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

To calculate the lattice energy of NaCl(s), we use the Born-Haber cycle, which includes several enthalpy changes to form the solid from its constituent elements in their standard states. The lattice energy can be calculated using the following steps:

1. The standard enthalpy change for the formation of NaCl(s) from its elements: \(\text{Na(s)} + \frac{1}{2} \text{Cl}_2 \text{(g)} \rightarrow \text{NaCl(s)} \quad \Delta H^\circ = -411 \, \text{kJ/mole}\).
2. The sublimation enthalpy of Na(s) to Na(g): \(\text{Na(s)} \rightarrow \text{Na(g)} \quad \Delta H^\circ = 107 \, \text{kJ/mole}\).
3. The bond dissociation enthalpy of Cl\(_2\)(g) to 2Cl(g): \(\text{Cl}_2 \text{(g)} \rightarrow 2\text{Cl}(g) \quad \Delta H^\circ = 242 \, \text{kJ/mole}\).
4. The first ionization energy of Na(g): \(\text{Na(g)} \rightarrow \text{Na}^+(g) + e^- \quad \Delta H^\circ = 502 \, \text{kJ/mole}\).
5. The electron affinity of Cl(g): \(\text{Cl(g)} + e^- \rightarrow \text{Cl}^-(g) \quad \Delta H^\circ = -355 \, \text{kJ/mole}\).

The equation for lattice energy \((U)\) is derived from the Born-Haber cycle and can be expressed as:

\(U = \Delta H_f(\text{NaCl}) - [\Delta H_{subl}(\text{Na}) + \frac{1}{2} \Delta H_{dissoc}(\text{Cl}_2) + \Delta H_{ion}(\text{Na}) + \Delta H_{ea}(\text{Cl})]\)

Substituting the given values:

\(U = (-411) - [107 + \frac{1}{2}(242) + 502 - 355]\)

Calculating step by step:

• Energy required for sublimation: 107 kJ/mol.
• Energy required to dissociate chlorine gas: \(\frac{1}{2} \times 242 = 121 \, \text{kJ/mole}\).
• Energy required for ionization: 502 kJ/mol.
• Energy released in electron affinity: -355 kJ/mol.

Plug these into the equation:

\(U = -411 - [107 + 121 + 502 - 355] = -411 - 375\)

Therefore:

\(U = -786 \, \text{kJ/mole}\)

The correct answer is \[ \boxed{-786 \, \text{kJ/mole}}. \].