Final answer:
The bond enthalpy of the Br-Cl bond can be calculated using the given reactions. By using the bond energies of the reactants and products in the reactions, we can set up an equation to solve for the bond enthalpy. The bond enthalpy of the Br-Cl bond is calculated to be 107.1 kJ/mol.
Step-by-step explanation:
The bond enthalpy of the Br-Cl bond can be calculated using the bond energies from the given reactions. First, we need to determine the bond energies for the reactants and products. From reaction a, we can see that one mole of Br2(g) requires 30.91 kJ/mol to be converted to Br2(g). From reaction b, we can determine that one mole of Br2(g) requires 192.9 kJ/mol to be converted to 2 moles of Br(g). From reaction c, we can determine that one mole of Cl2(g) requires 243.4 kJ/mol to be converted to 2 moles of Cl(g). Finally, from reaction d, we can determine that one mole of Br2(l) and one mole of Cl2(g) require 29.2 kJ/mol to form 2 moles of BrCl(g).
Using this information, we can set up an equation to solve for the bond enthalpy of the Br-Cl bond.
Let x represent the bond enthalpy of the Br-Cl bond. The reaction can be written as:
Br2(l) + Cl2(g) → 2BrCl(g)
The enthalpy change for this reaction can be calculated using the bond energies:
(2 * x) + 436 + 243 = 2 * 432 + 29.2
Simplifying the equation, we get:
2x + 679 = 864 + 29.2
2x + 679 = 893.2
2x = 893.2 - 679
2x = 214.2
x = 107.1 kJ/mol
Therefore, the bond enthalpy of the Br-Cl bond is 107.1 kJ/mol.