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 \(C\) to estimate the enthalpy change for this reaction. Would this reaction raise or lower the temperature of the stratosphere?

Short Answer

Expert verified
(a) The rate law for this reaction is of second order and can be written as: $$rate = (4.8 \times 10^{5} M^{-1}s^{-1})[\mathrm{O(g)}][\mathrm{O_{3}(g)}]$$ (b) The reaction likely occurs via a single elementary process since there are two reacting species (O and O3) colliding according to the rate law. (c) The enthalpy change for this reaction is \(\Delta H^{\circ}_{reaction} = -391.9\, kJ/mol\), indicating an exothermic reaction. This reaction would raise the temperature of the stratosphere.

Step by step solution

01

Identify the order of the reaction

Since the rate constant is given in units of \(M^{-1}s^{-1}\), we can determine the order of the reaction. The unit of rate constant for an \(n\) order reaction can be written as \((M)^{1-n}(s)^{-1}\). Comparing the given unit with this expression, we get \(1-n = -1\), which implies \(n = 2\). Consequently, the reaction is of second order.
02

Write the rate law

For a second-order reaction involving two reactants, the rate law is given by: $$rate = k[\mathrm{O(g)}][\mathrm{O_{3}(g)}]$$ Here, \(k = 4.8 \times 10^{5} M^{-1}s^{-1}\). Thus, the rate law for this reaction is: $$rate = (4.8 \times 10^{5} M^{-1}s^{-1})[\mathrm{O(g)}][\mathrm{O_{3}(g)}]$$ #b) Reaction through a single elementary process#
03

Analyze the molecularity of the reaction

The molecularity of a reaction refers to the number of colliding species in a given elementary process. As the rate law for this reaction is of second order, there are two reacting species colliding: O(g) and O3(g). This indicates that the reaction could occur via a single elementary process. #c) Calculate the enthalpy change#
04

Calculate the enthalpy change

We can estimate the enthalpy change for this reaction using \(\Delta H^{\circ}_{reaction} = \sum \Delta H^{\circ}_{products} - \sum \Delta H^{\circ}_{reactants}\). From Appendix C, we have: \(\Delta H^{\circ}_{f} (\mathrm{O(g)}) = 249.2\, kJ/mol\) \(\Delta H^{\circ}_{f} (\mathrm{O_{3}(g)}) = 142.7\, kJ/mol\) \(\Delta H^{\circ}_{f} (\mathrm{O_{2}(g)}) = 0\, kJ/mol\) Now, calculating \(\Delta H^{\circ}_{reaction}\): \(\Delta H^{\circ}_{reaction} = (2 \times \Delta H^{\circ}_{f} (\mathrm{O_{2}(g)})) - (\Delta H^{\circ}_{f} (\mathrm{O(g)}) + \Delta H^{\circ}_{f} (\mathrm{O_{3}(g)}))\) \(\Delta H^{\circ}_{reaction} = (2 \times 0) - (249.2 + 142.7) = -391.9\, kJ/mol\)
05

Effect of the reaction on the stratosphere temperature

Since the enthalpy change for this reaction is negative, the reaction is exothermic. This means that heat is released during the reaction. Such reactions lead to an overall increase in the temperature of their surroundings. Therefore, this reaction would raise the temperature of the stratosphere.

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