Question

Consider the reaction: $$ \mathrm{A}+\mathrm{B} \longrightarrow \text { products } $$ From the following data obtained a a certain temperature, determine the order of the reaction and calculate the rate constant. $$ \begin{array}{ccc} {[\mathbf{A}](\boldsymbol{M})} & {[\mathbf{B}](\boldsymbol{M})} & \text { Rate }(\boldsymbol{M} / \mathbf{s}) \\ \hline 1.50 & 1.50 & 3.20 \times 10^{-1} \\ 1.50 & 2.50 & 3.20 \times 10^{-1} \\ 3.00 & 1.50 & 6.40 \times 10^{-1} \end{array} $$

Short answer

The reaction is first order with respect to A and zeroth order with respect to B; rate constant \( k = 2.13 \times 10^{-1} \text{ s}^{-1} \).

Step-by-step solution

Understand the Reaction and Initial Data

The reaction given is \( A + B \rightarrow ext{products} \). The task is to determine the order of the reaction with respect to both \( A \) and \( B \) using the provided experimental data, and then calculate the rate constant of the reaction.

Analyze Experiment 1 and 2 - Determine Order with Respect to B

Compare experiments 1 and 2, where \([A]\) is constant at 1.50 M. The rate does not change despite increasing \([B]\) from 1.50 M to 2.50 M. This indicates that the rate does not depend on \([B]\), meaning the reaction is zeroth order with respect to \( B \).

Analyze Experiment 1 and 3 - Determine Order with Respect to A

Compare experiments 1 and 3, where \([B]\) is constant at 1.50 M. Doubling \([A]\) from 1.50 M to 3.00 M doubles the rate from \(3.20 \times 10^{-1} \text{ M/s}\) to \(6.40 \times 10^{-1} \text{ M/s}\). This indicates a direct proportionality, meaning the reaction is first order with respect to \( A \).

Write the Rate Law

Since the reaction is zeroth order with respect to \( B \) and first order with respect to \( A \), the rate law is: \[ ext{Rate} = k[A]^1[B]^0 = k[A] \]

Calculate the Rate Constant (k)

Using the rate law from Step 4 and any of the provided data points: - Choosing experiment 1 (\([A] = 1.50 \text{ M}, \text{Rate} = 3.20 \times 10^{-1} \text{ M/s}\)):\[ 3.20 \times 10^{-1} = k(1.50) \]Solve for \( k \): \[ k = \frac{3.20 \times 10^{-1}}{1.50} = 2.13 \times 10^{-1} \text{ s}^{-1} \]

Conclusion

The reaction is first order with respect to \( A \) and zeroth order with respect to \( B \). The rate constant is calculated to be \( 2.13 \times 10^{-1} \text{ s}^{-1} \).

Key concepts

Rate Constant

The rate constant, denoted as \( k \), is a proportionality factor in the rate law equation that links the rate of a chemical reaction to the concentrations of reactants. It is unique for each reaction and depends on the reaction conditions, particularly temperature. The units of the rate constant fluctuate depending on the overall order of the reaction, ensuring that the rate equation maintains consistent units.

• In a first-order reaction, like the one provided, the rate constant has units of \( ext{s}^{-1} \).
• It tells you how fast the reaction proceeds when all concentrations are at 1 M.
• An essential factor for determining reaction speed and comparison of different reactions.

The calculated rate constant, from the given problem, is \( 2.13 \times 10^{-1} \, \text{s}^{-1} \). This confirms how quickly the reaction occurs per second under the given conditions. By understanding the rate constant, you can predict how changing reactant concentrations may affect the overall reaction rate.

Kinetics

Kinetics is the study of the speed, or rate, of chemical reactions and the factors affecting them. It allows chemists to understand how different conditions alter reaction pathways.

• Important parameters include concentration, temperature, catalysts, and surface area.
• Studying kinetics helps in determining the mechanism of a reaction.
• Analyzing reaction data, as shown in the exercise, enables us to deduce reaction orders.

In the provided exercise, kinetics is used to determine the effect of reactant concentrations \([A]\) and \([B]\) on the rate. The experimental data shows that while \([A]\) affects the rate, \([B]\) does not, aiding in determining the reaction order with respect to each reactant. Understanding kinetics is crucial for optimizing chemical processes in industrial applications.

Rate Law

The rate law is an equation that summarises the relationship between the concentration of reactants and the rate of a reaction. It is expressed in the form: \[\text{Rate} = k[A]^m[B]^n\]where \( m \) and \( n \) are the orders of reaction with respect to each reactant. They reflect the degree to which the concentration of each reactant affects the rate.

• A rate law must be determined experimentally and cannot be inferred from the balanced equation alone.
• The rate law provides a clear representation of a reaction's kinetics.
• For complex reactions, the rate law can include more terms accounting for additional reactants or steps.

In your exercise, the rate law was derived to be \( \text{Rate} = k[A] \), indicating that the reaction is first order in \( A \) and zeroth order in \( B \). This means the rate is directly proportional to \([A]\), while \([B]\) does not influence the rate. Mastery of rate laws allows one to predict the effect of varying concentrations on the reaction speed.