Use the following data to estimate \(\Delta H_{\mathrm{f}}^{\circ}\) for barium bromide. $$ \mathrm{Ba}(s)+\mathrm{Br}_{2}(g) \longrightarrow \operatorname{BaBr}_{2}(s) $$ $$ \begin{array}{lr} \text { Lattice energy } & -1985 \mathrm{~kJ} / \mathrm{mol} \\ \text { First ionization energy of Ba } & 503 \mathrm{~kJ} / \mathrm{mol} \\ \text { Second ionization energy of Ba } & 965 \mathrm{~kJ} / \mathrm{mol} \\ \text { Electron affinity of } \mathrm{Br} & -325 \mathrm{~kJ} / \mathrm{mol} \\\ \text { Bond energy of } \mathrm{Br}_{2} & 193 \mathrm{~kJ} / \mathrm{mol} \\ \text { Enthalpy of sublimation of } \mathrm{Ba} & 178 \mathrm{~kJ} / \mathrm{mol} \end{array} $$

The Born-Haber cycle is a way to calculate the enthalpy of formation for an ionic compound by taking into account various energy changes involved in the reaction. In general, the cycle includes lattice energy, ionization energy, electron affinity, and bond energies. In our case, the Born-Haber cycle for barium bromide can be represented as follows: 1. Enthalpy of sublimation of Ba(s): \(Ba_{(s)} \rightarrow Ba_{(g)}\) 2. Bond energy of Br2(g): \(\frac{1}{2}Br_{2(g)} \rightarrow Br_{(g)}\) 3. First ionization energy of Ba(g): \(Ba_{(g)} \rightarrow Ba^{+}_{(g)} + e^-\) 4. Second ionization energy of Ba(g): \(Ba^{+}_{(g)} \rightarrow Ba^{2+}_{(g)} + e^-\) 5. Electron affinity of Br(g): \(Br_{(g)} + e^- \rightarrow Br^{-}_{(g)}\) 6. Formation of BaBr2(s) from its ions (Lattice energy): \(Ba^{2+}_{(g)} + 2Br^{-}_{(g)} \rightarrow BaBr_{2(s)}\) Now, let's calculate the enthalpy of formation using the given data.

To find the enthalpy of formation, we need to sum up the energy changes in each step of the Born-Haber cycle. ΔHf°(BaBr2) = ΔHsub(Ba) + 0.5 * Bond energy(Br2) + Ionization energy(Ba) + Electron affinity(Br) + Lattice energy

Now we'll insert the given values into the equation: ΔHf°(BaBr2) = 178 kJ/mol + 0.5 * 193 kJ/mol + (503 kJ/mol + 965 kJ/mol) - 2 * 325 kJ/mol - 1985 kJ/mol

Now calculate the sum: ΔHf°(BaBr2) = 178 + 96.5 + 1468 - 650 - 1985 ΔHf°(BaBr2) = -892.5 kJ/mol The standard enthalpy of formation for barium bromide (BaBr2) is approximately -892.5 kJ/mol.