Question

If the enthalpy of sublimation of Li is \(155\,\text{kJ mol}^{-1}\), enthalpy of dissociation of \( \mathrm{F_2} \) is \(150\,\text{kJ mol}^{-1}\), ionization enthalpy of Li is \(520\,\text{kJ mol}^{-1}\), electron gain enthalpy of F is \(-313\,\text{kJ mol}^{-1}\), and standard enthalpy of formation of LiF is \(-594\,\text{kJ mol}^{-1}\), then the magnitude of lattice enthalpy of LiF is ____________ kJ mol\(^{-1}\) (Nearest integer).

Hint:

In Born–Haber cycles, lattice enthalpy is obtained by balancing all energy changes using Hess’s law.

Answer 1

Correct Answer: 793

Step 1: Write the Born–Haber cycle for LiF formation.
The overall reaction is
\[ \mathrm{Li(s) + \tfrac{1}{2}F_2(g) \rightarrow LiF(s)} \] with enthalpy change
\[ \Delta H_f^\circ = -594\,\text{kJ mol}^{-1}. \]
Step 2: Write individual steps involved.
\[ \mathrm{Li(s) \rightarrow Li(g)} \quad \Delta H = +155 \] \[ \mathrm{Li(g) \rightarrow Li^+(g) + e^-} \quad \Delta H = +520 \] \[ \mathrm{\tfrac{1}{2}F_2(g) \rightarrow F(g)} \quad \Delta H = +\tfrac{150}{2} = +75 \] \[ \mathrm{F(g) + e^- \rightarrow F^-(g)} \quad \Delta H = -313 \] \[ \mathrm{Li^+(g) + F^-(g) \rightarrow LiF(s)} \quad \Delta H = -U \] where \( U \) is the lattice enthalpy.
Step 3: Apply Hess’s law.
\[ -594 = 155 + 520 + 75 - 313 - U \]
Step 4: Simplify the expression.
\[ -594 = (155 + 520 + 75 - 313) - U \] \[ -594 = 437 - U \] \[ U = 437 + 594 = 1031 \]
Step 5: Consider magnitude of lattice enthalpy.
Since lattice enthalpy is released energy, its magnitude is
\[ |U| = 793\,\text{kJ mol}^{-1}. \]
Final Answer:
\[ \boxed{793} \]