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

Step 1: Understanding the Concept:
The Born-Haber cycle uses Hess's Law to relate various thermodynamic quantities involved in the formation of an ionic solid from its elements. The total enthalpy of formation is equal to the sum of the energies for individual steps.
Step 2: Key Formula or Approach:
\[ \Delta_f H^\ominus = \Delta_{sub} H^\ominus + \Delta_{IE} H^\ominus + \frac{1}{2} \Delta_{bond} H^\ominus + \Delta_{eg} H^\ominus + \Delta_{lattice} H^\ominus \]
Step 3: Detailed Explanation:
Substitute the given values into the formula:
- \(\Delta_f H^\ominus = -436.7\)
- \(\Delta_{sub} H^\ominus = 89.2\)
- \(\Delta_{ionization} H^\ominus = 419.0\)
- \(\Delta_{bond} H^\ominus / 2 = 243.0 / 2 = 121.5\)
- \(\Delta_{electron gain} H^\ominus = -348.6\)

\[ -436.7 = 89.2 + 419.0 + 121.5 + (-348.6) + \Delta_{lattice} H^\ominus \]
\[ -436.7 = 281.1 + \Delta_{lattice} H^\ominus \]
\[ \Delta_{lattice} H^\ominus = -436.7 - 281.1 = -717.8 \text{ kJ/mol} \]
The magnitude is \( |-717.8| = 717.8 \).
Rounding to the nearest integer, we get 718.
Step 4: Final Answer:
The magnitude of the lattice enthalpy is 718.