Draw a born haber cycle of sodium chloride with the given values Calculate the lattice dissociation of sodium chloride Name the law applied State the law named

From the previous question and general chemical data:

• Heat of formation of $\text{NaCl}$, $\Delta H_f = -410 \text{ kJ/mol}$
• Sublimation energy of $\text{Na}$, $\Delta H_{sub} = +108 \text{ kJ/mol}$
• Atomization energy of $\text{Cl}$, $\Delta H_{atom} = +121 \text{ kJ/mol}$
• Ionization energy of $\text{Na}$, $\Delta H_{IE} = +496 \text{ kJ/mol}$ (Assumed standard value)
• Electron affinity of $\text{Cl}$, $\Delta H_{EA} = -349 \text{ kJ/mol}$ (Assumed standard value)

The Born-Haber cycle for $\text{NaCl}$ can be represented by the following series of reactions and their enthalpy changes:

1. Overall formation: $$\text{Na}_{(\text{s})} + \frac{1}{2}\text{Cl}_{2(\text{g})} \rightarrow \text{NaCl}_{(\text{s})} \quad \Delta H_f = -410 \text{ kJ/mol}$$
2. Sublimation of sodium: $$\text{Na}_{(\text{s})} \rightarrow \text{Na}_{(\text{g})} \quad \Delta H_{sub} = +108 \text{ kJ/mol}$$
3. Atomization of chlorine: $$\frac{1}{2}\text{Cl}_{2(\text{g})} \rightarrow \text{Cl}_{(\text{g})} \quad \Delta H_{atom} = +121 \text{ kJ/mol}$$
4. Ionization of gaseous sodium: $$\text{Na}_{(\text{g})} \rightarrow \text{Na}^{+}_{(\text{g})} + \text{e}^- \quad \Delta H_{IE} = +496 \text{ kJ/mol}$$
5. Electron affinity of gaseous chlorine: $$\text{Cl}_{(\text{g})} + \text{e}^- \rightarrow \text{Cl}^{-}_{(\text{g})} \quad \Delta H_{EA} = -349 \text{ kJ/mol}$$
6. Lattice formation of sodium chloride: $$\text{Na}^{+}_{(\text{g})} + \text{Cl}^{-}_{(\text{g})} \rightarrow \text{NaCl}_{(\text{s})} \quad \Delta H_{lattice, formation} = ?$$

According to Hess's Law, the overall enthalpy change for the formation of $\text{NaCl}$ is equal to the sum of the enthalpy changes of the individual steps:

$$\Delta H_f = \Delta H_{sub} + \Delta H_{atom} + \Delta H_{IE} + \Delta H_{EA} + \Delta H_{lattice, formation}$$

Substitute the known values:

$$-410 \text{ kJ/mol} = (+108 \text{ kJ/mol}) + (+121 \text{ kJ/mol}) + (+496 \text{ kJ/mol}) + (-349 \text{ kJ/mol}) + \Delta H_{lattice, formation}$$ $$-410 \text{ kJ/mol} = (108 + 121 + 496 - 349) \text{ kJ/mol} + \Delta H_{lattice, formation}$$ $$-410 \text{ kJ/mol} = 376 \text{ kJ/mol} + \Delta H_{lattice, formation}$$ $$\Delta H_{lattice, formation} = -410 \text{ kJ/mol} - 376 \text{ kJ/mol}$$ $$\Delta H_{lattice, formation} = -786 \text{ kJ/mol}$$

Lattice dissociation energy is the energy required to break one mole of an ionic solid into its constituent gaseous ions. It is the negative of the lattice formation energy.

$$\Delta H_{lattice, dissociation} = - \Delta H_{lattice, formation}$$ $$\Delta H_{lattice, dissociation} = - (-786 \text{ kJ/mol})$$ $$\Delta H_{lattice, dissociation} = \boxed{\text{+786 kJ/mol}}$$

The law applied is Hess's Law.

Hess's Law states that the total enthalpy change for a chemical reaction is the same, regardless of the pathway taken, as long as the initial and final conditions are the same.