Given the following data: $$ \begin{array}{ll}{\mathrm{NO}_{2}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{O}(g)} & {\Delta H=233 \mathrm{kJ}} \\ {2 \mathrm{O}_{3}(g) \longrightarrow 3 \mathrm{O}_{2}(g)} & {\Delta H=-427 \mathrm{kJ}}\end{array} $$ $$ \mathrm{NO}(g)+\mathrm{O}_{3}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) \quad \Delta H=-199 \mathrm{kJ} $$ Calculate the bond energy for the \(\mathrm{O}_{2}\) bond, that is, calculate \(\Delta H\) for: $$ \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{O}(g) \qquad \Delta H=? $$

The exercise asks for the enthalpy change of: \( O_{2}(g) \longrightarrow 2 O(g) \qquad \Delta H=? \)

First, we notice that in the given equations, we do not have O2 as a reactant and O as a product. However, we have O2 as a product and O as a reactant in some of them. To manipulate the given equations, we can reverse them to have the correct atoms on both reactant and product sides. By doing so, we will also need to change the sign of the enthalpy change for each reversed equation. We reverse the first two equations: 1. \( NO_{2}(g) \longrightarrow NO(g) + O(g) \quad \Delta H = 233 \, kJ \) (given equation) \( NO(g) + O(g) \longrightarrow NO_{2}(g) \quad \Delta H = -233 \, kJ \) (reversed equation) 2. \( 2 O_{3}(g) \longrightarrow 3 O_{2}(g) \quad \Delta H = -427 \, kJ \) (given equation) \( 3 O_{2}(g) \longrightarrow 2 O_{3}(g) \quad \Delta H = 427 \, kJ \) (reversed equation) The third equation remains the same: 3. \( NO(g) + O_{3}(g) \longrightarrow NO_{2}(g) + O_{2}(g) \quad \Delta H = -199 \, kJ \)

Now, let's add the reversed equations from Step 2 together: 1. \( NO(g) + O(g) \longrightarrow NO_{2}(g) \quad \Delta H = -233 \, kJ \) 2. \( 3 O_{2}(g) \longrightarrow 2 O_{3}(g) \quad \Delta H = 427 \, kJ \) 3. \( NO(g) + O_{3}(g) \longrightarrow NO_{2}(g) + O_{2}(g) \quad \Delta H = -199 \, kJ \) For better clarity, we rewrite these equations as: A. \( NO(g) + O(g) + 3 O_{2}(g) \longrightarrow NO_{2}(g) + 2 O_{3}(g) + NO(g) + O_{3}(g) \) By adding these equations, the following terms cancel out: NO(g), O(g), NO2(g), and O3(g) Resulting in: \( 3 O_{2}(g) \longrightarrow 2 O_{3}(g) + O_{2}(g) \) Which can be simplified as: \( 2 O_{2}(g) \longrightarrow 2 O_{3}(g) \)

We added the equations to get the simplified equation, so now we must also add their enthalpy changes: \(\Delta H_{total} = -233\, kJ + 427\, kJ - 199\, kJ = -5\, kJ \) Now we have the enthalpy change for \( 2 O_{2}(g) \longrightarrow 2 O_{3}(g) \), which is -5 kJ. To find the enthalpy change for the target equation, \( O_{2}(g) \longrightarrow 2 O(g) \), we will divide the enthalpy change we just calculated by 2: \(\Delta H = -\frac{5 \, kJ}{2} = -2.5 \, kJ\) So, the enthalpy change for the dissociation of \( O_{2}(g) \) to \( 2 O(g) \) is \(-2.5 \, kJ\).