Question

Consider the following data: \[ \text{Na(s)} + \frac{1}{2} \text{Cl}_2 \text{(g)} \rightarrow \text{NaCl(s)} \quad \Delta H^\circ = -411 \, \text{kJ/mole} \] \[ \text{Na(s)} \rightarrow \text{Na(g)} \quad \Delta H^\circ = 107 \, \text{kJ/mole} \] \[ \text{Cl}_2 \text{(g)} \rightarrow 2\text{Cl}(g) \quad \Delta H^\circ = 242 \, \text{kJ/mole} \] \[ \text{Cl(g)} + e^- \rightarrow \text{Cl}^-(g) \quad \Delta H^\circ = -355 \, \text{kJ/mole} \] \[ \text{Na(g)} \rightarrow \text{Na}^+(g) + e^- \quad \Delta H^\circ = 502 \, \text{kJ/mole} \] Find the lattice energy of NaCl(s).

• A. -786 kJ mole\(^{-1}\)
• B. -628 kJ mol\(^{-1}\)
• C. -428 kJ mole\(^{-1}\)
• D. -393 kJ mole\(^{-1}\)

Hint:

Use Hess's Law to sum up the enthalpy changes of various steps in a reaction to find the overall lattice energy.

Answer 1

The Correct Option is A

Step 1: Apply Hess’s Law.
The lattice energy of NaCl(s) is calculated using Hess’s law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes of the individual steps: \[ \text{Lattice Energy} = \Delta H_1 + \Delta H_2 + \Delta H_3 + \Delta H_4 + \Delta H_5. \] Substituting the given values: \[ \text{Lattice Energy} = -411 + 107 + 242 - 355 + 502 = -786 \, \text{kJ/mole}. \] Final Answer: \[ \boxed{-786 \, \text{kJ/mole}}. \]