Question

Dinitrogen pentoxide \(\left(\mathrm{N}_{2} \mathrm{O}_{5}\right)\) decomposes in chloroform as a solvent to yield \(\mathrm{NO}_{2}\) and \(\mathrm{O}_{2}\). The decomposition is first order with a rate constant at \(45^{\circ} \mathrm{C}\) of \(1.0 \times 10^{-5} \mathrm{~s}^{-1}\). Calculate the partial pressure of \(\mathrm{O}_{2}\) produced from \(1.00\) L of \(0.600 \mathrm{M} \mathrm{N}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of \(20.0 \mathrm{~h}\) if the gas is collected in a \(10.0-\mathrm{L}\) container. (Assume that the products do not dissolve in chloroform.)

Short answer

The partial pressure of O₂ produced in the container is approximately 0.0491 atm.

Step-by-step solution

Write the balanced chemical equation for the decomposition of N₂O₅

The balanced chemical equation for the decomposition of N₂O₅ is as follows: \[2\,\mathrm{N}_2\mathrm{O}_5 (\mathrm{liquid}) \rightarrow 4\,\mathrm{NO}_2 (\mathrm{gas}) + \mathrm{O}_2 (\mathrm{gas})\]

Use the first order reaction equation to find the amount of N₂O₅ decomposed

For a first order reaction: \[A = A_0 \mathrm{e}^{-kt}\] Where \(A\) is the concentration at time \(t\), \(A_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time (in seconds). Calculate the time in seconds: \[20.0 \,\text{h} = 20.0 \times 60 \times 60 \,\text{s} = 72000\,\text{s}\] Now, substitute the given values to find \(A\): \[A = 0.600 \,\mathrm{M} \times \mathrm{e}^{-1.0 \times 10^{-5} \mathrm{s}^{-1} \times 72000 \,\text{s}}\] \[A \approx 0.222\,\mathrm{M}\]

Calculate the amount of N₂O₅ decomposed and the corresponding amount of O₂ produced

Subtract the concentration at time \(t\) from the initial concentration to find the decomposed amount of N₂O₅: \[\Delta \mathrm{N}_2\mathrm{O}_5 = 0.600\,\mathrm{M} - 0.222\,\mathrm{M} = 0.378\,\mathrm{M}\] To find the amount of O₂ produced, we can see from the balanced chemical equation that 1 mol of O₂ is produced from 2 moles of N₂O₅. Thus: \[\Delta \mathrm{O}_2 = \frac{1}{2} \times \Delta \mathrm{N}_2\mathrm{O}_5 = \frac{1}{2} \times 0.378\,\mathrm{M} = 0.189\,\mathrm{M}\] Since the volume of the solution is 1.00 L, the number of moles of O₂ produced is \(0.189\,\mathrm{moles}\).

Calculate the partial pressure of O₂ using the ideal gas law

Using the ideal gas law formula, the partial pressure of O₂ can be calculated as follows: \[ P = \frac{nRT}{V}\] Where \(P\) is the partial pressure of O₂, \(n\) is the number of moles of O₂, \(R\) is the gas constant (0.08206 L atm/mol K), \(T\) is the temperature in Kelvin (45 + 273.15 = 318.15 K), and \(V\) is the volume of the container (10.0 L). Substitute the values to calculate \(P\): \[P = \frac{0.189\,\mathrm{moles} \times 0.08206\,\mathrm{L\,atm\,K}^{-1}\,\mathrm{mol}^{-1} \times 318.15\,\mathrm{K}}{10.0\,\mathrm{L}}\] \[P \approx 0.0491\,\mathrm{atm}\] Therefore, the partial pressure of O₂ produced in the container is approximately 0.0491 atm.

Key concepts

Chemical Kinetics

Chemical kinetics is the study of reaction rates and the steps involved in chemical processes. When we discuss a first-order reaction like the decomposition of dinitrogen pentoxide (\(\mathrm{N}_{2} \mathrm{O}_{5}\)), we refer to a process where the rate depends on the concentration of a single reactant. In this context, understanding the kinetics helps us predict how fast a reaction proceeds and how the concentration of reactants changes over time.

For a first-order reaction, the rate equation follows the expression \(A = A_0 \mathrm{e}^{-kt}\), indicating that the concentration of the reactant decreases exponentially over time due to its dependency on a constant rate (\(k\), also known as the rate constant) and time elapsed (\(t\)). This exponential decay is a hallmark of first-order kinetics.

This insight into chemical kinetics allows chemists to predict the behavior of chemical systems and design conditions to control reaction speeds. It's crucial in various fields such as pharmaceuticals and environmental science.

Reaction Rate Equations

Reaction rate equations describe how the concentration of reactants and products change over time. In the decomposition of \(\mathrm{N}_{2} \mathrm{O}_{5}\), the rate equation helps determine how much of the compound has decomposed after a certain period. By using the formula \(A = A_0 \mathrm{e}^{-kt}\), where \(A_0\) is the initial concentration, we estimate the concentration of reactants at any given time.

For the exercise in question, we start with an initial concentration of 0.600 M \(\mathrm{N}_{2} \mathrm{O}_{5}\). After 20 hours, or 72000 seconds, using the given rate constant of \(1.0 \times 10^{-5} \mathrm{~s}^{-1}\), we find the remaining concentration. The difference between the initial and remaining amounts of \(\mathrm{N}_{2} \mathrm{O}_{5}\) provides the extent of decomposition.

These calculations not only underline the insight we can gain from reaction rate equations but also demonstrate their utility in planning and executing reactions safely and efficiently.

Ideal Gas Law

The ideal gas law connects pressure, volume, temperature, and moles of gas in a system, formulated as \(PV = nRT\). This fundamental equation allows us to calculate the pressure of gases produced or consumed in a chemical reaction at any given temperature and volume.

In this exercise, after calculating the moles of oxygen \((\mathrm{O}_2)\) produced from \(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposition, we use the ideal gas law to find the oxygen's partial pressure in a 10.0- liter container. Here, \(P\) is pressure, \(V\) is the container volume, \(n\) is the number of moles of \(\mathrm{O}_2\), \(R\) is the gas constant (0.08206 L atm/mol K), and \(T\) is the temperature in Kelvin (45 + 273.15 = 318.15 K).

By substituting these values, we apply the ideal gas law to find the pressure contributed by the produced oxygen. This approach shows the practicality of the ideal gas law in solving real-life chemical problems, especially when considering gaseous reactions.