Use an enthalpy diagram to calculate the lattice energy of CaCl2 from the following information. Energy needed to vaporize one mole of Ca(s) is 192 kJ. For calcium, the first IE = 589.5 kJ mol-1, the second IE = 1146 kJ mol-1. The electron affinity of Cl is -348 kJ mol-1. The bond energy of Cl2 is 242.6 kJ per mole of Cl—Cl bonds. The standard heat of formation of CaCl2 is -795 kJ mol

Answer 1

Explain your trouble. You set up
Ca(s) --> Ca(g) dH = +192 kJ
Ca(g) --> Ca^+ + e dH = + 589.5 kJ
Ca^+(g) --> Ca^2+ + e dH = + 1146 kJ
Cl2(g)--> 2Cl(g) dH = 242.6 kJ
2*Cl(g) + 2e -->2*Cl^-(g) dH =2*-348 kJ
Ca^2+(g) + 2Cl^-(g) --> CaCl2 dH = lattice
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Ca(s) + Cl2(g) ==> CaCl2(s) dHformation = -795 kJ

Answer 2

-2302.8

Answer 3

Constructing the born-haber cycle and getting the relevant data like formation enthalpy of cacl2 , atomisation of ca, first and second ionisation energies of ca , atomisation of cl and then first electron affinity of cl. First equal the formation enthalpy to the sum of all the other enthalpy's and use letter X to represent the lattice enthalpy .