Question

\(\mathrm{NO}_{x^{\prime}}\) a mixture of \(\mathrm{NO}\) and \(\mathrm{NO}_{2},\) plays 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 \(\mathrm{NO}_{x}\) 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) 0.814 mg remains; (b) 46 hours are needed.

Step-by-step solution

Understanding Half-life and First-order Kinetics

The half-life of a first-order reaction is the time required for the concentration of a reactant to reduce to half its initial value. In this problem, the half-life (\( t_{1/2}\)) is given as 3.9 hours. We will use the formula for first-order kinetics: \( C_t = C_0 \times (1/2)^{(t/t_{1/2})} \). Where \( C_t \) is the concentration at time \( t \), and \( C_0 \) is the initial concentration.

Calculating Remaining \(\mathrm{NO}_x\) after 5.25 hours

Using the formula \( C_t = C_0 \times (1/2)^{(t/t_{1/2})} \), with \( C_0 = 1.50 \) mg, \( t = 5.25 \) hours, and \( t_{1/2} = 3.9 \) hours, we find the remaining \(\mathrm{NO}_x\):\[ C_{5.25} = 1.50 \times \left(\frac{1}{2}\right)^{(5.25/3.9)} \].Calculating this gives \( C_{5.25} \approx 0.814 \) mg.

Setting Up the Equations for Time Calculation

To find the time taken for \(1.50\) mg of \(\mathrm{NO}_x\) to decrease to \(2.50 \times 10^{-6}\) mg, use the rearranged formula: \( t = t_{1/2} \times \left(\frac{\log(C_t/C_0)}{\log(1/2)}\right) \). Substitute \( C_0 = 1.50 \) mg, \( C_t = 2.50 \times 10^{-6} \) mg, and \( t_{1/2} = 3.9 \) hours.

Calculating the Time Elapsed

Substitute the values into the rearranged formula: \[ t = 3.9 \times \left(\frac{\log(2.50 \times 10^{-6}/1.50)}{\log(1/2)}\right) \].This calculation gives \( t \approx 46 \) hours.

Key concepts

Understanding Half-Life

The concept of half-life is important in understanding how substances decrease over time. It refers to the time required for a reactant's concentration to decrease to half its original amount. For a first-order reaction, the half-life is constant and does not depend on the initial concentration.

When dealing with chemical reactions, knowing the half-life helps us predict how long a substance will persist in the environment or in a chemical process. In the case of atmospheric reactions, half-life can tell us how quickly pollutants like NOx break down.

In this exercise, the half-life of NOx is 3.9 hours. This means every 3.9 hours, the amount of NOx reduces by half. It's like a countdown that helps understand the longevity and impact of pollutants in the atmosphere.

Decoding First-Order Reactions

First-order reactions are characterized by a reaction rate that is directly proportional to the concentration of one reactant. In simple terms, the speed of the reaction depends only on the amount of one starting substance.

The mathematical expression for a first-order reaction shows a natural logarithmic relation: \[ k = \frac{\ln(2)}{t_{1/2}} \]where \( k \) is the rate constant and \( t_{1/2} \) is the half-life. This formula is crucial for calculating how reactants change over time.

In atmospheric chemistry, such as the breakdown of NOx to N2 and O2, understanding first-order kinetics is vital. It helps predict how these atmospheric reactions proceed and how long pollutants will last.

What is Photochemical Smog?

Photochemical smog is a type of air pollution that occurs when sunlight reacts with pollutants in the atmosphere. The main components that lead to this smog are nitrogen oxides (NOx) and volatile organic compounds (VOCs). These react with sunlight to form secondary pollutants like ozone, which contribute to the haze often seen in cities.

Photochemical smog is of great concern due to its health risks. It can irritate the eyes, throat, and lungs, and is harmful to both humans and the environment. Understanding the chemistry behind smog formation helps in developing strategies to mitigate its effects and improve air quality.

This exercise involving NOx plays into the larger story of how these substances contribute to smog, underlining the importance of managing emissions in urban areas.

Exploring Atmospheric Chemistry

Atmospheric chemistry focuses on the chemical processes that occur in the Earth's atmosphere. It examines compounds that exist from natural sources, such as volcanic eruptions, and human activities, like industrial emissions.

The study of atmospheric chemistry explains phenomena like the greenhouse effect, ozone depletion, and the creation of photochemical smog. It involves understanding reactions like the breakdown of pollutants and interactions with atmospheric conditions.

In the given exercise, exploring how NOx decomposes into N2 and O2 through a first-order reaction relates directly to atmospheric chemistry. By studying these processes, we better understand how different elements affect climate and air quality. This knowledge becomes crucial in making informed decisions about environmental policies.