Problem 221 - A Complete Resource Book in Chemistry for JEE Main

Question

For a first order reaction, (A) \(\rightarrow\) products, the concentration of A changes from \(0.10 \mathrm{M}\) to \(0.025\) Min 40 minutes. The rate of reaction when the concentration of \(\mathrm{A}\) is \(0.01 \mathrm{M}\), is: (a) \(3.47 \times 10^{-5} \mathrm{M} / \mathrm{min}\) (b) \(3.47 \times 10^{-4} \mathrm{M} / \mathrm{min}\) (c) \(1.73 \times 10^{-5} \mathrm{M} / \mathrm{min}\) (d) \(1.73 \times 10^{-4} \mathrm{M} / \mathrm{min}\)

Step-by-Step Solution

Answer

The rate of reaction is \( 3.47 \times 10^{-4} \text{ M/min} \), option (b).

Step 1: Understand the Reaction Order

The problem concerns a first order reaction characterized by the fact that the rate of reaction, \( R \), depends linearly on the concentration of one reactant, \( A \). The general rate equation for a first order reaction is \( R = k[A] \), where \( k \) is the rate constant.

Step 2: Use the First Order Kinetic Equation

For a first order reaction, the integrated rate law is given by \( \ln \left( \frac{[A]_0}{[A]} \right) = kt \), where \( [A]_0 \) is the initial concentration, \( [A] \) is the concentration at time \( t \), and \( t \) is the time.

Step 3: Calculate the Rate Constant \( k \)

Using the provided concentrations \([A]_0 = 0.10 \text{ M}\) and \([A] = 0.025 \text{ M}\) over \( t = 40 \text{ minutes}\), apply the first order kinetic equation: \[ \ln \left( \frac{0.10}{0.025} \right) = k \times 40 \]. Calculate \( k \).

Step 4: Solve for the Rate Constant \( k \)

Calculate the natural logarithm: \( \ln(4) \approx 1.386 \). Then, solve for \( k \): \[ k = \frac{1.386}{40} \approx 0.03465 \ \text{min}^{-1} \].

Step 5: Determine Rate at Given Concentration

Use the rate equation \( R = k[A] \) with \( k = 0.03465 \ \text{min}^{-1} \) and \([A] = 0.01 \text{ M}\): \[ R = 0.03465 \times 0.01 = 3.465 \times 10^{-4} \text{ M/min} \].

Step 6: Select the Correct Answer

Compare the calculated rate \( 3.465 \times 10^{-4} \text{ M/min} \) with the options. Round appropriately to match: the closest choice is \( 3.47 \times 10^{-4} \text{ M/min} \). Thus, the correct answer is (b) \( 3.47 \times 10^{-4} \text{ M/min} \).

Key Concepts

Rate of Reaction

Understanding how fast a reaction proceeds is crucial in chemistry. The rate of reaction tells us how quickly the concentration of a reactant decreases, or the concentration of a product increases, over time. For first-order reactions, the rate is directly proportional to the concentration of a single reactant.

This means if you double the concentration of the reactant, the rate of reaction also doubles. In mathematical terms, it's expressed with the formula: \[ R = -\frac{d[A]}{dt} = k[A] \]Here, \( R \) is the rate of the reaction, \( [A] \) is the concentration of the reactant, and \( k \) is the rate constant. Because the rate changes as the concentration changes, it's important to determine the conditions under which you measure it.

Rate Constant

The rate constant \( k \) is a factor that makes the math of rate of reaction work. It's unique to every reaction at a given temperature. It includes all the factors that affect the rate of the reaction, besides the concentration of the reactants.

For a first-order reaction, once you know the rate constant, you can predict how fast the reaction will proceed at different concentrations of the reactant. It's a bit like speed in a car - with the rate constant, you've got a sense of how fast things are moving. Calculations involving the rate constant for first-order reactions often use the formula: \[ k = \frac{1}{t} \ln \left(\frac{[A]_0}{[A]}\right) \]Here, \( [A]_0 \) is the initial concentration, and \([A]\) is the concentration at time \( t \). This constant helps us find out the rate when concentrations change.

Integrated Rate Law

The integrated rate law gives us a deeper understanding of how concentrations change over time for first-order reactions. It allows us to calculate the concentration of the reactant at any given time based on the initial concentration, without having to measure concentration every second.

For a first-order reaction, the integrated rate law is:\[ \ln([A]_0) - \ln([A]) = kt \] Simplified, this becomes:\[ \ln \left(\frac{[A]_0}{[A]} \right) = kt \]This formula links the concentration at the beginning, the concentration at time \( t \), and the time that has passed.

• \( [A]_0 \) is the starting concentration.
• \([A]\) is the concentration after time \( t \).
• \( k \) is the rate constant.

This understanding is essential for predicting how long it will take for a reaction to reach a certain point or to be considered complete.