15.2 Born-Haber Cycle 15.2.1 Define and apply the terms lattice enthalpy, and electron affinity 15.2.2 Explain how the relative sizes and the charges of ions affect the lattice enthalpies of different ionic compounds The relative value of the theoretical lattice enthalpy increases with higher ionic charge and smaller ionic radius due to increased attractive forces 15.2.3 Construct a Born-Haber cycle for group 1 and 2 oxides and chlorides and use it to calculate the enthalpy change 15.2.4 Discuss the difference between theoretical and experimental lattice enthalpy values of ionic compounds in terms of their covalent character.

Born-Haber Cycle A series of hypothetical steps and their enthalpy changes needed to convert elements to an ionic compound and devised to calculate the lattice energy. Using Hess’s law as a means to calculate the formation of ionic compounds

Born-Haber Cycle Steps Elements (standard state) converted into gaseous atoms Losing or gaining electrons to form cations and anions Combining gaseous anions and cations to form a solid ionic compound

Step 1: Atomisation The standard enthalpy change of atomisation is the ΔH required to produce one mole of gaseous atoms Na(s)  Na(g) ΔHoat = +109 kJmol-1

NOTE: for diatomic gaseous elements, Cl2, ΔHoat is equal to half the bond energy (enthalpy) Cl2(g)  Cl(g) ΔHoat = ½ E (Cl-Cl) ΔHoat = ½ (+242 ) ΔHoat = +121 kJmol-1

Step 2: Formation of gaseous ions Electron Affinity Enthalpy change when one mole of gaseous atoms or anions gains electrons to form a mole of negatively charged gaseous ions. Cl(g) + e-  Cl-(g) ΔHo = -364 kJmol-1 For most atoms = exothermic, but gaining a 2nd electron is endothermic due to the repulsion between the anion and the electron

Becoming cations Ionisation energy Enthalpy change for one mole of a gaseous element or cation to lose electrons to form a mole of positively charged gaseous ions Na(g)  Na+(g) + e- IE1= +494 kJmol-1

Lattice Enthalpy Energy required to convert one mole of the solid compound into gaseous ions. NaCl (s)  Na+(g) + Cl-(g) ΔHolat = +771kJmol-1 It is highly endothermic We cannot directly calculate ΔHolat , but values are obtained indirectly through Hess’s law for the formation of the ionic compound

Calculations Calculate the lattice energy of NaCl(s) using the following: (kJmol-1) Enthalpy of formation of NaCl = - 411 Enthalpy of atomisation of Na = +109 Enthalpy of atomisation of Cl = +121 Electron affinity of Cl = - 364 Ionisation energy of Na = + 494 Enthalpy of atomisation + electron affinity + ionisation = enthalpy of formation + lattice energy

Magnitude of Lattice enthalpy The greater the charge on the ions, the greater the electrostatic attraction and hence the greater the lattice enthalpy Ex: Mg2+ > Na+ The larger the ions, then the greater the separation of the charges and the lower the lattice enthalpy VICE VERSA

Trends ΔHolat Change from NaCl MgO 3889 Increased ionic charge NaCl 771 ------ KBr 670 Larger ions

Use of Born-Haber Cycles Empirical value of ΔHolat is found using Born-Haber cycle. Theoretical value of ΔHolat can be found by summing the electrostatic attractive and repulsive forces between the ions in the crystal lattice.

Compound Empirical value Theoretical value NaCl 771 766 KBr 670 667 KI 632 631 AgCl 905 770

Agreement Usually there is good agreement between empirical and theoretical values If there isn’t good agreement Implying that the description of the compound as ionic is inappropriate There could be a significant degree of covalent character in the bonding (EN difference less than 1.7) Presence of covalent character leads to an increase in ΔHolat