/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 The gas-phase reaction \(\mathrm... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The gas-phase reaction \(\mathrm{Cl}(g)+\operatorname{HBr}(g) \longrightarrow \mathrm{HCl}(g)+\) \(\mathrm{Br}(g)\) has an overall enthalpy change of \(-66 \mathrm{~kJ}\). The activation energy for the reaction is \(7 \mathrm{~kJ}\). (a) Sketch the energy profile for the reaction, and label \(E_{g}\) and \(\Delta E\). (b) What is the activation energy for the reverse reaction?

Short Answer

Expert verified
The activation energy for the reverse reaction is \(73 \mathrm{~kJ}\).

Step by step solution

01

Sketch the energy profile of the reaction

First, we need to plot the energy levels of the reactants and products in the energy profile diagram. Let's start with the reactant energy level where Cl(g) + HBr(g) are the reactants and set the energy level to zero as a reference. The overall enthalpy change (ΔE) for the reaction is -66 kJ, meaning the products have lower energy than the reactants. As the enthalpy change is negative, draw a horizontal line indicating the energy level of the products (HCl(g) + Br(g)) at -66 kJ. Now, we can show the activation energy (Eg) for the forward reaction. Eg is the difference between the energy of the transition state and the reactants' energy level. Since the activation energy is given to be 7 kJ, draw a peak at 7 kJ, representing the transition state.
02

Label Eg and ΔE

Now label the activation energy (Eg) and overall enthalpy change (ΔE) on the energy profile. The activation energy (Eg) is represented by the vertical distance between the energy level of the reactants and the transition state (7 kJ). ΔE is the difference in energy between the reactants and products and is indicated by the vertical distance between the two energy levels (-66 kJ).
03

Calculate the activation energy for the reverse reaction

Let Egr represent the activation energy for the reverse reaction. To find Egr, we should first determine the energy of the transition state for the reverse reaction. Since the reverse reaction is the opposite of the forward reaction, the overall enthalpy change for the reverse reaction would be +66 kJ, and the energy level for the reverse reaction's products would be at zero. Therefore, the transition state's energy doesn't change, and it remains at 7 kJ above the energy level of the products of the forward reaction. Thus, Egr would be the vertical distance between the energy level of HCl(g) + Br(g) and the transition state. This gives Egr = 7 kJ + 66 kJ = 73 kJ. In conclusion, the activation energy for the reverse reaction is 73 kJ.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Enthalpy Change
In a chemical reaction, the enthalpy change, denoted as \( \Delta H \), represents the difference in energy content between the reactants and products. It's a thermodynamic quantity that gives us an indication of the heat absorbed or released during a reaction. When it comes to sign conventions, a negative enthalpy change, as in the exercise \( -66 \mathrm{~kJ} \), signifies that the reaction is exothermic; energy is released to the surroundings. Conversely, a positive enthalpy change means the reaction is endothermic, drawing energy from the surroundings.

Understanding enthalpy is crucial for predicting the direction of a chemical reaction and for calculating the energy needs or yields of industrial processes. It's also important to remember that enthalpy change is a state function, meaning it is determined by the initial and final states of the system, regardless of the path the reaction takes.

For example, in the given problem, the negative enthalpy change indicates that forming HCl(g) and Br(g) from Cl(g) and HBr(g) will result in a release of energy, signalling a potential spontaneity of the forward reaction in certain conditions.
Energy Profile Diagram
An energy profile diagram visually represents the energy changes that occur during a chemical reaction, including the activation energy and the overall enthalpy change. When sketching such a diagram, as required in the exercise, one must first establish a baseline for the reactants' energy. Then, a peak is drawn to represent the activation energy needed to reach the transition state, the highest energy point along the reaction path.

Moreover, the products are shown at a lower position if the reaction is exothermic, as in the solved exercise, or higher if it's endothermic. This visual tool is incredibly beneficial for students as it aids in the conceptual understanding of the dynamic process of chemical reactions. It showcases why, even with a negative \( \Delta H \), not all reactions occur spontaneously due to the initial energy barrier represented by the activation energy.

Using energy profile diagrams, we gain insights into factors like the stability of reactants, the likelihood of a reaction proceeding, and the energy landscape the reactants must traverse to become products.
Chemical Kinetics
Chemical kinetics delves into the speed or rate of a chemical reaction and the mechanisms by which reactions occur. This field of study examines factors affecting these rates, such as temperature, concentration, surface area, and the presence of catalysts or inhibitors.

The activation energy is a critical component in chemical kinetics. It's the minimum energy required to initiate a reaction, influencing how fast the reaction will occur. A lower activation energy means the reaction can proceed faster under the same conditions. For the exercise provided, it is this energy hurdle of \( 7 \mathrm{~kJ} \) that must be overcome for the reaction between Cl(g) and HBr(g) to take place.

Moreover, the reaction's reverse activation energy, which amounts to \( 73 \mathrm{~kJ} \), shows us that more energy is required for the decomposition of HCl(g) and Br(g) back into Cl(g) and HBr(g), explaining why under ordinary conditions, the forward reaction is favored. The principles of chemical kinetics not only help us understand how fast a reaction will proceed but also how to control and manipulate reaction rates, which is incredibly useful in fields like pharmacology, environmental science, and materials engineering.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The reaction between ethyl bromide \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right)\) and hydroxide ion in ethyl alcohol at \(330 \mathrm{~K}, \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}(a l c)+\) \(\mathrm{OH}^{-}(a l c) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{Br}^{-}(a l c)\), is first order each in ethyl bromide and hydroxide ion. When \(\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}\right]\) is \(0.0477 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]\) is \(0.100 \mathrm{M}\), the rate of disappearance of ethyl bromide is \(1.7 \times 10^{-7} \mathrm{M} / \mathrm{s}\). (a) What is the value of the rate constant? (b) What are the units of the rate constant? (c) How would the rate of disappearance of ethyl bromide change if the solution were diluted by adding an equal volume of pure ethyl alcohol to the solution?

(a) Define the following symbols that are encountered in rate equations: \([\mathrm{A}]_{0}, t_{1 / 2}[\mathrm{~A}]_{t}, k .(\mathrm{b})\) What quantity, when graphed versus time, will yield a straight line for a firstorder reaction?

For each of the following gas-phase reactions, indicate how the rate of disappearance of each reactant is related to the rate of appearance of each product: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) (b) \(2 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{~N}_{2}(g)+\mathrm{O}_{2}(g)\) (c) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\)

The enzyme invertase catalyzes the conversion of sucrose, a disaccharide, to invert sugar, a mixture of glucose and fructose When the concentration of invertase is \(4.2 \times 10^{-7} M\) and the concentration of sucrose is \(0.0077 M\), invert sugar is formed at the rate of \(1.5 \times 10^{-4} \mathrm{M} / \mathrm{s} .\) When the sucrose concentration is doubled, the rate of formation of invert sugar is doubled also. (a) Assuming that the enzyme- substrate model is operative, is the fraction of enzyme tied up as a complex large or small? Explain. (b) Addition of inositol, another sugar, decreases the rate of formation of invert sugar. Suggest a mechanism by which this occurs.

As described in Exercise \(14.37\), the decomposition of sulfuryl chloride \(\left(\mathrm{SO}_{2} \mathrm{Cl}_{2}\right)\) is a first-order process. The rate constant for the decomposition at \(660 \mathrm{~K}\) is \(4.5 \times 10^{-2} \mathrm{~s}^{-1}\). (a) If we begin with an initial \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) pressure of 375 torr, what is the pressure of this substance after 65 s? (b) At what time will the pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decline to one-tenth its initial value?

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.