Problem 45 - Chemistry The Central Science In Si Units

Question

The reaction \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \longrightarrow \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)\) is first order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\). Using the following kinetic data, determine the magnitude and units of the first-order rate constant: \begin{tabular}{cc} \hline Time (s) & Pressure \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{kPa})\) \\ \hline 0 & \(101.3 \mathrm{kPa}\) \\ 2500 & \(95.95 \mathrm{kPa}\) \\ 5000 & \(90.69 \mathrm{kPa}\) \\ 7500 & \(85.92 \mathrm{kPa}\) \\ 10,000 & \(81.36 \mathrm{kPa}\) \end{tabular}

Step-by-Step Solution

Answer

The rate constant is \( 2.19 \times 10^{-5} \ \mathrm{s}^{-1} \).

Step 1: Understand First-Order Kinetics

A first-order reaction means that the rate of the reaction depends linearly on the concentration (or pressure, since this is a gas) of the reactant. The formula for a first-order reaction is \( k = \frac{-1}{t} \ln \left( \frac{[A]}{[A]_0}\right) \), where \( [A]_0 \) is the initial concentration or pressure, \( [A] \) is the concentration or pressure at time \( t \), and \( k \) is the rate constant.

Step 2: Use Initial Reaction Data

From the table, we know the initial pressure \( [SO_2Cl_2]_0 = 101.3 \ \mathrm{kPa} \) at \( t = 0 \).

Step 3: Use Reaction Data at 10,000 seconds

At \( t = 10,000 \) seconds, \( [SO_2Cl_2] = 81.36 \ \mathrm{kPa} \). Insert these values into the first-order rate formula: \[ k = \frac{-1}{10,000} \ln \left( \frac{81.36}{101.3} \right) \].

Step 4: Calculate the Natural Logarithm

First, calculate the fraction: \( \frac{81.36}{101.3} = 0.803 \). Then, calculate the natural logarithm: \( \ln(0.803) = -0.219 \).

Step 5: Compute the Rate Constant

Place the logarithm value in the rate formula: \[ k = \frac{-1}{10,000} \times (-0.219) = 2.19 \times 10^{-5} \].

Step 6: Determine Units for the Rate Constant

Since this is a first-order reaction, the units for \( k \) are inverse seconds \( \mathrm{s}^{-1} \).

Step 7: State Final Result

The magnitude and units of the first-order rate constant \( k \) are \( 2.19 \times 10^{-5} \ \mathrm{s}^{-1} \).

Key Concepts

Rate Constant Calculation

Understanding how to calculate the rate constant is crucial in working with first-order reactions. The rate constant, often symbolized as \( k \), is a critical value that indicates the speed of a chemical reaction. For first-order reactions, the relationship is straightforward: the reaction rate depends linearly on the concentration or, in the case of gaseous reactions, the pressure of the reactant. To compute the rate constant, we use the formula:

• \( k = \frac{-1}{t} \ln \left( \frac{[A]}{[A]_0}\right) \)

Here, \( [A]_0 \) refers to the initial concentration (or pressure), \( [A] \) is the concentration (or pressure) at time \( t \), and \( k \) is the rate constant we wish to calculate.
In our example, at time \( t = 0 \), the initial pressure is 101.3 kPa. At \( t = 10,000 \) seconds, the pressure decreases to 81.36 kPa. Plugging these values into our formula allows us to solve for \( k \). This straightforward formula highlights the essential role of time and concentration (or pressure) in determining the rate at which reactions progress.

Natural Logarithm

The concept of the natural logarithm is crucial for calculations in chemical kinetics, especially in first-order reactions. The natural logarithm is the logarithm to the base \( e \), where \( e \approx 2.718 \). It plays a vital role in transforming a ratio of concentrations or pressures into a linear form, which simplifies the process of solving for the rate constant.
In our calculation, we first find the fraction \( \frac{81.36}{101.3} = 0.803 \). The next step is to find its natural logarithm using the symbol \( \ln \):

• \( \ln(0.803) = -0.219 \)

This transformed value moves into the equation for determining \( k \). The negative sign indicates a decrease in concentration (or pressure) over time, highlighting the diminishing presence of the reactant as the reaction progresses.

Units of Rate Constant

Understanding the units of the rate constant \( k \) is vital for interpreting chemical kinetics accurately. Units provide critical insight into the nature of the reaction. For first-order reactions, the units of the rate constant are inverse seconds \( \mathrm{s}^{-1} \), reflecting how time directly affects the rate of change in concentration (or pressure).
To comprehend why, consider that in the formula \( k = \frac{-1}{t} \ln \left( \frac{[A]}{[A]_0}\right) \), the time \( t \) is in seconds. Thus, the \( k \) is a measure of the rate that reflects how many reaction events occur per second. The inverse characteristic \( \mathrm{s}^{-1} \) clearly indicates its relationship with time, telling us the exponential nature of the reaction decay. This unit is distinct from those of higher-order reactions, emphasizing the linear dependence characteristic of first-order kinetics.