Problem 86 - Chemistry The Central Science

Question

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?

Step-by-Step Solution

Answer

(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 1: 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.

Step 2: 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#

Step 3: 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#

Step 4: 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 \(\Delta H^{\circ}_{reaction}\): \(\Delta H^{\circ}_{reaction} = (2 \times \Delta H^{\circ}_{f} (\mathrm{O_{2}(g)})) - (\Delta H^{\circ}_{f} (\mathrm{O(g)}) + \Delta H^{\circ}_{f} (\mathrm{O_{3}(g)}))\) \(\Delta H^{\circ}_{reaction} = (2 \times 0) - (249.2 + 142.7) = -391.9\, kJ/mol\)

Step 5: 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.

Key Concepts

Reaction Order

In chemical kinetics, the reaction order is a crucial concept that describes how the rate of a chemical reaction is affected by the concentration of reactants. Reaction order helps us understand the dynamics of how reactant quantities can change the rate at which products form. This is often determined experimentally but can also be deduced from the units of the rate constant in some cases.
For the given reaction between oxygen atoms and ozone, the rate constant is presented with units of \(M^{-1}s^{-1}\). This indicates a second-order reaction. The reason is found in comparing this unit with the general formula for the rate constant of an \(n\)-order reaction, which follows \(M^{1-n}s^{-1}\). By determining the value of \(n\), we found it to be 2. Thus, both reactants affect the rate, making the reaction second-order.
Understanding reaction order is vital because it guides us in predicting how changes in concentration affect the reaction speed under constant temperature and pressure conditions.

Rate Law

The rate law is a mathematical expression that relates the rate of a reaction to the concentration of its reactants. It is a key component in understanding the kinetics of a chemical reaction, predicting the reaction's speed, and understanding its mechanism.
For our reaction, involving O(g) and O extsubscript{3}(g), we established earlier that it is second order. The rate law becomes particularly straightforward:

• Rate law: \( rate = k[ ext{O(g)}][ ext{O}_3 ext{(g)}] \)

The rate law states that the rate of this reaction is directly proportional to the concentration of both O(g) and O extsubscript{3}(g). Here, \(k\) is the rate constant, valued at \(4.8 \times 10^{5} M^{-1}s^{-1}\).
Knowing the rate law allows chemists to predict how quickly a reaction will proceed under different conditions. This is particularly useful when controlling reactions for chemical production or understanding how natural processes occur, such as in environmental chemistry.

Enthalpy Change

Enthalpy change, often annotated as \(\Delta H\), represents the heat absorbed or released during a chemical reaction at constant pressure. It can help determine if a reaction is exothermic or endothermic. Let's dive into this concept as applied to our reaction.
In the given exercise, the enthalpy change is calculated using the standard enthalpies of formation. By using the formula:

• \[ \Delta H^{\circ}_{\text{reaction}} = \sum \Delta H^{\circ}_{\text{products}} - \sum \Delta H^{\circ}_{\text{reactants}} \]

we substituted the known values for our reactants and products, leading to an estimated \(\Delta H^{\circ}_{reaction} = -391.9\, kJ/mol\).
This enthalpy change signified a negative value, indicating that the reaction releases energy in the form of heat as the products are formed. Understanding enthalpy change is crucial in predicting how much heat will be involved in a reaction, influencing how reactions are managed in both laboratory and industrial settings.

Exothermic Reactions

Chemical reactions are often classified based on heat exchange with their surroundings. Exothermic reactions are those that release heat, causing an increase in the temperature of the surroundings. This energy release happens because the energy required to break the reactants' bonds is less than the energy released when new bonds form in the products.
In our discussed reaction of oxygen atoms with ozone, the enthalpy change was calculated to be negative (\(-391.9 kJ/mol\)), indicating it is exothermic. Exothermic reactions like this one are critical to various natural and industrial processes. For example, they are essential in understanding atmospheric chemistry and its impact on global temperatures.
Recognizing a reaction as exothermic can have practical implications, such as adjusting reactant quantities or conditions to prevent overheating in a process, ensuring safe and efficient chemical production. In the case of atmospheric reactions, knowing if they are exothermic helps in understanding their role in climate dynamics.