Question

Given that a chlorine-oxygen bond has an enthalpy of 243 kJ/mol , an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the standard enthalpy of formation of ClO2 102.5 kJ/mol , calculate the value for the enthalpy of formation per mole of ClO(g). What is the value for the enthalpy of formation per mole of ClO(g)?

Answer 1

Explanation:

The chemical equation is as follows.

      $\frac{1}{2}Cl_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow ClO(g)$

And, the given enthalpy is as follows.

    $\frac{1}{2}Cl_{2}(g) + O_{2}(g) \rightarrow ClO_{2}(g)$;  $\Delta H$ = 102.5 kJ

    Cl-Cl = 243 kJ/mol,      O=O = 498 kJ/mol

Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.

   $\Delta H = \sum \text{bond broken in reactants} - \sum \text{bond broken in products}$

    102.5 = $[(\frac{1}{2})x + 498] - [(2)(243)]$

    102.5 = $(\frac{1}{2})x + 498 - 486$

     102.5 - 12 = $\frac{x}{2}$

           x = 181 kJ

Now, total bond enthalpy of per mole of ClO is calculated as follows.

       $\Delta H = \sum E(\text{bond broken in reactants}) - \sum (\text{bond broken in products})$

              x = $[(\frac{1}{2})181 + (\frac{1}{2})498] - 243$

                 = 339.5 - 243

                 = 96.5 kJ

Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.