Question

The Born-Haber cycle for KCl is evaluated with the following data:
\(\Delta_f H^\ominus\) for KCl = -436.7 kJ \(mol^{-1}\); \(\Delta_{sub} H^\ominus\) for K = 89.2 kJ \(mol^{-1}\);
\(\Delta_{ionization} H^\ominus\) for K = 419.0 kJ \(mol^{-1}\); \(\Delta_{electron gain} H^\ominus\) for \(Cl_{(g)\) = -348.6 kJ \(mol^{-1}\);
\(\Delta_{bond} H^\ominus\) for \(Cl_2\) = 243.0 kJ \(mol^{-1}\).
The magnitude of lattice enthalpy of KCl in kJ \(mol^{-1}\) is _________. (Nearest integer)}

Hint:

Be extremely careful with the bond enthalpy term; for \(Cl_2 \rightarrow KCl\), you only need half a mole of \(Cl_2\) to get one mole of \(Cl\) atoms, so always divide \(\Delta_{bond} H\) by 2.

Answer 1

Correct Answer: 718

To calculate the lattice enthalpy of KCl using the Born-Haber cycle, we follow these steps:

1. Identify known enthalpy changes:
   • Formation enthalpy of KCl, \(\Delta_f H^\ominus = -436.7 \, \text{kJ mol}^{-1}\)
   • Sublimation enthalpy of K, \(\Delta_{sub} H^\ominus = 89.2 \, \text{kJ mol}^{-1}\)
   • Ionization enthalpy of K, \(\Delta_{ionization} H^\ominus = 419.0 \, \text{kJ mol}^{-1}\)
   • Electron gain enthalpy of Cl, \(\Delta_{electron gain} H^\ominus = -348.6 \, \text{kJ mol}^{-1}\)
   • Bond dissociation enthalpy of \(Cl_2\), \(\Delta_{bond} H^\ominus = 243.0 \, \text{kJ mol}^{-1}\)
2. Write the Born-Haber cycle equation:
   \[ \Delta_f H^\ominus = \Delta_{sub} H^\ominus + \Delta_{ionization} H^\ominus + \frac{1}{2}\Delta_{bond} H^\ominus + \Delta_{lattice} H^\ominus + \Delta_{electron gain} H^\ominus \]
3. Substitute known values and solve for \(\Delta_{lattice} H^\ominus\):
   • \(-436.7 = 89.2 + 419.0 + \frac{1}{2}(243.0) + \Delta_{lattice} H^\ominus - 348.6\)
   • \(-436.7 = 89.2 + 419.0 + 121.5 + \Delta_{lattice} H^\ominus - 348.6\)
   • \(-436.7 = 281.1 + \Delta_{lattice} H^\ominus\)
   • \(\Delta_{lattice} H^\ominus = -436.7 - 281.1\)
   • \(\Delta_{lattice} H^\ominus = 717.8 \, \text{kJ mol}^{-1}\)
4. Round to the nearest integer: 718 kJ \(mol^{-1}\).
5. Verify the solution falls within the range: 718,718.

Thus, the lattice enthalpy of KCl is 718 kJ \(mol^{-1}\), and it is within the given range.