Question

Nitrogen oxides, \(\mathrm{NO}_{x}\) (a mixture of \(\mathrm{NO}\) and \(\mathrm{NO}_{2}\) collectively designated as \(\mathrm{NO}_{x}\) ), play an essential role in the production of pollutants found in photochemical smog. The \(\mathrm{NO}_{x}\) in the atmosphere is slowly broken down to \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) in a first-order reaction. The average half-life of \(\mathrm{NO}_{x}\) in the smokestack emissions in a large city during daylight is 3.9 hours. (a) Starting with \(1.50 \mathrm{mg}\) in an experiment, what quantity of NO, remains after 5.25 hours? (b) How many hours of daylight must have elapsed to decrease \(1.50 \mathrm{mg}\) of \(\mathrm{NO}_{x}\) to \(2.50 \times 10^{-6} \mathrm{mg} ?\)

Short answer

(a) 0.733 mg remains after 5.25 hours. (b) It takes 78.2 hours to reduce to 2.50 × 10⁻⁶ mg.

Step-by-step solution

Understanding the Problem and Given Data

We are given a first-order reaction in which \( \mathrm{NO}_{x} \) breaks down into \( \mathrm{N}_{2} \) and \( \mathrm{O}_{2}.\) The half-life of \( \mathrm{NO}_{x} \) is 3.9 hours. We need to calculate the remaining amount of \( \mathrm{NO}_{x} \) after 5.25 hours starting from 1.50 mg and also find out how long it takes to decompose to \( 2.50 \times 10^{-6} \) mg.

Calculate the Rate Constant

For a first-order reaction, the rate constant \( k \) can be calculated using the formula for half-life: \( k = \frac{\ln(2)}{t_{1/2}}, \) where \( t_{1/2} \) is the half-life of the reaction.\[ k = \frac{\ln(2)}{3.9 \text{ hours}} \approx 0.1777 \text{ hr}^{-1} \]

Use the First-Order Kinetics Formula for Part (a)

For the first-order reaction, the quantity remaining after time \( t \) can be calculated using the formula: \[ [\mathrm{NO}_x]_t = [\mathrm{NO}_x]_0 e^{-kt} \] where \( [\mathrm{NO}_x]_0 \) is the initial amount and \( [\mathrm{NO}_x]_t \) is the amount at time \( t. \)Here, \( t = 5.25 \) hours.\[[\mathrm{NO}_x]_{5.25} = 1.50 \times e^{-0.1777 \times 5.25} \text{ mg} \approx 0.733 \text{ mg} \]

Rearrange Formula for Part (b) to Solve for Time

To find the time required for \( 1.50 \text{ mg} \) to decrease to \( 2.50 \times 10^{-6} \text{ mg} \), we use the formula:\[ [\mathrm{NO}_x]_t = [\mathrm{NO}_x]_0 e^{-kt} \]Rearrange to solve for \( t \):\[ t = \frac{\ln([\mathrm{NO}_x]_0 / [\mathrm{NO}_x]_t)}{k} \] where \( [\mathrm{NO}_x]_t = 2.50 \times 10^{-6} \text{ mg}. \)Calculate \( t \):\[ t \approx \frac{\ln(1.50 / 2.50 \times 10^{-6})}{0.1777} \approx 78.2 \text{ hours} \]

Final Calculations and Results

For part (a), the remaining \( \mathrm{NO}_{x} \) after 5.25 hours is approximately 0.733 mg. For part (b), it takes approximately 78.2 hours of daylight to reduce \( \mathrm{NO}_{x} \) from 1.50 mg to \( 2.50 \times 10^{-6} \text{ mg}. \)

Key concepts

Half-Life Calculation

In a first-order reaction, substances diminish exponentially over time. This occurs in reactions such as the breakdown of nitrogen oxides, or \( \mathrm{NO}_{x} \), into \( \mathrm{N}_{2} \) and \( \mathrm{O}_{2} \).

A critical concept in these reactions is the half-life, the time required for half the amount of a substance to decompose. For a first-order reaction, the half-life is constant, meaning it doesn't depend on the initial amount.

To calculate the rate of decomposition, we use the first-order kinetics formula:

• Rate constant \( k = \frac{\ln(2)}{t_{1/2}} \)
• With \( t_{1/2} = 3.9 \) hours, we get \( k \approx 0.1777 \text{ hr}^{-1} \)

Once \( k \) is known, predict the amount of substance left after time \( t \) with:
\[ [\mathrm{NO}_x]_t = [\mathrm{NO}_x]_0 e^{-kt} \]
This formula helps solve part (a) of the exercise efficiently. One simply plugs in the values for \( t \), the initial amount, and the rate constant.

Nitrogen Oxides

Nitrogen oxides (\( \mathrm{NO} \) and \( \mathrm{NO}_2 \)) are crucial agents in atmospheric chemistry. Formed primarily through combustion processes, these gases contribute to air pollution.

In photochemical smog, \( \mathrm{NO}_x \) acts as a catalyst, reacting with sunlight and other compounds to form ozone and other pollutants. Their breakdown in the atmosphere involves reactions that are first-order in nature. This means that the rate of the reaction depends directly on the concentration of \( \mathrm{NO}_x \).

• Recent studies show that human activity significantly elevates levels of \( \mathrm{NO}_x \).
• This can increase the formation of smog, affecting air quality and visibility.

Understanding their behavior in the atmosphere, and specifically their half-life, is vital in developing strategies for air quality management.

Photochemical Smog

Photochemical smog is a type of air pollution created in sunny environments. It results from complex chemical reactions between nitrogen oxides, volatile organic compounds, and sunlight.

The presence of \( \mathrm{NO}_x \) is particularly significant, as these compounds facilitate the formation of ground-level ozone and other harmful pollutants.

• High ozone levels can lead to respiratory issues and other health problems in communities.
• Photochemical smog often leads to reduced visibility due to its formation of secondary pollutants.

Efforts to control smog mainly focus on reducing \( \mathrm{NO}_x \) emissions. Strategies include refining combustion processes and adopting cleaner technologies. Mitigating this pollution is essential to improving urban air quality.