91Ó°ÊÓ

The following data pertains to the reaction between A and \(B\)\begin{tabular}{llll} \hline S. & {\([\mathrm{A}]\)} & {\([\mathrm{B}]\)} & Rate \\ No. & \(\mathrm{mol} \mathrm{L}^{-1}\) & \(\mathrm{~mol} \mathrm{~L}^{-1}\) & \(\mathrm{Mol} \mathrm{L}^{-1} \mathrm{t}^{-1}\) \\ \hline 1 & \(1 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(2 \times 10^{-4}\) \\ 2 & \(2 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(4 \times 10^{-4}\) \\ 3 & \(2 \times 10^{-2}\) & \(4 \times 10^{-2}\) & \(8 \times 10^{-4}\) \\ \hline \end{tabular} Which of the following inferences are drawn from the above data? (1) Rate constant of the reaction is \(10^{-4}\) (2) Rate law of the reaction is \(\mathrm{k}[\mathrm{A}][\mathrm{B}]\) (3) Rate of reaction increases four times by doubling the concentration of each reactant. Select the correct answer the codes given below: a. 1 and 3 b. 2 and 3 c. 1 and 2 d. 1,2 and 3

Step by step solution

01

Analyze the Data

We are given three experiments with the following concentrations and rates: - Experiment 1: - [A] = 1 x 10^-2 M - [B] = 2 x 10^-2 M - Rate = 2 x 10^-4 M L^-1 t^-1 - Experiment 2: - [A] = 2 x 10^-2 M - [B] = 2 x 10^-2 M - Rate = 4 x 10^-4 M L^-1 t^-1 - Experiment 3: - [A] = 2 x 10^-2 M - [B] = 4 x 10^-2 M - Rate = 8 x 10^-4 M L^-1 t^-1

02

Determine the Rate Law

Look at Experiments 1 and 2, where [B] is constant and [A] is doubled. The rate also doubles, indicating the reaction is first-order with respect to [A]. Next, compare Experiments 2 and 3, where [A] is constant and [B] is doubled. The rate doubles again, indicating the reaction is also first-order with respect to [B]. Thus, the rate law is: \[ \text{Rate} = k[A][B] \]

03

Calculate the Rate Constant k

Using the rate law from Step 2 and data from any of the experiments, we can solve for the rate constant \( k \). Using Experiment 1: \[ 2 \times 10^{-4} = k (1 \times 10^{-2})(2 \times 10^{-2}) \] Solving for \( k \): \[ k = \frac{2 \times 10^{-4}}{(1 \times 10^{-2})(2 \times 10^{-2})} = 10^{-4} \] Thus, the rate constant is \( 10^{-4} \).

04

Check Effect of Doubling on Rate

Doubling both [A] and [B] quadruples the rate because the rate equation \( \text{Rate} = k[A][B] \) is first-order in both reactants. From Experiment 1 to 3, both reactants are effectively doubled, leading to a quadrupling of the rate, going from \( 2 \times 10^{-4} \) to \( 8 \times 10^{-4} \).

05

Identify Correct Inferences

From the analysis: Inference 1: Rate constant is \( 10^{-4} \) is correct. Inference 2: Rate law is \( k[A][B] \) is correct. Inference 3: Rate increases four times by doubling reactants is also correct. Thus the correct answer is option d. 1, 2, and 3.

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Key Concepts

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

Rate Law

Understanding the rate law is vital in chemical kinetics. A rate law expresses the reaction rate as a function of the concentration of the reactants. For example, in the problem concerning the reaction between substances A and B, the derived rate law is \( \text{Rate} = k[A][B] \). This law tells us that the rate of the reaction is dependent on the concentrations of A and B each raised to the power of one. This means that the reaction is first-order with respect to each reactant.

Rate laws are determined by experimentation. By observing how changes in concentration affect the reaction rate, we can deduce the order of each reactant. In this case, doubling A or B separately doubles the reaction rate, confirming their first-order status.

• This is a direct relationship, meaning changes in concentration have a predictable, proportional effect on rate.
• The resulting equation illustrates how rates can be calculated, given concentrations and the rate constant.

Fully grasping the rate law can help you predict how long a reaction will take, or how varying conditions might alter the rate.

Reaction Order

The reaction order in a chemical process provides insight into how the concentration of the reactants influences the rate. Knowing the reaction order is critical to understanding the dynamics of a reaction.

For the reaction between A and B, the exercise reveals that the reaction is first-order with respect to both A and B. This means:

• A one-to-one relationship exists between concentration and rate change. If the concentration of A is doubled, the rate doubles.
• Similarly, doubling B's concentration also doubles the rate.

This aspect confirms the overall reaction order as two, calculated as a sum of the individual orders of the reactants. The reaction order is key to understanding mechanisms and predicting the outcome of reactions with varying concentrations. It is vital to surveys across different times to ensure the observed kinetic effects are due to this fundamental kinetic law.

Rate Constant

The rate constant, denoted as \( k \), is a crucial parameter in the rate law equation. This constant provides the necessary conversion factor between reactant concentrations and reaction rate. Its value is determined under specific conditions, such as temperature, and is unique to every reaction.

Working through the exercise, we derived \( k = 10^{-4} \). This was found by using the experimental data and the formula: \[ k = \frac{\text{Rate}}{[A][B]} \]

The rate constant provides invaluable insight:

• It offers a bridge to calculate rates under different concentrations.
• It remains constant with the same reaction, unless changed by temperature or pressure.

Understanding the rate constant's role helps in not only predicting rate changes with varying concentrations but also in controlling reaction conditions to achieve desired reaction speeds. Mastery of the rate constant enables scientists and engineers to optimize reactions in industrial and laboratory settings.