Problem 8 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

From the data provided in the table, deduce the rate equation and the value of the rate constant for the following reaction. (Section \(9.4)\) \\[ \begin{aligned} \mathrm{CH}_{3} \mathrm{COCH}_{3}(\mathrm{aq})+\mathrm{Br}_{2}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \rightarrow & \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{Br}(\mathrm{aq}) \\ &+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Br}(\mathrm{aq}) \end{aligned} \\] $$\begin{array}{llll} \hline & \text { Initial } & \text { Initial } & \text { Initial } & \text { Initial } \\ & \text { concentration } & \text { concentration } & \text { concentration } & \text { rate of } \\ & \begin{array}{l} \text { of } \mathrm{CH}_{3} \mathrm{COCH}_{3} \\ \text { /moldm }^{-3} \end{array} & \begin{array}{l} \text { of } \mathrm{Br}_{2} \\ \text { /moldm }^{-3} \end{array} & \begin{array}{l} \text { of } \mathrm{H}^{+} \\ \text {/moldm }^{-3} \end{array} & \begin{array}{l} \text { reaction } \\ \text { /moldm }^{-3} \mathrm{s}^{-1} \end{array} \\ \hline 1 & 1.00 & 1.00 & 1.00 & 4.0 \times 10^{-3} \\ 2 & 2.00 & 1.00 & 1.00 & 8.0 \times 10^{3} \\ 3 & 2.00 & 2.00 & 1.00 & 8.0 \times 10^{3} \\ 4 & 1.00 & 1.00 & 2.00 & 8.0 \times 10^{-3} \end{array}$$

Step-by-Step Solution

Answer

Rate law: \( \text{Rate} = k[\text{CH}_3\text{COCH}_3][\text{H}^+] \) with \( k = 4.0 \times 10^{-3} \, \text{mol}^{-1}\text{dm}^3\text{s}^{-1} \). Reaction is first order with respect to \([\text{CH}_3\text{COCH}_3]\) and \([\text{H}^+]\), and zero order with respect to \([\text{Br}_2]\).

Step 1: Identify Variables and Rate Equation

We need to write the rate expression for the reaction \( \text{Rate} = k[\text{CH}_3\text{COCH}_3]^a[\text{Br}_2]^b[\text{H}^+]^c \), where \( k \) is the rate constant, and \( a, b, c \) are the orders of the reaction with respect to acetone, bromine, and hydrogen ions, respectively.

Step 2: Determine the Order with Respect to CH3COCH3

Compare experiments 1 and 2: Doubling the concentration of \([\text{CH}_3\text{COCH}_3]\) while keeping \([\text{Br}_2]\) and \([\text{H}^+]\) constant causes the rate to double. This implies that the reaction is first order with respect to \([\text{CH}_3\text{COCH}_3]\) (i.e., \( a = 1 \)).

Step 3: Determine the Order with Respect to Br2

Compare experiments 2 and 3: Doubling \([\text{Br}_2]\) while keeping \([\text{CH}_3\text{COCH}_3]\) and \([\text{H}^+]\) constant does not change the rate, indicating the reaction is zero order with respect to \([\text{Br}_2]\) (i.e., \( b = 0 \)).

Step 4: Determine the Order with Respect to H+

Compare experiments 1 and 4: Doubling \([\text{H}^+]\) while keeping \([\text{CH}_3\text{COCH}_3]\) and \([\text{Br}_2]\) constant causes the rate to double. This indicates the reaction is first order with respect to \([\text{H}^+]\) (i.e., \( c = 1 \)).

Step 5: Write the Rate Law

Based on our findings, the rate law is \( \text{Rate} = k[\text{CH}_3\text{COCH}_3]^1[\text{Br}_2]^0[\text{H}^+]^1 = k[\text{CH}_3\text{COCH}_3][\text{H}^+] \).

Step 6: Calculate the Rate Constant k

Using experiment 1, where \([\text{CH}_3\text{COCH}_3] = 1.00 \, \text{mol/dm}^3\), \([\text{H}^+] = 1.00 \, \text{mol/dm}^3\), and rate = \( 4.0 \times 10^{-3} \, \text{mol/dm}^3\text{s}^{-1} \), plug these into the rate law to solve for \( k \): \( 4.0 \times 10^{-3} = k(1.00)(1.00) \Rightarrow k = 4.0 \times 10^{-3} \, \text{mol}^{-1}\text{dm}^3\text{s}^{-1} \).

Key Concepts

Reaction Order

The term "reaction order" refers to the power to which the concentration of a reactant is raised in the rate equation. It helps us understand how the concentration of a reactant affects the rate of reaction.
Each reactant in a chemical reaction can have its own order, usually determined experimentally. The overall order of a reaction is the sum of these individual orders.
For instance, in the given reaction involving acetone, bromine, and hydrogen ions,

• it was found that the rate doubles when the concentration of acetone is doubled, without changing the others. This behavior indicates a first-order reaction concerning acetone.
• Similarly, a change in the concentration of bromine does not alter the rate, indicating a zero-order reaction concerning bromine.
• Finally, doubling the concentration of hydrogen ions also doubles the rate, showing a first-order reaction here as well.

By determining the order for each reactant, we can comprehend how changes in concentration can control the speed of a reaction.

Rate Constant

The rate constant, represented by the symbol \( k \), is a proportionality constant in the rate equation. It is crucial because it ties the reaction rate to the concentrations of reactants raised to their respective orders.
The value of the rate constant provides insights into the speed of a reaction under specific conditions.

• For example, in the provided exercise, using the established rate equation, we determined \( k = 4.0 \times 10^{-3} \text{ mol}^{-1}\text{dm}^3\text{s}^{-1} \).
• The units of \( k \) can vary based on the reaction's total order. Thus, analyzing \( k \)'s units becomes crucial for understanding the reaction characteristics.

It is important to note that the rate constant is not affected by changes in concentration, but it can be influenced by factors like temperature. Interpreting its value helps us understand the intrinsic reactivity and efficiency of a reaction.

Chemical Kinetics

Chemical kinetics is the branch of chemistry that explores the speed of chemical reactions and the factors influencing this speed. It examines how fast reactants transform into products.
In the context of our reaction, chemical kinetics helps identify the reaction mechanism and the steps involved by studying how quickly reactants decrease or products increase over time.
Important areas of focus in chemical kinetics include:

• Determining how reactant concentrations influence reaction rates, as observed through reaction orders and the rate constant.
• Investigating how variables like temperature and catalysts affect rates and mechanisms.
• Understanding the time evolution of reactions, enabling chemists to optimize conditions for desired rates.

By utilizing principles of chemical kinetics, we can not only predict and control reaction rates but also refine the synthesis processes and design efficient catalytic systems.