Problem 66 - Chemistry and Chemical Reactivity

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 NO in the smokestack emissions in a large city during daylight is \(3.9 \mathrm{h}\) (a) Starting with \(1.50 \mathrm{mg}\) in an experiment, what quantity of \(\mathrm{NO}_{x}\) remains after \(5.25 \mathrm{h} ?\) (b) How many hours of daylight must have elapsed to decrease \(1.50 \mathrm{mg}\) of \(\mathrm{NO}_{x}\) to \(2.50 \times 10^{-6} \mathrm{mg} ?\)

Step-by-Step Solution

Answer

(a) 0.591 mg NOx remains; (b) 73.38 hours are needed.

Step 1: Understand the First-Order Reaction Equation

The first-order reaction rate equation is given by \( [A] = [A]_0 \cdot e^{-kt} \), where \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time elapsed. For this problem, we'll use the half-life of NO to calculate \(k\).

Step 2: Calculate the Rate Constant (k) from Half-life

For a first-order reaction, the half-life \(t_{1/2}\) is related to the rate constant by the formula \( t_{1/2} = \frac{0.693}{k} \). Given that the half-life \( t_{1/2} = 3.9 \) hours, we solve for \(k\):\[k = \frac{0.693}{3.9} \approx 0.1777 \, \mathrm{h}^{-1}\]

Step 3: Determine Remaining NOx After 5.25 Hours

Using the first-order reaction formula \([A] = [A]_0 \cdot e^{-kt}\), we are to find \([A]\) after \(t = 5.25 \) hours:\[[A] = 1.50 \, \text{mg} \times e^{-0.1777 \times 5.25} \]\[[A] \approx 1.50 \, \text{mg} \times e^{-0.9323} \]\[[A] \approx 1.50 \, \text{mg} \times 0.394 \]\[[A] \approx 0.591 \, \text{mg}\]

Step 4: Calculate Time to Reach 2.50 × 10^-6 mg

To find the time required for \([A]\) to decrease from \(1.50 \, \text{mg}\) to \(2.50 \times 10^{-6} \, \text{mg}\), we use the same first-order formula:\[2.50 \times 10^{-6} = 1.50 \times e^{-kt}\]Solving for \(t\),\[e^{-kt} = \frac{2.50 \times 10^{-6}}{1.50}\]\[t = -\frac{\ln\left(\frac{2.50 \times 10^{-6}}{1.50}\right)}{k}\]\[t \approx -\frac{\ln(1.667 \times 10^{-6})}{0.1777}\]\[t \approx \frac{13.04}{0.1777} \approx 73.38 \, \text{hours}\]

Key Concepts

Nitrogen oxides

Nitrogen oxides, often abbreviated as \( \text{NO}_x \), comprise two gases, \( \text{NO} \) (nitric oxide) and \( \text{NO}_2 \) (nitrogen dioxide). These compounds are significant pollutants that play a crucial role in environmental issues like acid rain and photochemical smog. They are primarily produced from vehicle emissions and industrial processes.
Once released into the atmosphere, \( \text{NO}_x \) can undergo various chemical reactions. A key process is its breakdown or conversion into other compounds, which often occurs through a first-order reaction, where the rate of reaction is directly proportional to the concentration of the \( \text{NO}_x \) present. Understanding how \( \text{NO}_x \) behaves in the atmosphere is essential for making sense of their impact and how to potentially mitigate pollution.

Half-life

The concept of half-life is crucial in understanding how long it takes for a substance to reduce to half its original amount. For first-order reactions, this means that each half-life period reduces the amount of reactant by half, regardless of the amount initially present. In the case of \( \text{NO}_x \), the half-life in the atmosphere is given as 3.9 hours.
This specific half-life is telling us that every 3.9 hours, the concentration of \( \text{NO}_x \) will decrease to 50% of its original value. Using this value, we can calculate several important parameters such as the reaction rate constant -- a value which indicates how quickly the reaction occurs. This is necessary for determining how long it takes for \( \text{NO}_x \) levels to decline to safe standards, as observed in environmental cleanups.

Reaction rate constant

The reaction rate constant, represented by \( k \), is a fundamental part of chemical kinetics. It provides a measure of how fast a reaction takes place. For first-order reactions, \( k \) can be calculated using the formula \( t_{1/2} = \frac{0.693}{k} \), where \( t_{1/2} \) is the half-life.
In our scenario, with a half-life of 3.9 hours for \( \text{NO}_x \), we calculate \( k \) to be approximately 0.1777 \( \text{h}^{-1} \). This constant allows us to predict the concentration of \( \text{NO}_x \) at any given time, providing insights into how quickly \( \text{NO}_x \) pollutants can be mitigated in the atmosphere. This understanding is crucial for environmental policies and also for scientists working to control urban pollution levels.

Photochemical smog

Photochemical smog is a type of air pollution that's particularly problematic in urban areas with high vehicle emissions and sunlight. It forms when \( \text{NO}_x \) reacts with volatile organic compounds (VOCs) in the presence of sunlight, leading to a mix of airborne pollutants that can cause serious health and environmental effects.
This smog is characteristically brown in color and can significantly reduce air quality, affecting respiratory health and visibility. The primary reactions that lead to photochemical smog start with the oxidation of \( \text{NO} \) to \( \text{NO}_2 \), which then reacts further in sunlight to form harmful products like ozone, a key component of smog.
Understanding the roles of \( \text{NO}_x \) in the formation of photochemical smog is important for developing effective strategies to reduce air pollution in urban environments. It underscores the importance of regulating emissions and developing cleaner technologies to ensure healthier air quality.