Cyclopropane \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right)\) reacts to form propene \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right)\) in the gas phase. The reaction is first order in cyclopropane and has a rate constant of \(5.87 \times 10^{-4} / \mathrm{s}\) at \(485^{\circ} \mathrm{C}\). If a 2.5 - \(\mathrm{L}\) reaction vessel initially contains 722 torr of cyclopropane at \(485^{\circ} \mathrm{C}\), how long will it take for the partial pressure of cyclopropane to drop to below \(1,00 \times 10^{2}\) torr?

A first-order reaction is a kinetic model where the reaction rate depends linearly on the concentration of only one reactant. This means that if you were to double the concentration of the reactant, the rate of the reaction would also double. In practical terms, these reactions have a characteristic half-life, which remains constant regardless of the initial concentration. Understanding this concept helps in determining how quickly a reaction proceeds and is pivotal in fields like pharmacology, where the rate at which a drug is metabolized can be critical.

For a first-order reaction, the rate law can be expressed as \( \text{Rate} = k[\text{A}] \), where \( k \) is the rate constant and \( [\text{A}] \) is the concentration of the reactant. This implies that the rate at which a reactant is used up is directly proportional to its concentration at any given time during the reaction.

The rate constant, symbolized by \( k \) in chemical kinetics, is a proportionality factor that provides the relationship between the reactant concentration and the rate of the reaction. For a first-order reaction, it has units of \( \text{s}^{-1} \) indicating that it expresses the rate at which the concentration of a reactant changes per second.

The value of \( k \) is dependent on various factors such as temperature and the presence of a catalyst. In the specific exercise, the rate constant is given as \( 5.87 \times 10^{-4} / \text{s} \) at \( 485^\circ \text{C} \), and it is a crucial parameter for predicting the time it takes for the reaction to reach a certain point. It is through the rate constant that we can calculate how long it will take for the concentration of cyclopropane to reduce to the desired level.

The integrated rate law for a first-order reaction provides a relationship that connects the concentrations of a reactant at different times. It is mathematically represented as \( \ln([\text{A}]_{\text{initial}} / [\text{A}]_t) = kt \), where \( \ln \) denotes the natural logarithm. This equation allows one to determine the concentration of reactant \( [\text{A}]_t \) at any time \( t \), given the initial concentration \( [\text{A}]_{\text{initial}} \) and the rate constant \( k \).

In practice, this law helps us understand how the concentration of a reactant in a chemical reaction decreases over time. For the given problem, after calculating the initial and final concentrations of cyclopropane using the ideal gas law, the integrated rate law enables us to solve for the time taken for the concentration to reach the specified final value.

The ideal gas law is a fundamental equation in chemistry that relates the pressure \( (P) \) , volume \( (V) \) , temperature \( (T) \) and amount \( (n) \) in moles of gas through the equation \( PV = nRT \). The constant \( R \) is known as the ideal gas constant and is valued at \( 0.0821 \text{ L} \cdot \text{atm} / (\text{K} \cdot \text{mol}) \).

To solve for the initial concentration of cyclopropane in the given exercise, we convert pressure from torr to atmospheres and use the ideal gas law. The ability to switch between pressure and concentration is essential since both can be used to express amounts of gases in reactions under the assumption of ideal behavior. The ideal gas law comes into play when determining the amount of gas present at the start and end of the reaction, which is then used in conjunction with the integrated rate law to calculate the duration of the reaction.