Question

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

Short answer

The standard enthalpy of formation for barium bromide (BaBr2) is approximately -892.5 kJ/mol.

Step-by-step solution

Understand the Born-Haber Cycle

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.

Calculate the total energy change

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

Plug in the given values

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

Calculate the enthalpy of formation

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.

Key concepts

Enthalpy of Formation

The enthalpy of formation, (\(\Delta H_{\mathrm{f}}^{\circ}\) ), is the energy change that accompanies the formation of one mole of a compound from its elements in their standard states. This concept is integral to the Born-Haber Cycle because it allows us to estimate the overall energetic feasibility of forming specific ionic compounds.
When looking at the formation of barium bromide, we start with elemental barium and bromine.
- Barium (Ba) is a solid in its standard state.- Bromine (\(\text{Br}_2\)) is a diatomic gas.
The process of forming barium bromide involves several steps that contribute to the total energy required, including changing these elements into their respective gaseous and ionic forms.

Lattice Energy

Lattice energy is a crucial factor in understanding the stability of ionic compounds. It is the amount of energy released when ions come together to form a solid lattice, or conversely, the energy required to separate a mole of an ionic compound into its gaseous ions. For barium bromide, the lattice energy is given as \(-1985 \text{ kJ/mol}\).
This is a significant energy release, indicating a strong attraction between the barium cations (\(\text{Ba}^{2+}\)) and bromide anions (\(\text{Br}^-\)).
The magnitude of the lattice energy is influenced by the charge and size of the ions, where a higher charge and smaller ionic radii produce a stronger lattice.

Ionization Energy

Ionization energy refers to the energy required to remove an electron from an atom or ion in the gaseous state. For barium, this process happens in two steps:
- The first ionization energy is \(503 \text{ kJ/mol}\), required to remove the first electron and form \(\text{Ba}^+\), - The second ionization energy is \(965 \text{ kJ/mol}\), needed to remove a second electron and form \(\text{Ba}^{2+}\).
These two ionization steps are crucial as they convert barium into its stable ionic form, \(\text{Ba}^{2+}\), needed for properly forming barium bromide.Factors Influencing Ionization Energy- Atomic size: Smaller atoms have higher ionization energies.- Nuclear charge: A higher positive charge in the nucleus increases ionization energy.

Electron Affinity

Electron affinity is the energy change when an electron is added to a neutral atom in the gas phase to form a negatively charged ion. For bromine, it is \(-325 \text{ kJ/mol}\).
The negative sign signifies energy is released when bromine gains an electron to form \(\text{Br}^-\).
- Electron affinity reflects how strongly an atom attracts additional electrons.- In terms of the Born-Haber cycle for barium bromide, the electron affinity value helps us understand how bromine transitions its state to become part of an ionic compound.Importance in Chemical Bonding- A more negative electron affinity indicates a stronger tendency to gain electrons.- Halogens, like bromine, have high electron affinities compared to other elements.

Bond Energy

Bond energy is the measure of bond strength in a chemical bond. It represents the amount of energy required to break one mole of bonds in gaseous molecules.
For \(\text{Br}_2\), the bond energy is \(193 \text{ kJ/mol}\),indicating the energy needed to dissociate \(\text{Br}_2\) into two bromine atoms.
This step is essential in the Born-Haber Cycle, as we need individual bromine atoms to form ionic bonds with barium.
- Higher bond energy means stronger bonds.- In the Born-Haber process, breaking \(\text{Br}_2\) bonds is required, making bond energy a critical consideration.
Overall, each of these energies contributes to the comprehensive understanding of the energy requirements and outputs associated with compound formation.