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 \(20.0 \mathrm{~h}\) if the gas is collected in a \(10.0\)-L container. (Assume that the products do not dissolve in chloroform.)

Short answer

The partial pressure of O2 produced in the container after 20 hours is approximately 0.144 atm.

Step-by-step solution

Calculate the initial moles of N2O5

To find the initial moles of N2O5, we can use the formula: Moles of N2O5 = concentration × volume The concentration of N2O5 is given as 0.600 M, and the volume is 1.00 L. Moles of N2O5 = 0.600 M × 1.00 L = 0.600 moles

Calculate the remaining concentration of N2O5 after 20 hours

The first-order reaction equation is: \[A_t = A_0 e^{-kt}\] Where: - \(A_t\) is the remaining concentration of N2O5 at time t - \(A_0\) is the initial concentration of N2O5 - k is the rate constant (1.0 × 10⁻⁵ s⁻¹) - t is the time (20 hours or 72000 seconds) Plug in the known values: Frequency(N2O5) = \(A_0 e^{-kt}\) \(A_t = 0.600 e^{-(1.0 × 10^{-5})(72000)}\) \(A_t = 0.600 e^{-0.72}\) \(A_t ≈ 0.486\; M\)

Determine the moles of O2 produced in the reaction

Using stoichiometry, we know that for every 1 mole of N2O5 decomposed, 1/2 mole of O2 is produced. So, we need to find the moles of N2O5 that reacted: Initial moles of N2O5 = 0.600 moles Final moles of N2O5 = 0.486 moles × 1.00 L = 0.486 moles Moles of N2O5 reacted = 0.600 - 0.486 = 0.114 moles Moles of O2 produced = 0.114 moles × (1/2) = 0.057 moles

Calculate the partial pressure of O2 using the ideal gas law

The ideal gas law is given as: PV = nRT Where: - P is the pressure of the gas - V is the volume of the gas - n is the number of moles of the gas - R is the ideal gas constant (0.0821 L atm K⁻¹ mol⁻¹) - T is the temperature in Kelvins We want to find P (pressure of O2) and we are given the volume of the container as 10.0 L. Temperature is given as 45°C, which is equal to 318 K. Rearrange the equation to solve for P: P = nRT / V Plug in the known values: P = (0.057 moles)(0.0821 L atm K⁻¹ mol⁻¹)(318 K) / 10.0 L P ≈ 0.144 atm The partial pressure of O2 produced in the container after 20 hours is approximately 0.144 atm.

Key concepts

First-order reaction

In chemical kinetics, a first-order reaction is one where the rate of the reaction is directly proportional to the concentration of one reactant. This means if you double the concentration of the reactant, the rate of reaction also doubles. The mathematical representation of a first-order reaction is expressed by the exponential formula:\[ A_t = A_0 e^{-kt} \]Here, \(A_t\) is the concentration of the substance at time \(t\), \(A_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time.
First-order reactions often have a straight-line plot when you graph the natural logarithm of concentration vs. time, showing their exponential decay pattern.

• For the decomposition of dinitrogen pentoxide \( (N_2O_5) \), the rate constant \(k\) at \(45^{\circ}C\) is given as \(1.0 \times 10^{-5} \; s^{-1}\).
• The half-life of a first-order reaction is a constant value, independent of initial concentration, meaning it takes the same time for the concentration to halve at any point through the reaction.

Ideal gas law

The ideal gas law is a fundamental concept in chemistry that describes the behavior of an ideal gas, using the equation:\[ PV = nRT \]Where \(P\) is the pressure of the gas, \(V\) is the volume, \(n\) is the number of moles, \(R\) is the ideal gas constant (\(0.0821 \; \text{L atm K}^{-1} \text{mol}^{-1}\)), and \(T\) is the temperature in Kelvins.
This equation offers a simple way to relate these different properties of a gas and can be used to find any one variable if the others are known.

• In this exercise, we're using the ideal gas law to calculate the partial pressure of \(O_2\) gas given a container of fixed volume and temperature.
• Remember to always convert the temperature from Celsius to Kelvin when using the ideal gas law by adding \(273\) to the Celsius temperature.

Using the ideal gas law, you can determine how gases will behave under different conditions of pressure, temperature, and volume.

Stoichiometry

Stoichiometry is the section of chemistry that involves using relationships between reactants and products in chemical reactions to calculate quantities of substances. It allows chemists to predict the amounts of products formed in a reaction from known quantities of reactants.

• In the reaction provided, for every mole of \(N_2O_5\) that decomposes, half a mole of \(O_2\) is formed.
• Using stoichiometry, you can determine that if \(0.114\) moles of \(N_2O_5\) have reacted, then \(0.057\) moles of \(O_2\) are produced from the decomposition of \(N_2O_5\).

This proportionality is determined by the coefficients of the balanced equation for the decomposition reaction.
Stoichiometry is crucial in both laboratory settings and industrial processes for calculating yields and ensuring that reactions proceed efficiently.

Decomposition reaction

A decomposition reaction occurs when a single compound breaks down into two or more simpler substances. This type of reaction is often represented by the formula:\[ AB \rightarrow A + B \]In the context of the provided exercise, the compound \(N_2O_5\) decomposes to form nitrogen dioxide \(NO_2\) and oxygen \(O_2\).

• Decomposition reactions can be initiated by various factors, such as heat, light, or catalysts.
• For \(N_2O_5\), the reaction is first-order and is accelerated by an increase in temperature.

This particular decomposition is noteworthy because it produces gaseous products, which allows for the application of the ideal gas law in determining pressures formed during the reaction.
Understanding decomposition reactions is key in various fields, including environmental chemistry and materials science, where breaking down compounds is necessary for processing or eliminating substances.