Use the following data to calculate `Delta_("lattice")H^(@)` for NaBr. `Delta_("sub")H^(@)` for sodium metal = `108.4 kJ "mol"^(-1)`.Ionization enthalpy of sodium = 496`kJ "mol"^(-1)` Electron gain enthalpy of bromine = `-325 kJ "mol"^(-1)`.Bond dissociation enthalpy of bromine = 192 kJ ` "mol"^(-1)`. `Delta_(f)^(H^(@))` for NaBr (s) = `-360.1 kJ "mol"^(-1)`.

Explanation

To calculate the lattice enthalpy (Δ_lattice H) for NaBr using the provided data, we will apply Hess's law. The process involves the following steps:

### Step 1: Write the Hess's Law equation
According to Hess's law, the enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. The equation for the formation of NaBr from its elements can be expressed as:

\[ \Delta_f H^\circ (\text{NaBr}) = \Delta_{\text{sub}} H^\circ (\text{Na}) + \text{Ionization Energy of Na} + \Delta_{\text{diss}} H^\circ (\text{Br}_2) + \text{Electron Gain Enthalpy of Br} + \Delta_{\text{lattice}} H^\circ (\text{NaBr}) \]
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