Question

Part I. Use a Born-Haber cycle to calculate the lattice energy of KCl from the following data. (5 marks) Ionization energy of K(g) = 444.0 kJ mol-1 Electron Affinity of Cl(g) = -381.0 kJ mol-1 Energy to Sublime K(s) = 152.0 kJ mol-1 Bond energy of Cl2 = 201.0 kJ mol-1 ∆rH for K(s) + 1/2 Cl2(g) ↔ KCl(s) = -480.0 kJ mol-1 art

II. Using the lattice energy calculated in part I determine the enthalpy of solution potassium chloride is added to water. To solve this problem use ΔhydH(K+) = -326.00 kJ mol-1 and ΔhydH(Cl-) = -348.00 kJ mol-1.

Answer 1

Born-Haver cyele of kel is given by - K(s) + ₂ (8) > kacs)- Anth + Atsublimation + 1 Bond dissociation energy lattice energy) cca) 1 (9) - EA K cg) Ktego +IE Art = 4H sub +IE + 1 bond dissociation energy Arn= - 480KT /mol AH sub = + 1520 kJ/mol / Bond dissociation energy = 1 x 201.0 KJ/mor of uz 100.5 kJ/mol I E = 440.0 kJ /mol, EA= -38110 kJ/mol. [EA is given as a (-) ve value, so we do not use the (-) ve sign of the equation as we know the EAIS always (-ove I . From ea- . -480 = 152.0 + 440.0 + 100.5 - 381.0t u u= - 480-152 - 440 - 100.5 +381.0 10 =-791.5 kJ/mol. Therefore lattice energy of ka is - 791.5kJ/molie, - 791.5 KJ energy is übarated when 1 mol of ka crystal is formed. ka 420 K + (aq) + ci (ar) At solution - AHhydration a lattice energy = A = Taksoin = ke (nyd.) & Aha-Chyd.) - Lattice energy – 326 - 348 - (- 79115) kJ/mol. & 117.5 kJ/moll Therefore ive, this dissolve enthalpy of the solution will be + 1 17.5 kJ/mol amount of heat enthalpy is required to 1 mol kel into Hoo.