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{tabular}{llll} \hline S. No. & {\([\mathrm{A}] / \mathrm{mol} \mathrm{L}^{-1}\)} & {\([\mathrm{~B}] / \mathrm{mol} \mathrm{L}^{-1}\)} & Rate \(/ \mathrm{mol} \mathrm{L}^{-1} \mathrm{t}^{-1}\) \\ \hline 1 & \(1 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(2 \times 10^{-4}\) \\ II & \(2 \times 10^{-2}\) & \(2 \times 10^{-2}\) & \(4 \times 10^{-4}\) \\ \hline III & \(2 \times 10^{-2}\) & \(4 \times 10^{-2}\) & \(8 \times 10^{-4}\) \\ \hline \end{tabular} Which of the following inferences are drawn from the above data? 1\. Rate 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

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

Step 1: Analyze provided data

The data shows the concentrations of reactants A and B, and the rate of the reaction for different trials. The objective is to determine how changes in these concentrations affect the rate, and thus infer the rate law.

Step 2: Determine the effect of changing [A]

Compare experiments I and II where [B] remains constant. In experiment I, [A] is \(1 \times 10^{-2}\) and the rate is \(2 \times 10^{-4}\). In experiment II, [A] is doubled to \(2 \times 10^{-2}\), and the rate doubles to \(4 \times 10^{-4}\). This suggests that the rate is first order with respect to [A].

Step 3: Determine the effect of changing [B]

Compare experiments II and III where [A] remains constant. In experiment II, [B] is \(2 \times 10^{-2}\) and the rate is \(4 \times 10^{-4}\). In experiment III, [B] is doubled to \(4 \times 10^{-2}\), and the rate doubles to \(8 \times 10^{-4}\). This suggests that the rate is first order with respect to [B].

Step 4: Write the rate law expression

Since the reaction is first order with respect to both [A] and [B], the rate law can be expressed as \( \text{Rate} = k[A][B] \). This confirms inference 1.

Step 5: Calculate the rate constant \( k \)

From any experiment, use the rate law. For example, using experiment I where \([\text{A}] = 1 \times 10^{-2}\), \([\text{B}] = 2 \times 10^{-2}\), and rate = \(2 \times 10^{-4}\), solve \(2 \times 10^{-4} = k (1 \times 10^{-2})(2 \times 10^{-2})\) to find \(k = 10^{-4}\). This confirms inference 2.

Step 6: Assess the overall effect of doubling reactants

Doubling both [A] and [B] results in multiplying the rate by a factor of four, as seen from experiment I to III. Thus, quadrupling effect confirms inference 3.

Step 7: Evaluate given options

Based on the findings, all three inferences are valid. Therefore, the correct answer is (c) 1, 2, and 3.

Key Concepts

Rate Law

The rate law of a chemical reaction outlines how the rate depends on the concentration of reactants. In this exercise, we have a reaction involving reactants \(A\) and \(B\). By examining how changes in concentration affect the rate of reaction, we can deduce the rate law.

When comparing different trials, we noticed that doubling the concentration of \([A]\) while keeping \([B]\) constant results in the rate doubling. The same is true when doubling \([B]\) and keeping \([A]\) constant. This indicates a linear relationship between the concentration of each reactant and the rate, suggesting a first-order dependence.

• This allows us to formulate the rate law as \( \text{Rate} = k[A][B] \).
• Each reactant is first order, contributing linearly to the overall rate.

This rate law helps predict how changes in concentrations will influence the reaction speed, a core concept in chemical kinetics.

Reaction Order

The reaction order in a rate law determines how the concentration of each reactant affects the rate of reaction. It is expressed as an exponent in the rate law equation.

In our example, the exponents of both \([A]\) and \([B]\) are 1. This means that the reaction is first order with respect to both reactants. Observing the experiments, we saw:

• Doubling \([A]\) resulted in a doubled rate, and
• Doubling \([B]\) also doubled the rate.

These observations confirm that both reactants are involved linearly in determining the rate.
Summing the individual orders gives us the overall reaction order:

• The reaction is second order overall, as first order plus first order equals second order.

Understanding reaction order is crucial as it helps in recognizing the influence of each reactant on the reaction rate.

Rate Constant

The rate constant, \(k\), is a crucial factor in the rate law equation that relates the rate of reaction to the concentrations of the reactants. It is specific for a given reaction at a particular temperature.

In this exercise, we calculated \(k\) using the rate law expression: \( \text{Rate} = k[A][B] \). By substituting the known values from Experiment I,

• \(2 \times 10^{-4} = k \times (1 \times 10^{-2}) \times (2 \times 10^{-2})\)

Solving this equation gives a rate constant \(k\) of \(10^{-4}\).
The rate constant provides essential insights:

• It allows us to predict reaction rates under varied concentrations.
• Understanding \(k\) also helps in scaling reactions for practical applications.

Remember, \(k\) may change with temperature, leading to variations in reaction speed as conditions alter.