Question

The bond enthalpy of the oxygen-oxygen bond in O2 is 498 kJ/mol. Based on the enthalpy of the reaction represented above, what is the average bond enthalpy, in kJ/mol, of an oxygen-oxygen bond in O3?

Answer 1

The given question is incomplete. The complete question is as follows.

The enthalpy of formation of ozone is 142.7 kJ / mol. The bond energy of $O_{2}$ is 498 kJ / mol. What is the average O=O bond energy of the bent ozone molecule O=O=O?

Explanation:

The given reaction is as follows.

          $3O_{2}(g) \rightarrows 2O_{3}$

The value of $\Delta H$ for 2 moles of ozone is $2 \times 142.7 kJ/mol$ = 285.4 kJ/mol.

So, in this reaction three O=O bonds are broken down and four O-O bonds of ozone are formed.

Expression for the enthalpy of reaction is as follows.

              $\Delta H = B.E_{reactants} - B.E_{products}$

       285.4 kJ/mol = $(3 \times 498) - (2 \times O_{3})$

       285.4 kJ/mol = $1494 - 2 \times O_{3})$

           -1208.6 kJ/mol = $-2O_{3}$

               $O_{3}$ = 604.3 kJ/mol

Now, the average bond enthalpy of O-O bond in $O_{3}$ is as follows.

                          = $\frac{604.3}{3}$ kJ/mol

                          = 201.43 kJ/mol

Thus, we can conclude that for the given reaction average bond enthalpy of an oxygen-oxygen bond in $O_{3}$ is 201.43 kJ/mol.