The bond dissociation energies for the relevant bonds are given below each of the species involved in the following reaction.Calculate the overall $\Delta$H° for the reaction.

Question

Quizplus · Accepted Answer

The overall enthalpy change for the reaction can be calculated by subtracting the sum of the bond dissociation energies of the reactants from the sum of the bond dissociation energies of the products:
ΔH = Σ(bond dissociation energies of products) - Σ(bond dissociation energies of reactants)

Reactants:
F-F: 158 kJ/mol (two bonds)
S-H: 339 kJ/mol

Products:
F-S-F: 282 kJ/mol (one bond)
H-F: 568 kJ/mol (two bonds)

ΔH = [(282 kJ/mol) + (568 kJ/mol)] - [(158 kJ/mol x 2) + (339 kJ/mol)]
ΔH = -138 kJ/mol

Therefore, the overall enthalpy change for the reaction is -138 kJ/mol, which means that B is the correct choice.

The Bond Dissociation Energies for the Relevant Bonds Are Given Δ\DeltaΔ

The Bond Dissociation Energies for the Relevant Bonds Are Given $\Delta$