Problem 36 - Chemistry The Science in Context

Question

The chemistry of smog formation includes \(\mathrm{NO}_{3}\) as an intermediate in several reactions. a. If \(\Delta\left[\mathrm{NO}_{3}\right] / \Delta t\) is \(-2.2 \times 10^{5} \mathrm{m} M / \mathrm{min}\) in the following reaction, what is the rate of formation of \(\mathrm{NO}_{2} ?\) $$ \mathrm{NO}_{3}(g)+\mathrm{NO}(g) \rightarrow 2 \mathrm{NO}_{2}(g) $$ b. What is the rate of change of \(\left[\mathrm{NO}_{2}\right]\) in the following reaction if \(\Delta\left[\mathrm{NO}_{3}\right] / \Delta t\) is \(-2.3 \mathrm{m} M / \mathrm{min}\) ? $$ 2 \mathrm{NO}_{3}(g) \rightarrow 2 \mathrm{NO}_{2}(g)+\mathrm{O}_{2}(g) $$

Step-by-Step Solution

Answer

Answer: In the first reaction, the rate of formation of NO2 is 4.4 x 10^5 mM/min. In the second reaction, the rate of change of NO2 concentration is 1.15 x 10^5 mM/min.

Step 1: Analyze the stoichiometry of the reaction#a

Notice that for every 1 mole of \(\mathrm{NO}_{3}\) that reacts, 2 moles of \(\mathrm{NO}_{2}\) are formed. Thus, we can write the ratio of their rates as: $$ \frac{\Delta\left[\mathrm{NO}_{2}\right]}{\Delta t} = -2 \times \frac{\Delta\left[\mathrm{NO}_{3}\right]}{\Delta t} $$

Step 2: Substitute the rate of change of \(\mathrm{NO}_{3}\) given and calculate#a

The given rate of change of \(\mathrm{NO}_{3}\) is \(-2.2 \times 10^{5} \mathrm{m}M / \mathrm{min}\). Substitute this value in the formula we just derived to find the rate of formation of \(\mathrm{NO}_{2}\): $$ \frac{\Delta\left[\mathrm{NO}_{2}\right]}{\Delta t} = -2 \times (-2.2 \times 10^{5} \mathrm{m}M/\mathrm{min}) $$ Calculate the value: $$ \frac{\Delta\left[\mathrm{NO}_{2}\right]}{\Delta t} = 4.4 \times 10^{5} \mathrm{m} M / \mathrm{min} $$ The rate of formation of \(\mathrm{NO}_{2}\) is \(4.4 \times 10^{5} \mathrm{m} M / \mathrm{min}\). b. Calculate the rate of change of the concentration of \(\mathrm{NO}_{2}\)

Step 1: Analyze the stoichiometry of the reaction#b

Notice that for every 2 moles of \(\mathrm{NO}_{3}\) that reacts, 2 moles of \(\mathrm{NO}_{2}\) are formed. Thus, we can write the ratio of their rates as: $$ \frac{\Delta\left[\mathrm{NO}_{2}\right]}{\Delta t} = -\frac{1}{2} \times \frac{\Delta\left[\mathrm{NO}_{3}\right]}{\Delta t} $$

Step 2: Substitute the rate of change of \(\mathrm{NO}_{3}\) given and calculate#b

The given rate of change of \(\mathrm{NO}_{3}\) is \(-2.3 \times 10^{5} \mathrm{m}M / \mathrm{min}\). Substitute this value in the formula we just derived to find the rate of change of the concentration of \(\mathrm{NO}_{2}\): $$ \frac{\Delta\left[\mathrm{NO}_{2}\right]}{\Delta t} = -\frac{1}{2} \times (-2.3 \times 10^{5} \mathrm{m}M/\mathrm{min}) $$ Calculate the value: $$ \frac{\Delta\left[\mathrm{NO}_{2}\right]}{\Delta t} = 1.15 \times 10^{5} \mathrm{m} M / \mathrm{min} $$ The rate of change of the concentration of \(\mathrm{NO}_{2}\) is \(1.15 \times 10^{5} \mathrm{m} M / \mathrm{min}\).

Key Concepts

Stoichiometry

Stoichiometry plays a pivotal role in understanding chemical reactions. It allows us to quantify the relationships between reactants and products. The stoichiometry of a reaction tells us how the quantities of substances involved in a reaction relate to each other. This is essential when calculating the rate of reaction because it involves considering everything from initial moles to rates of formation or consumption.
In the context of the provided exercise, stoichiometry helps us determine the rate relationship between nitric oxide (\(\text{NO}_3\)) and nitrogen dioxide (\(\text{NO}_2\)). For the first reaction:

• One molecule of \(\text{NO}_3\) reacts to form two molecules of \(\text{NO}_2\).

For this scenario, the rate of formation of \(\text{NO}_2\) is twice the rate at which \(\text{NO}_3\) is being consumed. This relationship is derived from the stoichiometric coefficients in the balanced chemical equation. As chemical reactions are intrinsically tied to these ratios, stoichiometry acts as a fundamental tool in calculating and understanding rates of reaction.

NO3 Formation

The chemical component \(\text{NO}_3\), commonly referred to as nitrate, often acts as an oxidizing agent in various chemical reactions. In the scope of this exercise, \(\text{NO}_3\) functions as a reactant in the formation of nitrogen dioxide (\(\text{NO}_2\)). Understanding how \(\text{NO}_3\) forms and reacts in chemical processes is essential for predicting reaction behaviors and rates.
When \(\text{NO}_3\) serves as a reactant, its consumption rate directly influences product formation. Here, \(\text{NO}_3\) is primarily consumed to produce \(\text{NO}_2\). This is crucial for managing conditions where \(\text{NO}_3\) acts as a fleeting intermediate in air pollution and environmental chemistry.
Putting this into perspective, in environmental contexts like smog formation, monitoring the rate at which \(\text{NO}_3\) is consumed is vital to predict environmental impacts accurately. This knowledge aids in developing strategies for pollution control.

NO2 Formation

Nitrogen dioxide (\(\text{NO}_2\)) is a significant chemical species resulting from reactions involving nitrogen oxides, especially in atmospheric chemistry. Its formation is tightly interlinked with the consumption of other nitrogen oxides like \(\text{NO}_3\).
In reaction chemistry, tracking the formation of \(\text{NO}_2\) is crucial for understanding reaction kinetics and environmental implications. The rate at which \(\text{NO}_2\) is produced not only reflects the activity of preceding reactions but also impacts subsequent chemical transformations.
In the exercise presented, for every two moles of \(\text{NO}_2\) formed from \(\text{NO}_3\), you directly assess how quickly these products appear, which can be affected by factors such as temperature and pressure.
Given its role in environmental issues like air pollution, \(\text{NO}_2\) formation is extensively studied to mitigate its harmful effects and to design effective air quality management strategies.