Answer
The estimated change in energy (∆E) for the reaction Ba(s) + Br2(g) → BaBr2(s) is -2712 kJ/mol.
Work Step by Step
To estimate the change in energy (∆E) for the reaction Ba(s) + Br2(g) → BaBr2(s), we can use the given data and the Born-Haber cycle.
The Born-Haber cycle is a thermochemical cycle that allows us to calculate the lattice energy of an ionic compound by considering the various steps involved in its formation.
The steps involved in the formation of BaBr2 are:
1. Sublimation of Ba(s) to Ba(g)
2. Ionization of Ba(g) to Ba2+(g)
3. Dissociation of Br2(g) to 2Br(g)
4. Electron affinity of Br(g)
5. Formation of the ionic lattice BaBr2(s)
Using the given data, we can calculate the change in energy (∆E) for the reaction as follows:
∆E = Lattice energy of BaBr2(s) - (Enthalpy of sublimation of Ba(s) + 2 × First ionization energy of Ba(g) + Bond energy of Br2(g) + 2 × Electron affinity of Br(g))
Substituting the values:
∆E = -1985 kJ/mol - (178 kJ/mol + 2 × 503 kJ/mol + 193 kJ/mol + 2 × (-325 kJ/mol))
∆E = -1985 kJ/mol - (178 kJ/mol + 1006 kJ/mol + 193 kJ/mol - 650 kJ/mol)
∆E = -1985 kJ/mol - 727 kJ/mol
∆E = -2712 kJ/mol
Therefore, the estimated change in energy (∆E) for the reaction Ba(s) + Br2(g) → BaBr2(s) is -2712 kJ/mol.
