Problem 92 - A Complete Resource Book in Chemistry for JEE Main

Question

The rate constant of a first-order reaction, \(\mathrm{A} \longrightarrow\) products, is \(60 \times 10^{-4} \mathrm{~s}^{-1} .\) Its rate at \([\mathrm{A}]=\) \(0.01 \mathrm{~mol} \mathrm{~L}^{-1}\) would be (a) \(60 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (b) \(36 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (c) \(60 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\) (d) \(36 \times 10^{-1} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\)

Step-by-Step Solution

Answer

(a) \(60 \times 10^{-6} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}\).

Step 1: Understanding the Rate Law for First-Order Reactions

In a first-order reaction, the rate of the reaction is directly proportional to the concentration of the reactant. The rate can be expressed as \( \text{rate} = k[A] \), where \( k \) is the rate constant, and \( [A] \) is the concentration of the reactant.

Step 2: Using Given Data

We are given that the rate constant \( k = 60 \times 10^{-4} \text{ s}^{-1} \) and the concentration of \( A \) is \( [A] = 0.01 \text{ mol L}^{-1} \).

Step 3: Calculating the Reaction Rate

Substitute the given values into the rate equation: \[ \text{rate} = k[A] = 60 \times 10^{-4} \times 0.01 \]. Simplifying this, \( \text{rate} = 60 \times 10^{-6} \text{ mol L}^{-1} \text{ s}^{-1} \).

Step 4: Convert Rate from Seconds to Minutes

Since we need the rate in \( \text{mol L}^{-1} \text{ min}^{-1} \), convert from seconds to minutes: \( 60 \text{ s} = 1 \text{ min} \). Thus, \( 60 \times 10^{-6} \text{ mol L}^{-1} \text{ s}^{-1} \times 60 \text{ s min}^{-1} = 60 \times 10^{-4} \text{ mol L}^{-1} \text{ min}^{-1} \).

Key Concepts

Rate Constant

The rate constant, often represented by the symbol \( k \), is an essential component when dealing with reaction kinetics, especially in first-order reactions. For first-order reactions, the rate constant provides us with a quantitative measure of how fast the reaction is proceeding. It's specific to each reaction and depends on factors like temperature and the nature of the reactants.

In a first-order reaction, the rate constant has units of \( ext{s}^{-1} \), which signifies the change occurring per second. This is important as it helps differentiate between first-order and other orders of reactions since their constants will have different units.

The significance of the rate constant also extends to predicting the reaction's speed at any concentration of the reactant. For instance, if a reaction is known to have a high rate constant, it implies a fast reaction under the given conditions. Understanding and applying the rate constant effectively helps chemists control and optimize reactions in practical applications.

Reaction Rate Calculation

Reaction rate calculation is crucial for understanding how fast reactants are being converted into products. For first-order reactions, this calculation is straightforward, thanks to the relation: \( \text{rate} = k[A] \). Here, \( [A] \) represents the concentration of the reactant \( A \), and \( k \) is the rate constant.

By substituting the given values into the rate formula, we can accurately determine the reaction rate at any point. From the exercise, with \( k = 60 \times 10^{-4} \) and \( [A] = 0.01 \text{ mol L}^{-1} \), the calculation becomes very direct:

• Rate = \( 60 \times 10^{-4} \times 0.01 = 60 \times 10^{-6} \text{ mol L}^{-1} \text{ s}^{-1} \).

Calculating the reaction rate like this provides insights into how quickly the reaction proceeds at that specific concentration, guiding us in experimental setups and industry applications.

Conversion of Rate Units

In kinetics, converting units can sometimes be a necessity, especially when practical experimental data demands comparison in different time frames. Here, the rate was initially calculated in \( \text{mol L}^{-1} \text{ s}^{-1} \), but it was required in \( \text{mol L}^{-1} \text{ min}^{-1} \).

The conversion follows simple time arithmetic, where 1 minute is equivalent to 60 seconds. Thus, to convert from seconds to minutes, we multiply by 60:

• Rate = \( 60 \times 10^{-6} \text{ mol L}^{-1} \text{ s}^{-1} \times 60 \text{ s min}^{-1} = 60 \times 10^{-4} \text{ mol L}^{-1} \text{ min}^{-1} \).

This conversion is vital for aligning experimental data with theoretical expectations or industry standards, ensuring consistency across different analyses and applications.