Question

The Born-Haber cycle for KCI is evaluated with the following data:

`Delta "H"_"f"^"o"` for KCl = - 436.7 kJ mol^-1;

`Delta "H"_"f"^"o"` for K = 89.2 kJ mol^-1

`Delta "H"_"ionization"^"o"` for K = 419.0 kJ mol^-1 `Delta "H"_"ionization"^"o"` for Cl_(g) = - 348.6 kJ mol^-1 

`Delta "H"_"bond"^"o"` for Cl₂ = 243.0 kJ mol^-1

The magnitude of lattice enthalpy of KCl in kJ mol^-1 is ______. (Nearest integer)

Options

• 560 kJ/mol

• 718 kJ/mol

• 200 kJ/mol

• 652 kJ/mol

Answer 1

The magnitude of lattice enthalpy of KCl in kJ mol^-1 is 718 kJ/mol. 

Explanation:

From Born Haber Cycle

`Delta "H"_"f"^"o" "KCl" = Delta "H"_"formation"^"o"
("K") + Delta "H"_"ionization"^"o" ("K") + 1/2 Delta "H"_"bond"^"o" ("Cl"_2)`` + Delta "H"_"electron gain enthalpy"^"o" ("Cl") + Delta "H"_"lattice"^"o" ("KCl")`

`=> -436.7 " kJ/mol" = 89.2 + 419 + 1/2 (243) + (- 348.6) + Delta "H"_"lattice" ("KCl")`

`therefore Delta "H"_"lattice" = - 717.8 " kJ/mol"` ≈ 718  kJ/mol