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

Short Answer

Expert verified
The estimated standard enthalpy of formation for BaBr₂ is \(-796 \, \mathrm{kJ/mol}\).

Step by step solution

01

Ionization Energies of Barium

To remove one electron from a neutral barium atom (1st ionization energy) and another electron from the singly ionized barium atom (2nd ionization energy), we have to provide energy equal to the sum of the first and second ionization energies: IE_total = 503 kJ/mol + 965 kJ/mol = 1468 kJ/mol
02

Electron Affinity of Bromine

Two bromine atoms (in Br₂) each gain an electron to form two bromine anions. Since we are given the electron affinity per bromine atom, we need to multiply this value by 2 to account for both atoms in the reaction: EA_total = 2 × (-325 kJ/mol) = -650 kJ/mol (Note: The negative sign means that energy is released.)
03

Bond Energy of Bromine

The bond energy for Br₂ is the amount of energy required to break the bond between the two bromine atoms. This value is given as: BE_Br2 = 193 kJ/mol
04

Enthalpy of Sublimation of Barium

The reaction requires the sublimation of solid barium to gaseous barium. We are given the enthalpy of sublimation for this process: ΔH_sublimation_Ba = 178 kJ/mol
05

Lattice Energy of Barium Bromide

Next, consider the lattice energy of BaBr₂. This energy value is given as: LE_BaBr2 = -1985 kJ/mol (Note: The negative sign means that energy is released.)
06

Calculate the Standard Enthalpy of Formation

Finally, add all the above energy values to estimate the standard enthalpy of formation (∆Hf°) for barium bromide: ΔHf°(BaBr₂) = IE_total + EA_total + BE_Br2 + ΔH_sublimation_Ba + LE_BaBr2 ΔHf°(BaBr₂) = 1468 kJ/mol + (-650 kJ/mol) + 193 kJ/mol + 178 kJ/mol + (-1985 kJ/mol) ΔHf°(BaBr₂) = -796 kJ/mol So the estimated standard enthalpy of formation for BaBr₂ is -796 kJ/mol.

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