Problem 40 - Chemistry The Molecular Science

Question

The compound \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes in a first-order reaction $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{~g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) $$ that has a half-life of \(1.47 \times 10^{4} \mathrm{~s}\) at \(600 . \mathrm{K}\). If you begin with \(1.6 \times 10^{-3} \mathrm{~mol}\) of pure \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a \(2.0-\mathrm{L}\) flask, calculate at what time the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) will $$ \text { be } 1.2 \times 10^{-4} \mathrm{~mol} \text { . } $$

Step-by-Step Solution

Answer

Approximately 5.499 × 10⁴ s (or 54990 s) is needed for the compound to reach 1.2 × 10⁻⁴ mol.

Step 1: Use Half-Life to Find Rate Constant

For a first-order reaction, the relationship between the half-life (\(t_{1/2}\)) and the rate constant (\(k\)) is given by the formula:\[ t_{1/2} = \frac{0.693}{k} \]. Given \( t_{1/2} = 1.47 \times 10^4 \text{ s} \), we can solve for \( k \):\[ k = \frac{0.693}{1.47 \times 10^4} = 4.71 \times 10^{-5} \text{ s}^{-1} \].

Step 2: Write the First-Order Integrated Rate Equation

For a first-order reaction, the integrated rate equation is:\[ \ln \left( \frac{[A]_t}{[A]_0} \right) = -kt \],where \([A]_t\) is the concentration at time \( t \), \([A]_0\) is the initial concentration, and \( k \) is the rate constant.

Step 3: Calculate Initial Concentration

The initial mole concentration \([A]_0\) in the flask can be found by dividing the initial moles by the volume of the flask. Thus:\[ [A]_0 = \frac{1.6 \times 10^{-3} \text{ mol}}{2.0 \text{ L}} = 8.0 \times 10^{-4} \text{ mol L}^{-1} \].

Step 4: Calculate Subsequent Concentration

Similarly, the concentration at the desired state, \([A]_t\), is:\[ [A]_t = \frac{1.2 \times 10^{-4} \text{ mol}}{2.0 \text{ L}} = 6.0 \times 10^{-5} \text{ mol L}^{-1} \].

Step 5: Substitute Values into Integrated Rate Equation

Using the integrated rate equation:\[ \ln \left( \frac{6.0 \times 10^{-5}}{8.0 \times 10^{-4}} \right) = -4.71 \times 10^{-5} \times t \].

Step 6: Solve for Time \(t\)

First, calculate the natural logarithm:\[ \ln \left( \frac{6.0 \times 10^{-5}}{8.0 \times 10^{-4}} \right) = \ln (0.075) \approx -2.590 \].Next, solve for \( t \):\[ -2.590 = -4.71 \times 10^{-5} \times t \].Divide both sides by \(-4.71 \times 10^{-5}\) to find \( t \):\[ t \approx \frac{2.590}{4.71 \times 10^{-5}} \approx 5.499 \times 10^4 \text{ s} \].

Key Concepts

First-Order Reaction

In chemical kinetics, a **first-order reaction** is a reaction where the rate depends linearly on the concentration of only one reactant. For a first-order reaction, the rate law is expressed as \( ext{Rate} = k[A] \), where \( k \) is the rate constant and \( [A] \) is the concentration of the reactant. This means that the reaction rate will change proportionally if the concentration of \( A \) changes. First-order reactions are quite common in chemical processes.

• Depend only on the concentration of one reactant.
• Doubling the concentration of the reactant doubles the rate.

Understanding this concept is crucial because it forms the basis for determining how quickly a substance will transform over time under given conditions.

Rate Constant

The **rate constant**, denoted as \( k \), is a crucial parameter in the study of reaction kinetics that helps measure the speed of a chemical reaction. It is specific to each reaction and varies with temperature. For a first-order reaction, the units of \( k \) are \( s^{-1} \), indicating that the reaction rate is dependent on time.The rate constant is calculated using the half-life of the reaction with the formula \( t_{1/2} = \frac{0.693}{k} \). By rearranging the formula, you can solve for the rate constant as shown:\[ k = \frac{0.693}{t_{1/2}}\]This formula highlights the inverse relationship between the rate constant and half-life. As the rate constant increases, the half-life decreases, indicating a faster reaction.

Half-Life

The **half-life** of a reaction, represented as \( t_{1/2} \), is the time required for half of the reactant to be consumed. For first-order reactions, the half-life is particularly important because it is constant and does not depend on the initial concentration of the reactant.

• First-order reaction half-life: \( t_{1/2} = \frac{0.693}{k} \).
• It is a fixed value for a given reaction, making it useful for predictions.

In the sample problem, the given half-life is \( 1.47 \times 10^4 \) seconds, which helps calculate the rate constant, demonstrating how the concept of half-life is practically used in computations. Understanding the half-life helps predict when a reactant will decrease to any fractional amount over time.

Integrated Rate Law

The **integrated rate law** for a first-order reaction provides a way to connect the concentration of reactants over time. It is given by the equation:\[\ln \left( \frac{[A]_t}{[A]_0} \right) = -kt\]Here, \([A]_t\) is the concentration of the reactant at time \( t \), \([A]_0\) is the initial concentration, and \( k \) is the rate constant. This equation allows you to determine the concentration at any given time, provided you have the rate constant.Using this equation, and knowing the initial and desired concentrations of \( \text{SO}_2 \text{Cl}_2 \), you can calculate the time it will take for the reaction to reach a particular concentration. This application of integrated rate laws is significant in kinetics because it provides insight into how rapidly reactions proceed and helps in planning experiments and processes.