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For two parallel first-order reactions, what is the overall activation energy of reaction? The yields of \(\mathrm{B}\) and \(\mathrm{C}\) in products are \(40 \%\) and \(60 \%\), respectively. \(\mathrm{A} \stackrel{\mathrm{Ea}=20 \mathrm{kcal} / \mathrm{mol}}{\longrightarrow} \mathrm{B} \quad \mathrm{A} \stackrel{\mathrm{Ea}=40 \mathrm{kcal} / \mathrm{mol}}{\longrightarrow} \mathrm{C}\) (a) \(60 \mathrm{kcal} / \mathrm{mol}\) (b) \(32 \mathrm{kcal} / \mathrm{mol}\) (c) \(28 \mathrm{kcal} / \mathrm{mol}\) (d) \(20 \mathrm{kcal} / \mathrm{mol}\)

Short Answer

Expert verified
The overall activation energy of the reaction is 32 kcal/mol.

Step by step solution

01

Understand the concept of overall activation energy

For parallel reactions, the overall activation energy can be considered as a weighted average of the individual activation energies given the products formed. The weights are the yields of each product.
02

Calculate the weighted average of the activation energies

Use the formula for weighted average with the activation energies and yields as weights: Overall Activation Energy = (Yield of B * Activation Energy for B) + (Yield of C * Activation Energy for C).
03

Insert the given values and calculate the overall activation energy

Insert the given yields and activation energies: Overall Activation Energy = (0.40 * 20 kcal/mol) + (0.60 * 40 kcal/mol). Calculate the weighted average: Overall Activation Energy = (8 kcal/mol) + (24 kcal/mol) = 32 kcal/mol.

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Key Concepts

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

Understanding Parallel First-Order Reactions
Parallel first-order reactions occur when a single reactant can follow multiple pathways to form different products with each reaction pathway being a first-order reaction. This is characterized by the reactant's ability to transform into various products at rates that depend on the concentration of the reactant only, not on the concentrations of the products.

In the context of the exercise, reactant A has two pathways, one leading to product B and one to product C. Each of these pathways follows a first-order kinetic rate law. This implies that the rate of formation of each product is directly proportional to the concentration of A, without any influence from one path on the other. To predict the behavior and yield of these reactions, it's essential to understand both the kinetics of the reactions and the conditions under which they occur. These insights can help in optimizing the conditions to obtain the desired product more efficiently.
Activation Energies Explained
Activation energy, denoted as Ea, is the minimum energy that must be overcome for a chemical reaction to occur. Think of it like the energy needed to push a ball over a hill so that it can roll down the other side. The concept of activation energy is crucial in understanding how different conditions, like temperature, can affect the rate of a reaction.

In our exercise, the Ea for the formation of products B and C from A are different, being 20 kcal/mol and 40 kcal/mol, respectively. This difference explains why the reactions might occur at different rates under the same conditions. The higher the activation energy, the fewer molecules will have enough energy to reach the transition state, a high-energy state that precedes the formation of products, resulting in a slower reaction rate.
Chemical Kinetics Fundamentals
Chemical kinetics is the area of chemistry that studies the rates of chemical reactions, the factors that affect these rates, and the mechanisms by which the reactions proceed. It provides insights into the step-by-step process of a reaction at the molecular level, which is vital for controlling and predicting the outcomes of chemical processes.

Referring to the given exercise, kinetics helps us understand how the reaction rates of A turning into B and C are governed by their first-order nature. It explains how the reaction speed is influenced by the reactant concentration and the activation energy. The whole concept of calculating overall activation energy comes from the understanding of these kinetic principles, shedding light on which reaction will dominate under specific conditions and thus predicting the yields of B and C.
Solving Physical Chemistry Problems
Tackling physical chemistry problems, such as the one in our exercise, requires a methodical approach. Breaking down complex problems into manageable steps is key to understanding and solving them. In this case, we began by comprehensively understanding the concept of overall activation energy, followed by the calculation using a weighted average based on product yields.

Working through physical chemistry problems often involves a mix of conceptual understanding and mathematical calculations. By correctly applying the weighted average formula, we can calculate that the overall activation energy for the formation of products B and C from A is 32 kcal/mol. Developing strong problem-solving strategies can greatly aid students in not just answering textbook questions, but also in applying these concepts to real-world chemistry challenges.

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Most popular questions from this chapter

As the initial concentration increases from \(0.75\) to \(1.55 \mathrm{M}\) in a reaction, \(t_{1 / 2}\) decreases from 60 to \(29 \mathrm{~s}\). The order of the reaction is (a) zero (b) first (c) second (d) third

If \(I\) is the intensity of absorbed light and \(C\) is the concentration of \(\mathrm{AB}\) for the photochemical process: \(\mathrm{AB}+h \mathrm{v} \rightarrow \mathrm{AB}^{*}\), the rate of formation of \(\mathrm{AB}^{*}\) is directly proportional to (a) \(C\) (b) \(I\) (c) \(I^{2}\) (d) \(C \cdot I\)

The activation energy of a reaction can be lowered by (a) increasing temperature (b) lowering temperature (c) adding a catalyst (d) removing one or more products.

In \(80 \%\) ethanol at \(55^{\circ} \mathrm{C}\), isopropyl bromide reacts with hydroxide ion according to the following kinetics: $$ \begin{array}{l} -\frac{\mathrm{d}[\mathrm{RX}]}{\mathrm{d} t}=\left(4.8 \times 10^{-5} \mathrm{M}^{-1} \mathrm{~s}^{-1}\right) \\ {[\mathrm{RX}]\left[\mathrm{OH}^{-}\right]+2.4 \times 10^{-6} \mathrm{~s}^{-1}[\mathrm{RX}]} \end{array} $$ What percentage of isopropyl bromide reacts by the \(S_{\mathrm{N}_{2}}\) mechanism when \(\left[\mathrm{OH}^{-}\right]=0.01 \mathrm{M} ?\) (a) \(16.67 \%\) (b) \(83.33 \%\) (c) \(66.67 \%\) (d) \(33.33 \%\)

The suggested mechanism for the reaction: \(\mathrm{CHCl}_{3}(\mathrm{~g})+\mathrm{Cl}_{2}(\mathrm{~g}) \rightarrow \mathrm{CCl}_{4}(\mathrm{~g})\) \(+\mathrm{HCl}(\mathrm{g})\), is $$ \mathrm{Cl}_{2} \underset{K_{2}}{\stackrel{K_{1}}{\leftrightarrows}} 2 \mathrm{C} \mathrm{C} \text { (fast) } $$ \(\mathrm{CHCl}_{3}+\mathrm{Ci} \stackrel{K_{3}}{\longrightarrow} \mathrm{HCl}+\dot{\mathrm{C}} \mathrm{Cl}_{3}(\mathrm{slow})\) $$ \dot{\mathrm{C}} \mathrm{Cl}_{3}+\mathrm{C} \mathrm{\textrm{l }} \stackrel{K_{4}}{\longrightarrow} \mathrm{CCl}_{4} \text { (fast) } $$ The experimental rate law consistent with the mechanism is (a) rate \(=K_{3}\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]\) (b) rate \(=K_{4}\left[\mathrm{CCl}_{3}\right][\mathrm{Cl}]\) (c) rate \(=K_{\text {eq }}\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]\) (d) rate \(=K_{3} K_{\text {eq }}^{1 / 2}\left[\mathrm{CHCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]^{1 / 2}\)

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