Problem 20 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

For the non-stoichiometric reaction \(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}\), the following kinetic data were obtained in three separate experiments, all at \(298 \mathrm{~K}\). \begin{tabular}{|c|c|c|} \hline Initial Concentration (A) & Initial Concentration (B) & Initial rate of formation of C \(\left(\mathbf{m o l} \mathbf{L}^{-1} \mathbf{s}^{-1} \mathbf{)}\right.\) \\ \hline \(0.1 \mathrm{M}\) & \(0.1 \mathrm{M}\) & \(1.2 \times 10^{-3}\) \\ \hline \(0.1 \mathrm{M}\) & \(0.2 \mathrm{M}\) & \(1.2 \times 10^{-3}\) \\ \hline \(0.2 \mathrm{M}\) & \(0.1 \mathrm{M}\) & \(2.4 \times 10^{-3}\) \\ \hline \end{tabular}The rate law for the formation of \(\mathrm{C}\) is: [Main 2014] (a) \(\frac{d \mathrm{c}}{d t}=k[\mathrm{~A}][\mathrm{B}]\) (b) \(\frac{d \mathrm{c}}{d t}=k[\mathrm{~A}]^{2}[\mathrm{~B}]\)(c) \(\frac{d \mathrm{c}}{d t}=k[\mathrm{~A}][\mathrm{B}]^{2}\) (d) \(\frac{d \mathrm{c}}{d t}=k[\mathrm{~A}]\)

Step-by-Step Solution

Answer

The rate law is \( \frac{d[\text{C}]}{dt} = k[\text{A}] \) (option d).

Step 1: Understand the Problem

We want to determine the rate law for the formation of C in the reaction. The rate law can be expressed in the form \( \frac{d[\text{C}]}{dt} = k[\text{A}]^m[\text{B}]^n \), where \( k \) is the rate constant, and \( m \) and \( n \) represent the reaction orders with respect to A and B, respectively.

Step 2: Compare Experiments 1 and 2

In Experiments 1 and 2, the initial concentration of A is constant at 0.1 M, but the concentration of B changes from 0.1 M to 0.2 M. The rate of formation of C does not change, implying that the reaction rate is independent of the concentration of B. Therefore, \( n = 0 \).

Step 3: Compare Experiments 1 and 3

In Experiments 1 and 3, the initial concentration of B is constant at 0.1 M, but the concentration of A changes from 0.1 M to 0.2 M. The rate of formation of C doubles from \( 1.2 \times 10^{-3} \) to \( 2.4 \times 10^{-3} \), indicating that the rate is directly proportional to the concentration of A. Therefore, \( m = 1 \).

Step 4: Derive the Rate Law

Based on the findings from Steps 2 and 3, the rate law can be written as \( \frac{d[\text{C}]}{dt} = k[\text{A}] \). The reaction rate is first order with respect to A and zero order with respect to B.

Key Concepts

Rate Law

The rate law is a mathematical equation that describes the speed, or rate, at which a chemical reaction proceeds. It expresses the rate as a function of the concentration of its reactants. For a reaction where the rate law is described by the equation\( \frac{d[ ext{Product}]}{dt} = k[ ext{Reactant}_1]^m[ ext{Reactant}_2]^n \), we define several important variables:

• \( k \) is the rate constant.
• \( [ ext{Reactant}_1] \) and \( [ ext{Reactant}_2] \) are the concentrations of the reactants.
• \( m \) and \( n \) are the orders of the reaction with respect to each reactant.

The overall rate law helps us to understand how changes in the concentration or state of a reactant can affect the rate at which products are formed.
By studying the rate law, chemists can predict how the reaction will behave under different conditions, which is valuable in both academic and industrial settings.
In the exercise, we identified the rate law for the reaction as \( \frac{d[ ext{C}]}{dt} = k[ ext{A}] \), indicating a first-order reaction with respect to reactant A.

Reaction Order

Reaction order indicates the power to which the concentration of a reactant is raised in a rate law equation. Each reactant in a chemical reaction can have its own individual reaction order, and together they make up the overall order of the reaction. The reaction order tells us how the concentration of each reactant affects the rate of reaction. If a reaction is zero order with respect to a reactant, - the rate of reaction does not depend on the concentration of that reactant. - Doubling the concentration will have no effect on the rate. For a first-order reaction: - The rate is directly proportional to the concentration of the reactant. - Doubling the concentration of the reactant will double the rate of reaction. In the given exercise, we determined: - The reaction is first order with respect to A because, when concentration of A was doubled, the rate of formation of C also doubled. - The reaction is zero order with respect to B because changing its concentration did not change the rate.

Rate Constant

The rate constant, symbolized as \( k \), is an integral part of a reaction's rate law. It allows us to quantify how fast a reaction occurs. The value of \( k \) can vary significantly depending on the reaction conditions, such as temperature, and is determined experimentally for each specific reaction.Key characteristics of the rate constant:

• The unit of \( k \) depends on the overall order of the reaction. For example, for a first-order reaction, the unit is \( s^{-1} \).
• It remains constant given fixed temperature conditions.
• A larger \( k \) value signifies a faster reaction rate, assuming constant concentrations.

To find \( k \) in the exercise, the rate law is established first. Using the data from Experiment 1: \[ 1.2 \times 10^{-3} = k [0.1] \]Solving for \( k \), we find that the rate constant is the quotient of the initial rate and the concentration of A used in the experiments, elucidating how it defines the intrinsic speed of the reaction.