Problem 114 - Chemistry The Science in Context

Question

Ozone reacts completely with NO, producing \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). A \(10.0 \mathrm{L}\) vessel is filled with 0.280 mol of \(\mathrm{NO}\) and 0.280 mol of \(\mathrm{O}_{3}\) at \(350 \mathrm{K}\). Find the partial pressure of each product and the total pressure in the flask at the end of the reaction.

Step-by-Step Solution

Answer

The partial pressures of NO2 and O2 are both 8.0 atm, and the total pressure in the flask after the reaction is 16.0 atm.

Step 1: Write the balanced chemical equation

The balanced chemical equation for the reaction between \(\mathrm{NO}\) and \(\mathrm{O}_{3}\) is: \(\mathrm{NO} + \mathrm{O}_{3} \rightarrow \mathrm{NO}_{2} + \mathrm{O}_{2}\) Since one mole of \(\mathrm{NO}\) reacts with one mole of \(\mathrm{O}_{3}\), and both reactants are present in equal amounts, the reaction will proceed to completion.

Step 2: Determine the final amount of each substance

At the end of the reaction, the initial amount of \(\mathrm{NO}\) is converted to \(\mathrm{NO}_{2}\), and all the \(\mathrm{O}_{3}\) has reacted to produce \(\mathrm{O}_{2}\). Therefore, the final amounts of each substance in the vessel are: - \(\mathrm{NO}_{2}:\) 0.280 mol (from the consumed \(\mathrm{NO}\)) - \(\mathrm{O}_{2}:\) 0.280 mol (from the consumed \(\mathrm{O}_{3}\))

Step 3: Calculate the partial pressures of the products using the Ideal Gas Law

The Ideal Gas Law states that \(PV = nRT\), where \(P\) is the pressure, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature. We can use this equation to find the partial pressures of \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\): For \(\mathrm{NO}_{2}\): \(P_{\mathrm{NO}_{2}} = \frac{n_{\mathrm{NO}_{2}}RT}{V} = \frac{0.280 \times 0.0821 \times 350}{10.0} = 8.0 \ \mathrm{atm}\) For \(\mathrm{O}_{2}\): \(P_{\mathrm{O}_{2}} = \frac{n_{\mathrm{O}_{2}}RT}{V} = \frac{0.280 \times 0.0821 \times 350}{10.0} = 8.0 \ \mathrm{atm}\)

Step 4: Calculate the total pressure in the flask

We can calculate the total pressure in the flask by simply adding the partial pressures of the products: \(P_{total} = P_{\mathrm{NO}_{2}} + P_{\mathrm{O}_{2}} = 8.0 \ \mathrm{atm} + 8.0 \ \mathrm{atm} = 16.0 \ \mathrm{atm}\) The partial pressure of each product is 8.0 atm, and the total pressure in the flask at the end of the reaction is 16.0 atm.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a crucial equation in chemistry that relates the pressure, volume, temperature, and number of moles of a gas. It is represented as \( PV = nRT \), where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume of the gas container.
• \( n \) is the number of moles of gas.
• \( R \) is the universal gas constant, approximately \( 0.0821 \text{ L atm K}^{-1}\text{ mol}^{-1} \).
• \( T \) is the temperature in Kelvin.

This equation is vital for calculating the partial pressures of gases in a mixture. In the context of our problem, knowing the volume, temperature, and moles allowed us to determine the pressure of each gas produced by the reaction. Breakdowns like this help in understanding how the abstract formula applies in real situations, such as with gases in a chemical reaction within a fixed-volume flask.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products. Each reaction is represented by a balanced chemical equation, which must follow the law of conservation of mass. This means the number of atoms for each element is the same on both sides of the equation. In our case, the reaction between \( \mathrm{NO} \) and \( \mathrm{O}_{3} \) producing \( \mathrm{NO}_{2} \) and \( \mathrm{O}_{2} \) reflects this:
\( \mathrm{NO} + \mathrm{O}_{3} \rightarrow \mathrm{NO}_{2} + \mathrm{O}_{2} \)
For every mole of \( \mathrm{NO} \), one mole of \( \mathrm{O}_{3} \) is required to complete the reaction. This direct one-to-one ratio is straightforward, allowing us to predict the products' amounts with confidence.
Understanding these fundamentals is key in analyzing reactions and predicting how substances will interact under given conditions.

Ozone Decomposition

Ozone \( (\mathrm{O}_{3}) \) plays an important role in atmospheric chemistry and its reactions have significant implications. In this exercise, ozone decomposes by reacting with nitrogen monoxide \( (\mathrm{NO}) \), creating nitrogen dioxide \( (\mathrm{NO}_{2}) \) and oxygen \( (\mathrm{O}_{2}) \). This illustrates an interaction where ozone acts as an oxidizing agent, taking electrons from another molecule (\( \mathrm{NO} \) in this case).
This decomposition is crucial because it forms part of the atmospheric processes affecting air quality and environmental conditions. In simple terms, understanding these reactions helps us grasp how pollutants get transformed in the environment. It's a stark reminder of why controlling urban emissions is significant, as these reactions contribute to both ozone depletion and atmospheric balance.

Atmospheric Chemistry

Atmospheric chemistry is the study of the chemical compositions and reactions occurring in the Earth's atmosphere. This field explores how natural and human-caused compounds interact in the air around us. Our problem scenario involves a reaction relevant to atmospheric sciences: the interaction between airborne chemicals \( \mathrm{NO} \) and \( \mathrm{O}_{3} \).

Ozone's interaction with other gases like \( \mathrm{NO} \) showcases the dynamic nature of atmospheric chemistry. These reactions influence the concentration of atmospheric gases and the formation of secondary pollutants such as \( \mathrm{NO}_{2} \). Such processes affect everything from air quality to climate patterns.
Understanding atmospheric chemistry can aid in developing better environmental policies. By studying these interactions, scientists work towards mitigating pollution and protecting public health. This knowledge serves as the foundation for legislation aimed at limiting emissions that could worsen atmospheric conditions.