Problem 60 - Advanced Problems in Physical Chemistry for Competitive Examinations

Question

An organic compound A decomposes following two parallel first-order reactions: \(\mathrm{A} \stackrel{K_{1}}{\longrightarrow} \mathrm{B}\) and \(\mathrm{A} \stackrel{K_{2}}{\longrightarrow} \mathrm{C}\). If \(K_{1}\) is \(1.25 \times 10^{-5} \mathrm{~s}^{-1}\) and \(\frac{K_{1}}{K_{2}}=\frac{1}{9}\), the value of \(\frac{[\mathrm{C}]}{[\mathrm{A}]}\) after one hour of start of reaction by taking only \(A\), is \((\ln 1.568=0.45)\) (a) \(\frac{1}{9}\) (b) \(0.5112\) (c) \(1.4112\) (d) \(\frac{9}{20}\)

Step-by-Step Solution

Answer

(b) \(0.5112\)

Step 1 - Calculate the rate constant for the second reaction

Given the ratio of the rate constants, \(K_1/K_2 = 1/9\), we can solve for \(K_2\) as follows: \[K_2 = K_1 * 9 = (1.25 \times 10^{-5} \text{s}^{-1}) \times 9\]

Step 2 - Deduce the combined rate constant for the reactions

Since the reactions are parallel first-order reactions, the overall rate constant \(K\) will be the sum of the individual rate constants: \[K = K_1 + K_2\]

Step 3 - Calculate the concentration ratio \(\frac{[C]}{[A]}\) after one hour

Using the first-order integrated rate law for decomposition of A, \(\frac{[C]}{[A]} = \frac{k_2}{k} (1 - e^{-kt})\), where \([A]\) is the initial concentration of A, \([C]\) is the concentration of C at time \({t}\), and \(t\) is 3600 seconds (1 hour). Plug in the values for the rate constants and time to calculate the ratio.

Step 4 - Select the correct answer

Compare the calculated value with the provided options and select the one that matches your calculated value of \(\frac{[C]}{[A]}\).

Key Concepts

Rate Constant Calculation

Understanding how to calculate the rate constant for a chemical reaction is crucial in the study of reaction kinetics. The rate constant, typically symbolized by the letter 'k', is a quantitative measure of how quickly a reaction occurs. In the given exercise, an organic compound A decomposes through two parallel first-order reactions with rate constants 'k1' and 'k2'.

For first-order reactions, the rate of reaction is directly proportional to the concentration of the reactant. Thus, knowing 'k1', we can calculate 'k2' given the ratio \(\frac{k1}{k2}=\frac{1}{9}\). By multiplying both sides of this equation by 'k1' and rearranging, we derive 'k2' as \(k1 * 9\). This step is vital because it allows us to analyze the second reaction independently and relate it to the first.

Example of Calculating 'k2' from 'k1'

To calculate 'k2' using 'k1', simply multiply 'k1' by the given ratio. If 'k1' is \(1.25 \times 10^{-5} s^{-1}\), then:
\[k2 = 1.25 \times 10^{-5} \times 9 = 1.125 \times 10^{-4} s^{-1}\]This calculation is the cornerstone for further understanding the kinetics of both reactions occurring simultaneously.

Integrated Rate Law

The integrated rate law for first-order reactions is a key concept in chemical kinetics, relating the concentration of reactants to time and the rate constant. It allows us to track the change in concentration of reactants over time. For a first-order reaction, the law takes the form:\[\ln\left(\frac{[A]_0}{[A]}\right) = kt\]where \( [A]_0 \) is the initial concentration of the reactant, 'A', \( [A] \) is the concentration of 'A' after time 't', and 'k' is the first-order rate constant.

In the context of our exercise with compound A decomposing into products B and C, we can use the integrated rate law to determine the concentration ratio \(\frac{[C]}{[A]}\) after one hour. The formula we use in this scenario is a modification for parallel reactions:\[\frac{[C]}{[A]} = \frac{k2}{k} (1 - e^{-kt})\]where 'k' is the combined rate constant for the parallel reactions (the sum of 'k1' and 'k2').

Applying the Integrated Rate Law

With the known rate constants and the elapsed time, we can plug these values into the integrated rate law to find out how much of A has been converted to C after one hour. This application illustrates the practical use of the integrated rate law in analyzing the progression of a reaction over time, which is critical for researchers and professionals in the field of chemistry.

Reaction Kinetics

Reaction kinetics is the field of chemistry that studies the rates at which chemical reactions proceed and the factors that affect these rates. Kinetics provides insights into the speeds of reaction, the mechanism by which reactions occur, and how different variables—such as concentration, temperature, and the presence of a catalyst—impact these rates.

Understanding kinetics is essential when examining reactions such as the parallel decomposition of compound A. It enables chemists to predict how long a reaction will take to reach a certain point, or to determine the yield of products at any given time. In the case of our parallel first-order reactions, reaction kinetics allows us to analyze the simultaneous conversion of A into B and C through two different pathways, each characterized by their individual rate constants 'k1' and 'k2'.

Significance in Parallel Reactions

In parallel reactions, like those involved in the decomposition of compound A, reaction kinetics helps us understand how each pathway contributes to the overall reaction rate and product distribution. It guides the calculation of the overall rate constant and further allows us to use the integrated rate law to relate reaction progress to time. This fundamental understanding is essential for controlling reaction conditions in industrial processes and scientific research.