Problem 15 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

The addition of bromine to propene is an elementary reaction with a rate constant, \(k\) \\[ \mathrm{CH}_{2}=\mathrm{CHCH}_{3}+\mathrm{Br}_{2} \rightarrow \mathrm{CH}_{2} \mathrm{BrCHBrCH}_{3} \\] Kinetic studies were carried out at \(298 \mathrm{K}\) using excess \(\mathrm{Br}_{2}\). For \(\left[\mathrm{Br}_{2}\right]_{0}=0.20 \mathrm{mol} \mathrm{dm}^{-3},\) the pseudo-first order rate constant, \(k^{\prime}\) for the reaction was found to be \(900 \mathrm{s}^{-1}\). What is the value of \(k\) at \(298 \mathrm{K} ? \text { (Section } 9.4)\)

Step-by-Step Solution

Answer

The value of the second order rate constant, \( k \), is \( 4500 \, \text{dm}^3 \text{mol}^{-1} \text{s}^{-1} \).

Step 1: Understanding the Given Data

We are given the elementary reaction between propene and bromine, and the condition of excess bromine, allowing us to treat the reaction as pseudo-first order. The initial concentration of bromine is \( [\mathrm{Br}_2]_0 = 0.20 \, \text{mol dm}^{-3} \), and the pseudo-first order rate constant is \( k' = 900 \, \text{s}^{-1} \).

Step 2: Relating Pseudo-first Order to Second Order

In a pseudo-first order reaction, the rate constant \( k' \) is related to the true second order rate constant \( k \) and the excess reagent concentration. The relation is given by \( k' = k[\text{Br}_2]_0 \). This simplifies the rate expression under the condition of excess bromine.

Step 3: Solving for the True Rate Constant

Use the relation \( k' = k[\text{Br}_2]_0 \) to solve for \( k \). Substitute \( k' = 900 \, \text{s}^{-1} \) and \( [\mathrm{Br}_2]_0 = 0.20 \, \text{mol dm}^{-3} \) into the equation. This gives \( k = \frac{k'}{[\mathrm{Br}_2]_0} = \frac{900}{0.20} \).

Step 4: Calculating the Value of k

Calculate \( k \) from the equation \( k = \frac{900}{0.20} = 4500 \, \text{dm}^3 \text{mol}^{-1} \text{s}^{-1} \). This is the second order rate constant at \( 298 \, \text{K} \).

Key Concepts

Rate Constant

When examining chemical reactions, the rate constant, denoted as \( k \), is a crucial factor. It provides insight into how quickly a reaction proceeds under specific conditions. The rate constant is influential in understanding the time scale of reactions:

• A larger value of \( k \) indicates a faster reaction.
• A smaller value suggests a slower reaction.

The unit of the rate constant varies depending on the order of the reaction. In second order reactions, for example, as discussed in the original problem, \( k \) is typically expressed in \( \, \text{dm}^3 \, \text{mol}^{-1} \, \text{s}^{-1} \). This reflects the dependence of the reaction rate on the concentrations of two reactants. Recognizing the unit of \( k \) can help identify the reaction order.

Kinetic Studies

To understand the dynamics of chemical reactions, kinetic studies are essential. These studies involve experimenting and observing rates of reactions under varied conditions such as temperature, reactant concentrations, and pressure. By conducting kinetic studies, we can determine:

• The rate law of a reaction, which expresses the rate of a reaction in terms of the concentration of reactants.
• The activation energy, which is the minimum energy required for a reaction to occur.
• Reaction mechanisms, offering insight into the step-by-step sequence by which reactants turn into products.

Employing excess of one reactant (as done with bromine in the provided exercise) is a common tactic to simplify analysis, allowing researchers to focus on a single reactant's effect on the reaction rate.

Elementary Reaction

An elementary reaction is a fundamental step in a reaction mechanism that describes a single transformation where reactants directly convert to products. These reactions are simple enough to write a straightforward rate law. For an elementary reaction, the rate law can be directly written from the molecularity, which refers to the number of molecules coming together to react in the step.
Directly related to the original exercise, the reaction between propene and bromine is described as an elementary reaction. In such cases, the stoichiometric coefficients from the balanced equation indicate the order of the reaction with respect to each reactant. This simplicity allows for clear and accurate determination of the reaction order, crucial for calculating accurate rate constants.

Second Order Reaction

A second order reaction is characterized by the rate being proportional to the product of two reactant concentrations or the square of a single reactant concentration. In the given problem, the second order concerns the interplay between bromine and propene. The rate law for a second order reaction can be written as:\[\text{Rate} = k [A][B]\]where \( [A] \) and \( [B] \) are concentrations of the reactants. In the context of excess bromine, the reaction simplifies. It manifests as a pseudo-first order reaction, where the concentration of bromine is almost constant throughout the experiment. This adjustment simplifies kinetic analysis, and you're able to uncover the genuine second order rate constant using the relation between the pseudo-first order rate constant \( k' \) and the realistic second order rate constant \( k \) through known concentrations of the reactants. Understanding this concept is key in interpreting and analyzing kinetic data for reactions exhibiting multiple orders.