Question

Consider the following information. • The lattice energy of KCl is AHlattice = -701 kJ/mol. • The enthalpy of sublimation of K is AHsub = 89.0 kJ/mol. • The first ionization energy of K is IE1 = 419 kJ/mol. • The electron affinity of Cl is AHEA = -349 kJ/mol. • The bond energy of Cl, is BE = 243 kJ/mol. Determine the enthalpy of formation, AHf, for KCl(s). AHư= kJ/mol

Answer 1

For, KCl , the Born - Haber cycle is given below

H Kast ² l 2 (2) Alf kee(s) 489KB ausub / Argeſ K(8) ce (8) thig LIE, LEA (-39953) **(9) + Ci (8) – TU=-701 kJ 4H f = 4Hsub +IE+ / AHBE+EA TU

Then,

$\Delta H_f$ =
ΔΗρμή
+ IE + $\frac{1}{2}$
HBE
+ $\Delta H$ EA + U

Or,  $\Delta H_f$

= [89.0 + 419 + $\frac{1}{2}$ × (243) +( -349) - 701] KJ/mol

= - 420.5 KJ/mol