1. 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.

2. 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

4. Born-Haber Cycle Steps 1. Elements (standard state) converted into gaseous atoms 2. Losing or gaining electrons to form cations and anions 3. Combining gaseous anions and cations to form a solid ionic compound

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

6.  NOTE: for diatomic gaseous elements, Cl 2, ΔH o at is equal to half the bond energy (enthalpy)  Cl 2 (g)  Cl(g) ΔH o at = ½ E (Cl-Cl) ΔH o at = ½ (+242 ) ΔH o at = +121 kJmol -1

7. 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) ΔH o = -364 kJmol -1 For most atoms = exothermic, but gaining a 2 nd electron is endothermic due to the repulsion between the anion and the electron

8. 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- IE 1 = +494 kJmol -1

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

11. 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

12. Magnitude of Lattice enthalpy  The greater the charge on the ions, the greater the electrostatic attraction and hence the greater the lattice enthalpy  Ex: Mg 2+ > Na +  The larger the ions, then the greater the separation of the charges and the lower the lattice enthalpy  VICE VERSA

13. Trends ΔH o lat Change from NaCl MgO3889Increased ionic charge NaCl771------ KBr670Larger ions

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

15. CompoundEmpirical valueTheoretical value NaCl771766 KBr670667 KI632631 AgCl905770

16. 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 ΔH o lat