Question

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

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

Step-by-step solution

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

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.)

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

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

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.)

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.

Key concepts

Standard Enthalpy of Formation

The standard enthalpy of formation, often symbolized as \( \Delta H_{\mathrm{f}}^{\circ} \), is a critical concept in chemistry. It represents the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states. This value helps predict reaction energies and is crucial for understanding the stability of compounds.
The formation of barium bromide (\( \text{BaBr}_2 \)) from barium and bromine can be thought of as a multi-step process involving ionization, electron affinity, and lattice formation. Each of these steps contributes to the overall energy change. By summing the energies of ionization of barium, electron affinities of bromine, bond energy of \( \text{Br}_2 \), sublimation energy of barium, and the lattice energy of barium bromide, we arrive at the standard enthalpy of formation for barium bromide. This value provides insights into how much energy is released or absorbed when the compound is formed under standard conditions.

Lattice Energy

Lattice energy is the amount of energy released when gaseous ions form a crystalline solid. It provides insights into the strength of ionic bonds in a compound. For barium bromide, the lattice energy is large and negative, indicating a strong attraction between barium and bromide ions.
The high lattice energy of \( -1985 \ \text{kJ/mol} \) implies that a significant amount of energy is required to separate the ions. This factor contributes substantially to the determination of the compound's standard enthalpy of formation. As lattice energy involves forces between ions, it is an indicator of the compound's thermal stability.

• High lattice energy often results in a high melting point and stability.
• Compounds with high lattice energies are typically less soluble in water.

Ionization Energy

Ionization energy is the energy required to remove an electron from an atom or ion. For barium, two ionization energies are crucial. The first ionization energy is for removing one electron to create a positive ion, while the second is for removing another electron to attain a dipositive cation.
In barium bromide formation, the ionization energies total to \( 1468 \ \text{kJ/mol} \).
Key Points:

• First ionization energy for barium: \( 503 \ \text{kJ/mol} \).
• Second ionization energy for barium: \( 965 \ \text{kJ/mol} \).
• Necessary for converting barium atoms into ions to form ionic bonds with bromide ions.

Ionization energies help to understand the ease or difficulty with which an atom gives up electrons. Higher ionization energy generally indicates a stronger hold over electrons, influencing the formation of ionic bonds.

Electron Affinity

Electron affinity measures the energy change when an electron is added to a neutral atom in the gaseous state, turning it into an anion. It is typically expressed as a negative value since energy is often released during this process.
For bromine, the electron affinity is \( -325 \ \text{kJ/mol} \) per atom. Since \( \text{BaBr}_2 \) involves two bromine anions, the total electron affinity is \( -650 \ \text{kJ/mol} \).
This energy release helps offset the energy input required for the ionization of barium.

• Electron affinity is crucial in understanding how atoms become negatively charged ions.
• It provides insights into an element's tendency to gain electrons.

By studying electron affinity, we can better predict which elements are likely to participate in exothermic reactions when forming compounds like barium bromide.