Problem 71 - A Complete Resource Book in Chemistry for JEE Main

Question

In a first-order reaction \(\mathrm{A} \longrightarrow \mathrm{P}\), the ratio of \(\mathrm{a} /(\mathrm{a}-\mathrm{x})\) was found to be 8 after 60 minutes. If the concentration is \(0.1 \mathrm{M}\) then the rate of reaction in moles of A reacted per minutes is (a) \(2.226 \times 10^{-3} \mathrm{~mol}\) litre \(^{-1} \mathrm{~min}^{-1}\) (b) \(3.466 \times 10^{-3} \mathrm{~mol}\) litre \(^{-1} \mathrm{~min}^{-1}\) (c) \(4.455 \times 10^{-3} \mathrm{~mol}\) litre \(^{-1} \mathrm{~min}^{-1}\) (d) \(5.532 \times 10^{-3} \mathrm{~mol}\) litre \(^{-1} \mathrm{~min}^{-1}\)

Step-by-Step Solution

Answer

The correct answer is (b) \(3.466 \times 10^{-3} \, \text{mol} \, \text{litre}^{-1} \, \text{min}^{-1}\).

Step 1: Understanding the Reaction Order

The reaction given is a first-order reaction, which follows the rate law: \( \text{rate} = k [A] \), where \(k\) is the rate constant and \([A]\) is the concentration of \(\mathrm{A}\). The integrated rate equation for a first-order reaction is \( \ln \left(\frac{[A]_0}{[A]_t}\right) = kt \), where \([A]_0\) is the initial concentration, \([A]_t\) is the concentration at time \(t\), and \(k\) is the rate constant.

Step 2: Analyzing the Given Data

The problem states \( \frac{a}{a-x} = 8 \) at \( t = 60 \) minutes. "a" is the initial concentration and "x" is the concentration change. From the ratio, \([A]_0 = 0.1 \, M\), it follows that \( \frac{0.1}{0.1-x} = 8\). Simplifying, we find \(0.1 = 8(0.1-x)\).

Step 3: Solving for x

Rearrange and solve for \(x\):\[0.1 = 0.8 - 8x \8x = 0.8 - 0.1 \x = \frac{0.7}{8} = 0.0875 \, M\]

Step 4: Finding the Rate Constant

Substitute the values back into the rate equation: \(\ln\left(\frac{[A]_0}{[A]_t}\right) = kt \\ln\left(\frac{0.1}{0.1-0.0875}\right) = k \times 60\)After simplification, \(\ln(8) = 60k\). Calculate \(\ln(8)\) which gives approximately 2.079. Thus:\[k = \frac{2.079}{60} = 0.03465 \, \text{min}^{-1}\]

Step 5: Calculating the Rate of Reaction

The rate of reaction for first-order is calculated as: rate = \(k[A]_0\) So, the rate is:\[0.03465 \times 0.1 = 0.003465 \, \text{mol} \, \text{litre}^{-1} \, \text{min}^{-1}\]

Step 6: Final Answer

Comparing the calculated rate \(0.003465\) with the options, it corresponds to (b) \(3.466 \times 10^{-3} \, \text{mol} \, \text{litre}^{-1} \, \text{min}^{-1}\).

Key Concepts

Rate law

The rate law is a fundamental concept in understanding how fast a reaction proceeds. For a first-order reaction, such as the one in our exercise, the rate of the reaction is directly proportional to the concentration of a single reactant, in this case, substance A. This relationship is mathematically expressed through the rate law equation:

• Rate = \( k [A] \)

Where \(k\) is the rate constant that remains unchanged during the reaction, and \([A]\) is the concentration of reactant A at any point in time.
First-order reactions' rate laws imply that if the concentration of A is doubled, the reaction rate also doubles. This direct proportionality makes it easier to predict how changing the concentration of reactant A influences the reaction speed.
Understanding this principle is crucial for analyzing how chemical reactions proceed and for calculating kinetic parameters such as the rate constant and reaction rate.

Integrated rate equation

The integrated rate equation provides a means to relate the concentration of a reactant to time. For a first-order reaction, the concentration of the reactant \([A]\) at a given time \(t\) can be determined using the integrated rate equation:

• \( \ln \left(\frac{[A]_0}{[A]_t}\right) = kt \)

Where:

• \([A]_0\) is the initial concentration of reactant A,
• \([A]_t\) is the concentration of A at time \(t\), and
• \(k\) is the rate constant.

This equation helps in determining the time required for the concentration to reach a certain level or for calculating the rate constant when experimental data are available.
In our exercise, given a reaction time and a concentration ratio, this equation allows us to pin down the exact value of \(k\). It serves as a powerful tool for chemists to analyze reaction progress quantitatively over time.

Rate constant

The rate constant \(k\) is a crucial parameter in the kinetics of a reaction. It provides vital information about the speed of a reaction under specific conditions. For first-order reactions, \(k\) carries units of inverse time, typically expressed in \(\text{min}^{-1}\) or \(\text{s}^{-1}\).
The value of the rate constant depends on factors like temperature and the nature of the reactants but remains constant for a given reaction at fixed conditions.
In the context of the exercise, we calculated \(k\) by rearranging the integrated rate equation and using the known concentration change over a set period. This calculation of \( k = 0.03465 \, \text{min}^{-1} \) highlights how this constant forms the backbone of quantifying chemical kinetics and understanding how reactions proceed.

Concentration change

In a reaction, concentration changes help us understand how much of a reactant has been consumed over a time period. In the given first-order reaction exercise, "a" represents the initial concentration, and "x" is the change in concentration, resulting in a decreased level of reactant A.
Determining the concentration change is essential to understand the reaction's progress. Using the equation \(\frac{a}{a-x} = 8\), we derived the amount of reactant A that had reacted after 60 minutes.
Once "x" is known, it becomes easy to substitute back into the integrated rate equation to further analyze the reaction. This interplay of concentration changes and kinetic equations demonstrates the dynamic nature of chemical reactions and the importance of precise calculations in understanding these processes. By calculating \(x = 0.0875 \, M\), we quantify how much A is converted, enabling us to delve deeper into the reaction's kinetics.