Problem 59 - A Complete Resource Book in Chemistry for JEE Main

Question

\(75 \%\) of a first-order reaction was completed in 32 min. When was \(50 \%\) of the reaction completed? (a) \(24 \mathrm{~min}\) (b) \(16 \mathrm{~min}\) (c) \(8 \mathrm{~min}\) (d) \(64 \mathrm{~min}\)

Step-by-Step Solution

Answer

(b) 16 min

Step 1: Understand the First-Order Reaction

In a first-order reaction, the rate of reaction is directly proportional to the concentration of one reactant. The integrated rate law for a first-order reaction is given by: \[ [A]_t = [A]_0 e^{-kt} \] where \( [A]_t \) is the concentration at time \( t \), \( [A]_0 \) is the initial concentration, and \( k \) is the rate constant. The equation can also be expressed in terms of the initial concentration and the remaining concentration as ln \( \frac{[A]_0}{[A]_t} = kt. \)

Step 2: Relate Percentage Completion to Concentration

For a first-order reaction, the fraction of reactant remaining at time \( t \) is given by \([A]_t / [A]_0\). So, when the reaction is 75% complete, 25% of the reactant remains. This means: \( \frac{[A]_t}{[A]_0} = 0.25\).

Step 3: Calculate the Rate Constant

Using the first-order equation ln \( \frac{[A]_0}{[A]_t} = kt \), and substituting \( \frac{[A]_t}{[A]_0} = 0.25 \) and \( t = 32 \) min, we find \( k \): \[ \ln \left( \frac{1}{0.25} \right) = k \times 32 \] Solve for \( k \): \[ k = \frac{\ln(4)}{32} \approx 0.0433 \, \text{min}^{-1} \]

Step 4: Apply the Rate Constant to 50% Completion

When the reaction is 50% complete, 50% of the reactant remains, so \( \frac{[A]_t}{[A]_0} = 0.5 \). Using the equation: \[ \ln \left( \frac{1}{0.5} \right) = k \times t \] We already know \( k \), so substitute and solve for \( t \): \[ \ln(2) = 0.0433 \times t \] \[ t = \frac{\ln(2)}{0.0433} \approx 16 \text{ min} \]

Step 5: Verify the Solution

Verify the result by ensuring consistency with the equation used and the calculated rate constant. Since \( t \approx 16 \text{ min} \) provides a logical outcome for the time at which the reaction is 50% complete, the calculation is confirmed.

Key Concepts

Rate Constant

In the study of chemical kinetics, the rate constant, denoted as \( k \), is a crucial parameter. It reflects how fast or slow a reaction progresses. For first-order reactions, the rate constant is defined based on the time it takes for the concentration of a reactant to decrease. This relationship can be expressed simply using the integrated rate law.

• The rate constant \( k \) has units of \( ext{time}^{-1} \), such as \( ext{min}^{-1} \). This signifies that it determines how quickly the concentration of reactants changes over time.
• It remains consistent as long as the temperature is stable, making it a reliable measure for predicting reaction progress.

In our exercise, we calculated \( k \) by understanding that at 75% completion (where only 25% of the original reactant remains), the rate constant becomes \( k = \frac{\ln(4)}{32} \), roughly \( 0.0433 \text{ min}^{-1} \). This calculation showcases how precise the role of \( k \) can be in determining how long a reaction takes to progress to a certain point.

Integrated Rate Law

The integrated rate law is an essential component for tackling any first-order reaction problem. It provides a mathematical relationship that relates the concentration of reactants at a given time to the initial concentration, using the rate constant.

• The general form for a first-order reaction is given by \( [A]_t = [A]_0 e^{-kt} \), which simplifies to \( \ln \left( \frac{[A]_0}{[A]_t} \right) = kt \).

Using this formula, we can deduce how much of a reactant remains after a certain time. Thus, it becomes straightforward to solve for the unknowns if the concentration at a particular time or the rate constant is known.
For example, understanding that 75% completion means 25% of reactant remains allows us to plug values into the integrated rate law and solve for the time or rate constant, enhancing our ability to predict future concentrations.

Percentage Completion

Percentage completion in chemical reactions provides a practical way to gauge how far along a reaction has progressed. It helps in visualizing and calculating the extent of reactant transformation over time.

• A 50% completion means half of the starting reactant has undergone the reaction, which in terms of concentration means \( [A]_t = \frac{1}{2}[A]_0 \).
• Similarly, 75% completion implies that 25% of the initial reactant concentration remains.

In the given exercise, evaluating percentage completion allowed us to determine certain intervals of time when these completions occur. Utilizing the integrated rate law, we juxtaposed the percentages to find out at what time exactly 50% of the reaction was completed, leading to the use of the rate constant to back-calculate to a time of approximately 16 minutes. Understanding the concept of percentage completion enables us to anticipate and confirm the stages of a reaction accurately.