Problem 40 - Chemistry: An Atoms-Focused Approach

Question

Tropospheric Ozone Tropospheric (lower atmosphere) ozone is rapidly consumed in many reactions, including $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ Use the following data to calculate the instantaneous rate of the reaction at \(t=0.000 \mathrm{s}\) and \(t=0.052 \mathrm{s}\) $$\begin{array}{cc}\text { Time (s) } & {[\mathrm{NO}](\mathrm{M})} \\\0.000 & 2.0 \times 10^{-8} \\\\\hline 0.011 & 1.8 \times 10^{-8} \\\\\hline 0.027 & 1.6 \times 10^{-8} \\\\\hline 0.052 & 1.4 \times 10^{-8} \\\\\hline 0.102 & 1.2 \times 10^{-8} \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

Based on the given data and calculations, the approximate instantaneous rates of the reaction at \(t=0.000 \mathrm{s}\) and \(t=0.052 \mathrm{s}\) are \(-\frac{2.0 \times 10^{-9}}{0.011} \,\text{M/s}\) and \(-\frac{2.0 \times 10^{-9}}{0.025} \,\text{M/s}\), respectively.

Step 1: Determine the rate equations

According to the balanced chemical equation: $$\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow\mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g)$$ The relationship between the rates would look like this: \(-\frac{d[\mathrm{O}_3]}{dt} = -\frac{d[\mathrm{NO}]}{dt} = \frac{d[\mathrm{NO}_2]}{dt} = \frac{d[\mathrm{O}_2]}{dt}\) Since we only have information on the concentration for NO, in this case, we will focus on the change in concentration of NO.

Step 2: Calculate the instantaneous rate at \(t=0.000 s\) and \(t=0.052 s\)

To calculate the instantaneous rate at the given times, we find the average rate over the interval in which the given time lies and then use it to approximate the instantaneous rate at each required time. For \(t = 0.000 s\), we can calculate the average rate between \(t = 0.000 s\) and \(t = 0.011 s\) as follows: Average rate \( = \frac{\Delta[\mathrm{NO}]}{\Delta t} = \frac{1.8 \times 10^{-8} - 2.0 \times 10^{-8}}{0.011 - 0.000}\) Average rate \( = -\frac{2.0 \times 10^{-9}}{0.011}\) Now, we will approximate the instantaneous rate at \(t = 0.000 s\) as this average rate. Instantaneous rate at \(t = 0.000 s = -\frac{2.0 \times 10^{-9}}{0.011}\) For \(t = 0.052 s\), we can calculate the average rate between \(t = 0.027 s\) and \(t = 0.052 s\) as follows: Average rate \(=\frac{\Delta[\mathrm{NO}]}{\Delta t} = \frac{1.4 \times 10^{-8} - 1.6 \times 10^{-8}}{0.052 - 0.027}\) Average rate \( = -\frac{2.0 \times 10^{-9}}{0.025}\) Now, we will approximate the instantaneous rate at \(t = 0.052 s\) as this average rate. Instantaneous rate at \(t = 0.052 s = -\frac{2.0 \times 10^{-9}}{0.025}\)

Step 3: Write the final answer

The instantaneous rate of the reaction at: - \(t=0.000 \mathrm{s}\) is approximately \(-\frac{2.0 \times 10^{-9}}{0.011} \,\text{M/s}\) - \(t=0.052 \mathrm{s}\) is approximately \(-\frac{2.0 \times 10^{-9}}{0.025} \,\text{M/s}\)

Key Concepts

Tropospheric Ozone

Tropospheric ozone is a significant component of air pollution and is found in the lower atmosphere, or troposphere. It plays a crucial role in environmental processes.
Tropospheric ozone is not emitted directly into the air but forms through chemical reactions between nitrogen oxides (NOx) and volatile organic compounds (VOC) in the presence of sunlight.

• It is a harmful pollutant at ground level, contributing to smog and respiratory problems in humans.
• Despite its harmful effects, it is essential for absorbing some ultraviolet radiation, though it cannot replace the protective role of the ozone layer in the stratosphere.

Understanding the reactions involving ozone, such as its consumption in atmospheric processes, helps us manage its levels and mitigate its adverse effects.

Chemical Kinetics

Chemical kinetics is the study of the rate at which chemical reactions occur and the factors affecting them. It involves understanding how different conditions such as temperature, pressure, and concentration influence reactions.
Key concepts in chemical kinetics include:

• **Reaction Rate**: The change in concentration of reactants or products per unit time. Rates can be measured or calculated at different stages of a reaction.
• **Rate Law**: An equation that relates the rate of reaction to the concentration of reactants. It often takes the form: rate = k[A] ^{m}[B] ^{n}, where k is the rate constant, and m and n indicate how the rate depends on the concentration of reactants A and B.
• **Activation Energy**: The minimum energy required for a reaction to occur. This value determines how temperature affects the reaction rate.

Chemical kinetics provides valuable insights into reaction mechanisms, helping scientists and engineers optimize industrial processes and environmental strategies.

Instantaneous Rate Calculation

The instantaneous rate of a reaction is an important concept in chemical kinetics, representing the speed of a reaction at a specific moment in time. Unlike average rates, instantaneous rate gives the precise rate right at a particular time.
To calculate:

• Use the concentration changes and time intervals very close to the specific time point.
• For instance, in the exercise, we used a short time interval around 0.000 s and 0.052 s to determine the instantaneous rate at these moments.
• Mathematically, it involves taking the derivative of concentration with respect to time, but practically, it’s approximated by using closely spaced experimental data.

This methodical calculation helps in analyzing reaction speed accurately, which is crucial in both laboratory and real-world applications where reaction speed matters.