Title: Lesson 7 Lattice Enthalpies and Enthalpy Change of Solution Learning Objectives: Understand the term lattice enthalpy Identify and explain trends in lattice enthalpy Explain how lattice enthalpy and enthalpies of hydration are related to enthalpy change of solution

Theoretical lattice enthalpies can be calculated from the ionic model Assumption is the crystal is made up of perfectly spherical ions. Assumption is the only interaction is due to electrostatic forces between ions. The energy needed to separate the ions depends on the product of the ionic charges and the sum of the ionic radii. An increase in ionic radius decreases attraction. An increase in ionic charge increases the attraction. K is a constant that depends on the geometry of the lattice n and m are the magnitude of the charges on the ions. Ionic radii can be determined from X-ray diffraction measurements

Solutions

Lattice enthalpies depend on the size and charge of the ions What pattern can you see in the table? What is changing as you move down each group? Lattice enthalpy decreases as the size of the cation or anion increases Explain the difference in lattice enthalpies for these ionic structures...

Comparing Lattice Enthalpies Study Table 18 in the data booklet. What is the relationship between lattice enthalpy and the charge on an ion? Explain why this occurs Use suitable pairs of lattice enthalpies to illustrate your answer What is the relationship between lattice enthalpy and the size of an ion:

Key Points Lattice enthalpy is the equivalent of bond enthalpy in ionic compounds Born-Haber cycles are a specialist type of Hess cycle Each individual step in the reaction is included Lattice enthalpy: Increases with ionic charge Decreases with ionic size

Solutions

Enthalpies of solution Enthalpy of solution can be calculated by measuring the temperature change of the solution (q = mct) Enthalpy of solution can also be obtained by measuring enthalpy changes for solutions with increasing volumes of water until a limit is reached. Ionic compounds dissolved readily in water as the ions are strongly attracted to the polar solvent water. (See diagrams below) Ions separated and surrounded by water are said to be hydrated. The strength of interaction between the polar water molecules and the separated ions is given by their hydration enthalpies.

Enthalpy of hydration ‘The enthalpy of hydration of an ion is the enthalpy change that occurs when one mole of gaseous ions is dissolved to form an infinitely dilute solution of one mole of aqueous ions (under standard temperature and pressure).’ As there is a force of attraction between the ions and the polar water molecules, it is exothermic (negative enthalpy value) QUESTION: What do we notice about the enthalpies as we move down the groups? (Think attraction and ionic radius) Think: Solid is sublimed into gaseous ions, then plunged into water!

Hydration enthalpies of the ions are inversely proportional to the ionic radii A = Constant Down group 1 and 7 exothermic hydration enthalpy decreases Across period 3 exothermic hydration enthalpy increases due to increased ionic charge and decrease in ionic radii.  Increased attraction to polar water molecules. This leads to: B = Constant n = charge of ion

Enthalpy change of solution is related to the lattice enthalpy and hydration enthalpies

Solutions

Li+(g) + F(g) By Hess’s Law LEdiss = - (-616) Li+(g) + F-(g) Q1. Use the following data to construct a Born-Haber cycle for the formation of LiF and then calculate a value for its lattice association enthalpy. Enthalpy of atomisation of Li(s) = +160.0 kJ mol-1 Enthalpy of atomisation of F2(g) = +79.0 kJ mol-1 First ionisation energy of Li(g) = +520.0 kJ mol-1 First electron affinity of F(g) = -334.0 kJ mol-1 Enthalpy of formation of LiF(s) = -616.0 kJ mol-1 Li+(g) + F(g) I1[Li] EA1[F] By Hess’s Law = - 334 = + 520 LEdiss = - (-616) Li+(g) + F-(g) Li(g) + F(g) + 160 + 79 + 520 ΔHØat [Li(s)] + ΔHØat [½F2(l)] + (- 334) = + 160 + 79 = + 1041 kJ mole-1 LEdiss Li(s) + ½F2(l) ΔHØf [LiF] = - 616 LiF(s)

The Ca2+ ion is smaller and more highly charged than Li+ Q2 The values of the lattice association energies of potassium iodide and calcium iodide experimentally determined from Born-Haber cycles and theoretically calculated from an ionic model are shown below. Experimental Lattice Dissociation Energy / kJ mole-1 Theoretical Lattice Dissociation Energy / kJ mole-1 KI(s) -651 -636 CaI2(s) -2074 -1905 Explain why the experimental lattice energy of potassium iodide is less exothermic than the experimental lattice energy of calcium iodide. The Ca2+ ion is smaller and more highly charged than Li+ (i.e. Ca2+ more charge dense than Li+) forces of attraction between Ca2+ and I- ions > K+ and I- ions  more energy needed to overcome forces to separate Ca2+ & I-. Explain why the experimental and theoretical values of the lattice energy are almost the same for potassium iodide, but are significantly different for calcium iodide. Bonding in KI is pure ionic but CaI2 shows significant covalency because the I- ion is significantly polarised by the more charge dense Ca2+ but not by the low charge density K+ ion.

Click Here For DATA M2+(g) + 2Cl(g) By Hess’s Law M2+(g) + 2Cl-(g) Q3. Use a Born-Haber cycle for the formation of MCl2 to calculate the first electron affinity of chlorine, given that metal M has ΔHat = +150, I1 = +736 and I2 = +1450 kJ mole-1 and ΔHf of MCl2 is -642 kJ mole-1 and LEdiss of MCl2 is +2526 kJ mole-1. Click Here For DATA M2+(g) + 2Cl(g) 2EA1[Cl] I1[M] + I2[M] = + 736 + 1450 By Hess’s Law M2+(g) + 2Cl-(g) +2526 = - (-642) M(g) + 2Cl(g) + 150 + 2(121) + 736 + 1450 ΔHØat [M(s)] + 2ΔHØat [½Cl2(g)] + 2EA1[Cl] = + 150 + 2(+121) LEdiss M(s) + Cl2(g) EA1[Cl] = - 347 kJ mole-1 = + 2526 ΔHØf [MCl2] = - 642 MCl2(s)

DATA : ALL VALUES IN kJ mole-1 ELEMENT Hoat I1 I2 I3 EA1 EA2 Li(s) +159 Li(g) +520 Na(s) +107 Na(g) +496 K(s) +89 K(g) +419 Cs(s) +76 Cs(g) +376 Mg(s) +148 Mg(g) +738 +1451 Ca(s) +178 Ca(g) +590 +1145 Al(s) +326 Al(g) +578 +1817 +2745 1/2F2(g) +79 F(g) -328 1/2Cl2(g) +121 Cl(g) -349 1/2Br2(g) +112 Br(g) -325 1/2I2(g) I(g) -295 1/2O2(g) +249 O(g) -141 +798 1/2S8(g) +279 S(g) -200 +640