Chemical Energetics

Three new enthalpy changes, one new destination

Everything in AS energetics described molecules reacting. This topic describes ionic solids being built from, or pulled apart into, their ions — and that needs a precise vocabulary for each stage of the journey:

Each arrow gets its own named enthalpy change with an examiner-precise definition. Learn the definitions word for word — they are asked directly, almost on every paper, and each carries its own mark. Two things every one of these definitions must pin down: how many moles (almost always exactly one mole of something), and the physical states (gas, solid, aqueous). Behind the word standard sit the standard conditions themselves — $298\text{ K}$ and $101\text{ kPa}$ — which is why every one of these quantities carries the $\ominus$ symbol.

Enthalpy change of atomisation, ΔH_at

Definition (mark-scheme wording): the energy required when one mole of gaseous atoms is formed from the element (in its standard state).

You are pulling apart a lump of metal, or splitting a diatomic molecule, until only separate single atoms remain — bonds and attractions must be overcome, so $\Delta H_{\text{at}}$ is always endothermic, always positive.

The $\tfrac12$ matters: the definition fixes one mole of gaseous atoms, and one mole of chlorine atoms comes from only half a mole of $\text{Cl}_2$ molecules. Writing $\text{Cl}_2\text{(g)} \rightarrow 2\text{Cl(g)}$ describes two moles of atoms — a different quantity entirely.

Lattice energy, ΔH_latt

Definition (mark-scheme wording): the energy change when one mole of an ionic solid is formed from its gaseous ions (under standard conditions).

Oppositely charged ions attract, so packing gaseous ions into a lattice releases energy: by this definition $\Delta H_{\text{latt}}$ is always exothermic, always negative. Notice the direction built into the syllabus wording — gas phase ions → solid lattice — formation, not separation.

First electron affinity, EA₁

The metal side of the journey is now sorted (atomise, then remove electrons — ionisation energies from AS). The non-metal side needs electrons added, and that has its own name.

Definition (mark-scheme wording): the enthalpy change when one mole of gaseous atoms gains one electron each to form one mole of gaseous 1− ions.

A neutral atom pulls an incoming electron towards itself, so $\text{EA}_1$ is usually exothermic (negative) for Groups 16 and 17. But not always — which is exactly what the examiner wants you to explain.

What controls the size of an electron affinity

Two competing effects decide how exothermic adding an electron is:

• Nuclear attraction for the incoming electron — stronger with a higher nuclear charge, a smaller atomic radius, and less shielding by inner shells. More attraction → more energy released → more exothermic.
• Repulsion between the incoming electron and the electrons already there — a crowded atom resists the newcomer. More repulsion → less exothermic.

Down Group 17 ($\text{Cl} \rightarrow \text{Br} \rightarrow \text{I}$): each element adds a shell, so radius and shielding increase while the pull on the incoming electron weakens — $\text{EA}_1$ becomes less exothermic down the group.

Down Group 16 ($\text{O} \rightarrow \text{S} \rightarrow \text{Se}$): the same argument applies — radius and shielding increase, so $\text{EA}_1$ becomes less exothermic down this group too.

Across a period (Group 16 vs Group 17, e.g. S vs Cl): the Group 17 atom has a higher nuclear charge with the same number of inner shells, so it attracts the incoming electron more strongly — Group 17 $\text{EA}_1$ values are more exothermic than their Group 16 neighbours.

Second electron affinities are always endothermic

Adding a second electron means forcing it onto an ion that is already negative: $\text{O}^-\text{(g)} + \text{e}^- \rightarrow \text{O}^{2-}\text{(g)} \qquad \text{EA}_2 = +844\text{ kJ mol}^{-1}$

Repulsion between the negative ion and the incoming negative electron must be overcome, so $\text{EA}_2$ is always positive — no exceptions. This sign matters in Born–Haber cycles for −2 anions (§02).

A clean demonstration: connecting the "half-molecule" equation to EA₁

Questions often quote an enthalpy change for the whole half-molecule equation rather than for $\text{EA}_1$ alone. Suppose you are told:

and asked for $\Delta H_{\text{at}}$ of X. The quoted equation bundles together two different things: breaking one mole of X–X bonds ($\Delta H_{\text{at}}$, twice over) and adding two moles of electrons ($2\times\text{EA}_1$):

Positive, as every atomisation must be — a built-in sense check.

Why this diagram exists

You cannot measure the lattice energy of NaCl directly in any calorimeter — you cannot start with a flask of gaseous $\text{Na}^+$ and $\text{Cl}^-$ ions. But $\Delta H_f^\ominus(\text{NaCl})$ can be measured, and so can every step along an alternative route that happens to pass through those gaseous ions. A Born–Haber cycle is simply Hess's law applied to an ionic solid: the direct route from elements to solid, and an indirect route through gaseous atoms and gaseous ions, must have equal enthalpy changes.

Every term in the cycle is one of the definitions from §01 — nothing new needs memorising, only assembling.

Reading the cycle, arrow by arrow

Start at the bottom-left level: the elements in their standard states, $\text{M(s)} + \tfrac12\text{X}_2\text{(g)}$. The indirect route climbs from there:

1. Atomise the metal ($\Delta H_{\text{at}}\text{(M)}$, positive): $\text{M(s)} \rightarrow \text{M(g)}$.
2. Ionise the metal ($\text{IE}_1$, positive): $\text{M(g)} \rightarrow \text{M}^+\text{(g)} + \text{e}^-$.
3. Atomise the non-metal ($\Delta H_{\text{at}}\text{(X)}$, positive): $\tfrac12\text{X}_2\text{(g)} \rightarrow \text{X(g)}$.
4. Add an electron ($\text{EA}_1$, negative): $\text{X(g)} + \text{e}^- \rightarrow \text{X}^-\text{(g)}$.
5. Form the lattice ($\Delta H_{\text{latt}}$, negative): $\text{M}^+\text{(g)} + \text{X}^-\text{(g)} \rightarrow \text{MX(s)}$.

The direct route is a single arrow: $\Delta H_f^\ominus\text{(MX)}$, negative. Hess's law says direct = indirect:

A clean demonstration (+1/−1 solid, invented numbers)

Find $\Delta H_f^\ominus$ of MX given: $\Delta H_{\text{at}}\text{(M)} = +108$, $\text{IE}_1\text{(M)} = +496$, $\Delta H_{\text{at}}\text{(X)} = +121$, $\text{EA}_1\text{(X)} = -349$, $\Delta H_{\text{latt}} = -780$ (all kJ mol⁻¹).

Step 1 — sum the indirect route, one term at a time:

Step 2 — the sum IS $\Delta H_f^\ominus$ (direct route = indirect route):

Negative, as a formation enthalpy of a stable ionic solid should be. In a real question it is usually $\Delta H_f$ that is known and one of the five steps that is unknown — then the same equation is rearranged: unknown $= \Delta H_f - (\text{sum of the other four})$.

The +2/−1 case: where the ×2 multipliers appear

For a solid like $\text{MgCl}_2$ or $\text{ZnI}_2$, one formula unit contains one $\text{M}^{2+}$ but two $\text{X}^-$ ions. Two consequences ripple through the cycle:

• The non-metal side is needed two moles at a time: $2\times\Delta H_{\text{at}}\text{(X)}$ and $2\times\text{EA}_1$. (If the element's data are quoted per mole of $\text{X}_2$ molecules — a sublimation step or a bond energy — remember one mole of $\text{X}_2$ gives two moles of X atoms.)
• The metal needs both ionisation energies: $\text{IE}_1 + \text{IE}_2$, both positive.

The −2-anion cases: EA₁ is joined by EA₂

When the anion carries a 2− charge, two electrons must be loaded onto each anion atom — and §01 showed those two arrivals have opposite signs ($\text{EA}_1$ exothermic, $\text{EA}_2$ always endothermic). Both terms therefore appear side by side in the cycle.

M₂X solids (+1 cation, e.g. $\text{Ag}_2\text{O}$ or $\text{Na}_2\text{S}$): two moles of metal atoms are atomised and ionised, and each anion atom absorbs both electrons:

MX solids with 2+/2− ions (e.g. $\text{MgO}$ or $\text{CaO}$): the metal now needs $\text{IE}_1 + \text{IE}_2$ and the anion needs $\text{EA}_1 + \text{EA}_2$:

The electron bookkeeping ties the halves together: the two electrons the metal releases (via $2\times\text{IE}_1$, or via $\text{IE}_1+\text{IE}_2$) are exactly the two the non-metal absorbs ($\text{EA}_1$, then $\text{EA}_2$). And watch for data tables that bundle the whole non-metal side into one entry such as "$\tfrac12\text{O}_2\text{(g)} + 2\text{e}^- \rightarrow \text{O}^{2-}\text{(g)}$" — that single number is $\Delta H_{\text{at}}\text{(O)} + \text{EA}_1 + \text{EA}_2$ combined; the exercise below makes you unpick it.

Two dials control every lattice energy

Lattice energy is electrostatics: oppositely charged ions attracting across the distances between them. Only two things about the ions can change that attraction:

• Ionic charge — bigger charges on both ions → stronger attraction → more energy released on lattice formation → $\Delta H_{\text{latt}}$ more negative (bigger magnitude).
• Ionic radius — smaller ions → the charges sit closer together → stronger attraction → $\Delta H_{\text{latt}}$ more negative.

So the most exothermic lattice energies belong to small, highly charged ion pairs. Magnesium oxide ($\text{Mg}^{2+}\text{O}^{2-}$, both tiny doubly-charged ions, $\Delta H_{\text{latt}} \approx -3800$) sits far beyond sodium chloride ($\text{Na}^+\text{Cl}^-$, both singly charged and larger, $-787$).

A clean demonstration: ordering three solids

Order $\text{NaCl}$, $\text{MgO}$ and $\text{KCl}$ by lattice energy magnitude (most exothermic first), using the reasoning, not memorised values.

Step 1 — compare charges. $\text{MgO}$ has $2+/2-$ ions; the other two are $+/-$ only. Doubling both charges strengthens the attraction enormously — $\text{MgO}$ is the most exothermic of the three.

Step 2 — compare radii within the same charges. $\text{Na}^+$ is smaller than $\text{K}^+$, so $\text{NaCl}$ packs its charges closer than $\text{KCl}$ does: $\text{NaCl}$ more exothermic than $\text{KCl}$.

Result: $\text{MgO}$ (most exothermic) $>$ $\text{NaCl}$ $>$ $\text{KCl}$ (least).

Every ordering question in the bank is this two-step argument — charges first, then radii — with the final link always stated explicitly: stronger attraction between the ions → more exothermic lattice energy.

What has to happen, energetically, when a salt dissolves

Drop an ionic solid into water and two things must happen: the lattice is pulled apart into free gaseous ions (costly — it is the reverse of lattice formation), and each gaseous ion becomes surrounded by water molecules (hydrated — rewarding, because ion–dipole attractions form). Each stage has its own named enthalpy change.

Enthalpy change of hydration, $\Delta H_{\text{hyd}}$ — mark-scheme wording: the enthalpy change when one mole of gaseous ions dissolves in water to form one mole of aqueous ions.

Water molecules arrange themselves with their slightly negative oxygen end towards cations (and their slightly positive hydrogen ends towards anions), so attraction is always formed: $\Delta H_{\text{hyd}}$ is always exothermic, always negative.

Enthalpy change of solution, $\Delta H_{\text{sol}}$ — the enthalpy change when one mole of an ionic solid dissolves in water to form aqueous ions in an infinitely dilute solution. The "one mole" is pinned to the solid only — as the NaCl equation shows, the aqueous ions it produces may number more than one mole.

Unlike the other two, $\Delta H_{\text{sol}}$ can be either sign — it is the balance between the costly break-up and the rewarding hydration.

Deriving the working equation from the cycle

Reading the indirect route the way its arrows point:

Two details carry all the marks in this calculation:

• The sign flip on the lattice term. $\Delta H_{\text{latt}}$ is negative (formation); dissolving needs the reverse step, so the term entering the sum is $-\Delta H_{\text{latt}}$ — a positive quantity.
• The stoichiometry. A formula unit like $\text{MgF}_2$ hydrates one $\text{Mg}^{2+}$ and two $\text{F}^-$, so the hydration sum is $\Delta H_{\text{hyd}}(\text{Mg}^{2+}) + 2\times\Delta H_{\text{hyd}}(\text{F}^-)$. Forgetting the ×2 is the single most penalised slip in this topic.

A clean demonstration (+1/−1 solid, invented numbers)

Find $\Delta H_{\text{sol}}$ of KBr given: $\Delta H_{\text{latt}} = -680$, $\Delta H_{\text{hyd}}(\text{K}^+) = -310$, $\Delta H_{\text{hyd}}(\text{Br}^-) = -330$ (all kJ mol⁻¹).

Step 1 — hydration sum: $-310 + (-330) = -640$.

Step 2 — subtract the lattice energy:

Positive — dissolving KBr would be endothermic. Yet KBr dissolves freely. Hold that thought: §06's entropy is what resolves this apparent paradox, and questions love asking you to connect the two.

The same two dials, pointed at water

§03's argument was about ion attracting ion. For hydration it is ion attracting water: the ion's electric field pulls in the water dipoles (oxygen end first for cations), and the stronger that pull, the more energy released on hydrating.

• Higher charge → $\Delta H_{\text{hyd}}$ more exothermic.
• Smaller radius → the same charge packed tighter (higher charge density) → $\Delta H_{\text{hyd}}$ more exothermic.

A clean demonstration: ordering ions by hydration enthalpy

Arrange $\text{Na}^+$, $\text{Mg}^{2+}$, $\text{Al}^{3+}$ by $\Delta H_{\text{hyd}}$ (most exothermic first). They are isoelectronic — all have 10 electrons — so radius falls as charge rises across the three.

Step 1: charge rises $\text{Na}^+ < \text{Mg}^{2+} < \text{Al}^{3+}$; Step 2: radius shrinks in the same order. Both effects point the same way, so:

with the explicit link: higher charge density → stronger attraction to water's dipoles → more energy released on hydration. Real values show how dramatic this is: $-322$ for $\text{K}^+$ versus $-1650$ for $\text{Ca}^{2+}$ (M/J 2025 Q4).

The solubility argument: which enthalpy changes faster?

$\Delta H_{\text{sol}} = \Sigma\Delta H_{\text{hyd}} - \Delta H_{\text{latt}}$ ties everything together. Compare salts as the cation grows down a group:

• Both terms get less exothermic (bigger cation → weaker attractions everywhere).
• But they shrink at different rates. For the fluorides and hydroxides, $\Delta H_{\text{latt}}$ changes faster than $\Delta H_{\text{hyd}}$: F⁻ and OH⁻ are small anions, so growing the cation increases the inter-ionic distance by a large fraction — moving the lattice energy a long way — while the hydration sum (where only the cation's term changes) moves less.
• Subtracting the faster-shrinking term makes $\Delta H_{\text{sol}}$ more exothermic down the group — so solubility increases.

For the sulfates the argument reverses: the large $\text{SO}_4^{2-}$ anion dominates the lattice distance, so the cation's growth barely moves $\Delta H_{\text{latt}}$ — while $\Delta H_{\text{hyd}}$ of the cation still falls steeply. Now hydration shrinks faster, $\Delta H_{\text{sol}}$ becomes less exothermic, and solubility decreases down the group.

The "which changes faster" sentence is the heart of the 4-mark explanation — learn it as a sentence, not a formula.

The second driver: disorder

§04 ended on a paradox: dissolving $\text{MgF}_2$ is endothermic ($\Delta H_{\text{sol}} = +21$), and KBr's was $+40$ in our invented example — yet both dissolve. Enthalpy cannot be the only thing pushing processes along. The second driver is entropy, $S$:

Definition (syllabus wording, asked almost verbatim every year): the number of possible arrangements of the particles and their energy in a given system.

A gas spread through a flask can arrange its particles — and share out their energy — in astronomically more ways than the same particles locked in a crystal. Systems with more possible arrangements have higher entropy.

Predicting the sign of ΔS

Four patterns cover nearly every exam question:

• Change of state: melting, boiling, subliming → $\Delta S$ positive (solid < liquid < gas). Freezing, condensing → negative.
• Dissolving: an ionic solid's ordered lattice becomes mobile hydrated ions → $\Delta S$ usually positive. (Its reverse, crystallising, is negative.)
• Temperature change: heating anything lets its particles access more energy arrangements → $\Delta S$ positive.
• Reactions: count gas molecules on each side. Gases dominate entropy because they occupy enormously more volume. More gas molecules produced → $\Delta S$ positive; fewer → negative. Changes in moles of solids or liquids barely matter by comparison.

That last bullet is the one to internalise: $2\text{H}_2\text{(g)} + \text{O}_2\text{(g)} \rightarrow 2\text{H}_2\text{O(l)}$ has $\Delta S$ strongly negative — three moles of gas become none — even though a liquid replaces a gas on the right.

Calculating ΔS from standard entropies

Every substance has a tabulated standard entropy, $S^\ominus$, in $\text{J K}^{-1}\text{mol}^{-1}$. Unlike formation enthalpies, standard entropies are never zero — even elements have disorder (the third-law baseline of a perfect crystal at 0 K is the only zero, and nothing sits there at 298 K).

with each substance's value multiplied by its balancing coefficient — the same bookkeeping as the Hess's law formula from AS, but watch the units: entropies come in joules, enthalpies in kilojoules. That mismatch is about to become the classic trap of §07.

A clean demonstration (invented numbers)

Find $\Delta S^\ominus$ for $\text{N}_2\text{O}_4\text{(g)} \rightarrow 2\text{NO}_2\text{(g)}$, given $S^\ominus(\text{N}_2\text{O}_4) = +304$ and $S^\ominus(\text{NO}_2) = +240\text{ J K}^{-1}\text{mol}^{-1}$.

Step 1 — products, with the coefficient of 2: $2 \times 240 = +480$

Step 2 — reactants: $+304$ (coefficient 1).

Step 3 — subtract:

Sense-check: one mole of gas becomes two — more gas molecules — so positive, as calculated. The sign check costs five seconds and catches most slips.

One number that balances both drivers

A process can be pushed along by exothermicity ($\Delta H$ negative) or by rising disorder ($\Delta S$ positive) — and each driver fights the other when they disagree. The Gibbs free energy change is the single quantity that weighs them against each other:

with $T$ the absolute temperature in kelvin. A process is feasible (allowed by thermodynamics — note: feasible says nothing about how fast it happens; kinetics is a separate topic) when:

At $\Delta G = 0$ the process sits exactly on the boundary — this is what defines a melting point, for instance: solid and liquid coexist because neither direction wins.

Reading the temperature switchover off a graph

Because $\Delta G^\ominus = \Delta H^\ominus - T\Delta S^\ominus$ is linear in $T$, plotting $\Delta G^\ominus$ against $T$ gives a straight line — and every feature of that line has a name:

• the y-intercept ($T=0$ kills the $T\Delta S$ term) is $\Delta H^\ominus$;
• the gradient is $-\Delta S^\ominus$ (falling line ⇒ $\Delta S^\ominus$ positive);
• the x-intercept, where $\Delta G^\ominus = 0$, is the temperature where feasibility flips: $T_{\text{min}} = \dfrac{\Delta H^\ominus}{\Delta S^\ominus}$.

A clean demonstration (+/+, invented numbers)

A reaction has $\Delta H^\ominus = +50\text{ kJ mol}^{-1}$ and $\Delta S^\ominus = +100\text{ J K}^{-1}\text{mol}^{-1}$. At what temperature does it become feasible?