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 \(2.8 \times\) \(10^{-3} \min ^{-1}\) 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 about 495 minutes for the concentration to drop to the given level.

Step 1: Understanding the Reaction Type

The reaction given is a first-order reaction, which means the rate of the reaction is directly proportional to the concentration of the reactant, \ \( \text{SO}_2\text{Cl}_2 \). First-order reactions follow the kinetics equation: \ \( [A] = [A]_0 e^{-kt} \), where \ \([A]_0\) is the initial concentration, \ \([A]\) is the concentration at time \ \(t\), \ \(k\) is the rate constant, and \ \(t\) is the time.

Step 2: Applying the First-Order Kinetics Formula

To find the time \ \(t\), we use the first-order equation \ \( [A] = [A]_0 e^{-kt} \). For this problem, \ \([A]_0 = 1.24 \times 10^{-3} \text{ mol/L} \), \ \([A] = 0.31 \times 10^{-3} \text{ mol/L} \), and \ \(k = 2.8 \times 10^{-3} \text{ min}^{-1} \). Plug these values into the equation.\ We need to solve for \ \(t\). Rearranging gives, \ \ e^{-kt} = \frac{[A]}{[A]_0}. \

Step 3: Solving for Time

Isolate the exponential term by calculating the ratio: \ \(\frac{[A]}{[A]_0} = \frac{0.31 \times 10^{-3}}{1.24 \times 10^{-3}} = 0.25\). Therefore, \ \(e^{-kt} = 0.25\). \ Take the natural logarithm (ln) on both sides to solve for \ \(-kt\): \ \ -kt = \ln(0.25).\ Next, solve for \ \(t\) by substituting the known value of \ \(k\) and simplifying: \ \ t = -\frac{\ln(0.25)}{k} = -\frac{\ln(0.25)}{2.8 \times 10^{-3}}. \

Step 4: Calculating the Time Required

Evaluate \ \(-\ln(0.25) = 1.3863\). Now divide by the rate constant to find \ \(t\): \ \ t = \frac{1.3863}{2.8 \times 10^{-3}} = 495.11 \ So, it will take approximately 495 minutes for the concentration to decrease from \ 1.24 \times 10^{-3} \ to \ 0.31 \times 10^{-3} \ mol/L.

Key Concepts

rate constant

In the study of chemical kinetics, the rate constant is a crucial factor that determines the speed at which a reaction occurs. Specifically, for a first-order reaction, the rate constant, denoted as \(k\), provides a direct relationship between the concentration of the reactant and the rate of the reaction. This is presented in the equation:

• \([A] = [A]_0 e^{-kt}\)

where \([A]_0\) is the initial concentration and \([A]\) is the concentration at time \(t\). The rate constant has unique dimensions that reflect the order of the reaction. For first-order reactions, it's expressed in terms of time, such as \(\text{min}^{-1}\).

As you can see, in this reaction, the rate constant is \(2.8 \times 10^{-3} \text{ min}^{-1}\). This means that as time progresses, the concentration of \(\mathrm{SO}_2 \mathrm{Cl}_2\) decreases at a rate influenced by this constant. Understanding the rate constant is key to predicting how long the reaction will take under specified conditions.

exponential decay

In first-order reactions, the concentration of a reactant decreases over time in a process known as exponential decay. This term describes how the concentration of \(\mathrm{SO}_2 \mathrm{Cl}_2\) decreases exponentially rather than linearly.

The governing formula is:

• \([A] = [A]_0 e^{-kt}\)

The exponential function \(e^{-kt}\) captures how the reactant's concentration decreases by a specific factor over equal time intervals. This decay reflects the natural diminishing of the reactant as the reaction proceeds.

Every time interval sees the concentration reduce by approximately the same percentage, not the same amount, which is why we call it exponential. Understanding this concept is vital for engineers and scientists as they can predict how much of a reactant will remain after a given time.

natural logarithm

The natural logarithm, abbreviated as \(\ln\), is an important mathematical function used in handling exponential equations, specifically when solving for time in first-order reaction kinetics.

When you rearrange the first-order kinetics equation to solve for \(t\):

• \(e^{-kt} = \frac{[A]}{[A]_0}\)

Taking the natural logarithm of both sides allows us to solve for \(-kt\):

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

In our example, \(\ln(0.25)\) helped determine how long it takes for the reactant's concentration to fall to a specified level. Using \(\ln\) allows the conversion of an exponential decay equation into a linear form, making calculations straightforward.

reaction kinetics

Reaction kinetics is the branch of chemistry that focuses on the rates of chemical reactions and the factors affecting those rates. For the decomposition of \(\mathrm{SO}_2 \mathrm{Cl}_2\), understanding the kinetics involves calculating how reactant concentrations change over time.
Several critical aspects influence reaction kinetics:

• Concentration of reactants
• Temperature
• Presence of a catalyst
• Specific rate constant for the reaction

In the given problem, the initial and final concentrations are essential to determine how long the reaction takes to proceed. Also, using kinetic principles, such as applying the relevant formulas and understanding the reaction type (first-order), allows for precise calculations. Reaction kinetics not only helps comprehend how reactions occur but also facilitates the prediction and control of reaction conditions in industrial and laboratory settings.