Problem 206 - A Complete Resource Book in Chemistry for JEE Main

Question

In a first-order reaction, the concentration of the reactant, decreases from \(0.8 \mathrm{M}\) to \(0.4 \mathrm{M}\) in 15 minutes. The time taken for the concentration to change from \(0.1 \mathrm{M}\) to \(0.025 \mathrm{M}\) is (a) 30 minutes (b) 60 minutes (c) \(7.5\) minutes (d) 15 minutes

Step-by-Step Solution

Answer

The time taken is 30 minutes (option a).

Step 1: Understanding First-Order Reactions

First-order reactions have a rate that is directly dependent on the concentration of a single reactant. The rate law for a first-order reaction is typically expressed as \(\text{rate} = k [A]\), where \([A]\) is the concentration and \(k\) is the rate constant.

Step 2: Using the First-Order Integrated Rate Law

The integrated rate law for a first-order reaction is \( \ln([A]_t) = \ln([A]_0) - kt \), where \([A]_0\) is the initial concentration, \([A]_t\) is the concentration at time \(t\), and \(k\) is the rate constant.

Step 3: Calculate the Rate Constant (k)

Given the initial concentration \([A]_0 = 0.8\) M and \([A]_t = 0.4\) M after 15 minutes, use the formula \( t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_t}\right) \) to find \(k\).\[ 15 = \frac{1}{k} \ln\left(\frac{0.8}{0.4}\right) \]Solving for \(k\),\[ k = \frac{\ln(2)}{15} \]

Step 4: Calculate Time for New Concentration Change

To find the time \( t \) for the concentration change from \(0.1\ M\) to \(0.025\ M\), substitute into \( t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_t}\right) \):\[ t = \frac{15}{\ln(2)} \ln\left(\frac{0.1}{0.025}\right) \]Calculate \( \ln\left(\frac{0.1}{0.025}\right) = \ln(4) \). Thus,\[ t = \frac{15}{\ln(2)} \cdot \ln(4) = 30 \]

Step 5: Compare with Given Options

Based on the calculated time, the time for the concentration change from \(0.1\ M\) to \(0.025\ M\) is 30 minutes. Thus, the correct answer is option (a) 30 minutes.

Key Concepts

Rate Constant (k)

In first-order reactions, the rate of reaction is directly proportional to the concentration of one reactant. The rate constant, denoted by \( k \), plays a pivotal role here. It's a measure of how quickly a reaction proceeds. Think of it as a speed dial for the reaction — a higher \( k \) means a faster reaction.
First-order reactions have the simplest rate equation, which looks like this:

• Rate = \(k[A]\)

The symbol \([A]\) represents the concentration of the reactant. The unit for \( k \) in first-order reactions is typically \( s^{-1} \) or \( min^{-1} \), varying based on what time unit you prefer.
Understanding \( k \) helps chemists and students predict how long a reaction will take under specific conditions. It's the key springboard for using further calculations in kinetic studies.

Integrated Rate Law

The integrated rate law offers a way to link concentration with time for first-order reactions. When dealing with kinetics, this form of the rate law is especially useful. It is represented by the equation:

• \( \ln([A]_t) = \ln([A]_0) - kt \)

Here, \([A]_t\) is the concentration at time \( t \), \([A]_0\) represents the initial concentration, and \( k \) is the trusty rate constant.
This equation can be rearranged to find time, initial concentration, or \( k \). It is incredibly handy for calculating the time required to reach a certain concentration, as showcased in the exercise. The equation illustrates how reactant concentration decreases over time logarithmically rather than linearly.

Concentration Change Calculation

In the exercise, we looked at how the concentrations drop from different starting points down to new levels within a specific timeframe. Calculating this time, or assessing changes like these, hinges on using the integrated rate law.
By inserting different values for initial and final concentrations, and leveraging the known rate constant \( k \), it is straightforward to predict when a particular concentration will be reached. This is what we tackled through the equation:

• \( t = \frac{1}{k} \ln\left(\frac{[A]_0}{[A]_t}\right) \)

This formula helps calculate how long it takes for the concentration of \( A \) to reduce from \(0.1 \ M\) to \(0.025 \ M \).
Concentration changes like this serve as a practical example of how reactions progress over time and inform decision-making in both industrial and research settings.