/*! 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 66 The activation energy for some r... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The activation energy for some reaction $$\mathrm{X}_{2}(g)+\mathrm{Y}_{2}(g) \longrightarrow 2 \mathrm{XY}(g)$$ is \(167 \mathrm{kJ} / \mathrm{mol},\) and \(\Delta E\) for the reaction is \(+28 \mathrm{kJ} / \mathrm{mol} .\) What is the activation energy for the decomposition of XY?

Short Answer

Expert verified
The activation energy for the decomposition of XY is \(195\,\mathrm{kJ/mol}\).

Step by step solution

01

Understand the energy diagram

In an energy diagram, the activation energy is represented as the difference in energy between the reactants and the transition state in the forward reaction. The change in energy, denoted as \(\Delta E\), is the difference in energy between the reactants and products. The activation energy for the reverse reaction (decomposition of XY) can be found by adding the change in energy for the reaction to the activation energy of the forward reaction.
02

Calculate the activation energy for the reverse reaction (decomposition of XY)

To calculate the activation energy for the decomposition of XY, we will add the change in energy for the reaction to the activation energy of the forward reaction: Activation energy for the reverse reaction = Activation energy for the forward reaction + Change in energy for the reaction \(E_\text{reverse} = E_\text{forward} + \Delta E\) Using the given values, \(E_\text{reverse} = 167\,\text{kJ/mol} + 28\,\text{kJ/mol}\)
03

Compute the answer

Now, calculate the activation energy for the reverse reaction: \(E_\text{reverse} = 167\,\text{kJ/mol} + 28\,\text{kJ/mol} = 195\,\text{kJ/mol}\) The activation energy for the decomposition of XY is \(195\,\mathrm{kJ/mol}\).

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.

Chemical Kinetics
Chemical kinetics is the branch of chemistry that deals with the speed or rate of a chemical reaction, as well as the mechanism by which the reaction occurs. To understand chemical kinetics, one must first grasp the concept of the reaction rate, which is a measure of how quickly reactants are converted into products.

Several factors influence reaction rates, including temperature, concentration of reactants, surface area, and the presence of catalysts. The rate of a reaction is proportionally related to the concentration of the reactants raised to a power (usually determined experimentally). This relationship is expressed in the rate law equation, which can be mathematically represented as:
\[ \text{Rate} = k [\text{Reactant}]^n \]
where \( k \) is the rate constant, and \( n \) is the order of the reaction with respect to a particular reactant.

A key part of chemical kinetics is understanding the energy changes that occur during a reaction. Every chemical reaction requires a certain amount of energy to proceed, known as the activation energy. This is the barrier that must be overcome for reactants to transform into products, which brings us to the next core concept - energy diagrams.
Energy Diagrams
Energy diagrams visually represent the energy changes that occur during chemical reactions. They are essential tools for understanding the energetics and kinetics of reactions. An energy diagram plots the energy of the system on the y-axis, with the reaction's progress on the x-axis.

The diagram usually features a curve peaking at the transition state - the point at which the system reaches its highest energy level before proceeding to product formation. The energy difference between the reactants and the peak of this curve signifies the activation energy (\(E_{\text{forward}}\)) needed to initiate the reaction.

The change in energy (\(\text{Delta E}\)) of a reaction is depicted as the difference in energy between the starting reactants and the final products. If the products are at a higher energy level than the reactants, the reaction is endothermic (absorbing energy), and if lower, the reaction is exothermic (releasing energy).

For the reverse reaction, such as the decomposition of \(XY(g)\), the activation energy will be equal to the activation energy of the forward reaction plus the energy difference between products and reactants (\(E_{\text{reverse}} = E_{\text{forward}} + \text{Delta E}\)), reflecting the energy needed to revert to the original reactants.
Reaction Mechanisms
A reaction mechanism is a step-by-step sequence of elementary reactions by which overall chemical change occurs. An elementary reaction is a single step with its own transition state and activation energy, and the overall mechanism is the combined sequence of these steps.

Understanding a reaction's mechanism allows chemists to piece together the entire puzzle of how reactants are transformed into products. Mechanisms help explain not only the speed and order of a reaction but also the role of intermediates (species that appear in the mechanism but not in the overall balanced equation) and the influence of different conditions on the reaction rate.

In the context of our exercise, the reaction between \(X_2(g)\) and \(Y_2(g)\) to form \(XY(g)\) would have its own mechanism, which would dictate the activation energy needed for each step. By studying each elementary reaction, chemists can also better understand the kinetics and thermodynamics - including energy profiles of both the forward and reverse reactions - associated with the overall process.

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

Experiments during a recent summer on a number of fireflies (small beetles, Lampyridaes photinus) showed that the average interval between flashes of individual insects was \(16.3 \mathrm{s}\) at \(21.0^{\circ} \mathrm{C}\) and \(13.0 \mathrm{s}\) at \(27.8^{\circ} \mathrm{C}\). a. What is the apparent activation energy of the reaction that controls the flashing? b. What would be the average interval between flashes of an individual firefly at \(30.0^{\circ} \mathrm{C} ?\) c. Compare the observed intervals and the one you calculated in part b to the rule of thumb that the Celsius temperature is 54 minus twice the interval between flashes.

Two isomers \((A \text { and } B)\) of a given compound dimerize as follows: $$\begin{aligned} &2 \mathrm{A} \stackrel{k_{1}}{\longrightarrow} \mathrm{A}_{2}\\\ &2 \mathrm{B} \stackrel{k_{2}}{\longrightarrow} \mathrm{B}_{2} \end{aligned}$$ Both processes are known to be second order in reactant, and \(k_{1}\) is known to be 0.250 \(\mathrm{L} / \mathrm{mol} \cdot \mathrm{s}\) at \(25^{\circ} \mathrm{C} .\) In a particular experiment \(\mathrm{A}\) and \(\mathrm{B}\) were placed in separate containers at \(25^{\circ} \mathrm{C}\) where \([\mathrm{A}]_{0}=1.00 \times 10^{-2} \mathrm{M}\) and \([\mathrm{B}]_{0}=2.50 \times 10^{-2} \mathrm{M} .\) It was found that after each reaction had progressed for 3.00 min, \([\mathrm{A}]=3.00[\mathrm{B}] .\) In this case the rate laws are defined as $$\begin{array}{l} \text { Rate }=-\frac{\Delta[\mathrm{A}]}{\Delta t}=k_{1}[\mathrm{A}]^{2} \\ \text { Rate }=-\frac{\Delta[\mathrm{B}]}{\Delta t}=k_{2}[\mathrm{B}]^{2} \end{array}$$ a. Calculate the concentration of \(\mathrm{A}_{2}\) after 3.00 min. b. Calculate the value of \(k_{2}\) c. Calculate the half-life for the experiment involving A.

In the Haber process for the production of ammonia, $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)$$ what is the relationship between the rate of production of ammonia and the rate of consumption of hydrogen?

Upon dissolving \(\operatorname{InCl}(s)\) in \(\mathrm{HCl}, \operatorname{In}^{+}(a q)\) undergoes a disproportionation reaction according to the following unbalanced equation: $$\operatorname{In}^{+}(a q) \longrightarrow \operatorname{In}(s)+\operatorname{In}^{3+}(a q)$$ This disproportionation follows first-order kinetics with a half-life of 667 s. What is the concentration of \(\operatorname{In}^{+}(a q)\) after \(1.25 \mathrm{h}\) if the initial solution of \(\operatorname{In}^{+}(a q)\) was prepared by dissolving \(2.38 \mathrm{g}\) InCl \((s)\) in dilute HCl to make \(5.00 \times 10^{2} \mathrm{mL}\) of solution? What mass of \(\operatorname{In}(s)\) is formed after \(1.25 \mathrm{h} ?\)

Which of the following statement(s) is(are) true? a. The half-life for a zero-order reaction increases as the reaction proceeds. b. A catalyst does not change the value of the rate constant. c. The half-life for a reaction, a \(A \longrightarrow\) products, that is first order in A increases with increasing \([\mathrm{A}]_{0}\) d. The half-life for a second-order reaction increases as the reaction proceeds.

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.