adds a complete inner shell. The incoming electron enters a shell further from the nucleus and is better shielded, so it is attracted less strongly and less energy is released when it is captured.

Down Group 16

Sulfur, selenium and tellurium follow the same pattern for exactly the same reason: from S to Te the first electron affinity becomes less exothermic as the atoms get larger and the incoming electron is further from, and better shielded from, the nucleus.

The exceptions at the top of each group

Fluorine's first electron affinity (−328 kJ mol⁻¹) is less exothermic than chlorine's, and oxygen's (−141 kJ mol⁻¹) is less exothermic than sulfur's. Fluorine and oxygen atoms are very small, so their outer shells are crowded. An electron added to such a small shell is close to the electrons already there and is strongly repelled by them; that repulsion offsets part of the attraction of the nucleus. From the second member of each group onwards, the larger outer shell relieves this crowding, and the ordinary trend takes over.

Across a period

Comparing elements in the same period isolates the effect of nuclear charge. Carbon's first electron affinity is only −120 kJ mol⁻¹, fluorine's −328 kJ mol⁻¹. The electron is added to the same shell in each case, with the same shielding, but fluorine has three more protons, so the attraction for the incoming electron is much greater. Examiners report that candidates often try to explain this difference in terms of electronegativity; the answer that earns the mark is the greater nuclear charge of fluorine and its greater attraction for the added electron.

Constructing a Born–Haber cycle23.1.3

A Born–Haber cycle is a Hess's law cycle, usually drawn as an energy-level diagram, that links the enthalpy change of formation of an ionic solid to the lattice energy through a series of steps, each of which can be measured. The formation of the solid from its elements can happen by two routes that start and finish in the same place:

• directly: elements in their standard states → ionic solid, with enthalpy change ΔH_f;
• indirectly: elements → gaseous atoms → gaseous ions → ionic solid, via atomisation, ionisation energy, electron affinity and finally the lattice energy.

By Hess's law the two routes have the same total enthalpy change. For a compound M^a+X^b−:

with each term multiplied by the number of moles the formula requires.

Drawing the energy-level diagram

The diagram is built from the bottom-left. A horizontal line represents the elements in their standard states, at zero. Every endothermic step is drawn as an arrow pointing up to a new, higher line; every exothermic step as an arrow pointing down. On every line you write the species present at that stage, with state symbols, so that the atoms and electrons balance on every level. The last arrow, the lattice energy, runs a long way down from the gaseous ions to the ionic solid, and the formation arrow joins the elements directly to the solid.

Cycles with doubly charged ions

For compounds of a Group 2 metal the metal must lose two electrons, so both the first and second ionisation energies appear. For an oxide or sulfide the non-metal must gain two electrons, so both electron affinities appear — the first exothermic, the second endothermic. The magnesium oxide cycle below shows both features. Notice how far above the elements the gaseous ions sit: 3239 kJ mol⁻¹ must be put in to make Mg²⁺(g) and O²⁻(g) from the elements. The lattice energy repays all of that and a further 602 kJ mol⁻¹, which is why MgO is stable.

Calculations with Born–Haber cycles23.1.4

Because the cycle is closed, any one unknown term can be calculated from the others. The method is always the same: write the Hess's law sum along the two routes, substitute every value with its own sign, multiply by the number of moles each step involves, and solve for the unknown.

The same cycle can be run backwards to find a quantity that is hard to measure. Electron affinities, in particular, are often obtained this way from a lattice energy that has been calculated independently.

What controls the size of a lattice energy23.1.5

Lattice energy comes from electrostatic attraction between oppositely charged ions. The force between two charges is proportional to the product of the charges and falls off with the distance between their centres. Two properties of the ions therefore control the size of a lattice energy:

• Ionic charge. The larger the charges on the ions, the stronger the attraction between them and the more exothermic the lattice energy. Doubling the charge on both ions quadruples the charge product.
• Ionic radius. The smaller the ions, the closer together their centres can approach and the stronger the attraction. Smaller ions give a more exothermic lattice energy.

The two effects are often combined in the phrase charge density: a small, highly charged ion has a high charge density and forms strong ionic bonds.

Reading the data

Compare NaCl (−787) with MgO. Both have the same arrangement of ions, and Mg²⁺ and O²⁻ are similar in size to Na⁺ and Cl⁻, but the charge product rises from 1 to 4 and the lattice energy rises nearly five-fold. Within compounds of the same charges the differences are smaller and come from size: LiF is more exothermic than KF because Li⁺ is smaller than K⁺; KF is more exothermic than KCl because F⁻ is smaller than Cl⁻; CaO is more exothermic than SrO because Ca²⁺ is smaller than Sr²⁺.

Past-paper practice · Set 23A · Lattice energy, electron affinity and Born–Haber cycles

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 23A.1[4]

Answer and marking guidance

(a)(i) Energy released when one mole of an ionic compound is formed ✔ from its gaseous ions (under standard conditions) ✔. (ii) Ca²⁺ and O²⁻ have higher charges / higher charge density than Li⁺ and F⁻ ✔ — the answer must refer to the charges of both ions. (iii) MgO −3600 or more negative AND BaO −3200 or less negative ✔ (both needed). Examiner insight: the definition was generally well answered; the errors were omitting "1 mole" of compound and writing "1 mole of gaseous ions". Part (ii) was found difficult because answers did not address the charges of both ions.

Question 23A.2[9]

Answer and marking guidance

(a)(i) Enthalpy (energy) change ✔ when one mole of electrons ✔ is gained by one mole of gaseous atoms ✔ — two points for one mark, all three for two marks. (ii) Energy is needed to overcome the repulsion between the electron and the negative ion ✔. (iii) Becomes less negative (less exothermic) down the group ✔; the added electron is further from the nucleus / atomic radius larger / more shielding, so less attraction between the nucleus and the incoming electron ✔. (b) −208 = 131 + 906 + 1733 + 62 + 151 + 2x − 2605 ✔ (the seven correct values); 2x = −586 ✔ (EA used twice); x = −293 kJ mol⁻¹ ✔. (c) First box: less negative ✔, because Cd²⁺ is larger / has lower charge density, so the attraction between the ions is weaker. Examiner insight: common errors in (b) were −586 (no division by 2), −262 (sublimation enthalpy of iodine omitted), and sign errors such as +293 or −85 (using +208). In (c) most chose the right box but gave no reasoning about the larger Cd²⁺ attracting I⁻ less strongly; in (a)(iii) many did not state that the attraction between the nucleus and the incoming electron decreases.

Question 23A.3[8]

Answer and marking guidance

(a) Bond energy: always positive; atomisation: always positive; formation: can be either ✔. (b) Enthalpy change when one mole of gaseous atoms is formed ✔ from the element in its standard state under standard conditions ✔. (c) −31 = 2(285) + 2(731) + (−141) + 798 + ½(496) + x ✔ (the six correct values, with 2 × for silver and ½ × for O=O) ✔; x = −2968 kJ mol⁻¹ ✔. Ag's second ionisation energy, the O–O bond energy and the first ionisation energy of oxygen are not used. (d) least exothermic Ag₂Se, then Ag₂S, most exothermic Ag₂O ✔; the anion radius increases (charge density decreases) from O²⁻ to Se²⁻, so attraction between the ions decreases ✔. Examiner insight: a common error in (a) was claiming that the enthalpy change of formation is always negative. In (b) errors included "one mole of gaseous ions" and omitting "one mole". In (d) many identified that O²⁻ is smallest but did not say this gives greater attraction between the ions in Ag₂O.

Question 23A.4[7]

Answer and marking guidance

(ii) Each species on the correct line, with state symbols: Zn(s) + S(s) on the level from which ΔH_f starts; Zn(g) + S(s) after ΔH_at(Zn); Zn(g) + S(g) after ΔH_at(S); Zn⁺(g) + S(g) + e⁻ after IE₁. That is eight species plus the electron: any three for one mark, any six for two, all nine for three. (iii) First EA becomes less negative (less exothermic) from S to Te ✔; atomic radius increases / more shielding ✔; so less attraction for the incoming electron ✔ — two points one mark, three points two marks. (iv) O²⁻ has the same charge as S²⁻ but is smaller ✔, so there is greater attraction between Zn²⁺ and O²⁻ / stronger ionic bonds ✔. Examiner insight: in (ii) common errors were writing S₂(g) and omitting an electron after IE₁; in (iii) some used the ambiguous word "decreases"; in (iv) some wrongly referred to stronger nuclear attraction instead of attraction between the ions.

Question 23A.5[6]

Answer and marking guidance

(c) Use of 2 × (−348) for EA(F) and +158 for the F–F bond energy ✔; use of +147, +736 and +1450 for magnesium ✔; ΔH_latt = −1102 − (147 + 158 + 736 + 1450 − 696) = −2897 kJ mol⁻¹ ✔. Note that two F atoms need one F–F bond broken, so 158 is used once. (d)(i) Energy change when one electron is added ✔ to each atom in one mole ✔ of gaseous atoms. (ii) F has the greater nuclear charge (more protons) AND so a greater attraction between the nucleus and the added electron ✔. Examiner insight: errors in (c) came from wrong multipliers: −3245 (1 × 348), −3166 (1 × 348 and ½ × 158), −3055 (2 × 158). In (d)(ii) a common wrong answer was an explanation in terms of electronegativity.

Enthalpy change of solution and enthalpy change of hydration23.2.1

Drop anhydrous calcium chloride into water and the beaker becomes noticeably warm; dissolve ammonium nitrate and it becomes cold enough to feel through the glass. In both cases an ionic lattice is being taken apart, which on its own would absorb a very large amount of energy — the reverse of the lattice energy. The ions do not end up as free gaseous ions, however. Each one becomes surrounded by water molecules, and that process releases energy. Whether dissolving is exothermic or endothermic depends on which of these two large quantities is bigger.

Water molecules are polar. The oxygen atom carries a partial negative charge and lone pairs; the hydrogen atoms carry partial positive charges. Around a cation the water molecules orient with their oxygen atoms inwards; around an anion they orient with hydrogen atoms inwards. These ion–dipole attractions are what release energy when an ion is hydrated.

"Infinitely dilute" means that so much water is used that adding more produces no further enthalpy change: every ion is fully surrounded by water and the ions are too far apart to interact with one another. Both definitions involve dissolving in water; published marking required that idea in both, and examiners noted that answers which left it out of either definition lost the mark.

• Hydration is always exothermic: ion–dipole attractions form and nothing is broken.
• Solution can be exothermic or endothermic, because it is the balance between breaking up the lattice (endothermic) and hydrating the ions (exothermic).

The solution energy cycle23.2.2, 23.2.3

Dissolving an ionic solid can be imagined as taking place in two stages, which together start and finish in the same place as the direct process:

1. The lattice is separated into gaseous ions. This is the reverse of the lattice energy, so its enthalpy change is −ΔH_latt, a large positive number.
2. The gaseous ions are hydrated. The enthalpy change is the sum of the hydration enthalpies of all the ions in the formula — negative.

By Hess's law:

which is often written in the equivalent form ΔH_latt + ΔH_sol = ΣΔH_hyd, obtained by going round the cycle the other way: gaseous ions → solid → solution gives the same total as gaseous ions → solution directly.

The NaCl cycle makes a point that recurs: ΔH_sol is usually the small difference between two very large numbers. A modest change in either the lattice energy or the hydration enthalpies can switch dissolving from exothermic to endothermic. That is why the sign of ΔH_sol cannot be predicted by inspection, and why it alone does not decide whether a salt dissolves (the entropy change, discussed later in this chapter, matters as well).

What controls the enthalpy change of hydration23.2.4

Hydration releases energy because water molecules are attracted to the ion. The stronger that ion–dipole attraction, the more exothermic the hydration enthalpy. The same two properties that control lattice energy control it:

• Ionic charge. A more highly charged ion attracts the polar water molecules more strongly.
• Ionic radius. A smaller ion lets the water molecules approach its centre more closely, so the attraction is stronger.

Why ions of higher charge density dominate solubility trends

Both lattice energy and hydration enthalpy become less exothermic as ions get larger, so when you move down a group the two changes pull ΔH_sol in opposite directions. Which one wins depends on the particular compounds. For the Group 2 hydroxides and sulfates this balance explains their opposite solubility trends — the full argument belongs to Group 2 at A Level and is developed there. The skill here is the cycle itself: once ΔH_latt and ΔH_hyd are known, ΔH_sol follows.

Past-paper practice · Set 23B · Enthalpies of solution and hydration

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 23B.1[9]

Answer and marking guidance

(a) Lattice energy always negative; hydration always negative; solution either ✔ (all three for the mark). (b) The energy (enthalpy) change when one mole of gaseous ions is dissolved in water ✔. (c)(i) Correct six values only: −682.8, 178.2, 590, 1145, 111.9, −324.6 ✔; 2 × used with bromine (2 × 111.9 and 2 × −324.6) ✔; −682.8 = 178.2 + 590 + 1145 + 2(111.9) + 2(−324.6) + ΔH_latt, so ΔH_latt = −2170.6 kJ mol⁻¹ ✔. The Br–Br bond energy, the ionisation energy of bromine and the solution/hydration data are not used here. (ii) Correct three numbers: −2170.6, −103.1, −1579 ✔; ΔH_hyd(Br⁻) = (−2170.6 − 103.1 + 1579) ÷ 2 = −347 kJ mol⁻¹ ✔. (iii) Br⁻ has a smaller ionic radius ✔ so it has stronger attractive forces with water molecules ✔. Examiner insight: hydration was the tick most often misplaced in (a). In (c)(ii) many did not realise the calculation first gives the hydration enthalpy of two moles of Br⁻. In (iii) the attraction to water molecules was awarded much less often than the size comparison. Show working: credit is available for the method.

Question 23B.2[6]

Answer and marking guidance

(a) K⁺(g) + Cl⁻(g) on the upper box AND K⁺(aq) + Cl⁻(aq) on the other ✔; three correct arrows: ΔH_latt from gaseous ions to KCl(s), ΔH_sol from KCl(s) to the aqueous ions, ΣΔH_hyd from gaseous to aqueous ions ✔. (b) Use of −155, −2493 and 2 × (−364) ✔; ΔH_hyd(Mg²⁺) = −155 + (−2493) − 2(−364) = −1920 kJ mol⁻¹ ✔ (minimum 3 s.f.). (c) Mg²⁺ is smaller than K⁺; Mg²⁺ has a greater charge; so greater attraction between Mg²⁺ and Cl⁻ (stronger ionic bonds) — two points one mark, all three two marks. Examiner insight: completing the cycle was found harder than identifying the species. In (c) many gave the charge and size but did not state that this leads to greater attraction between the ions in MgCl₂.

Question 23B.3[9]

Answer and marking guidance

(a) Energy change when one mole of ionic solid is formed from gaseous ions ✔. (b) ΔH_f = −2237 + 193 + 590 + 1150 + 2(121) + 2(−364) ✔ = −790 kJ mol⁻¹ ✔. (c) −342 kJ mol⁻¹, because the Br atom has a larger radius than Cl, so less attraction for the added electron ✔. (d)(i) Solution: energy change when one mole of solid dissolves in water ✔; hydration: energy change when one mole of gaseous ions dissolves in water ✔. (ii) ΔH_hyd(Cl⁻) = (−2237 − 83 + 1650) ÷ 2 ✔ = −335 kJ mol⁻¹ ✔. Examiner insight: in (a), marks were lost by omitting "one mole" or "gaseous ions". In (b) many did not multiply 121 and/or −364 by two. In (c) answers had to state clearly that the Br atom is larger. In (d)(i) both definitions must say the substance dissolves in water; in (d)(ii) some used their answer to (b) instead of the lattice energy, and others forgot to divide by two and gave −670.

What entropy measures23.3.1

A spontaneous change is one that tends to happen of its own accord, without being driven by a continuous input of energy from outside. Many spontaneous reactions are exothermic, and it is tempting to conclude that releasing energy is what makes a change spontaneous. The evidence says otherwise. Ice melts at room temperature, an endothermic change. Solid hydrated barium hydroxide mixed with solid ammonium chloride reacts readily on mixing and the mixture becomes very cold. Potassium chloride dissolves readily in water even though ΔH_sol is positive. Enthalpy alone cannot be the whole story.

What these changes have in common is that the particles, and the energy they carry, end up spread out in more ways than before. Ice becomes a liquid whose molecules are free to move past one another; an ionic lattice becomes ions dispersed through the water; solids react to give ammonia gas, whose molecules fill the whole available volume. The quantity that measures this is entropy.

Why more arrangements means "more likely"

Imagine four gas particles free to move between the two halves of a container. There is only one way for all four to be in the left-hand half, but there are six ways of having two in each half (particles 1 and 2 on the left, or 1 and 3, or 1 and 4, and so on). If every individual arrangement is equally likely, the evenly spread state turns up six times as often as the all-on-one-side state. With four particles the preference is modest; with a mole of particles it is overwhelming, and the particles are essentially certain to be found spread through the whole container. The same argument applies to the energy: there are far more ways to share a quantity of energy among many particles than among a few.

This gives a statistical picture of why spontaneous changes happen: a system tends to move towards the state that can be achieved in the most ways — the state of higher entropy — simply because that state is overwhelmingly more probable. The description of entropy as "disorder" is a loose shorthand for this; the syllabus definition in terms of arrangements of particles and energy is the one that earns credit, and examiners have commented that stronger answers used it.

Predicting the sign of an entropy change23.3.2

The syllabus asks for the sign of ΔS in three kinds of change. In each, the reasoning comes back to counting arrangements.

(a) Changes of state

In a solid the particles vibrate about fixed positions; in a liquid they move past one another while remaining in contact; in a gas they move independently through the whole volume. Each step from solid to liquid to gas gives the particles more possible positions and more ways of distributing their energy, so

Melting and boiling therefore have positive ΔS; freezing and condensing negative ΔS. The jump on boiling is far larger than on melting, because separating the particles completely increases the number of possible arrangements enormously. Dissolving an ionic solid in water also has a positive ΔS: the ordered lattice becomes ions dispersed through the solvent.

(b) Temperature changes

Heating a substance gives its particles more energy. There are more ways to share a larger amount of energy among the particles, and the particles also occupy a wider range of energy levels, so the entropy of a substance increases with temperature even without a change of state.

(c) Reactions that change the number of gaseous molecules

Because gases have by far the largest entropies, the sign of ΔS for a reaction is almost always decided by the change in the number of moles of gas:

• more moles of gas in the products than the reactants → ΔS positive (CaCO₃(s) → CaO(s) + CO₂(g));
• fewer moles of gas in the products → ΔS negative (N₂(g) + 3H₂(g) → 2NH₃(g); 2Mg(s) + O₂(g) → 2MgO(s)).

Calculating the entropy change of a reaction23.3.3

Unlike enthalpy, entropy has an absolute zero — a perfect crystal at 0 K — so every substance has a measurable standard molar entropy, S^⊖, at 298 K. Elements have non-zero standard entropies (compare ΔH_f, which is zero for an element in its standard state). The entropy change of a reaction is then simply the difference between the total entropy of the products and that of the reactants:

Each S^⊖ is multiplied by the number of moles in the balanced equation. The result is in J K⁻¹ mol⁻¹, per mole of the reaction as written.

Notice the pattern in the table: the gases all lie between about 130 and 310 J K⁻¹ mol⁻¹, while the solids are mostly below 100. Within the solids, the hard, strongly bonded ones (graphite, MgO) have the smallest entropies, because their particles are held most rigidly and the energy can be arranged in the fewest ways.

The Gibbs equation23.4.1

A change is driven in two ways: by releasing energy to the surroundings (negative ΔH) and by increasing the number of possible arrangements (positive ΔS). The Gibbs free energy change combines the two into a single quantity:

The TΔS term has units of energy, because entropy is measured per kelvin. It represents the part of the energy change associated with rearranging the particles and their energy. A negative ΔH makes ΔG more negative; so does a positive ΔS, through the −TΔS term. The temperature sets how much weight the entropy term carries: at low temperature ΔG is dominated by ΔH, at high temperature by TΔS.

Calculating ΔG23.4.2

Feasibility and the sign of ΔG23.4.3

"Feasible" is a statement about whether a reaction can happen, not about how fast it happens. The Gibbs equation contains no information about activation energy. Carbon burns in oxygen with a very negative ΔG, yet a diamond left in air is not seen to change: the activation energy is high, and the rate at room temperature is immeasurably slow. A feasible reaction may need heating, a catalyst or a spark to get it going at a useful rate. A thermodynamic prediction and a kinetic one answer different questions, and a complete answer about whether a reaction is observed needs both.

How temperature affects feasibility23.4.4

For most reactions ΔH^⊖ and ΔS^⊖ change only slightly with temperature, and at this level they are treated as constant. The Gibbs equation then has the form of a straight line:

A graph of ΔG against T is a straight line with intercept ΔH (at T = 0) and gradient −ΔS. The sign of ΔS decides whether ΔG rises or falls as the temperature increases, and there are four cases.

The temperature at which feasibility changes

For the two cases where the signs of ΔH and ΔS are the same, there is a temperature at which ΔG = 0. Setting ΔG = 0 in the Gibbs equation:

For an endothermic reaction with positive ΔS this is the minimum temperature for feasibility; for an exothermic reaction with negative ΔS it is the maximum.

Past-paper practice · Set 23C · Entropy and Gibbs free energy

Examination questions on this part of the unit. Try each one on paper before opening the answer.

Question 23C.1[3]

Answer and marking guidance

(i) Negative AND the number of gas molecules is reduced (gaseous F₂ is used up to form a solid) ✔. (ii) TΔS becomes more negative as T increases ✔; so the synthesis becomes less feasible AND ΔG becomes (less negative, eventually) positive ✔. Examiner insight: many imprecise answers to (i) omitted the key fact that gas molecules are used up. Part (ii) was found very difficult; strong answers used the Gibbs equation as the basis of a logical explanation, stating that TΔS becomes more negative, or −TΔS more positive, as temperature increases.

Question 23C.2[4]

Answer and marking guidance

(i) All five points plotted correctly ✔; best-fit straight line with a negative gradient ✔. (ii) Gradient ≈ (−10.3 − 9.9) ÷ (1176 − 1050) = −0.160 kJ K⁻¹ mol⁻¹ ✔ (working seen); gradient = −ΔS^⊖, so ΔS^⊖ = +0.160 kJ K⁻¹ mol⁻¹ = +160 J K⁻¹ mol⁻¹ (± 10) ✔. Examiner insight: most could calculate the gradient, but many could not turn it into ΔS in J K⁻¹ mol⁻¹: the common errors were not multiplying by 1000 and not converting the negative gradient into a positive ΔS.

Question 23C.3[5]

Answer and marking guidance

(i) ΔG^⊖ = 0, so T = ΔH^⊖ ÷ ΔS^⊖ ✔; T = 132 ÷ 0.616 = 214.3 K = −58.7 °C ✔ (the answer was asked for in °C). (ii) ΔS^⊖ = [203 + 8(70) + 2(192)] − [427 + 2(95)] = 1147 − 617 = +530 J K⁻¹ mol⁻¹ ✔; ΔG^⊖ = ΔH^⊖ − TΔS^⊖ ✔; ΔG^⊖ = 133 − 298 × 0.530 = −24.9 kJ mol⁻¹ ✔ (ecf). The reaction is feasible at room temperature, consistent with the observation that it happens on mixing.

Question 23C.4[5]

Answer and marking guidance

(i) The number of possible arrangements of the particles and the energy in a system (or a measure of the disorder of a system) ✔. (ii) ΔS^⊖ = 192.8 + 213.8 − 238.2 − 188.8 = −20.4 J K⁻¹ mol⁻¹ ✔; ΔH^⊖ = (−45.9) + (−393.5) − (−101.7) − (−241.8) = −95.9 kJ mol⁻¹ ✔; ΔG^⊖ = ΔH^⊖ − TΔS^⊖ ✔; ΔG^⊖ = −95.9 − 298 × (−0.0204) = −89.8 kJ mol⁻¹ ✔. Examiner insight: better answers to (i) stated "the number of possible arrangements of particles and energy in a system". Part (ii) was usually well answered; candidates who wrote down their calculated ΔH and ΔS were much more likely to gain partial credit when the final answer was wrong.

Question 23C.5[4]

Answer and marking guidance

(i) ΔS negative AND more moles of gaseous reactants than gaseous products (3 mol O₂ → 2 mol SO₂) ✔. (ii) ΔS^⊖ = 50.8 + 197.7 − 43.7 − 5.7 = +199.1 J K⁻¹ mol⁻¹ ✔. (iii) ΔG = ΔH − TΔS ✔; ΔG = +733 − (800 + 273) × 0.218 = +499 kJ mol⁻¹ ✔ (min 3 s.f.): positive, so not feasible at 800 °C. Examiner insight: in (i) a common error was to omit any reference to the gaseous state. In (iii) the common wrong answers were +558.6 (using T = 800 instead of 1073 K) and 2.33 × 10⁵ (ΔS not divided by 1000).

Question 23C.6[2]

Answer and marking guidance

Feasibility increases as temperature increases; ΔS is positive (dissolving; ΔH is positive); so TΔS becomes more positive / −TΔS more negative as T rises, and ΔG becomes negative — two points one mark, all three two marks. Examiner insight: most recognised that feasibility increases with temperature, but few explained it in terms of ΔS (TΔS becoming more positive) and ΔG becoming more negative.

Question 23C.7[3]

Answer and marking guidance

(i) The number of possible arrangements of particles and energy in a system ✔. (ii) ΔS is positive AND the ionic lattice (solid) forms aqueous ions, KCl(s) → K⁺(aq) + Cl⁻(aq); or ΔS positive AND so ΔG is negative, TΔS being greater than ΔH_sol ✔. (iii) More soluble at 80 °C AND ΔG is more negative at higher T (TΔS more positive) ✔. Examiner insight: the definition in (i) was well known. Part (ii) was found difficult: many stated that ΔS was positive without adequate explanation, and a common error was to explain it in terms of melting rather than dissolving.

Misconceptions and the examiner's view23.1–23.4

How the topic is assessed

Self-test

Fourteen questions across the whole unit. Each explanation says why the right answer is right and what the wrong ones get wrong.

Definitions to learn

Data used in this chapter

The values used in the text, figures and models come from three places: the data printed in the examination questions reproduced here, the worked examples' own stated data, and a small table of standard entropies and enthalpies of formation given with the entropy section. They are reference values for learning the methods, not the official data booklet. Different sources quote slightly different values for the same quantity — the first electron affinity of chlorine appears as −349 in one question and −364 in another, and the hydration enthalpy of Ca²⁺ as −1579 and −1650 — so in an examination always use the value the question gives.

The Born–Haber and solution-cycle figures are drawn with heights roughly proportional to the enthalpy changes, but with every step given a minimum height so that small steps remain legible; their vertical scales are approximate. The entropy–temperature graph is schematic, anchored at the standard molar entropies of ice, liquid water and steam.

Summary

23.1 Lattice energy and Born–Haber cycles

• ΔH_at forms one mole of gaseous atoms and is always endothermic; for a diatomic element it is half the bond energy.
• Lattice energy forms one mole of ionic compound from gaseous ions and is always exothermic.
• First EA adds one mole of electrons to one mole of gaseous atoms, usually exothermic; the second EA is endothermic because the electron is repelled by a negative ion.
• EA₁ becomes less exothermic down Groups 16 and 17 (greater distance and shielding); F and O are anomalously low because their small outer shells are crowded.
• Born–Haber cycle: ΔH_f = ΣΔH_at + ΣIE + ΣEA + ΔH_latt, with the formula deciding every multiplier.
• Lattice energy is more exothermic for higher ionic charges and smaller ions (higher charge density); charge dominates.

23.2 Enthalpies of solution and hydration

• ΔH_hyd: one mole of gaseous ions → aqueous ions; always exothermic.
• ΔH_sol: one mole of solid → aqueous ions; either sign.
• ΔH_sol = −ΔH_latt + ΣΔH_hyd; divide by the number of ions when finding one ion's ΔH_hyd.
• ΔH_hyd is more exothermic for smaller, more highly charged ions, which attract water molecules more strongly.

23.3 Entropy change

• Entropy is the number of possible arrangements of particles and their energy; S(solid) < S(liquid) << S(gas).
• ΔS is positive for melting, boiling, dissolving, heating, and reactions that increase the moles of gas.
• ΔS^⊖ = ΣS^⊖(products) − ΣS^⊖(reactants), with coefficients; elements have non-zero S^⊖.

23.4 Gibbs free energy change

• ΔG^⊖ = ΔH^⊖ − TΔS^⊖, with T in K and ΔS in kJ K⁻¹ mol⁻¹; feasible when ΔG is negative.
• ΔG against T is a straight line: intercept ΔH, gradient −ΔS.
• When ΔH and ΔS have the same sign, feasibility changes at T = ΔH/ΔS — a minimum temperature if both are positive, a maximum if both are negative.
• Feasibility says nothing about rate: a feasible reaction can be too slow to observe.

Examination checklist

• Can I write, with state symbols, the equation for every defined enthalpy change in this unit?
• Do my definitions contain "one mole", the right species (atoms, ions, compound) and "gaseous" where needed?
• In a Born–Haber cycle, have I multiplied each term by the moles the formula requires, and halved bond energies for one mole of atoms?
• Have I shown atoms (not molecules) after atomisation and the electrons after each ionisation?
• When comparing lattice or hydration enthalpies, have I named the ions, compared charge and radius, and stated the stronger attraction between ions or to water molecules?
• When finding ΔH_hyd of one ion from a 1:2 salt, have I divided by two?
• For the sign of ΔS, have I referred to the change in moles of gas (or the change of state)?
• In ΔG calculations, is T in kelvin and ΔS in kJ K⁻¹ mol⁻¹? Have I written ΔH and ΔS separately first?
• When explaining the effect of temperature, have I said how TΔS changes and what that does to the sign of ΔG?

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