Problem 67 - Chemistry Matter and Change

Question

Use the data in Table 16.4 to calculate the value of the specific rate constant, \(k\). $$ \begin{array}{|c|c|c|} \hline \begin{array}{c} \text { Experiment } \\ \text { Number } \end{array} & \begin{array}{c} \text { Initial } \\ {\left[\mathrm{CH}_{3} \mathrm{~N}_{2} \mathrm{CH}_{3}\right]} \end{array} & \begin{array}{c} \text { Initial } \\ \text { Reaction Rate } \end{array} \\ \hline 1 & 0.012 M & 2.5 \times 10^{-6} \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s}) \\ \hline 2 & 0.024 M & 5.0 \times 10^{-6} \mathrm{~mol} /(\mathrm{L} \cdot \mathrm{s}) \\ \hline \end{array} $$

Step-by-Step Solution

Answer

The specific rate constant, \( k \), is \( 2.083 \times 10^{-4} \text{ s}^{-1} \).

Step 1: Analyze the Data

The table provides data from two experiments, showing the initial concentration of \( \text{CH}_3 \text{N}_2 \text{CH}_3 \) and the initial reaction rate for each experiment. The initial concentrations are 0.012 M and 0.024 M, with corresponding reaction rates of \( 2.5 \times 10^{-6} \text{ mol/(L} \cdot \text{s)} \) and \( 5.0 \times 10^{-6} \text{ mol/(L} \cdot \text{s)} \) respectively.

Step 2: Write the Rate Law

For a simple reaction where the concentration of one reactant is changing, the rate law can be written as: \[ \text{Rate} = k [\text{CH}_3 \text{N}_2 \text{CH}_3]^n \] where \( k \) is the rate constant and \( n \) is the order of the reaction with respect to \( \text{CH}_3 \text{N}_2 \text{CH}_3 \).

Step 3: Determine the Reaction Order

To find \( n \), we compare the two experiments. When the concentration doubles from 0.012 M to 0.024 M, the rate also doubles from \( 2.5 \times 10^{-6} \) to \( 5.0 \times 10^{-6} \). This suggests that the reaction is first order with respect to \( \text{CH}_3 \text{N}_2 \text{CH}_3 \) since the rate change corresponds directly to the change in concentration.

Step 4: Solve for the Rate Constant \( k \)

Using experiment 1, substitute the values into the rate law. For a first-order reaction, it becomes: \[ 2.5 \times 10^{-6} = k (0.012) \] Solve for \( k \): \[ k = \frac{2.5 \times 10^{-6}}{0.012} = 2.083 \times 10^{-4} \text{ s}^{-1} \].

Step 5: Verify Consistency

Use the second experiment to confirm \( k \): \[ 5.0 \times 10^{-6} = k (0.024) \] Solving gives \( k = \frac{5.0 \times 10^{-6}}{0.024} = 2.083 \times 10^{-4} \text{ s}^{-1} \), consistent with the calculation from experiment 1.

Key Concepts

Rate Law

To understand the rate law, let's first look at what it describes. A rate law is an equation that links the reaction rate to the concentration of its reactants. In simple terms, it shows how fast a reaction happens based on the concentrations of the reactants involved. Here's how it works:

Consider a reaction with a single reactant:

• The general form of the rate law is \(\text{Rate} = k[\text{Reactant}]^n\), where:
• \(k\) is the rate constant — a value that indicates the speed of the reaction at a given temperature.
• [Reactant] is the concentration of the reactant.
• \(n\) is the order of the reaction concerning that specific reactant.

In our example with \(\text{CH}_3\text{N}_2\text{CH}_3\), the reaction rate changes with the concentration of the reactant, and we express it through the rate law formula.

Reaction Order

The reaction order is a vital aspect of reaction kinetics. It tells us how the concentration of a reactant affects the rate of the reaction. Determining the reaction order can sometimes be straightforward, as exemplified below.

To find the reaction order with respect to a given reactant:

• Compare how the rate changes when the concentration varies, keeping all other conditions constant.
• If doubling the concentration doubles the rate, the reaction is first order with respect to that reactant. Mathematically, this appears in the exponent — here, we observe \(n = 1\).

In the problem, the concentration of \(\text{CH}_3\text{N}_2\text{CH}_3\) is doubled from 0.012 M to 0.024 M, and the reaction rate doubles. This simple observation confirms that the reaction is first order in \(\text{CH}_3\text{N}_2\text{CH}_3\). First-order kinetics like this mean the reaction rate is directly proportional to the concentration of \(\text{CH}_3\text{N}_2\text{CH}_3\).

Rate Constant Calculation

Calculating the rate constant is an essential step in analyzing reaction kinetics. The rate constant, denoted as \(k\), provides insight into the rate of a reaction independent of the concentration of reactants. Here's how you do it:

Once the rate law is established and the reaction order is known, plug the values for the rate and concentration into the rate law equation:

• For the first experiment: \[2.5 \times 10^{-6} = k \times (0.012)^{1}\]
• Solve for \(k\): \[k = \frac{2.5 \times 10^{-6}}{0.012} = 2.083 \times 10^{-4} \, \text{s}^{-1}\]

To ensure accuracy, validate \(k\) using the second experiment:

• Use the equation: \[5.0 \times 10^{-6} = k \times (0.024)^{1}\]
• Solving for \(k\) gives the same result: \[k = \frac{5.0 \times 10^{-6}}{0.024} = 2.083 \times 10^{-4} \, \text{s}^{-1}\]

Consistency across experiments indicates that \(k\) is accurately calculated, providing a cornerstone measurement for predicting reaction rates under similar conditions.