Problem 150 - A Complete Resource Book in Chemistry for JEE Main

Question

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

Step-by-Step Solution

Answer

(c) 1, 2, and 3

Step 1: Determine the Reaction Order with Respect to \( [A] \)

Examine experiments I and II where \([B]\) is constant. Between these two experiments, \([A]\) is doubled from \(1 \times 10^{-2}\) to \(2 \times 10^{-2}\), and the rate also doubles from \(2 \times 10^{-4}\) to \(4 \times 10^{-4}\). This suggests that the rate is directly proportional to \([A]\), indicating a first-order dependence on \([A]\): \(m = 1\).

Step 2: Determine the Reaction Order with Respect to \( [B] \)

Next, examine experiments II and III, where \([A]\) is constant. Here, \([B]\) is doubled from \(2 \times 10^{-2}\) to \(4 \times 10^{-2}\), and the rate also doubles from \(4 \times 10^{-4}\) to \(8 \times 10^{-4}\). This suggests that the rate is also directly proportional to \([B]\), indicating a first-order dependence on \([B]\): \(n = 1\).

Step 3: Write the Rate Law

From Steps 1 and 2, we have identified that the reaction is first-order with respect to both \([A]\) and \([B]\). The rate law can thus be expressed as: \( \text{Rate} = k [A]^1 [B]^1 = k [A][B] \).

Step 4: Calculate the Rate Constant \( k \)

Utilize the data from any experiment to solve for \( k \). Using Experiment I: \( 2 \times 10^{-4} = k (1 \times 10^{-2})(2 \times 10^{-2}) \). Solving for \( k \), we find \( k = \frac{2 \times 10^{-4}}{2 \times 10^{-4}} = 1 \times 10^{-4} \text{ mol}^{-1} \text{L}^{1} \text{t}^{-1} \).

Step 5: Evaluate Each Statement

1. The rate law is consistent with result from Step 3. 2. The calculated rate constant from Step 4 is indeed \(1 \times 10^{-4}\). 3. Doubling each reactant increases the rate by four times, which corresponds to a first-order dependence on each reactant, supporting the statement.

Key Concepts

Reaction Order

Reaction order is a key concept in chemical kinetics. It tells us how the rate of a reaction depends on the concentration of its reactants. In this exercise, we observe the reaction between A and B using experimental data. By examining the rate of reaction when concentrations of A and B are altered, we determine their individual reaction orders.

• When you double the concentration of 1. A from Experiment I to II (keeping B constant), the rate also doubles. This indicates a first-order reaction with respect to A.
• Similarly, when doubling 2. B from Experiment II to III (keeping A constant), the rate doubles again. This means the reaction is first-order with respect to B.

Overall, the reaction is observed to be first-order with respect to both A and B, making the total reaction order 2.

Rate Law

The rate law is an expression that quantifies the rate of a chemical reaction using the concentration of reactants. It is crucial for predicting how reaction rates will change. From the given data, once you have determined that each reactant is first-order, you can write the rate law.
The general form of a rate law for two reactants is:

• \[ ext{Rate} = k [A]^m [B]^n \]

In this exercise:

• The rate law becomes:\[ ext{Rate} = k [A]^1 [B]^1 = k [A][B] \]

This equation shows that the reaction rate is proportional to both [A] and [B]. You can use the rate law to predict how changes in concentration affect the rate.

Rate Constant

The rate constant, denoted as \( k \), is an essential factor in the rate law, reflecting the speed of a reaction. It is unique for every reaction and determined experimentally. In this exercise, after establishing the rate law, we can calculate k.
By using data from Experiment I:

• Substitute the given values:\[ 2 imes 10^{-4} = k(1 imes 10^{-2})(2 imes 10^{-2}) \]
• Solving this equation gives:\[ k = rac{2 imes 10^{-4}}{2 imes 10^{-4}} = 1 imes 10^{-4} ext{ mol}^{-1} ext{L}^{1} ext{t}^{-1} \]

The calculated rate constant matches the information derived from the data. It indicates how quickly the reaction proceeds under defined conditions. Understanding the rate constant assists in controlling and optimizing reactions in various chemical processes.