Use the following data to calculate `Delta_("lattice") H^(Θ) " for " NaBr. Delta_("Sub")H^(Θ)` for sodium metal `= 108.4 kJ mol^(-1)`, ionisation 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^(Θ) " for " NaBr(s) = - 360.1 kJ mol^(-1)`

Explanation

To calculate the lattice energy (Δ_lattice H°) of NaBr using the provided data, we can follow these steps:

### Step 1: Write the formation reaction of NaBr
The formation reaction for NaBr can be represented as:
\[ \text{Na (s)} + \frac{1}{2} \text{Br}_2 (g) \rightarrow \text{NaBr (s)} \]

### Step 2: Identify the given data
We have the following data:
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