Question

Given that the hydration energies of Ca²⁺_(g) and Cl^-_(g) are -1562KJ/Mole and -364KJ/Mole respectively. The heat of solution (Hsoln) for one Mole of CaCl₂ is -82.9 KJ/Mole. Determine the lattice energy for CaCl₂.

Answer 1

∆Hydr of CaCl₂ = ∆Hydr Ca₂⁺ +∆Hydr Cl^-
=-1562+(-364x2)
=-1562+-728
=-2290Kjmol-1
∆H_sol = ∆Hl_aq+∆Hydr
∆Hlath = ∆Hsol-∆Hydr
= -82.9 = (-2290)
= -82.9+2290
= +2207.1Kj/mole^-1