Question

The following data pertains to the reaction between \(\mathrm{A}\) and \(\underline{B}\) \begin{tabular}{llll} \multicolumn{4}{c} { Table \(10.5\)} \\ \hline S. No. & {\([\mathrm{A}] \mathrm{mol} \mathrm{L}^{-1}\)} & {\([\mathrm{~B}] \mathrm{mol} \mathrm{L}^{-1}\)} & Rate mol \(\mathrm{L}^{-1} \mathrm{~S}^{-1}\) \\ \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 \([\mathrm{A}][\mathrm{B}]\) (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) land 2 (d) 1,2 and 3

Short answer

The correct answer is (d) 1, 2 and 3.

Step-by-step solution

Determine the Order with Respect to A

To find the order with respect to \(\mathrm{A}\), compare Experiments 1 and 2. The concentration of \(\mathrm{A}\) doubles (from \(1 \times 10^{-2}\) to \(2 \times 10^{-2}\)) while \([\mathrm{B}]\) remains constant. The rate also doubles from \(2 \times 10^{-4}\) to \(4 \times 10^{-4}\). This indicates that the reaction is first order in \(\mathrm{A}\).

Determine the Order with Respect to B

Next, compare Experiments 2 and 3. The concentration of \(\mathrm{B}\) doubles (from \(2 \times 10^{-2}\) to \(4 \times 10^{-2}\)) while \([\mathrm{A}]\) remains constant. The rate also doubles from \(4 \times 10^{-4}\) to \(8 \times 10^{-4}\). This shows that the reaction is first order in \(\mathrm{B}\).

Write the Rate Law Expression

Since the reaction is first order in both \(\mathrm{A}\) and \(\mathrm{B}\), the overall rate law is given by \(\text{Rate} = k[\mathrm{A}][\mathrm{B}]\). This confirms option (2).

Calculate the Rate Constant

Using the rate law \(\text{Rate} = k[\mathrm{A}][\mathrm{B}]\), substitute the values from any experiment, for example Experiment 1: \(2 \times 10^{-4} = k(1 \times 10^{-2})(2 \times 10^{-2})\). Solving for \(k\), we get \(k = 10^{-4}\), confirming option (1).

Evaluate Changes in Rate with Doubling of Reactants' Concentration

Since both reactants are first order, doubling the concentration of each reactant independently results in a double increase in rate. If both are doubled simultaneously, the rate increases fourfold. This validates option (3).

Key concepts

Rate Law

In chemical kinetics, the rate law gives us a mathematical expression that relates the rate of a chemical reaction to the concentration of its reactants. The rate law helps predict how the reaction rate changes as the concentration of the reactants changes. For the reaction between substances \( \mathrm{A} \) and \( \underline{B} \), we observed that the rate law expression could be written as \( \text{Rate} = k[\mathrm{A}][\mathrm{B}] \).

• The rate law is determined experimentally, and it is specific to a given reaction at a certain temperature.
• The expression \([\mathrm{A}][\mathrm{B}]\) indicates that the reaction is directly proportional to the concentrations of both \( \mathrm{A} \) and \( \mathrm{B} \).
• The constant \( k \) is the rate constant, which quantifies the reaction's speed at a given temperature.

Understanding the rate law allows scientists to control and manipulate reaction conditions, optimizing outcomes in industrial and laboratory settings.

Reaction Order

Reaction order refers to the power to which the concentration of a reactant is raised in the rate law. It indicates how the concentration of reactants affects the rate of the chemical reaction. From the provided data, we determined the reaction orders with respect to reactants \( \mathrm{A} \) and \( \underline{B} \) are both first order.

• First order in \( \mathrm{A} \) means the rate of reaction doubles when the concentration of \( \mathrm{A} \) is doubled.
• First order in \( \underline{B} \) means similarly, doubling \([\mathrm{B}]\) results in doubling the rate.
• The overall reaction order is the sum of the orders with respect to each reactant: first in \( \mathrm{A} \) plus first in \( \underline{B} \), resulting in an overall second order.

Knowing the reaction order is crucial for understanding reaction mechanics and predicting how different factors affect the reaction.

Rate Constant

The rate constant \( k \) is a vital component of the rate law, providing a measure of the intrinsic speed of a reaction under specified conditions. Calculated as part of the rate law, the rate constant embodies how quickly a reaction proceeds independent of reactant concentrations;

• It has a specific value for each reaction at a certain temperature.
• In our reaction example, we calculated \( k = 10^{-4} \) using experiment data and the rate law \( \text{Rate} = k[\mathrm{A}][\mathrm{B}] \).
• The units of \( k \) vary depending on the overall reaction order. For a second-order reaction like ours, the units are \( \mathrm{L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1} \).

Understanding the rate constant allows chemists to compare the speed of different reactions and adjust conditions for desired reaction rates.