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.

Answer 2

Since the chlorine-oxygen bond has an enthalpy of 243 kJ/mol, and the oxygen-oxygen bond has an enthalpy of 498 kJ/mol, the enthalpy of formation per mole of ClO(g) is 96.5 kJ/mol.

Let's consider the following thermochemical equation.

0.5 Cl₂(g) + O₂(g) ⇒ ClO₂(g)     ΔH°rxn = 102.5 kJ/mol

Since we know the enthalpy of the reaction and the enthalpy of some bonds, we can calculate the enthalpy of the Cl-Cl bond.

$\Delta Hrxn = \Sigma \Delta H(bonds\ broken ) - \Sigma \Delta H(bonds\ formed) \\\\\Delta Hrxn = 0.5 \Delta H(Cl-Cl) + \Delta H(O=O) - 2\Delta H(Cl-O)\\\\-0.5 \Delta H(Cl-Cl) = \Delta H(O=O) - 2\Delta H(Cl-O)-Delta Hrxn\\\\-0.5 \Delta H(Cl-Cl) = (498kJ/mol) - 2(243kJ/mol)-(102.5kJ/mol)\\\\\Delta H(Cl-Cl) = 181kJ/mol$

Now, let's consider the reaction for the formation of ClO(g).

0.5 Cl₂(g) + 0.5 O₂(g) ⇒ ClO(g)

We will use the same expression to calculate the enthalpy of this reaction.

$\Delta Hrxn = \Sigma \Delta H(bonds\ broken ) - \Sigma \Delta H(bonds\ formed)\\\\\Delta Hrxn = 0.5 \Delta H(Cl-Cl) + 0.5 \Delta H(O=O) - 1 \Delta H(Cl-O)\\\\\Delta Hrxn = 0.5 (181kJ/mol) + 0.5 (498kJ/mol) - 1 (243kJ/mol) = 96.5 kJ/mol$

Since the chlorine-oxygen bond has an enthalpy of 243 kJ/mol, and the oxygen-oxygen bond has an enthalpy of 498 kJ/mol, the enthalpy of formation per mole of ClO(g) is 96.5 kJ/mol.

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