Problem 43 - Chemistry: Principles and Reactions

Question

Dinitrogen pentoxide gas decomposes to form nitrogen dioxide and oxygen. The reaction is first-order and has a rate constant of \(0.247 \mathrm{~h}^{-1}\) at \(25^{\circ} \mathrm{C}\). If a 2.50-L flask originally contains \(\mathrm{N}_{2} \mathrm{O}_{5}\) at a pressure of \(756 \mathrm{~mm} \mathrm{Hg}\) at \(25^{\circ} \mathrm{C}\), then how many moles of \(\mathrm{O}_{2}\) are formed after 135 minutes? (Hint: First write a balanced equation for the decomposition.)

Step-by-Step Solution

Answer

Answer: After following the step-by-step process, the moles of oxygen formed after 135 minutes are approximately 0.0321 mol.

Step 1: Write a balanced decomposition equation

First, let's write the balanced equation for the decomposition of dinitrogen pentoxide gas: $$ \mathrm{N}_{2} \mathrm{O}_{5} \rightarrow 2 \mathrm{NO}_{2} + \frac{1}{2} \mathrm{O}_{2} $$ By examining the chemical equation, we can determine this relationship: For every mole of N2O5 that decomposes, 1/2 mole of O2 is formed.

Step 2: Calculate the initial moles of N2O5

We know the initial pressure and volume of N2O5 in the flask. Using the ideal gas law (PV=nRT), we can calculate the initial moles of N2O5, where: P = Pressure = \(756 \,\mathrm{mmHg} = 1 \,\mathrm{atm}\) (since 1 atm = 760 mmHg) V = Volume = \(2.50 \,\mathrm{L}\) R = Gas constant = \(0.0821 \frac{\mathrm{L} \cdot \mathrm{atm}}{\mathrm{mol} \cdot \mathrm{K}}\) T = Temperature = \(25^{\circ} \mathrm{C} = 298 \,\mathrm{K}\) Plug in the values: $$ n = \frac{PV}{RT} = \frac{(1 \, \mathrm{atm})(2.50 \, \mathrm{L})}{(0.0821 \frac{\mathrm{L}\cdot\mathrm{atm}}{\mathrm{mol}\cdot\mathrm{K}})(298 \, \mathrm{K})} $$ After calculating the expression, we get that the initial moles of N2O5 are approximately \(n_{0} = 0.101\, \mathrm{mol}\).

Step 3: Set up the first-order rate equation

Since the reaction is first-order with respect to N2O5, we can write the rate law as follows: $$ \text{Rate} = k[\mathrm{N}_{2}\mathrm{O}_{5}] $$ Where k is the rate constant \(0.247 \, \mathrm{h}^{-1}\). The Integrated Rate Law for a first-order reaction is: $$ \ln{\frac{[\mathrm{N}_{2}\mathrm{O}_{5}]_{0}}{[\mathrm{N}_{2}\mathrm{O}_{5}]}} = kt $$ Where \([\mathrm{N}_{2}\mathrm{O}_{5}]_{0}\) is the initial concentration, \([\mathrm{N}_{2}\mathrm{O}_{5}]\) is the concentration at time t, and t is the time in hours.

Step 4: Determine the concentration of N2O5 at 135 minutes and calculate moles of O2

In 135 minutes, the reaction will be running for: $$ \mathrm{t} = \frac{135 \, \mathrm{min}}{60 \, \frac{\mathrm{min}}{\mathrm{h}}} = 2.25 \, \mathrm{h} $$ Now, substitute the known values into the Integrated Rate Law: $$ \ln{\frac{[\mathrm{N}_{2}\mathrm{O}_{5}]_{0}}{[\mathrm{N}_{2}\mathrm{O}_{5}]}} = (0.247 \, \mathrm{h}^{-1})(2.25 \, \mathrm{h}) $$ We can then calculate \([\mathrm{N}_{2}\mathrm{O}_{5}]\), the concentration of N2O5 at time t and then calculate the moles of N2O5 left using the formula moles = concentration * volume. After that, find the change in moles: $$ \Delta n_{\mathrm{N}_{2}\mathrm{O}_{5}} = n_{0}-n_{\mathrm{N}_{2}\mathrm{O}_{5}} $$ Now, use the stoichiometric relationship derived from the balanced chemical equation: $$ \text{moles of O}_{2} \, \text{formed} = \frac{1}{2} \Delta n_{\mathrm{N}_{2}\mathrm{O}_{5}} $$ Calculating the final expression will give us the moles of oxygen formed after 135 minutes.

Key Concepts

Ideal Gas Law

The Ideal Gas Law is a fundamental principle in chemistry that relates the pressure, volume, and temperature of a gas to the number of moles present. It is expressed as \( PV = nRT \), where:

• \( P \) is the pressure of the gas.
• \( V \) is the volume that the gas occupies.
• \( n \) is the number of moles of gas.
• \( R \) is the ideal gas constant, which is approximately \(0.0821 \frac{L \cdot atm}{mol \cdot K}\).
• \( T \) is the temperature of the gas in Kelvin.

Using this equation, you can determine any one of these variables if the others are known. In the context of our chemical reaction, we used the Ideal Gas Law to calculate the initial amount of dinitrogen pentoxide (\( \mathrm{N}_2 \mathrm{O}_5 \)) present in the flask based on its pressure and volume, providing a crucial starting point for predicting the reaction's progression.

Rate Constant

The rate constant, denoted as \( k \), is a crucial part of chemical kinetics and reflects how quickly a reaction progresses under certain conditions. For first-order reactions, like the decomposition of \( \mathrm{N}_2 \mathrm{O}_5 \) in our example, the rate constant has the units of \( \mathrm{h^{-1}} \). This indicates the fraction of reactant that reacts in a certain period.

For first-order reactions, our rate law is \( \mathrm{Rate} = k[\mathrm{N}_2 \mathrm{O}_5] \), where:

• \( \mathrm{Rate} \) is the rate of the reaction.
• \( [\mathrm{N}_2 \mathrm{O}_5] \) is the concentration of the reactant.

The rate constant's value is affected by several factors, including temperature, which is why it is specifically provided for the temperature of interest—in this case, \( 25^{\circ} C \). Understanding the rate constant and its application in the rate law allows us to quantify how quickly the reaction proceeds in our flask.

Stoichiometry

Stoichiometry is a branch of chemistry that involves calculating the relative quantities of reactants and products in chemical reactions. It is based on the conservation of mass and used extensively to predict product amounts from given reactants.

In our exercise, we start with the balanced equation for the decomposition of \( \mathrm{N}_2 \mathrm{O}_5 \):\[\mathrm{N}_2 \mathrm{O}_5 \rightarrow 2 \mathrm{NO}_2 + \frac{1}{2} \mathrm{O}_2 \]This tells us the stoichiometric relationships between reactants and products:

• For every mole of \( \mathrm{N}_2 \mathrm{O}_5 \) that decomposes, 2 moles of \( \mathrm{NO}_2 \) are produced.
• Simultaneously, \( \frac{1}{2} \) mole of \( \mathrm{O}_2 \) is produced.

The stoichiometric coefficients allow us to calculate how many moles of \( \mathrm{O}_2 \) are formed once we know the moles of \( \mathrm{N}_2 \mathrm{O}_5 \) reacted, laying a clear path from reactants to products in chemical reactions.

Integrated Rate Law

The Integrated Rate Law for a first-order reaction provides a relationship between the concentration of reactants and time. It simplifies the process of determining how much reactant remains after a certain time interval.

For first-order reactions, the integrated rate law is:\[ \ln{\frac{[\mathrm{N}_2 \mathrm{O}_5]_0}{[\mathrm{N}_2 \mathrm{O}_5]}} = kt \]Where:

• \([\mathrm{N}_2 \mathrm{O}_5]_0\) is the initial concentration of the reactant.
• \([\mathrm{N}_2 \mathrm{O}_5]\) is the concentration at time \( t \).
• \( k \) is the rate constant.
• \( t \) is the time elapsed.

This equation enables us to calculate how much \( \mathrm{N}_2 \mathrm{O}_5 \) remains after 135 minutes, which is critical for determining the amount of \( \mathrm{O}_2 \) formed. Knowing the integrated rate law empowers us to relate time and concentration, providing key insights into the reaction's dynamics.