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

The rate is \( 36 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \), option (b).

Step 1: Identify the Formula

For a first-order reaction, the rate is given by the formula \( r = k imes [A] \), where \( r \) is the rate, \( k \) is the rate constant, and \( [A] \) is the concentration of reactant A.

Step 2: Plug in the Known Values

Substitute the known values into the rate equation: \( k = 60 \times 10^{-4} \mathrm{~s}^{-1} \) and \( [A] = 0.01 \mathrm{~mol} \mathrm{~L}^{-1} \), thus \( r = 60 \times 10^{-4} \times 0.01 \).

Step 3: Calculate the Rate in \( \, \mathrm{mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \)

Perform the multiplication: \( r = 60 \times 10^{-4} \times 0.01 = 0.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \).

Step 4: Convert Units to \( \, \mathrm{mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \)

To convert \( 0.6 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} \) to \( \, \mathrm{mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \), recognize there are 60 seconds in a minute: \( r = 0.6 \times 10^{-4} \times 60 = 36 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \).

Step 5: Choose the Correct Answer

Now, compare the calculated result with the given options. The closest match to \( 36 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1} \) is option (b).

Key Concepts

Rate Constant

The rate constant, often represented as \( k \), is a key concept in chemical kinetics, especially for a first-order reaction. It is a proportionality factor in the rate equation that gives the rate of a reaction at a certain concentration of reactants.

For a first-order reaction, the equation is typically written as \( r = k \times [A] \), where \( r \) is the reaction rate and \([A]\) is the concentration of reactant A.

The units of the rate constant can provide insight into the reaction order. For a first-order reaction like the one we're discussing, the units of \( k \) are \( \text{s}^{-1} \). This indicates that the reaction's progress is directly related to the reciprocal of time. This reflects the nature of first-order reactions, where the rate is directly proportional to the concentration of one reactant.

Understanding the unit of the rate constant and its role in the rate equation is crucial because it helps in calculating the reaction rate and comparing different reactions.

Reaction Rate

The reaction rate describes how fast or slow a reaction proceeds. In the context of first-order reactions, the rate can be determined using the simple equation \( r = k \times [A] \).

This equation tells us that the reaction rate is directly proportional to the concentration of the reactant, \([A]\). For example, if you double the concentration of \(A\), the reaction rate will also double.

Reaction rates are usually expressed in \( \text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1} \), which translates to how many moles of reactant disappear per liter every second.

Understanding reaction rate is essential because it tells you how quickly products will be formed, which is important for controlling industrial processes or predicting the progress of a reaction under different conditions. In the given exercise, using the rate constant and concentration of \( A \), we calculated an initial reaction rate before performing any unit conversions.

Unit Conversion

Converting units is crucial in chemistry to ensure that the results are precise and applicable in real-world scenarios. Particularly in kinetic problems, it is essential to convert the rates into the desired units for practical use.

In this exercise, the initial calculation provided the reaction rate in \( \text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1} \). However, the options were given in \( \text{mol} \cdot \text{L}^{-1} \cdot \text{min}^{-1} \). To convert, you need to recognize that one minute equals 60 seconds.

• Multiply the rate by 60 to change seconds into minutes.

For instance, if you have a rate of \( 0.6 \times 10^{-4} \text{mol} \cdot \text{L}^{-1} \cdot \text{s}^{-1} \), multiplying by 60 changes it to the desired units: \(36 \times 10^{-4} \text{mol} \cdot \text{L}^{-1} \cdot \text{min}^{-1} \).

Correctly converting units ensures that your results can be accurately compared with other data or utilized in further calculations.