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.)

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.

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 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.

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.