Problem 17 - Chemistry and Chemical Reactivity

Question

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is a first-order reaction: $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ 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 \(1.24 \times 10^{-3} \mathrm{mol} / \mathrm{L},\) 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: Identify the Formula

For a first-order reaction, we use the formula for time, \( t \), which is \( t = \frac{1}{k} \ln \left( \frac{[A_0]}{[A]} \right) \), where \([A_0]\) is the initial concentration, \([A]\) is the final concentration, and \(k\) is the rate constant.

Step 2: Substitute Given Values

Substitute the known values into the formula: \( k = 2.8 \times 10^{-3} \text{ min}^{-1} \), \([A_0] = 1.24 \times 10^{-3} \text{ mol/L} \), and \([A] = 0.31 \times 10^{-3} \text{ mol/L} \). The equation becomes \( t = \frac{1}{2.8 \times 10^{-3}} \ln \left( \frac{1.24 \times 10^{-3}}{0.31 \times 10^{-3}} \right) \).

Step 3: Simplify the Expression

Calculate the inside of the logarithm: \( \frac{1.24 \times 10^{-3}}{0.31 \times 10^{-3}} = 4 \). So the expression becomes \( t = \frac{1}{2.8 \times 10^{-3}} \ln 4 \).

Step 4: Calculate the Logarithm

Find the natural logarithm of 4: \( \ln 4 \approx 1.386 \).

Step 5: Compute the Time

Now calculate the value of \( t \): \( t = \frac{1}{2.8 \times 10^{-3}} \times 1.386 \approx 495 \) minutes.

Key Concepts

Decomposition reaction

In a decomposition reaction, a complex chemical compound breaks down into simpler substances. These reactions are typically driven by heat or light. The decomposition of \( \mathrm{SO}_2 \mathrm{Cl}_2 \) is an example, where one compound splits into two products: \( \mathrm{SO}_2 \) and \( \mathrm{Cl}_2 \). This process can be categorized as a first-order reaction because the rate of the reaction is directly proportional to the concentration of the reactant.

In the context of first-order kinetics, the rate at which these reactions occur decreases as the concentration of the reactant decreases. This is crucial for predicting how quickly a compound will decompose under certain conditions, using precise mathematical calculations based on concentration changes over time.

Rate constant

The rate constant, denoted by \( k \), is a fundamental parameter in the study of reaction kinetics. It provides insight into how fast a reaction progresses. For a first-order reaction, the unit of the rate constant is typically expressed in reciprocal time, such as \( \text{min}^{-1} \). In our specific example, the rate constant of the decomposition of \( \mathrm{SO}_2 \mathrm{Cl}_2 \) is \( 2.8 \times 10^{-3} \ \text{min}^{-1} \).

This value indicates how swiftly the reaction occurs over time. It is essential for calculating the time required for the reactant concentration to reach a specific level. The rate constant links the concentration of the decomposing reactant to the time it will take to decompose, allowing precise predictions of reaction timelines.

Natural logarithm

A natural logarithm, often denoted as \( \ln \), is a mathematical function used in various scientific calculations. It is particularly useful in reaction kinetics, where it helps describe the logarithmic relationship between the concentration of reactants and the rate of reaction. For a first-order reaction, the formula \( t = \frac{1}{k} \ln \left( \frac{[A_0]}{[A]} \right) \) explains how the concentration ratio transforms over time.

The use of natural logarithms simplifies the computation of time by converting multiplicative changes in concentration into additive terms, which are easier to manage mathematically. This is crucial in predicting how quickly a reaction progresses, as seen when calculating the time required for the decomposition of \( \mathrm{SO}_2 \mathrm{Cl}_2 \) to a lower concentration.

Concentration change

Concentration change is at the heart of reaction kinetics, particularly in decomposition reactions. It refers to the variation in the amount of reactant over time. Understanding how concentration changes allow us to predict the behavior of reactions. In our case, the concentration of \( \mathrm{SO}_2 \mathrm{Cl}_2 \) decreases from \( 1.24 \times 10^{-3} \ \mathrm{mol}/\mathrm{L} \) to \( 0.31 \times 10^{-3} \ \mathrm{mol}/\mathrm{L} \) as it decomposes.

This process is crucial for applying the first-order rate equation. By measuring the decrease in concentration, we can calculate how long the decomposition takes using the rate constant and natural logarithm. The ability to relate concentration changes to time is a powerful tool for chemists, enabling them to predict and control the reaction under various conditions.