Chemical energetics · ⁨الديناميكا الحرارية الكيميائية⁩

These cycles use several enthalpy changes 焓变 ($\Delta H$). Two new ones are:

• the enthalpy change of atomisation 原子化焓变, $\Delta H_{\text{at}}$ — the energy to make one mole of gaseous atoms from an element. It is always positive (bonds must break).
• the lattice energy 晶格能, $\Delta H_{\text{latt}}$ — the energy change when one mole of a solid ionic lattice forms from its gaseous ions. It is always negative (strong bonds form).

Electron affinity

The first electron affinity 电子亲和能 (EA) is the energy change when one mole of gaseous atoms each gain one electron to form one mole of $1-$ ions. The first EA is usually negative.

The same factors as ionisation energy apply (nuclear charge, atomic radius, shielding). Going down Group 16 or 17, the EA becomes less exothermic, because the atom is larger and pulls the extra electron in less strongly. (The very top element is an exception: its atom is so small that electron repulsion makes its EA less exothermic than the one below it.)

Born–Haber cycles

A Born–Haber cycle 玻恩哈伯循环 is an energy cycle that links the enthalpy change of formation of an ionic solid with its atomisation, ionisation energy, electron affinity and lattice energy. Using Hess's law, you go round the cycle to find any one unknown step. (You only need $+1$ and $+2$ cations and $-1$ and $-2$ anions.)

Worked example. Find the lattice energy of sodium chloride from these data (kJ mol⁻¹): enthalpy of formation $\Delta H_f = -411$; atomisation $\Delta H_{\text{at}}(\text{Na}) = +107$ and $\Delta H_{\text{at}}(\text{Cl}) = +122$; first ionisation energy of Na $= +496$; electron affinity of Cl $= -349$.

By Hess's law the direct formation route equals the route up and round the cycle:

so $\Delta H_{\text{latt}} = -411 - (107 + 122 + 496 - 349) = -787\ \text{kJ mol}^{-1}.$

What controls the size of a lattice energy

The lattice energy is more negative (stronger) when:

• the ionic charge is higher (e.g. $\text{Mg}^{2+}$ beats $\text{Na}^+$).
• the ionic radius is smaller.

Both make the attraction between the ions stronger.

• the enthalpy change of hydration 水合焓变, $\Delta H_{\text{hyd}}$ — the energy change when one mole of gaseous ions is surrounded by water to form aqueous ions. It is exothermic.
• the enthalpy change of solution 溶解焓变, $\Delta H_{\text{sol}}$ — the energy change when one mole of solute dissolves fully in water.

An energy cycle links the three:

(To dissolve, you first pull the lattice apart, then hydrate the ions.) Like lattice energy, $\Delta H_{\text{hyd}}$ is more exothermic for ions with a higher charge and a smaller radius.

Entropy 熵 ($S$) measures the number of ways the particles and their energy can be arranged in a system. More ways means more "disorder".

The entropy change 熵变 ($\Delta S$) is positive when disorder increases, and negative when it falls:

You can calculate it from standard entropies:

The Gibbs free energy 吉布斯自由能 change decides whether a reaction can happen on its own:

Here $T$ is the temperature in kelvin. A reaction is feasible (it can happen) when $\Delta G$ is negative or zero. The feasibility 可行性 therefore depends on temperature:

To find the changeover temperature, set $\Delta G = 0$, which gives $T = \Delta H / \Delta S$.

Worked example. A reaction has $\Delta H = +120\ \text{kJ mol}^{-1}$ and $\Delta S = +200\ \text{J K}^{-1}\,\text{mol}^{-1}$. Find $\Delta G$ at $298\ \text{K}$, and the temperature above which the reaction becomes feasible.

First match the units: $\Delta S = +0.200\ \text{kJ K}^{-1}\,\text{mol}^{-1}$. Then

Setting $\Delta G = 0$ gives $T = \Delta H/\Delta S = 120/0.200 = 600\ \text{K}$, so the reaction is feasible above $600\ \text{K}$.

• In a Born-Haber cycle get each step's direction and sign right (atomisation, ionisation $+$; electron affinity, lattice formation $-$) and apply Hess's law around it.
• Lattice energy is more exothermic for smaller, more highly charged ions (higher charge density).
• Predict the sign of $\Delta S$ from the state changes (more gas moles = more disorder).
• Use $\Delta G = \Delta H - T\Delta S$; feasible when $\Delta G \le 0$ — convert $\Delta S$ from $\text{J K}^{-1}\,\text{mol}^{-1}$ ($\div 1000$).