Problem 5 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

\begin{tabular}{|c|c|c|c|} \hline Experiment & \([\mathrm{A}] /\) \(\mathrm{molL}^{-1}\) & \([\mathrm{B}] /\) \(\mathrm{molL}^{-1}\) & Initial rate/ \(\mathrm{moL}^{-1} \mathrm{~min}^{-1}\) \\\ \hline I & \(0.1\) & \(0.1\) & \(6.00 \times 10^{-3}\) \\ \hline II & \(0.1\) & \(0.2\) & \(2.40 \times 10^{-2}\) \\ \hline II & \(0.2\) & \(0.1\) & \(1.20 \times 10^{-2}\) \\ \hline IV & \(\mathrm{X}\) & \(0.2\) & \(7.20 \times 10^{-2}\) \\ \hline V & \(0.3\) & \(\mathrm{Y}\) & \(2.88 \times 10^{-1}\) \\ \hline \end{tabular} \(\mathrm{X}\) and \(\mathrm{Y}\) in the given table are respectively: (a) \(0.4,0.4\) (b) \(0.4,0.3\) (c) \(0.3,0.4\) (d) \(0.3,0.3\)The results given in the below table were obtained during kinetic studies of the following reaction: \(2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C}+\mathrm{D}\) [Main Sep. 02, 2020 (II)]

Step-by-Step Solution

Answer

Option (a) \(0.4, 0.4\): \(X = 0.4\), \(Y = 0.4\).

Step 1: Identify the reaction rate law

The rate law expression for the reaction \(2A + B \rightarrow C + D\) can be written as: \(rate = k[A]^m[B]^n\), where \(k\) is the rate constant, and \(m\) and \(n\) are the reaction orders with respect to \(A\) and \(B\) respectively.

Step 2: Determine the order with respect to \(B\)

Compare Experiments I and II, where \([A]\) is constant and \([B]\) changes from \(0.1\) to \(0.2\). The rate of reaction increases from \(6.00 \times 10^{-3}\) to \(2.40 \times 10^{-2}\), which is four times. This suggests that \(n = 2\) because \(2^n = 4\).

Step 3: Determine the order with respect to \(A\)

Compare Experiments I and III, where \([B]\) is constant and \([A]\) changes from \(0.1\) to \(0.2\). The rate of reaction increases from \(6.00 \times 10^{-3}\) to \(1.20 \times 10^{-2}\), which is two times. This suggests that \(m = 1\) because \(2^m = 2\).

Step 4: Find the rate constant \(k\)

Using the rate law expression \(rate = k[A]^1[B]^2\) from any known experiment (e.g., Experiment I: \(6.00 \times 10^{-3} = k(0.1)^1(0.1)^2\)), solve for \(k\). The resulting rate constant \(k = 6\).

Step 5: Solve for \(X\) using Experiment IV

In Experiment IV, use the rate law \(rate = k[A]^1[B]^2\) with \(rate = 7.20 \times 10^{-2}\), \([B] = 0.2\), and \(k = 6\) to solve for \(A\): \(7.20 \times 10^{-2} = 6[X](0.2)^2\). Solving gives \(X = 0.4\).

Step 6: Solve for \(Y\) using Experiment V

In Experiment V, use the rate law \(rate = k[A]^1[B]^2\) with \(rate = 2.88 \times 10^{-1}\), \([A] = 0.3\), and \(k = 6\) to solve for \(B\): \(2.88 \times 10^{-1} = 6(0.3)[Y]^2\). Solving gives \(Y = 0.4\).

Step 7: Choose the correct option

From the calculations, \(X = 0.4\) and \(Y = 0.4\). Thus, the correct option is (a) \(0.4, 0.4\).

Key Concepts

Rate Law

In chemical kinetics, understanding the rate law is essential. The rate law expresses the relationship between the rate of a chemical reaction and the concentration of its reactants. For a generic reaction where substances A and B produce products, the rate law can be formulated as:

• Rate = k[A]^m[B]^n

Here, "k" is a constant known as the rate constant, while "m" and "n" represent the reaction orders with respect to A and B respectively.
This equation shows how changes in the concentration of each reactant influence the rate of reaction.
While the rate law helps predict how fast a reaction will proceed under specific concentrations, it is determined experimentally rather than from the balanced equation. Understanding this concept can guide predictions and inform experiments, helping chemists control and optimize reactions.

Reaction Order

The reaction order is a crucial concept to grasp when studying reaction kinetics. It indicates the power to which the concentration of a reactant is raised in the rate law. In our specific example, for the reaction 2A + B → C + D, understand the step-by-step determination of reaction orders:

• First, observe what happens to the rate as the concentration of one reactant changes, keeping the others constant.
• In this case, observing experiments I and II shows how the order with respect to B is found: increasing [B] from 0.1 to 0.2 molL⁻¹ quadruples the rate, suggesting a second-order reaction in B.
• Comparing experiments I and III illustrates the order with respect to A: doubling [A] doubles the rate, indicating a first-order reaction in A.

This detailed understanding enables precise reaction predictions and forms a foundation for exploring more complex kinetic behavior.

Rate Constant

The rate constant, denoted as "k", is a pivotal factor in the rate law equation. While its value is specific to each reaction, the rate constant remains consistent under the same conditions of temperature and catalyst concentration. To find "k", use a rate law equation derived from experiment data. For instance, using data from Experiment I for the equation rate = k[A]^1[B]^2:

• Plug in the rate as 6.00 × 10⁻³ molL⁻¹ min⁻¹, [A] as 0.1 molL⁻¹, and [B] as 0.1 molL⁻¹.
• Solve for "k" to find it equals 6.

Understanding and calculating the rate constant provides insight into the reaction's speed and allows for comparisons across different reactions or conditions. Also, it integrates into larger models, predicting how variables like temperature influence the rate with established theories, like the Arrhenius equation.

Chemical Kinetics

Chemical kinetics is the overarching study focused on rates of chemical processes. It encompasses understanding not just the speed of reactions, but also the mechanisms by which they occur. By examining reactant concentrations, reaction orders, and rate constants, chemical kinetics provides a framework to describe reactions in an empirical and mathematical way. Some key areas within chemical kinetics include:

• The influence of factors such as temperature, pressure, and catalysts on reaction rates.
• The use of different methods to determine reaction rates and propose mechanisms.

This field bridges the gap between theoretical chemistry and practical applications, allowing scientists to improve industrial processes and develop new technologies. As students delve into kinetics, they unlock the tools needed to control reactions efficiently, an essential skill in both academic and industrial environments.