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 \(1.00 \mathrm{~L}\) of \(0.600 \mathrm{M} \mathrm{N}_{2} \mathrm{O}_{5}\) solution at \(45^{\circ} \mathrm{C}\) over a period of \(20.0 \mathrm{hr}\) 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 gas in the 10-L container can be calculated as follows: 1. Calculate the final concentration of N2O5 using the first-order integrated rate law: \( \ln \frac{[\mathrm{N}_{2}\mathrm{O}_{5}]_{t}}{[\mathrm{N}_{2}\mathrm{O}_{5}]_{0}} = -kt \). 2. Calculate the amount of N2O5 decomposed: \(\mathrm{Moles\;Decomposed\;N}_{2}\mathrm{O}_{5} = \mathrm{Decomposed\;N}_{2}\mathrm{O}_{5} \times V\). 3. Calculate the amount of O2 produced from decomposed N2O5: \(\mathrm{Moles\;Produced\;O}_{2} = \mathrm{Moles\;Decomposed\;N}_{2}\mathrm{O}_{5}\). 4. Determine the partial pressure of O2 gas using the ideal gas law formula: \(P_\mathrm{O_2} = \frac{\mathrm{Moles\;Produced\;O}_{2} \times R \times T}{V_\mathrm{container}}\).

Step-by-step solution

Calculate the final concentration of N2O5 using the first-order integrated rate law.

For a first-order reaction, we can use the integrated rate law to find the final concentration of N2O5, given the initial concentration and the elapsed time. The first-order integrated rate law is: \[ \ln \frac{[\mathrm{N}_{2}\mathrm{O}_{5}]_{t}}{[\mathrm{N}_{2}\mathrm{O}_{5}]_{0}} = -kt \] We have the initial concentration \([\mathrm{N}_{2}\mathrm{O}_{5}]_{0} = 0.600 \;\mathrm{M}\), the rate constant \(k = 1.0 \times 10^{-5} \;\mathrm{s^{-1}}\), and the elapsed time \(t = 20.0 \;\mathrm{hr}\). First, we need to convert the time from hours to seconds: \[t_\mathrm{sec}= t_\mathrm{hr} \times 3600\] Calculate \(t_\mathrm{sec}\) and plug the values into the rate law to find the final concentration of N2O5.

Calculate the amount of N2O5 decomposed.

We can find the amount of N2O5 decomposed by subtracting the final concentration from the initial concentration: \[\mathrm{Decomposed\;N}_{2}\mathrm{O}_{5} = [\mathrm{N}_{2}\mathrm{O}_{5}]_{0} - [\mathrm{N}_{2}\mathrm{O}_{5}]_{t}\] Next, we need to multiply the decomposed N2O5 concentration by the volume of the solution (1 L) to find the number of moles of N2O5 decomposed: \[\mathrm{Moles\;Decomposed\;N}_{2}\mathrm{O}_{5} = \mathrm{Decomposed\;N}_{2}\mathrm{O}_{5} \times V\]

Calculate the amount of O2 produced from decomposed N2O5.

In the reaction, for every mole of N2O5 decomposed, one mole of O2 is produced. Thus, the moles of O2 produced can be found as: \[\mathrm{Moles\;Produced\;O}_{2} = \mathrm{Moles\;Decomposed\;N}_{2}\mathrm{O}_{5}\]

Determine the partial pressure of O2 gas.

We now need to determine the partial pressure of O2 gas using the ideal gas law formula, which is: \[PV = nRT\] The pressure of the gas in atmospheres can be found by rearranging the formula: \[P = \frac{nRT}{V}\] Use the universal gas constant \(R = 0.0821\frac{\mathrm{L}\cdot \mathrm{atm}}{\mathrm{K}\cdot \mathrm{mol}}\), temperature in Kelvin \(T = (45 + 273.15)\;\mathrm{K}\), moles of O2 gas, and the volume of the container (10 L). Calculate the partial pressure of O2 gas: \[P_\mathrm{O_2} = \frac{\mathrm{Moles\;Produced\;O}_{2} \times R \times T}{V_\mathrm{container}}\] The final result is the partial pressure of O2 gas in the 10-L container.

Key concepts

First-Order Integrated Rate Law

Understanding the first-order integrated rate law is essential for solving problems involving first-order reactions. In such reactions, the rate at which reactants are converted into products is directly proportional to the concentration of one reactant.

The mathematical expression for this law is: \[ln \frac{{[\mathrm{A}]_t}}{{[\mathrm{A}]_0}} = -kt\]Here, \( [\mathrm{A}]_0 \) is the initial concentration of the reactant, \( [\mathrm{A}]_t \) is the concentration at time \( t \), \( k \) is the rate constant, and \( t \) is the time. By rearranging and solving this equation, we can predict concentrations at any given time during the reaction's progress.

This law is particularly useful when dealing with gases that decompose or react over time, as in the example problem.

Rate Constant

The rate constant, denoted by \( k \), is a critical parameter in the kinetics of a chemical reaction. It is unique for each chemical reaction at a given temperature and is independent of the concentrations of the reactants.

For a first-order reaction, the rate constant can be used with the integrated rate law to calculate how the concentration of a reactant decreases over time. In the exercise provided, the rate constant is given by \( 1.0 \times 10^{-5} \mathrm{~s}^{-1} \), and it is used to determine how much dinitrogen pentoxide (\( \mathrm{N}_{2} \mathrm{O}_{5} \)) decomposes over a specific period. It essentially provides a measure of the reaction's speed.

Partial Pressure

Partial pressure is a concept important in chemistry, especially when discussing reactions involving gases. It refers to the pressure exerted by a single gas in a mixture of gases. Each gas in the mixture exerts a pressure independently as if the other gases were not present.

The partial pressure of a gas can be calculated using the moles of the gas, the temperature, and the volume based on the ideal gas law. In the context of our exercise, we're interested in the pressure contributed by the oxygen gas (\( \mathrm{O}_{2} \)) produced from the decomposition of dinitrogen pentoxide in a given volume.

Ideal Gas Law

The ideal gas law is a fundamental equation that allows us to relate the pressure, volume, number of moles, and temperature of a gas in a closed system. The equation is given by: \[ PV = nRT \]Where \( P \) represents the pressure, \( V \) is the volume of the gas, \( n \) refers to the number of moles, \( R \) is the ideal or universal gas constant, and \( T \) is the temperature in Kelvin.

For our exercise example, one would use this law to calculate the partial pressure of oxygen gas produced in a container after dinitrogen pentoxide decomposes. By doing this, you can directly link the chemical reaction happening at the molecular level with the observable physical properties of the gas, such as pressure.