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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.

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Presentation on theme: "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."— Presentation transcript:

1 15.2 Born-Haber Cycle Define and apply the terms lattice enthalpy, and electron affinity 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 Construct a Born-Haber cycle for group 1 and 2 oxides and chlorides and use it to calculate the enthalpy change 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

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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

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11 Calculations  Calculate the lattice energy of NaCl(s) using the following: (kJmol -1 ) Enthalpy of formation of NaCl = Enthalpy of atomisation of Na = +109 Enthalpy of atomisation of Cl = +121 Electron affinity of Cl = Ionisation energy of Na =  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 NaCl 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 NaCl KBr KI 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


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