A reaction that contributes to the depletion of ozone in the stratosphere is the direct reaction of oxygen atoms with ozone: $$ \mathrm{O}(g)+\mathrm{O}_{3}(g) \longrightarrow 2 \mathrm{O}_{2}(g) $$ At \(298 \mathrm{~K}\) the rate constant for this reaction is $4.8 \times 10^{5}\( \)M^{-1} \mathrm{~s}^{-1}$. (a) Based on the units of the rate constant, write the likely rate law for this reaction. (b) Would you expect this reaction to occur via a single elementary process? Explain why or why not. (c) Use \(\Delta H_{f}^{\circ}\) values from Appendix \(\mathrm{C}\) to estimate the enthalpy change for this reaction. Would this reaction raise or lower the temperature of the stratosphere?

The rate constant for the reaction is given as \(4.8 \times 10^5 M^{-1}s^{-1}\). The units of the rate constant can help us determine the order of the reaction. Since the units are \(M^{-1}s^{-1}\), it suggests that the reaction is second order, i.e., the overall order is 2, with one reactant molecule participating in the reaction. The rate law for this reaction can be written as: $$ Rate = k[\mathrm{O(g)}][\mathrm{O}_{3}(g)] $$ Where \(k\) is the rate constant and \([\mathrm{O(g)}]\) and \([\mathrm{O}_{3}(g)]\) are the concentrations of the reactants.

The reaction has a rate law of order 2. An elementary process has a reaction order equal to its molecularity (the number of molecules directly involved in the elementary step). In this reaction, there are two molecules directly involved (O and O3). As the reaction has an order equal to its molecularity, it is an elementary process. Therefore, this reaction likely occurs via a single elementary process.

To calculate the enthalpy change for the reaction, we need to use the given ΔH°f values from Appendix C. We can calculate the enthalpy change for the reaction (ΔH°rxn) by taking the sum of the enthalpies of formation of the products and subtracting the sum of the enthalpies of formation of the reactants: $$ \Delta H_{rxn}^{\circ} = \Sigma \Delta H_{f, products}^{\circ} - \Sigma \Delta H_{f, reactants}^{\circ} $$ From Appendix C, we get the following values: ΔH°f (O(g)) = 249.2 kJ/mol ΔH°f (O3(g)) = 142.7 kJ/mol ΔH°f (O2(g)) = 0 kJ/mol Now we can calculate the enthalpy change: $$ \Delta H_{rxn}^{\circ} = (2 \times \Delta H_{f, O_{2}(g)}^{\circ}) - (\Delta H_{f, O_{3}(g)}^{\circ} + \Delta H_{f, O(g)}^{\circ}) $$ $$ \Delta H_{rxn}^{\circ} = (2 \times 0) - (142.7 + 249.2) = -391.9 \mathrm{kJ/mol} $$ Since the enthalpy change is negative, the reaction is exothermic, meaning it releases heat. As a result, this reaction would raise the temperature of the stratosphere.