Problem 19 - Chemistry and Chemical Reactivity

Question

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is a first-order reaction: $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ The rate constant for the reaction is \(2.8 \times 10^{-3} \mathrm{min}^{-1}\) at \(600 \mathrm{K} .\) If the initial concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is \(1.24 \times 10^{-3}\) mol/L, how long will it take for the concentration to drop to \(0.31 \times 10^{-3} \mathrm{mol} / \mathrm{L} ?\)

Step-by-Step Solution

Answer

It takes approximately 495 minutes.

Step 1: Understand the Problem

We need to determine the time required for the concentration of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease from \(1.24 \times 10^{-3}\ \text{mol/L}\) to \(0.31 \times 10^{-3}\ \text{mol/L}\) for a first-order reaction with a given rate constant.

Step 2: Recall the First-Order Kinetics Formula

For a first-order reaction, the formula to calculate time \(t\) based on initial concentration \([A]_0\), final concentration \([A]\), and rate constant \(k\) is: \[t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]}\right)\]

Step 3: Identify Given Values

From the problem, we identify \([A]_0 = 1.24 \times 10^{-3}\ \text{mol/L}\), \([A] = 0.31 \times 10^{-3}\ \text{mol/L}\), and \(k = 2.8 \times 10^{-3}\ \text{min}^{-1}\).

Step 4: Calculate the Natural Logarithm Ratio

Calculate the ratio of initial to final concentration:\[ \frac{[A]_0}{[A]} = \frac{1.24 \times 10^{-3}}{0.31 \times 10^{-3}} = 4.0 \]Then find the logarithm:\[ \ln(4.0) \approx 1.386 \]

Step 5: Solve for Time using the Formula

Substitute the values into the formula:\[ t = \frac{1}{2.8 \times 10^{-3}} \times 1.386 \]Calculate \(t\):\[ t \approx 495 \ \text{minutes} \]

Key Concepts

Understanding the Rate Constant in First-Order Reactions

The rate constant, often denoted as \(k\), is a crucial element in understanding the speed of a chemical reaction. For first-order reactions, like the decomposition of \(\text{SO}_2\text{Cl}_2\), it provides the rate at which the concentration of a reactant decreases over time. In this exercise, the rate constant is given as \(2.8 \times 10^{-3}\, \text{min}^{-1}\). This means for every minute, a constant fraction of the original concentration decomposes. The unit \(\text{min}^{-1}\) suggests a reaction progress that is linear with time on a logarithmic scale. Understanding the rate constant allows chemists to predict how quickly a reaction will proceed under specific conditions. Key points to remember include that the rate constant is dependent on temperature, thus changing if the temperature varies.

The Role of the Natural Logarithm in Kinetics

When dealing with first-order reactions, the natural logarithm, \(\ln\), is central to calculating the time required for a change in concentration. This is because the kinetics of first-order reactions can be described using logarithmic relationships. In our case, to find out how long it takes for the concentration to change, we use \(\ln\) to determine the ratio of the initial and final concentrations. The reason \(\ln\) is used here is because it transforms the rapidly changing exponential decay into a linear format which makes manipulating and solving equations more straightforward. In this example, we calculated \( \ln(4.0) \approx 1.386 \), which represents the natural logarithm of the ratio of initial to final concentration, translating a relationship that might seem complex into a simpler arithmetic problem.

Applying the First-Order Kinetics Formula

The core concept to solving problems like this one lies in the first-order kinetics formula. For any first-order reaction, the time \(t\) can be determined using the formula \[ t = \frac{1}{k} \ln \left( \frac{[A]_0}{[A]} \right) \], where \([A]_0\) and \([A]\) are the initial and final concentrations, respectively, and \(k\) is the rate constant. This formula allows us to connect the change in concentration with time directly. To use the formula effectively, ensure that all the concentrations, the rate constant, and time units are consistent. As we've seen in the solution, by substituting \([A]_0 = 1.24 \times 10^{-3} \) mol/L, \([A] = 0.31 \times 10^{-3} \) mol/L, and \(k = 2.8 \times 10^{-3} \ \text{min}^{-1}\), we can solve for \(t\), resulting in a time of approximately 495 minutes. This is a practical application of kinetic theory allowing prediction of reaction timelines.