Problem 26 - Chemistry and Chemical Reactivity

Question

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$\mathrm{sO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})$$ is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2},\) and the reaction has a half-life of 245 minutes at \(600 \mathrm{K}\). If you begin with \(3.6 \times 10^{-3}\) mol of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a 1.0 -L. flask, how long will it take for the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to decrease to \(2.00 \times 10^{-4}\) mol?

Step-by-Step Solution

Answer

It takes approximately 1021 minutes.

Step 1: Understanding First-Order Kinetics

For a first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant. The equation that describes a first-order reaction is the integrated rate law: \[[A] = [A]_0 e^{-kt}\]where \([A]\) is the concentration at time \(t\), \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is time.

Step 2: Calculate the Rate Constant

We know the reaction is first order and the half-life \(t_{1/2}\) for a first-order reaction is:\[t_{1/2} = \frac{0.693}{k}\]Given \(t_{1/2} = 245\) minutes, we can solve for the rate constant \(k\):\[k = \frac{0.693}{245} \approx 0.00283\, ext{min}^{-1}\]

Step 3: Apply the Integrated Rate Law

Using the integrated rate law equation, we can solve for \(t\). The initial concentration \([A]_0\) is \(3.6 \times 10^{-3}\, ext{mol/L}\), and \([A]\) is \(2.00 \times 10^{-4}\, ext{mol/L}\):\[2.00 \times 10^{-4} = 3.6 \times 10^{-3} \times e^{-0.00283t}\]To isolate \(e^{-0.00283t}\), divide both sides by \(3.6 \times 10^{-3}\):\[e^{-0.00283t} = \frac{2.00 \times 10^{-4}}{3.6 \times 10^{-3}} \approx 0.0556\]

Step 4: Solve for Time \(t\)

To solve for \(t\), take the natural logarithm of both sides:\[-0.00283t = \ln(0.0556)\]Calculate \(\ln(0.0556)\), which is approximately \(-2.89\). Then:\[-0.00283t = -2.89\]Solving for \(t\):\[t = \frac{-2.89}{-0.00283} \approx 1021\, ext{minutes}\]

Step 5: Final Calculation and Result

Therefore, it will take approximately 1021 minutes for the concentration of \( ext{SO}_2 ext{Cl}_2\) to decrease from \(3.6 \times 10^{-3}\, ext{mol/L}\) to \(2.00 \times 10^{-4}\, ext{mol/L}\). This is the time required for this first-order reaction under the given conditions.

Key Concepts

Half-Life Calculation

In first-order reaction kinetics, the concept of half-life is a key element, which pertains to the time required for the concentration of a reactant to decrease to half of its initial value. This is especially important because, for first-order reactions, the half-life is constant and does not depend on the initial concentration of the reactant. This characteristic allows chemists to predict how quickly a reaction proceeds based on the half-life value.

For a first-order reaction, you can calculate the half-life (\(t_{1/2}\)) using the equation:

• \(t_{1/2} = \frac{0.693}{k}\)

Here, \(k\) is the rate constant. The constant value 0.693 comes from the natural logarithm of 2. In the example problem, where the half-life of SO\(_2\)Cl\(_2\) is given as 245 minutes, this formula can be used to find the rate constant. By rearranging the formula, you solve for \(k\):

• \(k = \frac{0.693}{245}\)

Integrated Rate Law

The integrated rate law for a first-order reaction elegantly describes how the concentration of a reactant decreases over time. This formula is crucial for determining the amount of reactant remaining after a given period has elapsed. By integrating the differential rate law for first-order reactions, we arrive at the formula:

• \([A] = [A]_0 e^{-kt}\)

In this equation, \([A]_0\) represents the initial concentration, while \([A]\) is the concentration at time \(t\). The term \(e^{-kt}\) reflects the exponential decay nature of first-order reactions. With the problem's given values, the initial concentration of \(3.6 \times 10^{-3}\) mol/L eventually reduces to \(2.00 \times 10^{-4}\) mol/L.

This formula allows for flexible calculations depending on the unknown variable. Whether solving for final concentration, initial concentration, rate constant, or time elapsed, rearranging the equation reveals any necessary parameter with precision. The ease of applying this equation is a significant reason why integrated rate laws are fundamental in chemical kinetics.

Rate Constant

The rate constant, symbolized as \(k\), is an intrinsic factor of a chemical reaction signifying how fast the reaction occurs. Importantly, it functions independently of the reactant concentrations, specifically in first-order reactions. Instead, its value depends on the nature of the reaction and the conditions under which it takes place, such as temperature and the presence of a catalyst.

The units of the rate constant in first-order kinetics are s\(^{-1}\) or min\(^{-1}\), which illustrate that the reaction rate is proportional to the concentration of a single reactant. To find \(k\), you can rearrange the half-life equation for a first-order reaction. With a known half-life, like in our example (245 minutes), we use:

• \(k = \frac{0.693}{t_{1/2}}\)

This provides the rate constant as approximately 0.00283 min\(^{-1}\). Understanding \(k\) enables predictions about how fast the concentration of reactants decreases and is vital for comparing reaction speeds between different reactions or conditions.