The Born-Haber cycle for KCl is evaluated with the following data: Δ H for KCl = -436.7 kJ mol ; Δ H — Chemical Bonding Chemistry Question
Question
The Born-Haber cycle for KCl is evaluated with the following data: Δ H for KCl = -436.7 kJ mol ; Δ H for K = 89.2 kJ mol ; Δ H for K = 419.0 kJ mol ; Δ H for Cl = -348.6 kJ mol ; Δ H for Cl = 243.3 kJ mol The magnitude of lattice enthalpy of KCl in kJ mol is ________ (Nearest integer) f ⊖ -1 sub ⊖ -1 ionisation ⊖ -1 electron gain ⊖ (g) -1 bond ⊖ 2 -1 -1
💡 Solution & Explanation
**Step 1: Set up the Born-Haber cycle equation** The Born-Haber cycle relates formation enthalpy to lattice enthalpy: ΔH_f = ΔH_sub + ΔH_ionisation + ½ΔH_bond + ΔH_electron gain + ΔH_lattice **Step 2: Identify all given values** - ΔH_f(KCl) = -436.7 kJ/mol - ΔH_sub(K) = 89.2 kJ/mol - ΔH_ionisation(K) = 419.0 kJ/mol - ΔH_electron gain(Cl) = -348.6 kJ/mol - ΔH_bond(Cl₂) = 243.3 kJ/mol **Step 3: Calculate the bond energy contribution** ½ΔH_bond(Cl₂) = ½(243.3) = 121.65 kJ/mol **Step 4: Rearrange to solve for lattice enthalpy** ΔH_lattice = ΔH_f - ΔH_sub - ΔH_ionisation - ½ΔH_bond - ΔH_electron gain **Step 5: Substitute values** ΔH_lattice = -436.7 - 89.2 - 419.0 - 121.65 - (-348.6) ΔH_lattice = -436.7 - 89.2 - 419.0 - 121.65 + 348.6 ΔH_lattice = -717.95 kJ/mol **Step 6: Report the magnitude** |ΔH_lattice| = 717.95 ≈ 718 kJ/mol Therefore, the answer is 718.00.