Problem 146 - A Complete Resource Book in Chemistry for JEE Main

Question

A follows parallel path Ist order reactions giving B and C as shown: If initial concentration of \(\mathrm{A}\) is \(0.25 \mathrm{M}\), calculate the concentration of \(\mathrm{C}\) after 5 hour of reaction. Given, \(\lambda_{1}=1.5 \times 10^{-5} \mathrm{~s}^{-1}, \lambda_{2}=5 \times 10^{-6} \mathrm{~s}^{-1}\) (a) \(7.55 \times 10^{-3} \mathrm{M}\) (b) \(1.89 \times 10^{-2} \mathrm{M}\) (c) \(5.53 \times 10^{-3} \mathrm{M}\) (d) \(3.51 \times 10^{-3} \mathrm{M}\)

Step-by-Step Solution

Answer

The concentration of C after 5 hours is approximately \( 1.89 \times 10^{-2} \ M \). Option (b) is correct.

Step 1: Understand the Problem

We are given that substance A decomposes into B and C through parallel first-order reactions. We have two rate constants \( \lambda_1 \) and \( \lambda_2 \) and need to calculate the concentration of C after 5 hours, given the initial concentration of A.

Step 2: Convert Time Units

Convert the time from hours to seconds, because the rate constants are given in seconds. Since 1 hour = 3600 seconds, 5 hours = \( 5 \times 3600 = 18000 \) seconds.

Step 3: First-Order Reaction Formula

For a first-order reaction, the formula for concentration at time \( t \) is \( [A]_t = [A]_0 e^{-\lambda t} \). We will find the concentration of \( A \) after 18000 seconds using this formula.

Step 4: Calculate Remaining A

The remaining concentration of \( A \) after time \( t \) considering both reactions is \( [A]_t = [A]_0 e^{-(\lambda_1 + \lambda_2)t} \). Substitute to find \( [A]_t = 0.25 e^{-(1.5 \times 10^{-5} + 5 \times 10^{-6}) \times 18000} \).

Step 5: Evaluate the Exponential Expression

Calculate the expression: \( \lambda_1 + \lambda_2 = 2 \times 10^{-5} \ s^{-1} \) and \( (\lambda_1 + \lambda_2) \times 18000 = 0.36 \). Thus, \( [A]_t = 0.25 e^{-0.36} \).

Step 6: Calculate Remaining A

Compute \( e^{-0.36} \approx 0.6977 \), therefore, \( [A]_t \approx 0.25 \times 0.6977 = 0.174425 \ M \). This is the concentration of A remaining after 5 hours.

Step 7: Calculate the Concentration of C

The concentration of \( C \) formed is \( [C] = [A]_0 \frac{\lambda_2}{\lambda_1 + \lambda_2}(1 - e^{-(\lambda_1 + \lambda_2)t}) \). Substitute \( \lambda_2 = 5 \times 10^{-6}, \Lambda = 2 \times 10^{-5} \) and evaluate: \( [C] = 0.25 \times \frac{5 \times 10^{-6}}{2 \times 10^{-5}} (1 - 0.6977) \).

Step 8: Calculate Final Result

Perform the calculation to find \( [C] = 0.0625 \times 0.3023 = 0.01889375 \ M \). Expressing it in scientific notation gives approximately \( 1.89 \times 10^{-2} \ M \).

Step 9: Select the Correct Option

Compare calculated concentration with given options. We find that the concentration closest to our calculation is option (b) \( 1.89 \times 10^{-2} \ M \).

Key Concepts

Parallel reactions

In chemistry, parallel reactions occur when a single reactant can undergo two or more reactions simultaneously, producing different products. For instance, in the given exercise, substance A decomposes into B and C via two separate reaction pathways. This is a classic example of parallel first-order reactions. Understanding these reactions is crucial, as it demonstrates how a single reactant can affect multiple outcomes in a chemical reaction. By considering the different paths and their respective rates, chemists can predict the amounts of various products formed over time, helping in the design and control of chemical processes.

Rate constant

The rate constant, often denoted by the Greek letter lambda (\( \lambda \)), is a crucial factor in determining the speed of a reaction. It quantifies the rate at which a reaction occurs. In first-order reactions, the rate constant has units of time inverse, commonly in seconds (\( s^{-1} \)). The rate constant varies for different reactions, explaining why some reactions occur faster than others, even when they involve the same reactant. Understanding and calculating the rate constants for parallel reactions allows chemists to estimate the concentration of products formed at various times during the reaction.

Exponential decay

Exponential decay in chemical reactions describes how the concentration of a reactant decreases over time. For first-order reactions, this relationship is expressed by the formula \( [A]_t = [A]_0 e^{-\lambda t} \). This equation shows that the concentration of the reactant decreases exponentially as time progresses. Exponential decay is vital in predicting how much of a reactant remains after a certain period, which is crucial for reactions like those in the exercise where multiple pathways affect the total reaction progress. This concept is widely applicable, encompassing scenarios from radioactive decay in nuclear chemistry to the clearance of drugs in pharmacokinetics.

Chemical kinetics

Chemical kinetics is the study of rates of chemical processes and the factors affecting them. It involves the examination of how different variables—such as temperature, concentration, and pressure—influence the speed of a chemical reaction. Understanding kinetics helps in determining reaction mechanisms, which can explain the steps and intermediates the reactants go through before forming the final products. The exercise highlights first-order kinetics, often characterized by their simplicity, making them a good starting point for studying more complex reactions involving multiple steps or competing reaction paths. Learning chemical kinetics allows scientists to optimize conditions for industrial processes, predict side effects in drug development, and elucidate natural processes occurring in the environment.