Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 3

A reaction has a \(\Delta E_{\mathrm{rxn}}\) of \(-450 \mathrm{~kJ} / \mathrm{mol}\). The products have an energy of \(20 \mathrm{~kJ} / \mathrm{mol}\). (a) Is the energy of the reactants higher or lower than that of the products? Explain. (b) What is the energy of the reactants? (c) Is the reaction exothermic or endothermic? How do you know? (d) Does the reaction go uphill or downhill in energy? (e) Draw a reaction-energy profile for the reaction.

Short Answer

Expert verified
(a) The energy of the reactants is higher than that of the products because the reaction releases energy (\(\Delta E_{\mathrm{rxn}} = -450 \mathrm{~kJ/mol}\)). (b) The energy of the reactants is -430 kJ/mol. (c) The reaction is exothermic because \(\Delta E_{\mathrm{rxn}}\) is negative. (d) The reaction goes downhill in energy because it is exothermic. (e) Draw a reaction-energy profile with reactants at -430 kJ/mol and products at 20 kJ/mol, connected by a downward line, indicating a downhill direction in energy.

Step by step solution

01

(a) Determining the energy of reactants compared to products

To find if the energy of reactants is higher or lower than that of the products, we can consider the value of \(\Delta E_{\mathrm{rxn}}\). If the value is negative, it indicates that the reactants have a higher energy than the products because energy was released during the reaction. If the value is positive, it indicates that the reactants have a lower energy than the products because energy was absorbed during the reaction. In this case, since \(\Delta E_{\mathrm{rxn}} = -450 \mathrm{~kJ/mol}\), the energy of the reactants is higher than the products because the reaction releases energy.
02

(b) Calculating the energy of the reactants

To find the energy of the reactants, we can use the given information about the energy of the products and the \(\Delta E_{\mathrm{rxn}}\). We can use the formula: Reactant Energy = Product Energy + \(\Delta E_{\mathrm{rxn}}\) Plugging in the given values, we get: Reactant Energy = \(20 \mathrm{~kJ/mol} - 450 \mathrm{~kJ/mol}\) Reactant Energy = \( -430 \mathrm{~kJ/mol}\)
03

(c) Identifying the reaction type

We can identify if the reaction is exothermic or endothermic based on the value of \(\Delta E_{\mathrm{rxn}}\). If it is negative, the reaction is exothermic, meaning it releases energy during the reaction. If it is positive, the reaction is endothermic, meaning it absorbs energy during the reaction. In this case, since \(\Delta E_{\mathrm{rxn}} = -450 \mathrm{~kJ/mol}\), we can conclude that the reaction is exothermic.
04

(d) Determining if the reaction goes uphill or downhill in energy

A reaction that goes uphill in energy means it requires energy to proceed and is endothermic, while a reaction that goes downhill in energy releases energy and is exothermic. Since we established in part (c) that the reaction is exothermic, the reaction must go downhill in energy.
05

(e) Drawing a reaction-energy profile

To draw a reaction-energy profile, plot the energy levels on the y-axis and the reaction progress on the x-axis. Begin by drawing the reactants' energy level at -430 kJ/mol and the products' energy level at 20 kJ/mol. As the reaction is exothermic, the energy level of the products should be lower than that of the reactants. Therefore, connect the two points with a downward line, indicating that the reaction goes downhill in energy.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

Consider the reaction \(\mathrm{A}_{2}+\mathrm{B}_{2} \rightarrow 2 \mathrm{AB}\), for which \(\Delta E_{\mathrm{rxn}}=-100 \mathrm{~kJ}\). Forming 1 mole of A \(-B\) bonds releases \(150 \mathrm{~kJ}\). How much energy does it take to break the reactant bonds?

The reaction \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightarrow 2 \mathrm{NH}_{3}\) is exothermic. Draw a reaction-energy profile for the reaction. Label the gap that represents \(\Delta E_{\mathrm{rxn}} .\)

A reaction occurs in which 1 mole of \(\mathrm{A}\) is converted to 1 mole of \(B\). If 1 mole of \(A\) has an energy content of \(20 \mathrm{~kJ}\) and 1 mole of \(\mathrm{B}\) has an energy content of \(60 \mathrm{~kJ}\), is this reaction exothermic or endothermic? Calculate \(\Delta E_{\mathrm{rxn}} .\)

Consider the reaction. Kinetics studies reveal a first-order rate dependence on the concentration of the \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}\) and a zero-order dependence on the concentration of \(\mathrm{H}_{2} \mathrm{O}\). (a) What happens to the reaction rate as the \(\left(\mathrm{CH}_{3}\right)_{3} \mathrm{C}-\mathrm{Br}\) concentration is changed? What happens to the reaction rate as the \(\mathrm{H}_{2} \mathrm{O}\) concentration is changed? (b) Two mechanisms for this reaction are offered below. Can you rule out either of them? Is either mechanism plausible, given the overall balanced equation and kinetic data? Explain your answer fully.

Is it wise to postulate a three-molecule collision as an elementary step in a reaction mechanism? Explain your answer.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemistry Branches

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation
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.

Sign-up for free