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Chapter 16: Problem 67

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

Short Answer

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

Step by step solution

01

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.
02

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 \( \text{CH}_3 \text{N}_2 \text{CH}_3 \).
03

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.
04

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 \( k \): \[ k = \frac{2.5 \times 10^{-6}}{0.012} = 2.083 \times 10^{-4} \text{ s}^{-1} \].
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 鈥� 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 鈥� 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.

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Most popular questions from this chapter

Explain the function of the specific rate constant in a rate-law equation.

Explain why a crushed solid reacts with a gas more quickly than a large chunk of the same solid.

Calculate A reaction between \(A\) and \(B\) to form AB is first order in \(A\) and first order in B. The rate constant, \(k,\) equals 0.500 \(\mathrm{mol} /(\mathrm{L} \cdot \mathrm{s}) .\) What is the rate of the reaction when \([\mathrm{A}]=2.00 \times 10^{-2} M\) and \([\mathrm{B}]=1.50 \times 10^{-2} \mathrm{M?}\)

In the method of initial rates used to determine the rate law for a chemical reaction, what is the significance of the word initial?

Visualize the reaction energy diagram for a one-step, endothermic chemical reaction. Compare the heights of the activation energies for the forward and reverse reactions.
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