For KCl(s): $Δ_{f} H = - 436.7$ , enthalpy of sublimation of K $= 89.2$ , ionisation enthalpy of K $= 418.8$ , $\frac{1}{2}$ bond dissociation enthalpy of $Cl X_{2} = 121.7$ , electron gain enthalpy of Cl $= - 348.6$ (all in $kJ mo l^{- 1}$ ). Using a Born-Haber cycle, calculate the lattice enthalpy, defined as $Δ H$ for $KCl (s) K X^{+} (g) + Cl X^{-} (g)$ , in $kJ mo l^{- 1}$ (one decimal place).

Numerical answer type, answer in kJ/mol, tolerance 0.1.

Answer: 717.8 kJ/mol

By Hess's law,

Δ_{f} H = Δ_{s u b} + I E + \frac{1}{2} D + Δ_{e g} H - U

, where

U

is the (positive) lattice enthalpy. The sum of the gas-phase steps is

89.2 + 418.8 + 121.7 - 348.6 = 281.1

. Thus

U = 281.1 - (- 436.7) = 717.8 kJ mo l^{- 1}