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A substance undergoes first-order decomposition, it follows two parallel reactions \(k_{1}=1.26 \times 10^{-4} \mathrm{~s}^{-1}\) and \(k_{2}=3.8 \times 10^{-5} \mathrm{~s}^{-1}\) The percentage distribution of \(\mathrm{Y}\) and \(\mathrm{Z}\) are (a) \(80 \% \mathrm{Y}\) and \(20 \% \mathrm{Z}\) (b) \(72.83 \% \mathrm{Y}\) and \(32.71 \% \mathrm{Z}\) (c) \(76.83 \% \mathrm{Y}\) and \(23.17 \% \mathrm{Z}\) (d) \(62.4 \% \mathrm{Y}\) and \(90.5 \% \mathrm{Z}\)

Short Answer

Expert verified
Option (c) is correct: 76.83% Y and 23.17% Z.

Step by step solution

01

Understanding Parallel Reactions

In this problem, a substance decomposes through two parallel first-order reactions. The rates of these reactions are given by the rate constants \( k_1 = 1.26 \times 10^{-4} \, \text{s}^{-1} \) for Reaction 1 and \( k_2 = 3.8 \times 10^{-5} \, \text{s}^{-1} \) for Reaction 2. These rate constants represent the proportion of the substance turning into products Y and Z per second.
02

Calculate Total Rate Constant

For parallel reactions, the total rate constant \( k_{\text{total}} \) is the sum of the rate constants for each reaction: \[ k_{\text{total}} = k_1 + k_2 = 1.26 \times 10^{-4} + 3.8 \times 10^{-5} = 1.64 \times 10^{-4} \, \text{s}^{-1} \]
03

Calculate Fraction of Product Y

The fraction of the decomposition leading to product Y is determined by the ratio of \( k_1 \) to the total rate constant: \[ \text{Fraction of Y} = \frac{k_1}{k_{\text{total}}} = \frac{1.26 \times 10^{-4}}{1.64 \times 10^{-4}} \approx 0.7683 \] This implies approximately 76.83% of the substance decomposes to form Y.
04

Calculate Fraction of Product Z

Similarly, the fraction leading to product Z is determined by the ratio of \( k_2 \) to the total rate constant: \[ \text{Fraction of Z} = \frac{k_2}{k_{\text{total}}} = \frac{3.8 \times 10^{-5}}{1.64 \times 10^{-4}} \approx 0.2317 \] Thus, approximately 23.17% of the substance decomposes to form Z.
05

Compare with Given Options

The calculated percentages of 76.83% for Y and 23.17% for Z match option (c). Therefore, option (c) is the correct distribution.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding First-Order Reactions
First-order reactions are a fundamental concept in chemical kinetics, where the rate of reaction is directly proportional to the concentration of a single reactant. This means that as the concentration of the reactant decreases, the rate of reaction decreases as well. In a first-order reaction, the rate law is expressed as:\[\text{Rate} = k[A]\]Where:- \(k\) is the rate constant specific to the reaction.- \([A]\) is the concentration of the reactant at any given time.An important feature of first-order reactions is that they have a constant half-life, irrespective of the initial concentration. The half-life \(t_{1/2}\) is calculated using the formula:\[t_{1/2} = \frac{0.693}{k}\]Understanding this concept is crucial when analyzing reactions that proceed through parallel pathways, as each pathway may have its own distinct rate constant, influencing the reaction's outcome.
Role of Rate Constants in Parallel Reactions
Rate constants, denoted as "\(k\)", are essential parameters in the study of reaction kinetics. They provide valuable information on the speed and dynamics of a chemical reaction. In the context of parallel reactions, rate constants determine how much of a reactant is converted into various products.For parallel reactions, multiple pathways exist for a single reactant, each with its own rate constant. The overall behavior of these reactions can be comprehended by calculating an effective rate constant:- If a reactant can form two different products via parallel first-order reactions with rate constants \(k_1\) and \(k_2\), the total rate constant \(k_{\text{total}}\) for the reactant's consumption can be calculated as: \[k_{\text{total}} = k_1 + k_2\]The fraction of each product formed in parallel reactions is proportional to the ratio of its pathway's rate constant to the total rate constant. Hence, analyzing these ratios unveils which pathways are more dominant.- \[\text{Fraction of Product} = \frac{k_n}{k_{\text{total}}}\]where \(k_n\) is the rate constant for a specific product's formation pathway. This insight allows prediction of the product distribution in the reaction.
Chemical Decomposition in Reactions
Chemical decomposition is a process where a single compound breaks down into two or more simpler substances. This transformation can occur through various mechanisms, with first-order and parallel reactions being prime examples. Decomposition often results in multiple reaction pathways, particularly when a substance converts into different products via parallel reactions. In such processes, the substance initially decomposes according to first-order kinetics, with separate pathways contributing to distinct products. These pathways are characterized by their individual rate constants, each indicating how quickly the decomposition occurs along that path. Key aspects of chemical decomposition: - **Parallel Pathways**: Multiple simultaneous reactions occurring from the same reactant, leading to different products. - **Rate Determination**: Each pathway is governed by its rate constant, affecting the proportion of products formed. - **Product Analysis**: The ratios of products formed provide insights into the dominant decomposition pathways. A sound understanding of chemical decomposition is essential for predicting outcomes in chemical reactions, particularly in industrial applications, where efficiency and product yield are crucial. By analyzing rate constants and reaction mechanisms, chemists can optimize conditions to favor the desired reaction pathways.

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Most popular questions from this chapter

The reaction \(\mathrm{X} \longrightarrow\) Product follows first-order kinetics, in 40 minutes, the concentration of \(\mathrm{X}\) changes from \(0.1 \mathrm{M}\) to \(0.025 \mathrm{M}\), then the rate of reaction when concentration of \(\mathrm{X}\) is \(0.01 \mathrm{M}\) is? (a) \(3.47 \times 10^{-5} \mathrm{M} / \mathrm{min}\) (b) \(1.73 \times 10^{-4} \mathrm{M} / \mathrm{min}\) (c) \(1.73 \times 10^{-5} \mathrm{M} / \mathrm{min}\) (d) \(3.47 \times 10^{-4} \mathrm{M} / \mathrm{min}\)

The basic theory of Arrhenius equation is that (1) activation energy and pre-exponential factors are always temperature independent (2) the number of effective collisions is proportional to the number of molecule above a certain thresh old energy. (3) as the temperature increases, the number of molecules with energies exceeding the threshold energy increases. (4) the rate constant in a function of temperature (a) 2,3 and 4 (b) 1,2 and 3 (c) 2 and 3 (d) 1 and 3

In a hypothetical reaction \(\mathrm{X} \rightarrow \mathrm{Y}\), the activation energy for the forward and backward reaction is 15 and \(9 \mathrm{~kJ}\) mol \(^{-1}\) respectively. The potential energy of \(X\) is \(10 \mathrm{~kJ}\) \(\mathrm{mol}^{-1}\), Identify the correct statement(s). (a) The threshold energy of the reaction is \(25 \mathrm{~kJ}\). (b) The potential energy \(\mathrm{fY}\) is \(16 \mathrm{~kJ}\) (c) Heat of reaction is \(6 \mathrm{~kJ}\). (d) The reaction is endothermic.

A first-order reaction is \(50 \%\) completed in 30 minutes at \(27^{\circ} \mathrm{C}\). Its rate constant is (a) \(2.31 \times 10^{-2} \mathrm{~min}^{-1}\) (b) \(3.21 \times 10^{-2} \mathrm{~min}^{-1}\) (c) \(4.75 \times 10^{-2} \mathrm{~min}^{1}\) (d) \(1.33 \times 10^{-3} \mathrm{~min}^{-1}\)

An aqueous solution of sugar undergoes acid catalysed hydrolysis. \(50 \mathrm{~g}\) sugar in \(125 \mathrm{~mL}\) water rotates the plane of plane polarized light by \(+13.1^{\circ}\) at \(\mathrm{t}=0 .\) After complete hydrolysis, it shows a rotation of \(-3.75^{\circ} .\) The percentage hydrolysis of sugar at time ' \(t\) ' in the same solution having a rotation of \(5^{\circ}\) is (a) \(42 \%\) (b) \(58 \%\) (c) \(48 \%\) (d) \(55 \%\)

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