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Chapter 14: Problem 56

For the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) the activation energy \(\left(E_{a}\right)\) and overall \(\Delta E\) are 154 \(\mathrm{kJ} / \mathrm{mol}\) and 136 \(\mathrm{kJ} / \mathrm{mol}\) , respectively. (a) Sketch the energy profile for this reaction, and label \(E_{a}\) and \(\Delta E\) . (b) What is the activation energy for the reverse reaction?

Short Answer

Expert verified
The activation energy for the reverse reaction of the elementary process \(\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g)\) is 18 kJ/mol.

Step by step solution

01

Sketch the energy profile for the forward reaction

To sketch the energy profile, we will plot energy on the y-axis and the reaction progress on the x-axis. 1. Start by drawing the y-axis representing energy and the x-axis representing the reaction progress. 2. Locate the initial energy level of the reactants, N2O5, somewhere on the y-axis. 3. Then, draw an upward curve from the reactants' energy level, representing the activation energy for the forward reaction, Ea. The highest point on this curve represents the transition state. 4. Locate the final energy level of products, NO2 and NO3 somewhere lower on the y-axis (since \(\Delta E = -136 kJ/mol\) is negative, the energy of products is less than the energy of reactants). 5. Now, draw a downward curve from the transition state, representing the release of energy in the formation of products. 6. Label the initial energy level of N2O5 as E1, the transition state's energy as E2, and the final energy level of products as E3. 7. Label the activation energy, Ea, as the difference between E2 (transition state's energy) and E1 (N2O5 energy level), and the overall energy change, ∆E, as the difference between E3 (products' energy) and E1 (N2O5 energy level).
02

Determine the activation energy for the reverse reaction

To find the activation energy for the reverse reaction (Ea_reverse), we will use the relationship between activation energy, reaction energy change, and activation energy for the reverse reaction: Ea_reverse = Ea - ∆E Given that Ea = 154 kJ/mol and ∆E = 136 kJ/mol, we can calculate Ea_reverse Ea_reverse = 154 kJ/mol - 136 kJ/mol = 18 kJ/mol The activation energy for the reverse reaction is 18 kJ/mol.

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Key Concepts

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

Chemical Kinetics
Chemical kinetics is the branch of chemistry that deals with the rates of chemical reactions and the factors that affect these rates. It’s critical for students to understand that the rate at which a chemical reaction proceeds can depend on various factors, including concentration of reactants, temperature, and the presence of catalysts.

For instance, an increase in temperature usually accelerates a reaction because it provides the particles with more energy to overcome the activation energy barrier. Similarly, higher concentrations of reactants can lead to more frequent collisions, which might increase the rate of reaction. Catalysts, on the other hand, lower the activation energy needed for the reaction to take place without being consumed themselves, hence increasing the reaction rate.

Understanding the principles of chemical kinetics can help students predict and control the outcomes of chemical processes, which is essential in various industries such as pharmaceuticals, energy, and materials science.
Reaction Progress
When we talk about reaction progress in chemistry, we refer to the course a reaction takes from reactants to products. This is not just the physical transformation but also the energy changes that occur during the process.

During a reaction progress, the system goes through a high energy state called the transition state. This state is not the same as intermediates, which may occur in multi-step reactions. The transition state is the peak of energy that reactants must achieve before transforming into products. The difference in energy between reactants and the transition state is known as the activation energy, and it's a critical concept because it determines the rate at which the reaction will occur.

Understanding reaction progress helps students grasp the concept of reaction mechanisms and the detailed sequence of events that lead to the formation of products.
Energy Profile Diagram
An energy profile diagram is a graphical representation used to visualize the energy changes during a chemical reaction. It plots the energy on the y-axis against the reaction progress on the x-axis.

The initial energy level corresponds to the energy of the reactants, and the final level to the products. The activation energy is represented by the peak of the curve from the reactants to the transition state. The overall energy change, represented by ∆E, is the difference in energy between the reactants and products. If ∆E is negative, it indicates an exothermic reaction where energy is released to the surroundings. Conversely, a positive ∆E indicates an endothermic reaction where energy is absorbed from the surroundings.

These diagrams are an invaluable tool for students to quickly understand the energy landscape of a reaction and judge whether a reaction is likely to occur spontaneously.
Endothermic and Exothermic Reactions

Understanding Heat Flow

In chemistry, reactions can generally be classified as endothermic or exothermic, depending on whether they absorb or release energy in the form of heat. This concept is crucially tied both to reaction progress and energy profiles.

  • Endothermic Reactions: These reactions require an input of energy to proceed. The energy is absorbed from the surroundings, typically as heat, hence resulting in a cooling effect. On an energy profile diagram, endothermic reactions are depicted with a final energy level that is higher than the initial energy level, indicating that energy has been absorbed.
  • Exothermic Reactions: These release energy to the surroundings, often in the form of heat, which can be felt as warmth. On an energy profile diagram, exothermic reactions will have a final energy level lower than the initial energy level, showing that energy has been released into the surroundings.

The direction of heat flow not only determines the thermal characteristic of the reaction but also influences the reaction rate and can be a deciding factor in the spontaneity of the reaction's forward or reverse processes.

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Most popular questions from this chapter

(a) What are the units usually used to express the rates of reactions occurring in solution? (b) As the temperature increases, does the reaction rate increase or decrease? (c) As a reaction proceeds, does the instantaneous reaction rate increase or decrease?

Platinum nanoparticles of diameter \(\sim 2 \mathrm{nm}\) are important catalysts in carbon monoxide oxidation to carbondioxide. Platinum crystallizes in a face-centered cubic arrangement with an edge length of 3.924 A. (a) Estimate how many platinum atoms would fit into a 2.0 -nm sphere; the volume of a sphere is \((4 / 3) \pi r^{3} .\) Recall that \(1 \hat{\mathrm{A}}=1 \times 10^{-10} \mathrm{m}\) and \(1 \mathrm{nm}=1 \times 10^{-9} \mathrm{m} .\) (b) Estimate how many platinum atoms are on the surface of a \(2.0-\mathrm{nm}\) Pt sphere, using the surface area of a sphere \(\left(4 \pi r^{2}\right)\) and assuming that the "footprint" of one Pt atom can be estimated from its atomic diameter of 2.8 A. (c) Using your results from (a) and (b), calculate the percentage of Pt atoms that are on the surface of a 2.0 -nm nanoparticle. (d) Repeat these calculations for a 5.0 -nm platinum nanoparticle. (e) Which size of nanoparticle would you expect to be more catalytically active and why?

(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?

(a) For a generic second-order reaction \(\mathrm{A} \longrightarrow \mathrm{B}\) , what quantity, when graphed versus time, will yield a straight line? (b) What is the slope of the straight line from part line? (b) What is the slope of the straight line from part (a)? (c) Does the half-life of a second-order reaction increase, decrease, or remain the same as the reaction proceeds?

The activation energy of an uncatalyzed reaction is 95 \(\mathrm{kJ} / \mathrm{mol} .\) The addition of a catalyst lowers the activation energy to 55 \(\mathrm{kJ} / \mathrm{mol}\) . Assuming that the collision factor remains the same, by what factor will the catalyst increase the rate of the reaction at (a) \(25^{\circ} \mathrm{C},\) (b) \(125^{\circ} \mathrm{C} ?\)
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