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23.1
Lattice energy and Born–Haber cycles · พลังงานแลตทิซและวัฏจักร Born–Haber
Syllabus · หลักสูตร
English
define and use the terms: (a) enthalpy change of atomisation, $\Delta H_{\text{at}}$ (b) lattice energy, $\Delta H_{\text{latt}}$ (the change from gas phase ions to solid lattice)
(a) define and use the term first electron affinity, EA (b) explain the factors affecting the electron affinities of elements (c) describe and explain the trends in the electron affinities of the Group 16 and Group 17 elements
construct and use Born–Haber cycles for ionic solids (limited to +1 and +2 cations, –1 and –2 anions)
carry out calculations involving Born–Haber cycles
explain, in qualitative terms, the effect of ionic charge and of ionic radius on the numerical magnitude of a lattice energy
Source: Cambridge International syllabus · แหล่งที่มา: หลักสูตร Cambridge International
English
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:
The Born–Haber cycle · วัฏจักรโบน-เฮ伯 (The Born–Haber cycle)
Lattice energy can't be measured directly, so it is found from a cycle of measurable steps. · พลังงานแลตทิซไม่สามารถวัดได้โดยตรง จึงคำนวณได้จากวัฏจักรของขั้นตอนที่วัดได้
Source: Cambridge International syllabus · แหล่งที่มา: หลักสูตร Cambridge International
English
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.
(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.
define the term entropy, $S$, as the number of possible arrangements of the particles and their energy in a given system
predict and explain the sign of the entropy changes that occur: (a) during a change in state, e.g. melting, boiling and dissolving (and their reverse) (b) during a temperature change (c) during a reaction in which there is a change in the number of gaseous molecules
calculate the entropy change for a reaction, $\Delta S$, given the standard entropies, $S^\ominus$, of the reactants and products, $\Delta S^\ominus = \Sigma S^\ominus \text{(products)} - \Sigma S^\ominus \text{(reactants)}$ (use of $\Delta S^\ominus = \Delta S^\ominus_{\text{surr}} + \Delta S^\ominus_{\text{sys}}$ is not required)
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:
$\Delta H$
$\Delta S$
When feasible
negative
positive
at all temperatures
positive
positive
only at high temperature
negative
negative
only at low temperature
positive
negative
never
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
$$\Delta G = \Delta H - T\Delta S = 120 - 298 \times 0.200 = +60\ \text{kJ mol}^{-1}\quad(\text{positive, so not yet feasible}).$$
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 $\text{J K}^{-1}\,\text{mol}^{-1}$ ($\div 1000$).
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