Chapter 9: Problem 11
For the decomposition reaction: \(\mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})} \longrightarrow 2 \mathrm{NO}_{2(\mathrm{~g})} ;\) the initial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4}\) falls from \(0.46\) atm to \(0.28\) atm in 30 minute. What is the rate of appearance of \(\mathrm{NO}_{2}\) ?
Short Answer
Expert verified
The rate of appearance of \(\mathrm{NO_2}\) is \(0.72\,\text{atm}/\text{hour}\).
Step by step solution
01
Understanding the Reaction
The given chemical equation is a decomposition reaction where dinitrogen tetroxide \(\mathrm{N_2O_4}\) decomposes into nitrogen dioxide \(\mathrm{NO_2}\). For every 1 mole of \(\mathrm{N_2O_4}\) that decomposes, 2 moles of \(\mathrm{NO_2}\) are produced. The stoichiometry of the reaction is 1:2.
02
Calculate the Change in Pressure of \(\mathrm{N_2O_4}\)
The change in pressure (\(\Delta P_{N_2O_4}\)) of \(\mathrm{N_2O_4}\) is the difference between the initial and final pressures. Calculate \(\Delta P_{N_2O_4}\) by subtracting the final pressure from the initial pressure: \(0.46\,\text{atm} - 0.28\,\text{atm} = 0.18\,\text{atm}\).
03
Calculate the Change in Moles of \(\mathrm{N_2O_4}\)
Under constant volume and temperature, the change in moles is directly proportional to the change in pressure. Hence, \(\Delta n_{N_2O_4}\) is also 0.18 moles (since the pressure is given in atm and we assume 1 atm equals 1 mole as a simplification for the rate calculation).
04
Calculate the Change in Moles of \(\mathrm{NO_2}\)
Since 1 mole of \(\mathrm{N_2O_4}\) gives 2 moles of \(\mathrm{NO_2}\), the change in moles of \(\mathrm{NO_2}\) (\(\Delta n_{NO_2}\)) will be twice that of \(\mathrm{N_2O_4}\). Therefore, calculate \(\Delta n_{NO_2}\) as \(2 \times 0.18\,\text{moles} = 0.36\,\text{moles}\).
05
Calculate the Rate of Appearance of \(\mathrm{NO_2}\)
Rate of appearance is defined as the change in concentration (or moles) over time. Given that the change occurs in 30 minutes (or 0.5 hours), divide the change in moles of \(\mathrm{NO_2}\) by the time interval to find the rate: \(\frac{0.36\,\text{moles}}{0.5\,\text{hours}} = 0.72\,\text{moles}/\text{hour}\).
06
Convert the Rate to Correspond to Pressure Change
Since the problem gives and asks for pressures, the rate should also be in terms of pressure. Assuming the ideal gas law applies (which states that 1 mole corresponds to the same volume at a fixed temperature and pressure), the rate of appearance of \(\mathrm{NO_2}\) by pressure is also \(0.72\,\text{atm}/\text{hour}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Decomposition Reaction
Decomposition reactions are a type of chemical reaction where a single compound splits into two or more simpler substances. These reactions often require an input of energy, such as heat, light, or electricity, to occur. The general formula for a decomposition reaction is:
\( AB \to A + B \).
In the context of the textbook problem, dinitrogen tetroxide (\( \mathrm{N_2O_4} \)) decomposes into nitrogen dioxide (\( \mathrm{NO_2} \)), following the stoichiometry of 1:2. This means that each molecule of \( \mathrm{N_2O_4} \) breaks down to form two molecules of \( \mathrm{NO_2} \). Understanding this reaction is crucial, as it lays the foundation for calculating the stoichiometry, and the rate of the reaction.
\( AB \to A + B \).
In the context of the textbook problem, dinitrogen tetroxide (\( \mathrm{N_2O_4} \)) decomposes into nitrogen dioxide (\( \mathrm{NO_2} \)), following the stoichiometry of 1:2. This means that each molecule of \( \mathrm{N_2O_4} \) breaks down to form two molecules of \( \mathrm{NO_2} \). Understanding this reaction is crucial, as it lays the foundation for calculating the stoichiometry, and the rate of the reaction.
Stoichiometry
Stoichiometry is the quantitative relationship between the reactants and products in a chemical reaction. It is based on the conservation of mass and the balanced chemical equation. For the problem at hand, the stoichiometry is derived from the balanced equation:
\( \mathrm{N_2O_4_{(g)}} \rightarrow 2 NO_{2_{(g)}} \).
This indicates that for every mole of \( \mathrm{N_2O_4} \) that decomposes, two moles of \( \mathrm{NO_2} \) are produced. When calculating stoichiometric quantities, it's important to use a consistent set of units—moles in this case—and to understand how these relate to measurable quantities like pressure, especially under ideal gas behavior conditions. When students master stoichiometric calculations, they can predict the amounts of products formed or reactants needed in a chemical reaction, which is essential for laboratory work and industrial processes.
\( \mathrm{N_2O_4_{(g)}} \rightarrow 2 NO_{2_{(g)}} \).
This indicates that for every mole of \( \mathrm{N_2O_4} \) that decomposes, two moles of \( \mathrm{NO_2} \) are produced. When calculating stoichiometric quantities, it's important to use a consistent set of units—moles in this case—and to understand how these relate to measurable quantities like pressure, especially under ideal gas behavior conditions. When students master stoichiometric calculations, they can predict the amounts of products formed or reactants needed in a chemical reaction, which is essential for laboratory work and industrial processes.
Rate of Reaction
The rate of reaction is a key concept in chemical kinetics, referring to the speed at which a chemical reaction proceeds. It is typically expressed as the change in concentration of a reactant or product per unit time. In the case of gases, where concentration is often related to pressure under the assumption of ideal gas behavior, the rate can be given in terms of pressure change per unit time.
For a decomposition reaction like \( \mathrm{N_2O_4} \rightarrow 2 \mathrm{NO_2} \), the rate of appearance of \( \mathrm{NO_2} \) is linked to the rate of decomposition of \( \mathrm{N_2O_4} \). The rate can be determined by measuring the pressure change of \( \mathrm{NO_2} \) over a specific time interval, given the stoichiometric relationships. In the exercise, the decrease in pressure of \( \mathrm{N_2O_4} \) directly translates to the creation of \( \mathrm{NO_2} \) pressure, allowing students to calculate the rate of appearance with the formula:
\( \text{Rate of appearance} = \frac{\Delta \text{Pressure}_{NO_2}}{\Delta \text{time}} \).
Understanding the rate of reaction not only helps in academic settings but is also fundamental in industries that depend on controlled reaction speeds, such as pharmaceuticals, manufacturing, and materials science.
For a decomposition reaction like \( \mathrm{N_2O_4} \rightarrow 2 \mathrm{NO_2} \), the rate of appearance of \( \mathrm{NO_2} \) is linked to the rate of decomposition of \( \mathrm{N_2O_4} \). The rate can be determined by measuring the pressure change of \( \mathrm{NO_2} \) over a specific time interval, given the stoichiometric relationships. In the exercise, the decrease in pressure of \( \mathrm{N_2O_4} \) directly translates to the creation of \( \mathrm{NO_2} \) pressure, allowing students to calculate the rate of appearance with the formula:
\( \text{Rate of appearance} = \frac{\Delta \text{Pressure}_{NO_2}}{\Delta \text{time}} \).
Understanding the rate of reaction not only helps in academic settings but is also fundamental in industries that depend on controlled reaction speeds, such as pharmaceuticals, manufacturing, and materials science.