The enthalpy change for the following reaction is -81.1 kj. using bond energies, estimate the h-br bond energy in: hbr(g) * 2hbr(g) + cl2(g) > 2hcl(g) + br2(g)

The Bond energy of H-Br is +366 KJ

Explanation:

Bond energy is the energy required to break one mole of a covalent bond in a gaseous molecule. It is denoted as 'E'. The bond energy for Chlorine E(Cl-Cl) molecule is +242 KJ. This means that it takes 242 KJ of energy to break the Cl-Cl bond. It also takes the same amount of energy to form Chlorine molecule from chlorine atoms.

       Bond energies can be used to calculate the enthalpy of reactions

Using the example from the question

2HBr(g) + Cl2(g) > 2HCl(g) + Br2(g)

Two H-Br  and one Cl-Cl bonds will be broken and two H-Cl bonds with one Br-Br bonds will be formed

E(Cl-Cl) = +242 KJ

E(H-Cl) = +431 KJ

E (Br-Br) = +193 KJ

E (H-Br) =

Therefore, Enthalpy (ΔH) = bond breaking - bond forming

                                   -81.1 = [2x + 242] - [(2 x 431) + 193)]

                                   -81.1 =  [2x + 242]  - [862 + 193]

                                   -81.1 =  [2x + 242]  - 1055

                                   -81.1 + 1055 = 2x + 242

                                   973.9 = 2x + 242

                                   973.9 - 242 = 2x

                                   731.9 = 2x

                                   x = 731.9 / 2

                                   x = 365.95

                                   x = 366 KJ

x is the bond energy for H-Br which is +366 KJ

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