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) \(3.51 \times 10^{-3} \mathrm{M}\)

Short answer

The concentration of C after 5 hours is approximately \(1.89 \times 10^{-2} \text{ M}\), matching option (b).

Step-by-step solution

Understand the Reaction Pathway

The reaction involves parallel first-order reactions where A decomposes to form products B and C independently. The rate constants for these reactions are given by \(\lambda_{1}\) and \(\lambda_{2}\), respectively.

Identify the Relevant Equation

For a first-order reaction, the concentration at any time \(t\) is given by the equation \([A] = [A]_0 e^{-\lambda t}\), where \([A]_0\) is the initial concentration. We use this equation separately for products B and C.

Calculate Decomposition for C

The concentration of C formed is given by \([C] = [A]_0 (1 - e^{-\lambda_2 t})\), where \([A]_0 = 0.25 \text{ M}\) and \(\lambda_2 = 5 \times 10^{-6} \text{ s}^{-1}\). First, calculate \(t\) in seconds: \(t = 5 \times 3600 = 18000 \text{ s}\).

Compute the Exponential Term

Calculate \( e^{-\lambda_2 t} = e^{-5 \times 10^{-6} \times 18000} \). Use a calculator to find \(e^{-0.09}\).

Calculate the Concentration of C

Substitute back into the formula: \([C] = 0.25 (1 - e^{-0.09})\). Compute \(e^{-0.09}\) approximately as \(0.9139\) and substitute to get \([C] = 0.25 (1 - 0.9139) = 0.25 \times 0.0861\).

Compute Final Value

Multiply to get the final concentration: \([C] = 0.25 \times 0.0861 = 0.021525 \text{ M}\). Compare this result to the given options.

Key concepts

Parallel Reactions

Parallel reactions occur when a single reactant can decompose along two or more pathways, resulting in different products. In the exercise provided, the initial reactant \(A\) undergoes decomposition through two parallel reactions. One part of \(A\) converts into product \(B\) while the other part transforms into product \(C\). This independent pathway mechanism means that each reaction proceeds at its own rate, as determined by its specific reaction rate constant. Parallel reactions are common in complex reaction systems, where different products form simultaneously from the same reactant.Understanding parallel reactions is crucial, especially since each pathway has its own independent rate constant, symbolized by different \(\lambda\) values. These constants, \(\lambda_1\) and \(\lambda_2\), help determine the rate at which \(A\) decomposes into either \(B\) or \(C\). It's essential to understand this distinction, as it shows how diverse products can emerge from one substrate under specific conditions.

Reaction Rate Constants

Reaction rate constants are essential in determining the speed of reaction pathways. These constants are fixed values that are determined experimentally and are distinct for different reactions. In first-order reactions, where the rate of reaction is directly proportional to the concentration of a single reactant, the rate constant is denoted by the Greek letter \(\lambda\).In the example problem, the rate constants \(\lambda_1\) and \(\lambda_2\) are given as \(1.5 \times 10^{-5} \text{ s}^{-1}\) and \(5 \times 10^{-6} \text{ s}^{-1}\) respectively. These values indicate the probability per unit time that the reactant \(A\) will convert to \(B\) or \(C\). A higher rate constant suggests a faster reaction; therefore, \(A\) is more likely to become \(B\) than \(C\) over the same time period, assuming other conditions remain constant.Understanding the role of the reaction rate constant can help predict how a reaction progresses and how the concentration of products varies over time. It is a fundamental concept in kinetics that informs the speed at which chemical changes occur.

Exponential Decay Formula

The exponential decay formula is crucial in analyzing first-order reactions. It describes how the concentration of a reactant decreases over time. For our specific scenario, this formula is expressed as:\[ [A] = [A]_0 e^{-\lambda t} \]where \([A]_0\) is the initial concentration, \([A]\) is the concentration at time \(t\), \(\lambda\) is the reaction rate constant, and \(t\) is the time elapsed.This formula shows that the concentration decreases exponentially, rather than linearly, over time. The use of the exponential constant \(e\) reflects nature’s tendency towards randomness and dissipation over time. This approach applies to both decomposition pathways for \(B\) and \(C\).In the exercise, the calculation for the product \(C\) involves determining \(e^{-\lambda_2 t}\), illustrating how much of the reactant \(A\) has converted after 5 hours. This exponential reduction highlights the typical loss of reactant concentration as the reaction progresses, one of the critical characteristics of first-order kinetic behavior.

Product Concentration Calculation

Calculating the concentration of a product in parallel reactions involves applying the exponential decay formula to focus on the specific decomposition pathway. For the product \(C\), the concentration at a given time \(t\) can be calculated using:\[ [C] = [A]_0 (1 - e^{-\lambda_2 t}) \]This equation isolates the formation of \(C\) based on the pathway related to \(\lambda_2\), considering how much initial reactant \(A\) hasn't decomposed yet into \(B\). The term \(1 - e^{-\lambda_2 t}\) represents the fraction of \(A\) converted to \(C\) after time \(t\).In our scenario, with a starting concentration of \(0.25 \text{ M}\), and given that \(\lambda_2 = 5 \times 10^{-6} \text{ s}^{-1}\), the solution involves converting 5 hours into seconds as 18000 s, calculating the exponential term \(e^{-0.09}\) and finally applying it to find \([C]\) as \(0.021525 \text{ M}\). Understanding these steps is key to accurately determining how much of each product is formed over time in systems involving parallel first-order reactions.