Problem 123 - A Complete Resource Book in Chemistry for JEE Main

Question

The following data pertains to the reaction between \(\mathrm{A}\) and \(\mathrm{B}\) $$ \begin{array}{llll} \hline \text { S. No. } & {[\mathrm{A}] \mathrm{mol} \mathrm{L}^{-1}} & {[\mathrm{~B}] \mathrm{mol} \mathrm{L}^{-1}} & \begin{array}{l} \text { Rate mol } \\ \mathrm{L}-\mathbf{S}^{-1} \end{array} \\ \hline \text { 1. } & 1 \times 10^{-2} & 2 \times 10^{-2} & 2 \times 10^{-4} \\\ 2 . & 2 \times 10^{-2} & 2 \times 10^{-2} & 4 \times 10^{-4} \\ 3 . & 2 \times 10^{-2} & 4 \times 10^{-2} & 8 \times 10^{-8} \\ \hline \end{array} $$ 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

Step-by-Step Solution

Answer

Option (b) 2 and 3 are correct.

Step 1: Determine the Order with Respect to A

We compare experiments 1 and 2, where the concentration of \( [B] \) is constant and \( [A] \) doubles from \( 1 \times 10^{-2} \text{ mol L}^{-1} \) to \( 2 \times 10^{-2} \text{ mol L}^{-1} \). The rate also doubles from \( 2 \times 10^{-4} \) to \( 4 \times 10^{-4} \), indicating a first-order reaction with respect to \( A \).

Step 2: Determine the Order with Respect to B

We compare experiments 2 and 3, where \( [A] \) is constant and \( [B] \) doubles from \( 2 \times 10^{-2} \text{ mol L}^{-1} \) to \( 4 \times 10^{-2} \text{ mol L}^{-1} \). However, the rate unexpectedly changes from \( 4 \times 10^{-4} \) to \( 8 \times 10^{-8} \), suggesting a calculation error. Assuming the rate was intended to double like \( [B] \) from \( 4 \times 10^{-4} \) to \( 8 \times 10^{-4} \), we infer a first-order dependence on \( B \).

Step 3: Formulate the Rate Law

Given the reaction orders from Steps 1 and 2, the rate law can be expressed as the product of \( A \) and \( B \): \( ext{Rate} = k[A][B] \).

Step 4: Calculate the Rate Constant

Using the rate law \( ext{Rate} = k[A][B] \) and data from experiment 1: \( 2 \times 10^{-4} = k(1 \times 10^{-2})(2 \times 10^{-2}) \), we find \( k = 10^{-1} \text{ L mol}^{-1} \text{ s}^{-1} \).

Step 5: Evaluate Statement 3

If both \( A \) and \( B \) concentrations are doubled, the rate equation \( ext{Rate} = k[A][B] \), predicts the rate increasing by a factor of 4 (since doubling both reactants increases the rate by a factor of 2 for each reactant). This supports statement 3.

Step 6: Select the Correct Answer

With the rate law \( [A][B] \) and the quadrupling effect in rate confirmed by doubling reactant concentrations, statement 2 and 3 are correct. Despite a discrepancy in calculating \( k \), our final selection is (b) 2 and 3.

Key Concepts

Rate Law

When studying chemical reactions, understanding the rate law is crucial. The rate law equation showcases how the concentration of reactants influences the overall rate of the reaction. Specifically, for any given reaction, the rate can be expressed as a product of a rate constant and the concentrations of the reactants each raised to a power.
The general form is given by:

• Rate = k[A]\(^m\)[B]\(^n\)

Here,

• \(k\) is the rate constant
• \([A]\) and \([B]\) are concentrations of the reactants
• \(m\) and \(n\) are the orders of the reaction with respect to each reactant.

Understanding and determining the correct form of the rate law is essential because it gives insights into how changes in concentration affect the reaction speed. You determine it by examining experimental data, often by looking at how alterations in reactant concentrations change the rate.

Reaction Order

Reaction order conveys how the concentration of each reactant affects the overall rate of the reaction. From the rate law, those exponents \(m\) and \(n\) are specifically known as the reaction orders with respect to their respective reactants.
To determine these, scientists commonly conduct experiments where the concentration of one reactant is varied while others are held constant. Examining how the rate adjusts gives clues about the order:

• If doubling a reactant causes the rate to double, the reaction is first order with respect to that reactant.
• If the rate increases fourfold, the reaction is second order.

Understanding the reaction order is vital because it predicts how changes in concentrations of the reactants influence the overall reaction rate. In the exercise, it is noted that both reactants A and B hold a first-order relationship with their rate of reaction.

Rate Constant

The rate constant, denoted as \(k\), is a crucial parameter in the rate law expression. It provides the proportionality factor that correlates the concentrations and the observed reaction rate. The unit of the rate constant depends on the overall order of the reaction, ensuring that the rate law equation remains dimensionally consistent.
Calculating \(k\) involves inputting known values from experimental data into the rate law equation. Using the information from the original exercise, \(k\) can be deduced using values from a specific experiment:

• Using experiment 1:
• Rate = \(2 \times 10^{-4}\), \([A] = 1 \times 10^{-2}\), and \([B] = 2 \times 10^{-2}\).
• The equation solves to \(k = 10^{-1} \text{ L mol}^{-1} \text{ s}^{-1}\).

The significance of the rate constant is that it remains unchanged regardless of the concentrations, provided the temperature is constant. This makes \(k\) a reliable metric for comparing reaction speeds under varying conditions.