Problem 20 - Chemistry and Chemical Reactivity

Question

The conversion of cyclopropane to propene, described in Example \(15.5,\) occurs with a first-order rate constant of \(5.4 \times 10^{-2} \mathrm{h}^{-1} .\) How long will it take for the concentration of cyclopropane to decrease from an initial concentration \(0.080 \mathrm{mol} / \mathrm{L}\) to \(0.020 \mathrm{mol} / \mathrm{L} ?\)

Step-by-Step Solution

Answer

Approximately 25.67 hours.

Step 1: Understanding First-Order Reactions

First-order reactions have a rate that is directly proportional to the concentration of one reactant. The rate law can be expressed as \( \frac{d[A]}{dt} = -k[A] \). For first-order reactions, the integrated rate law is \( \ln[A]_t = -kt + \ln[A]_0 \), where \([A]_t\) is the concentration at time \(t\), \( [A]_0 \) is the initial concentration, and \( k \) is the rate constant.

Step 2: Inserting Given Values

In this problem, the initial concentration \([A]_0\) is 0.080 mol/L, the final concentration \([A]_t\) is 0.020 mol/L, and the rate constant \(k\) is \(5.4 \times 10^{-2} \text{ h}^{-1} \). We need to find the time \( t \) it takes for the reaction to reach \([A]_t\).

Step 3: Rearranging the Integrated Rate Equation

Using the integrated rate law \( \ln[A]_t = -kt + \ln[A]_0 \), rearrange to solve for \( t \):\(t = \frac{\ln[A]_0 - \ln[A]_t}{k}.\)

Step 4: Solving for Time

Substitute the given values into the rearranged equation:\[t = \frac{\ln(0.080) - \ln(0.020)}{5.4 \times 10^{-2}} = \frac{\ln\left(\frac{0.080}{0.020}\right)}{5.4 \times 10^{-2}} = \frac{\ln(4)}{5.4 \times 10^{-2}}.\]Calculate \( \ln(4) \approx 1.3863 \), then divide by the rate constant:\[ t \approx \frac{1.3863}{5.4 \times 10^{-2}} \approx 25.67 \text{ hours}. \]

Step 5: Conclusion

The time required for the concentration of cyclopropane to decrease from 0.080 mol/L to 0.020 mol/L is approximately 25.67 hours.

Key Concepts

Rate constant

The rate constant, often denoted as \(k\), is a crucial aspect of chemical kinetics. It is a fixed value that quantifies the speed of a reaction under specific conditions. For first-order reactions, the rate constant has units of inverse time, such as \(\text{h}^{-1}\). In our example, the rate constant \(k\) is \(5.4 \times 10^{-2} \text{ h}^{-1}\). This means that the transformation of cyclopropane to propene happens at this steady pace when measured against time.

Determining the rate constant is essential for predicting how fast reactions will proceed. It also helps chemists understand whether a reaction is feasible within certain timeframes.

• The higher the rate constant, the faster the reaction occurs.
• Environmental factors like temperature and pressure could influence \(k\).

Grasping the concept of the rate constant allows students to foresee reaction duration, which is pivotal in practical applications, such as chemical manufacturing and pharmacokinetics.

Integrated rate law

The integrated rate law for first-order reactions connects concentration with time, providing a straightforward method to calculate how reactant concentrations change over time. This equation is expressed as \(\ln[A]_t = -kt + \ln[A]_0\). Here, \([A]_t\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, and \(k\) is the rate constant.

Using this form of the rate law, we can determine how long it takes for a reaction to reach a certain concentration. In our example, by rearranging the integrated rate law, we calculate time \(t\) using the formula \(t = \frac{\ln[A]_0 - \ln[A]_t}{k}\).

• It helps convert a nonlinear concentration-time relationship into a linear equation.
• This simplification is especially user-friendly when graphing the data.

The integrated rate law is a valuable tool in kinetics, making it possible to predict how a system evolves over time.

Concentration changes

Understanding changes in concentration is fundamental to analyzing reaction progress. In a first-order reaction, the rate depends on the concentration of one reactant. This means that as the reactant is consumed, its concentration decreases exponentially over time.

In our scenario, cyclopropane's concentration drops from \(0.080 \text{ mol/L}\) to \(0.020 \text{ mol/L}\). The integrated rate law equation gives us the means to calculate the time it takes for this change.

• The process is exponentially proportional due to the nature of the logarithmic relationship.
• The rate of concentration change decreases as the reactant is consumed.

By closely examining concentration changes, students can better predict how long a reaction will reach a desired endpoint, facilitating efficient planning and resource allocation in chemical processes.

Kinetics calculations

Kinetics calculations involve determining how variables such as concentration and time impact the speed of a chemical reaction. These calculations are crucial for understanding the reaction mechanisms and predicting the behavior of chemical processes in real-world scenarios.

To solve our example problem, we used kinetics calculations to find the time needed for cyclopropane's concentration to reach a certain level. By applying the rearranged integrated rate law, \(t = \frac{\ln(0.080) - \ln(0.020)}{5.4 \times 10^{-2}}\), we derived that the reaction time is approximately 25.67 hours.

• The application of kinetics calculations can help in quality control and optimization in industries.
• Mastering these calculations can aid in experimental design and hypothesis testing.

Through kinetics calculations, students gain analytical skills necessary for various fields, enabling them to tackle complex chemical problems with confidence.