Problem 78 - Chemistry: An Atoms-Focused Approach

Question

Hydroperoxyl radicals react rapidly with ozone to produce oxygen and OH radicals: $$\mathrm{HO}_{2}(g)+\mathrm{O}_{3}(g) \rightarrow \mathrm{OH}(g)+2 \mathrm{O}_{2}(g)$$ The rate of this reaction was studied in the presence of a large excess of ozone. Determine the pseudo-first-order rate constant and the second-order rate constant for the reaction from the following data: $$\begin{array}{cll} \text { Time (ms) } & {\left[\mathrm{HO}_{2}\right](\mathrm{M})} & {\left[\mathrm{O}_{3}\right](\mathrm{M})} \\ \hline 0 & 3.2 \times 10^{-5} & 1.0 \times 10^{-3} \\ \hline 10 & 2.9 \times 10^{-5} & 1.0 \times 10^{-3} \\ \hline 20 & 2.6 \times 10^{-6} & 1.0 \times 10^{-3} \\ \hline 30 & 2.4 \times 10^{-6} & 1.0 \times 10^{-2} \\ \hline 80 & 1.4 \times 10^{-6} & 1.0 \times 10^{-3} \\ \hline \end{array}$$

Step-by-Step Solution

Answer

Answer: The pseudo-first-order rate constant is \(9.24 \times 10^{-4}\ M^{-1}s^{-1}\) and the second-order rate constant is \(9.24\times 10^{-1}\ M^{-1}s^{-1}\).

Step 1: Find the rate of decrease in concentration of hydroperoxyl radicals

We are given the concentrations of \(\mathrm{HO}_{2}\) at different times. Calculate the change in concentration over the given time intervals: \(\Delta [\mathrm{HO}_{2}] = [\mathrm{HO}_{2}]_{t+\Delta t} - [\mathrm{HO}_{2}]_{t}\) For the interval 0 to 10 ms: \(\Delta [\mathrm{HO}_{2}] = 2.9\times 10^{-5} M - 3.2\times 10^{-5} M =-0.3\times 10^{-5} M\) \(\Delta t = 10 ms - 0 ms = 10 ms\) For the interval 20 to 30 ms: \(\Delta [\mathrm{HO}_{2}] = 2.4\times 10^{-6} M - 2.6\times 10^{-6} M =-0.2\times 10^{-6} M\) \(\Delta t = 30 ms - 20 ms = 10 ms\) For the interval 30 to 80 ms: \(\Delta [\mathrm{HO}_{2}] = 1.4\times 10^{-6} M - 2.4\times 10^{-6} M =-1.0\times 10^{-6} M\) \(\Delta t = 80 ms - 30 ms = 50 ms\)

Step 2: Calculate the pseudo-first-order rate constant

Find the average rate of decrease of \(\mathrm{HO}_{2}\) concentration for each time interval by dividing the change in concentration by the time taken: Average Rate = \(\dfrac{\Delta [\mathrm{HO}_{2}]}{\Delta t}\) For the interval 0 to 10 ms: \(Average\ Rate = -\dfrac{0.3\times 10^{-5} M}{10 ms}\) For the interval 20 to 30 ms: \(Average\ Rate = -\dfrac{0.2\times 10^{-6} M}{10 ms}\) For the interval 30 to 80 ms: \(Average\ Rate = -\dfrac{1.0\times 10^{-6} M}{50 ms}\) Now, to find the pseudo-first-order rate constant (k'), divide the average rate by the concentration of \(\mathrm{HO}_{2}\) at the beginning of that time interval: For the interval 0 to 10 ms: \(k' = \dfrac{-\frac{0.3\times 10^{-5} M}{10 ms}}{3.2\times 10^{-5} M} = 9.38 \times 10^{-4} M^{-1}s^{-1}\) For the interval 20 to 30 ms: \(k' = \dfrac{-\frac{0.2\times 10^{-6} M}{10 ms}}{2.6\times 10^{-6} M} = 10.0 \times 10^{-4} M^{-1}s^{-1}\) For the interval 30 to 80 ms: \(k' = \dfrac{-\frac{1.0\times 10^{-6} M}{50 ms}}{2.4\times 10^{-6} M} = 8.33 \times 10^{-4} M^{-1}s^{-1}\) Average pseudo-first-order rate constant, k': \(k'= \dfrac{9.38\times 10^{-4} + 10.0\times 10^{-4} + 8.33\times 10^{-4}}{3} = 9.24 \times 10^{-4}\ M^{-1}s^{-1}\)

Step 3: Calculate the second-order rate constant

Now, we can find the second-order rate constant (k) by dividing the pseudo-first-order constant(found in step 2) by the concentration of ozone: \(k = \dfrac{k'}{[\mathrm{O}_{3}]}=\frac{9.24\times 10^{-4}\ M^{-1}s^{-1}}{1.0\times 10^{-3} M} = 9.24\times 10^{-1}\ M^{-1}s^{-1}\) The pseudo-first-order rate constant is \(9.24 \times 10^{-4}\ M^{-1}s^{-1}\) and the second-order rate constant is \(9.24\times 10^{-1}\ M^{-1}s^{-1}\).

Key Concepts

Rate Constant

In chemical kinetics, the rate constant is a crucial parameter. It provides insight into the speed of a reaction. The rate constant, often represented by the symbol \(k\), is part of the rate equation. This equation expresses how the concentration of a reactant changes with time. For a reaction like \( ext{HO}_2 + ext{O}_3 \rightarrow ext{OH} + 2 ext{O}_2\), the rate constant depends on several factors:

• The nature of the reactants and the mechanism of the reaction.
• The temperature at which the reaction occurs.

In our study, we calculated both a pseudo-first-order rate constant, \(k'\), and a second-order rate constant, \(k\). While \(k'\) considers the influence of excess ozone, \(k\) gives a complete picture of the reaction's dependence on both reactants at a given temperature.

Pseudo-First-Order Reaction

A pseudo-first-order reaction simplifies complex kinetics by making it look like a first-order reaction. This is achieved by having one reactant in large excess. Thus, its concentration remains nearly constant throughout the reaction.
For the hydroperoxyl radicals and ozone reaction, ozone is in excess. It allows us to treat it as a pseudo-first-order reaction concerning \( ext{HO}_2\). We use the formula:\[Rate ext{ } = k' imes [ ext{HO}_2]\]Here, \(k'\) is the pseudo-first-order rate constant for this simplified reaction. The beauty of this approach lies in its simplicity, making calculations more straightforward while still accurately predicting reaction dynamics.

Concentration Change

Tracking the concentration change of reactants provides valuable insights into the progression of a chemical reaction. As time progresses, the concentration of \( ext{HO}_2\) decreases, illustrating how the reaction proceeds over time.
Through measurements at different times, such as 0 ms and 10 ms, the change in concentration can be calculated using the formula:\[\Delta [ ext{HO}_2] = [ ext{HO}_2]_{t + \Delta t} - [ ext{HO}_2]_t\]The rate of change of concentration of \( ext{HO}_2\) in each interval can indicate how fast the reaction occurs. Such a quantitative approach helps chemists understand the reaction's phase and predict future behavior.

Hydroperoxyl Radicals

Hydroperoxyl radicals, denoted \( ext{HO}_2\), are key players in atmospheric chemistry. They participate in various reactions that influence air quality and the ozone layer.
In our specific reaction,\[\text{HO}_2 + \text{O}_3 \rightarrow \text{OH} + 2\text{O}_2\]these radicals react with ozone, another crucial atmospheric component. The resulting products include hydroxyl radicals (\( ext{OH}\)) and oxygen molecules. This reaction sheds light not just on the behavior of \( ext{HO}_2\), but also on larger atmospheric processes impacting climate and environmental health.

Ozone Reaction

Ozone (\( ext{O}_3\)) features prominently in atmospheric chemistry due to its role in absorbing harmful ultraviolet radiation. It can also participate in reactions that may deplete its concentration in the stratosphere.
In our example, ozone reacts with hydroperoxyl radicals, forming new species that can further interact in the atmosphere. Such reactions are part of the complex network of chemical transformations impacting air quality and climate.
Understanding this reaction helps us comprehend how pollutants influence the ozone layer and environmental health. By studying such interactions, scientists can develop strategies to mitigate adverse environmental effects, crucial in addressing global air quality challenges.