Question

Using the bond energies below determine the value and sign of delta H for the
reaction: 2H-Br + F-F → 2H-F + Br-Br
Bond energies in kJ/mol: F-F 154.8 Br-Br 190, H-F 565, H-Br 362.3

Answer 1

How is reaction enthalpy calculated when using bond enthalpies?

Answer 2

When you're given bond energies, the equation to find deltaH is

deltaH = sum of the deltaH of bonds broken - sum of the deltaH of bonds formed

In other words, you add up the enthalpy of bonds broken in the reactants and subtract them with the sum of the enthalpy of bonds broken in the products

Answer 3

deltaH = (2 x 362.3 + 154.8) - (2 x 565 + 190) = -440.6 kJ

Thus this reaction is an exothermic process b/c deltaH is < 0

Answer 4

Edit: subtract them with the sum of the enthalpy of bonds FORMED (not broken) in the products