Unit 7 Chemical Processes Study Guide

8/14/2019 Unit 7 Chemical Processes Study Guide

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Chapter 14 1

Chapter 14 - The Process of Chemical Reactions

Review Skills14.1 Collision Theory: A Model for theReaction Process

The Basics of Collision Theory Endergonic Reactions Summary of Collision Theory

14.2 Rates of Chemical Reactions

Temperature and Rates ofChemical Reactions

Concentration and Rates ofChemical Reactions

Catalysts Homogeneous and Heterogeneous CatalystsSpecial Topic 14.1: Green Chemistry

- The Development of New and Better

Catalysts

14.3 Reversible Reactions and Chemical

Equilibrium

Reversible Reactions andDynamic Equilibrium

Equilibrium Constants Determination of Equilibrium

Constant Values

Equilibrium Constants and Extentof Reaction

Heterogeneous Equilibria Equilibrium Constants and

Temperature

Internet: Calculating Concentrationsand Gas Pressures

Internet: pH and pH Calculations

Internet: Weak Acids and Equilibrium

Constants

14.4 Disruption of Equilibrium

The Effect of Changes inConcentrations on Equilibrium

Systems

Internet: Changing Volume and Gas

Phase Equilibrium

Le Chteliers Principle The Effect of Catalysts on

Equilibria

Special Topic 14.2: The Big

QuestionHow Did We Get Here?

Chapter GlossaryInternet: Glossary Quiz

Chapter ObjectivesReview Questions

Key IdeasChapter Problems


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2 Stud y Guid e forAn Introduction to Chemistry

Section Goals and Introductions

Section 14.1 Collision Theory: A Model for the Reaction Process

Goals

To describe a model, called collision theory, that helps us to visualize the process ofmany chemical reactions.

To use collision theory to explain why not all collisions between possible reactants leadto products.

To use collision theory to explain why possible reactants must collide with an energyequal to or above a certain amount to have the possibility of reacting and forming

products.

To show how the energy changes in chemical reactions can be described with diagrams. To use collision theory to explain why possible reactants must collide with a specific

orientation to have the possibility of reacting and forming products.

Once again, this chapter emphasizes that if you develop the ability to visualize changes on theparticle level, it will help you understand and explain many different things. This section

introduces you to a model for chemical change that is called collision theory, which helps you

explain the factors that affect the rates of chemical reactions. These factors are described inSection 14.2.

Section 14.2 Rates of Chemical Reactions

Goals

To show how rates of chemical reactions are described. To explain why increased temperature increases the rates of most chemical reactions. To explain why increased concentration of reactants increases the rates of chemical

reactions.

To describe how catalysts increase the rates of certain chemical reactions.This section shows how collision theory helps you explain the factors that affect rates of

chemical changes. These factors include amounts of reactants and products, temperature, andcatalysts.

Section 14.3 Reversible Reactions and Chemical Equilibrium

Goals

To explain why chemical reactions that are reversible come to a dynamic equilibriumwith equal forward and reverse rates of reaction.

To show what equilibrium constants are and how they can be determined. To describe how equilibrium constants can be used to show the relative amounts of

reactants and products in the system at equilibrium.

To explain the effect of temperature on equilibrium systems and equilibrium constants.This section takes the basic ideas of dynamic equilibrium introduced in Chapter 12 and

applies them to reversible chemical changes. This is a very important topic, so plan to spend

some extra time on this section, if necessary. You will also learn how equilibrium constantsare used to describe the relative amounts of reactants and products for a chemical reaction at


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Chapter 14 Map


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Chapter 14 5

Chapter Checklist

Read the Review Skills section. If there is any skill mentioned that you have not yet

mastered, review the material on that topic before reading this chapter.

Read the chapter quickly before the lecture that describes it.

Attend class meetings, take notes, and participate in class discussions.

Work the Chapter Exercises, perhaps using the Chapter Examples as guides.

Study the Chapter Glossary and test yourself on our Web site:

Internet: Glossary Quiz

Study all of the Chapter Objectives. You might want to write a description of how you

will meet each objective.

This chapter has logic sequences in Figures 14.11, 14.13, 14.15, 14.22, and 14.25.Convince yourself that each of the statements in these sequences logically leads to the

next statement.

To get a review of the most important topics in the chapter, fill in the blanks in the Key

Ideas section.

Work all of the selected problems at the end of the chapter, and check your answers with

the solutions provided in this chapter of the study guide.

Ask for help if you need it.

Web Resources

Internet: Calculating Concentrations and Gas Pressures

Internet: pH and pH Calculations

Internet: Weak Acids and Equilibrium Constants

Internet: Changing Volume and Gas Phase Equilibrium

Internet: Glossary Quiz

Exercises Key

Exercise 14.1 Writing Equilibrium Constant Expressions: Sulfur dioxide, SO2, oneof the intermediates in the production of sulfuric acid, can be made from the reaction of

hydrogen sulfide gas with oxygen gas. Write the equilibrium constant expressions forKCand KP

for the following equation for this reaction. (Objs 24 & 25)2H2S(g) + 3O2(g) 2SO2(g) + 2H2O(g)

K =[SO ] [H O]

[H S] [O ]K =

P P

P PC

22

22

22

23 P

SO2

H O2

H S2

O3

2 2

2 2


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Exercise 14.2 Equilibrium Constant Calculation:Ethanol, C2H5OH, can be made fromthe reaction of ethylene gas, C2H4, and water vapor. A mixture of C2H4(g) and H2O(g) is allowed

to come to equilibrium in a container at 110 C, and the partial pressures of the gases are foundto be 0.35 atm for C2H4(g), 0.75 atm for H2O(g), and 0.11 atm for C2H5OH(g). What is KP for

this reaction at 110 C? (Obj 26)C2H4(g) + H2O(g) C2H5OH(g)

K =P

P P=

0 .11at m

0 .35 at m 0 .75 at mP

C H OH

C H H O

2 5

2 4 2a f

= 0 .42 1/at m or 0 .42

Exercise 14.3 Predicting the Extent of Reaction:Using the information in Table 14.1,predict whether each of the following reversible reactions favors reactants, products, or neither at

25 C. (Obj 27)

a. This reaction is partially responsible for the release of pollutants from automobiles.2NO(g) O2(g) 2NO2(g)

According t o Table 14.1, t he KPfor t his react ion is 2.2 10 12, so it favorsproducts.

b. The NO2(g) molecules formed in the reaction in part (a) can combine to form N2O4.2NO2(g) N2O4(g)

According t o Table 14.1, t he KPfor t his react ion is 6 .7. Neither react ant s

nor product s are favored.

Exercise 14.4 Writing Equilibrium Constants for Heterogeneous Equilibria:Thefollowing equation describes one of the steps in the purification of titanium dioxide, which is

used as a white pigment in paints. Liquid titanium(IV) chloride reacts with oxygen gas to form

solid titanium oxide and chlorine gas. Write KCand KP expressions for this reaction. (Objs 24 &

25)

TiCl4(l) + O2(g) TiO2(s) + 2Cl2(g)

K =[Cl ]

[O ]K =

P

PC

22

2

P

Cl2

O

2

2


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Chapter 14 7

Exercise 14.5 Predicting the Effect of Disruptions on Equilibrium: Nitric acidcan be made from the exothermic reaction of nitrogen dioxide gas and water vapor in the

presence of a rhodium and platinum catalyst at 700-900 C and 5-8 atm. Predict whether each ofthe following changes in the equilibrium system will shift the system to more products, to more

reactants, or to neither. Explain each answer in two ways: (1) by applying Le Chteliersprinciple and (2) by describing the effect of the change on the forward and reverse reaction rates.(Objs 40- 42 & 44- 46)

Rh/Pt

3NO2(g) + H2O(g) 2HNO3(g) + NO(g) + 37.6 kJ

750-920 C

5-8 atm

a. The concentration of H2O is increased by the addition of more H2O. (1) Using Le Cht elier's Principle, we predict t hat t he syst em will shift to more

products t o part ially count eract t he increase in H2O.

(2) The increase in t he concent rat ion of wat er vapor speeds t he forwardreact ion wit hout init ially affect ing t he rat e of t he reverse react ion. Theequilibrium is disrupt ed, and t he syst em shifts to more products because t he

forward rat e is great er t han t he reverse rat e.

b. The concentration of NO2 is decreased. (1) Using Le Cht elier's Principle, we predict t hat t he syst em will shift to

more reactants t o part ially count eract t he decrease in NO2.

(2) The decrease in t he concent rat ion of NO2(g) slows t he forward react ionwit hout init ially affect ing t he rat e of t he reverse react ion. The equilibrium is

disrupt ed, and t he syst em shifts toward more reactants because t he

reverse rat e is great er t han t he forward rat e.c. The concentration of HNO3(g) is decreased by removing the nitric acid as it forms.

(1) Using Le Cht elier's Principle, we predict t hat t he syst em will shift tomore productst o part ially count eract t he decrease in HNO3.

(2) The decrease in t he concent rat ion of HNO3(g) s lows t he reverse react ionwit hout init ially affect ing t he rat e of t he forward react ion. The equilibrium is

disrupt ed, and t he syst em shifts toward more products because t he


d. The temperature is decreased from 1000 C to 800 C. (1) Using Le Cht elier's Principle, we predict t hat t he syst em shift s in t he

exot hermic direct ion t o part ially count eract t he decrease in t emperat ure. As

t he syst em shifts toward more products, energy is released, and t he

t emperat ure increases.

(2) The decreased t emperat ure decreases t he rat es of bot h t he forward andreverse react ions, but it has a great er effect on t he endot hermic react ion.

Because t he forward react ion is exot hermic, t he reverse react ion must be

endot hermic. Therefore, t he reverse react ion is slowed more t han t he forward


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react ion. The syst em shifts toward more products because t he forward r at e

becomes great er t han t he reverse rat e.

e. The Rh/Pt catalyst is added to the equilibrium system. (1) Le Cht elier's Principle does not apply here. (2) The cat alyst speeds bot h t he forward and t he reverse rat es equally. Thus

t here is no shift in t he equilibrium. The purpose of t he cat alyst is t o bring t he

syst em t o equilibrium fast er.

Review Questions Key

1. Describe what you visualize occurring inside a container of oxygen gas, O2, at roomtemperature and pressure.

The gas is composed of O2 molecules t hat are moving const ant ly in t he cont ainer. For a

t ypical gas, t he average dist ance bet ween part icles is about t en t imes t he diamet er ofeach part icle. This leads t o t he gas part icles t hemselves t aking up only about 0 .1% of t he

t ot al volume. The ot her 9 9 .9% of t he t ot al volume is empt y space. According t o our

model, each O2 molecule moves f reely in a st raight -line pat h unt il it collides wit h anot her

O2 molecule or one of t he walls of t he cont ainer. The part icles are moving fast enough t o

break any at t ract ion t hat might form bet ween t hem, so aft er t wo part icles collide, t hey

bounce off each ot her and cont inue on alone. Due t o collisions, each part icle is

const ant ly speeding up and slowing down, but it s average velocit y st ays const ant as

long as t he t emperat ure st ays const ant .

2. Write in each blank the word that best fits the definition.a. Energyis the capacity to do work.b. Kinetic energy is the capacity to do work due to the motion of an object.c. A(n) endergonicchangeis a change that absorbs energy.d. A(n) exergonicchangeis a change that releases energy.e. Thermalenergyis the energy associated with the random motion of particles.f. Heatis thermal energy that is transferred from a region of higher temperature to a

region of lower temperature as a result of the collisions of particles.

g. A(n) exothermicchangeis a change that leads to heatenergy being evolved from thesystem to the surroundings.

h. A(n) endothermicchangeis a change that leads the system to absorb heatenergyfrom the surroundings.

i. A(n) catalystis a substance that speeds a chemical reaction without beingpermanently altered itself.

3. When the temperature of the air changes from 62 C at 4:00 A.M. to 84 C at noon on asummer day, does the average kinetic energy of the particles in the air increase, decrease, or

stay the same?

Increased t emperat ure meansincreased average kinetic energy.

4. Explain why it takes energy to break an OO bond in an O3 molecule.Separat e at oms are less st able, and t herefore, higher pot ent ial energy t han at oms in a

bond. The Law of Conservat ion of Energy st at es t hat energy cannot be creat ed or


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Chapter 14 9

dest royed, so energy must be added t o t he syst em. It always t akes energy t o break

at t ract ions bet ween part icles.

5. Explain why energy is released when two oxygen atoms come together to form an O2molecule.

At oms in a bond are more st able, and t herefore, lower pot ent ial energy. The Law ofConservat ion of Energy st at es t hat energy cannot be creat ed or dest royed, so energy is

released from t he syst em. Energy is always released when new at t ract ions bet ween

part icles are formed.

6. Explain why some chemical reactions releaseheatto their surroundings.If t he bonds in t he product s are st ronger and lower pot ent ial energy t han in t he

react ant s, energy will be released from t he syst em. If t he energy released is due t o t he

conversion of pot ent ial energy t o kinet ic energy, t he t emperat ure of t he product s will be

higher t han t he original react ant s. The higher t emperat ure product s are able t o t ransfer

heat t o t he surroundings, and t he t emperat ure of t he surroundings increases.

7. Explain why some chemical reactions absorbheatfrom their surroundings.If t he bonds in t he product s are weaker and higher pot ent ial energy t han in t he

react ant s, energy must be absorbed. If t he energy absorbed is due t o t he conversion of

kinet ic energy t o pot ent ial energy, t he t emperat ure of t he product s will be lower t han t he

original react ant s. The lower t emperat ure product s are able t o absorb heat from t he

surroundings, and t he t emperat ure of t he surroundings decreases.

8. What are the general characteristics of any dynamic equilibrium system?The syst em must have t wo opposing changes, from st at e A t o st at e B and from st at e B

t o st at e A. For a dynamic equilibrium t o exist , t he rat es of t he t wo opposing changes

must be equal, so t here are const ant changes bet ween st at e A and st at e B but no net

change in t he component s of t he syst em.

Key Ideas Answers

9. At a certain stage in the progress of a reaction, bond breaking and bond making are of equal

importance. In other words, the energy necessary for bond breaking is balancedby theenergy supplied by bond making. At this turning point, the particles involved in the reaction

are joined in a structure known as the activated complex, ortransition state.

11. In a chemical reaction, the minimum energy necessary for reaching the activated complex

and proceeding to products is called the activation energy. Only the collisions that provide a

net kinetic energy equal to orgreater than the activation energy can lead to products.

13. The energies associated with endergonic (or endothermic) changes are described with

positive values.

15. Because the formation of the new bondsprovides some of the energy necessary to break the

old bonds, the making and breaking of bonds must occur more or less simultaneously. This

is possible only when the particles collide in such a way that the bond-forming atoms are

close to each other.


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17. Increased temperature means an increase in the average kinetic energy of the collisions

between the particles in a system. This leads to an increase in the fraction of the collisions

that have enough energy to reach the activated complex (the activation energy).

19. One of the ways in which catalysts accelerate chemical reactions is by providing a(n)

alternative pathwaybetween reactants and products that has a(n) lower activation energy.21. If the catalyst is not in the same state as the reactants, the catalyst is called a(n)

heterogeneous catalyst.

23. The extent to which reversible reactions proceed toward products before reaching

equilibrium can be described with a(n) equilibrium constant, which is derived from the ratio

of the concentrations of products to the concentrations of reactants at equilibrium. Forhomogeneous equilibria, the concentrations of all reactants and products can be described in

moles per liter, and the concentration of each is raised to a power equal to its coefficient in a

balanced equation for the reaction.

25. The larger a value for K, the farther the reaction shifts toward products before the rates of

the forward and reverse reactions become equal and the concentrations of reactants and

products stop changing.

27. Changing temperature always causes a shift in equilibrium systemssometimes toward more

products and sometimes toward more reactants.

29. If the forward reaction in a reversible reaction is endergonic, increased temperature will shift

the system toward more products.

31. Le Chtelier's principle states that if a system at equilibrium is altered in a way that disrupts

the equilibrium, the system will shift in such a way as to counter the change.

Problems Key

Section 14.1 Collision Theory: A Model for the Reaction Process

33. Assume that the following reaction is a single-step reaction in which one of the OO bonds

in O3 is broken and a new NO bond is formed. The heat of reaction is 226 kJ/mol.

NO(g) + O3(g) NO2(g) + O2(g) + 226 kJ

a. With reference to collision theory, describe the general process that takes place as thisreaction moves from reactants to products. (Obj 2)

NO and O3 molecules are const ant ly moving in t he cont ainer, somet imes wit h a

high velocit y and somet imes more slowly. The part icles are const ant ly colliding,

changing t heir direct ion of mot ion, and speeding up or s lowing down. If t he

molecules collide in a way t hat put s t he nit rogen at om in NO near one of t he

out er oxygen at oms in O3, one of t he OO bonds in t he O3 molecule begins t obreak, and a new bond bet ween one of t he oxygen at oms in t he ozone molecule

and t he nit rogen at om in NO begins t o form. If t he collision yields enough energy

t o reach t he act ivat ed complex, it proceeds on t o product s. If t he molecules do

not have t he correct orient at ion, or if t hey do not have enough energy, t hey

separat e wit hout a react ion t aking place.


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Chapter 14 11

b. List the three requirements that must be met before a reaction between NO(g) andO3(g) is likely to take place. (Obj 11)

NO and O3 molecules must collide, t hey must collide wit h t he correct orient at ion

t o form an NO bond at t he same t ime t hat an OO bond is broken, and t hey

must have t he minimum energy necessar y t o reach t he act ivat ed complex (t heact ivat ion energy).

c. Explain why NO(g) and O3(g) must collide before a reaction can take place.(Obj 3)

The collision brings t he at oms t hat will form t he new bonds close, and t he net

kinet ic energy in t he collision provides t he energy necessary t o reach t he

act ivat ed complex and proceed t o product s.

d. Explain why it is usually necessary for the new NO bonds to form at the same timethat the OO bonds are broken. (Obj 4)

It t akes a significant amount of energy t o break OO bonds, and collisions

bet ween part icles ar e not likely t o provide enough. As NO bonds f orm, t heyrelease energy, so t he format ion of t he new bonds can provide energy t o

supplement t he energy provided by t he collisions. The sum of t he energy of

collision and t he energy released in bond format ion is more likely t o provide

enough energy for t he react ion.

e. Draw a rough sketch of the activated complex.

f. Explain why a collision between NO(g) and O3(g) must have a certain minimumenergy (activation energy) in order to proceed to products. (Obj 5)

In t he init ial st age of t he react ion, t he energy released in bond making is less

t han t he energy absorbed by bond breaking. Therefore, energy must be available

from t he colliding part icles t o allow t he react ion t o proceed. At some point in t he

change, t he energy released in bond format ion becomes equal t o t he energy

absorbed in bond breaking. If t he colliding part icles have enough energy t o reach

t his point (in ot her words, if t hey have t he act ivat ion energy), t he react ion

proceeds t o product s.


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g. The activation energy for this reaction is 132 kJ/mol. Draw an energy diagram for thisreaction, showing the relative energies of the reactants, the activated complex, and the

products. Using arrows show the activation energy and heat of reaction. (Obj 7)

h. Is this reaction exothermic or endothermic? (Objs 6 & 8)The negat ive sign for t he heat of react ion shows t hat energy is r eleased overall,

so t he react ion is exothermic.i. Explain why NO(g) and O3(g) molecules must collide with the correct orientation if a

reaction between them is likely to take place. (Obj 10)

For a react ion t o be likely, new bonds must be made at t he same t ime as ot her

bonds are broken. Therefore, t he nit rogen at om in NO must collide wit h one of t he

out er oxygen at oms in O3.

Section 14.2 Rates of Chemical Reactions

35. Consider the following general reaction for which gases A and B are mixed in a constant

volume container.

A(g) + B(g) C(g) + D(g)

What happens to the rate of this reaction when

a. more gas A is added to the container?Increased concent rat ion of react ant A leads t o increased rat e of collision

bet ween A and B and t herefore leads t o increased rate of reaction.

b. the temperature is decreased?Decreased t emperat ure leads t o decreased average kinet ic energy of collisions

bet ween A and B. This leads t o a decrease in t he percent age of collisions wit h t he


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Chapter 14 13

minimum energy necessar y for t he react ion and t hereforeleads t o decreased

rate of reaction.

c. a catalyst is added that lowers the activation energy?Wit h a lower act ivat ion energy, t here is a great er percent age of collisions wit h t he

minimum energy necessar y for t he react ion and t hereforean increased rate ofreaction.

37. The reactions listed below are run at the same temperature. The activation energy for the firstreaction is 132 kJ/mol. The activation energy for the second reaction is 76 kJ/mol. In which

of these reactions would a higher fraction of collisions between reactants have the minimum

energy necessary to react (the activation energy)? Explain your answer.

NO(g) + O3(g) NO2(g) + O2(g) Activation energy = 132 kJ

I

(aq) + CH3Br(aq) CH3I(aq) + Br(aq) Activation energy = 76 kJ

At a part icular t emperat ure, t he lower t he act ivat ion energy is, t he higher t he

percent age of collisions wit h at least t hat energy or more will be. Thus t he secondreact ion would have t he higher fract ion of collisions wit h t he act ivat ion energy.

39. Two reactions can be described by the energy diagrams below. What is the approximate

activation energy for each reaction? Which reaction is exothermic and which is endothermic?

The approximat e act ivat ion energy for react ion 1is 30 kJ and for react ion 2 is 6 0 kJ.

React ion 1is endot hermic, and react ion 2 is exot hermic.

41. Explain why chlorine atoms speed the conversion of ozone molecules, O3, and oxygen atoms,O, into oxygen molecules, O2. (Obj 14)

In part , chlorine at oms are a t hreat t o t he ozone layer just because t hey provide anot her

pat hway for t he conversion of O3 and O t o O2, but t here is anot her r eason. The react ion

bet ween O3 and Cl t hat forms ClO and O2 has an act ivat ion energy of 2.1kJ/mole. At 25

C, about t hree of every seven collisions (or 4 3%) have enough energy t o reach t he

act ivat ed complex. The react ion bet ween O and ClO t o form Cl and O2 has an act ivat ion

energy of only 0.4 kJ/mole. At 25 C, about 8 5% of t he collisions have at least t his

energy. The uncat alyzed react ion has an act ivat ion energy of about 17 kJ/mole. At 25 C


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(298 K), about one of every one t housand collisions (or 0 .1%) bet ween O3 molecules and

O at oms has a net kinet ic energy large enough t o form t he act ivat ed complex and

proceed t o product s. Thus a much higher fract ion of t he collisions have t he minimum

energy necessary t o react for t he cat alyzed react ion t han for t he direct react ion

bet ween O3 and O. Thus a much great er fract ion of t he collisions has t he minimum energynecessary for t he react ion t o proceed for t he cat alyzed react ion t han for t he

uncat alyzed react ion. Figures 14.14 and 14.15 of t he t ext book illust rat e t his.

43. Using the proposed mechanism for the conversion of NO(g) into N2(g) and O2(g) as anexample, write a description of the four steps thought to occur in heterogeneous catalysis.(Obj 16)

Step 1: The react ant s (NO molecules) collide wit h t he surface of t he cat alyst where

t hey bind t o t he cat alyst . This st ep is called adsorpt ion. The bonds wit hin t he react ant

molecules ar e weakened or even broken as t he react ant s are adsorbed. (NO bonds are

broken.)

Step 2: The adsorbed part icles (separat e N and O at oms) move over t he surface of t hecat alyst .

Step 3: The adsorbed part icles combine t o form product s (N2 and O2).

Step 4: The product s (N2 and O2) leave t he cat alyst .

See Figure 14.16 of t he t ext book.

Section 14.3 Reversible Reactions and Chemical Equilibrium

45. Equilibrium systems have two opposing rates of change that are equal. For each of the

following equilibrium systems that were mentioned in earlier chapters, describe what is

changing in the two opposing rates.

a. a solution of the weak acid acetic acid, HC2H3O2 (Chapter 6)Acet ic acid molecules react wit h wat er t o form hydronium ions and acet at e ions,

and at t he same t ime, hydronium ions react wit h acet at e ions t o ret urn t o acet ic

acid molecules and wat er.

HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2

(aq)

b. pure liquid in a closed container (Chapter 12)Liquids evaporat e t o form vapor at a rat e t hat is balanced by t he ret urn of vapor

t o liquid.

c. a closed bottle of carbonated water with 4 atm of CO2 in the gas space above theliquid (Chapter 13)

Carbon dioxide escapes fr om t he solut ion at a rat e t hat is balanced by t he

ret urn of CO2 t o t he solut ion.


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Chapter 14 15

47. Two gases, A and B, are added to an empty container. They react in the following reversible

reaction.

A(g) + B(g) C(g) + D(g)

a. When is the forward reaction rate greatest: (1) when A and B are first mixed, (2)when the reaction reaches equilibrium, or (3) sometime between these two events?

The forward react ion rat e is at it s peak when A and B are first mixed. Because A

and B concent rat ions are diminishing as t hey form C and D, t he rat e of t he

forward r eact ion declines st eadily unt il equilibrium is reached.

b. When is the reverse reaction rate greatest: (1) when A and B are first mixed, (2) whenthe reaction reaches equilibrium, or (3) sometime between these two events?

The reverse react ion rat e is at it s peak when t he react ion reaches equilibrium.

Because C and D concent rat ions are increasing as t hey form f rom A and B, t he

rat e of t he reverse react ion increases st eadily unt il equilibrium is reached.

49. Assume that in the following reversible reaction both the forward and the reverse reactions

take place in a single step.I

(aq) + CH3Br(aq) CH3I(aq) + Br

(aq)

a. With reference to the changing forward and reverse reaction rates, explain why thisreaction moves toward a dynamic equilibrium with equal forward and reverse

reaction rates. (Obj 20)

When I ions and CH3Br molecules are added t o a cont ainer, t hey begin t o collide

and react . As t he react ion proceeds, t he concent rat ions of I and CH3Br

diminish, so t he rat e of t he forward react ion decreases. Init ially, t here are no

CH3I molecules or Br ions in t he cont ainer, so t he rat e of t he reverse react ion is

init ially zero. As t he concent rat ions of CH3I and Br increase, t he rat e of t he

reverse react ion increases.

As long as t he rat e of t he forward react ion is great er t han t he rat e of

t he reverse react ion, t he concent rat ions of t he react ant s (I and CH3Br) will

st eadily decrease, and t he concent rat ions of product s (CH3I and Br) will

const ant ly increase. This leads t o a decrease in t he forward rat e of t he react ion

and an increase in t he rat e of t he reverse react ion. This cont inues unt il t he t wo

rat es become equal. At t his point , our syst em has reached a dynamic equilibrium.

b. Describe the changes that take place once the reaction reaches an equilibrium state.Are there changes in the concentrations of reactants and products at equilibrium?

Explain your answer. (Obj 21)

In a dynamic equilibrium for r eversible chemical react ions, t he forward and reverse

react ion rat es are equal, so alt hough t here are const ant changes bet ween

react ant s and product s, t here is no net change in t he amount s of each. I and

CH3Br are const ant ly react ing t o form CH3I and Br, but CH3I and Br

are

react ing t o reform CH3Br and I at t he same rat e. Thus t here is no net change in

t he amount s of I, CH3Br, CH3I, or Br.


16/20


50. Write KCand KP expressions for each of the following equations. (Objs 22 & 23)

a. 2CH4(g) C2H2(g) + 3H2(g)

K =[C H ][H ]

[CH ]

K =P P

P

C2 2 2

3

4

2 P

C H H3

CH

2

2 2 2

4

b. 2N2O(g) + O2(g) 4NO(g)

K =[NO]

[N O] [O ]K =

P

P PC

4

22

2

PNO4

N O2

O2 2

c. Sb2S3(s) + 3H2(g) 2Sb(s) + 3H2S(g)

K =[H S]

[H ]K =

P

PC

23

23 P

H S3

H3

2

2

52. A mixture of nitrogen dioxide and dinitrogen tetroxide is allowed to come to equilibrium at

30 C, and the gases partial pressures are found to be 1.69 atm N2O4 and 0.60 atm NO2.(Obj 24)

a. On the basis of these data, what is KP for the following equation?NO2(g) N2O4(g)

K =P

P=

(1.6 9)

0.60P

N1/ 2

NO

1/ 22

2

O4 = 2.2

b. On the basis of these data, what is KP for the following equation?2NO2(g) N2O4(g)

K =P

P

=1.6 9

(0.60)

PN

NO

2

2

2

O4

2= 4.7

c. Table 14.1 lists the KP for the equation in Part (b) as 6.7 at 25 C. Explain why youranswer to Part (b) is not 6.7.

Changing t emperat ure leads t o a change in t he value for an equilibrium const ant .

(Because KPfor t his r eact ion decreases wit h increasing t emperat ure, t he

react ion must be exot hermic.)

54. Predict whether each of the following reactions favors reactants, products, or neither at the

temperature for which the equilibrium constant is given. (Obj 25)

a. CH3OH(g) + CO(g) CH3CO2H(g) KP = 1.2 1022

at 25 C

KP < 102 so react ant s favored

b. CH4(g) + 4Cl2(g) CCl4(g) + 4HCl(g) KP = 3.3 1068 at 25 C

KP > 102 so product s favored

c. CO(g) + Cl2(g) COCl2(g) KP = 0.20 at 600 C

102 < KP < 102 so neit her favored


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Chapter 14 17

56. Write the KCexpression for the following equation. Explain why the concentration of

CH3OH is left out of the expression. (Objs 22 & 26)

CO(g) + 2H2(g) CH3OH(l)

C 22

1

K = [CO][H ]

If t he number of moles of CH3OH(l) in t he cont ainer is doubled, it s volume doubles t oo,

leaving t he concent rat ion (mol/L) of t he met hanol const ant . Increasing or decreasing t he

t ot al volume of t he cont ainer will not change t he volume occupied by t he liquid met hanol,

so t he concent rat ion (mol/L) of t he CH3OH(l) also remains const ant wit h changes in t he

volume of t he cont ainer. The const ant concent rat ion of met hanol can be incorporat ed

int o t he equilibrium const ant it self and left out of t he equilibrium const ant expression.

3C2 2

2 3 2

[CH OH] K' 1K' = = = K

[CO] [H ] [CH OH] [CO][H ]

58. Ethylene, C2H4, is one of the organic substances found in the air we breathe. It reacts withozone in an endothermic reaction to form formaldehyde, CH2O, which is one of the

substances in smoggy air that cause eye irritation.

2C2H4 (g) + 2O3(g) + energy 4CH2O(g) + O2(g)

a. Why does the forward reaction take place more rapidly in Los Angeles than in awilderness area of Montana with the same air temperature?

Los Angeles has a much higher ozone concent rat ion t han in t he Mont ana

wilderness.

b. For a variety of reasons, natural systems rarely reach equilibrium, but if this reactionwas run in the laboratory, would increased temperature for the reaction at equilibrium

shift the reaction to more reactants or more products?Toward more products (Increased t emperat ure favors t he endot hermic

direct ion of reversible react ions.)

60. When the temperature of an equilibrium system for the following reaction is increased, the

reaction shifts toward more reactants. Is the reaction endothermic or exothermic?

H2(g) + Br2(g) 2HBr(g)

Increased t emperat ure favors t he endot hermic direct ion of reversible react ions,

so t his react ion is endot hermic in t he reverse direct ion and exot hermic in t heforward direct ion.

62. Assume that you are picking up a few extra dollars to pay for textbooks by acting as a

trainers assistant for a heavyweight boxer. One of your jobs is to wave smelling salts underthe nose of the fighter to clear his head between rounds. The smelling salts are ammonium

carbonate, which decomposes in the following reaction to yield ammonia. The ammonia does

the wakeup job. Suppose the fighter gets a particularly nasty punch to the head and needs anextra jolt to be brought back to his senses. How could you shift the following equilibrium to

the right to yield more ammonia?

(NH4)2CO3(s) + energy 2NH3(g) + CO2(g) + H2O(g)

Increased t emperat ure will drive t his endot hermic react ion t oward product s, so warming

t he smelling salt cont ainer in your hands will increase t he amount of ammonia released.


18/20


64. Formaldehyde, CH2O, is one of the components of embalming fluids and has been used to

make foam insulation and plywood. It can be made from methanol, CH3OH (often called

wood alcohol). The heat of reaction for the combination of gaseous methanol and oxygen gasto form gaseous formaldehyde and water vapor is 199.32 kJ per mole of CH2O formed, so

the reaction is exothermic.2CH3OH(g) + O2(g) 2CH2O(g) + 2H2O(g)

a. Increased temperature drives the reaction toward reactants and lowers the value forthe equilibrium constant. Explain why this is true. (Objs 27 & 28)

Increased t emperat ure increases t he rat e of bot h t he forward and t he reverse

react ions, but it increases t he rat e of t he endergonic react ion more t han it

increases t he rat e of t he exergonic react ion. Therefore, changing t he t emperat ure

of a chemical syst em at equilibrium will disrupt t he balance of t he forward and

reverse rat es of react ion and shift t he syst em in t he direct ion of t he endergonic

react ion. Because t his react ion is exot hermic in t he forward direct ion, it must be

endot hermic in t he reverse direct ion. Increased t emperat ure shift s t he syst emt oward more react ant s, decreasing t he rat io of product s t o react ant s and,

t herefore, decreasing t he equilibrium const ant .

b. This reaction is run by the chemical industry at 450-900 C, even though theequilibrium ratio of product to reactant concentrations is lower than at roomtemperature. Explain why this exothermic chemical reaction is run at high

temperature despite this fact. (Obj 29)

To maximize t he percent yield at equilibrium, t he react ion should be run at as low

a t emperat ure as possible, but at low t emperat ure, t he rat es of t he forward and

reverse react ions are bot h very low, so it t akes a long t ime for t he syst em t o

come t o equilibrium. In t his case, it is best t o run t he react ion at hight emperat ure t o get t o equilibrium quickly. (The unreact ed met hanol can be

recycled back int o t he original react ion vessel aft er t he formaldehyde has been

removed fr om t he product mixt ure.)

Section 14.4 Disruption of Equilibrium

66. Urea, NH2CONH2, is an important substance in the production of fertilizers. The equation

shown below describes an industrial reaction that produces urea. The heat of reaction is

135.7 kJ per mole of urea formed. Predict whether each of the following changes in theequilibrium system will shift the system to more products, to more reactants, or to neither.

Explain each answer in two ways: (1) by applying Le Chteliers principle and (2) by

describing the effect of the change on the forward and reverse reaction rates. (Objs 30- 33)

2NH3(g) + CO2(g) NH2CONH2(s) + H2O(g) + 135.7 kJ

a. The concentration of NH3 is increased by the addition of more NH3. (In the industrialproduction of urea, an excess of ammonia is added so that the ratio of NH 3 to CO2 is

3:1.)

Using Le Cht elier's Principle, we predict t hat t he syst em will shift to moreproducts t o part ially count eract t he increase in NH3.


19/20

Chapter 14 19

The increase in t he concent rat ion of ammonia speeds t he forward react ionwit hout init ially affect ing t he rat e of t he reverse react ion. The equilibrium is

disrupt ed, and t he syst em shifts to more products because t he forward r at e

is great er t han t he reverse rat e.

b. The concentration of H2O(g) is decreased by removing water vapor. Using Le Cht elier's Principle, we predict t hat t he syst em will shift to more

products t o part ially count eract t he decrease in H2O.

The decrease in t he concent rat ion of H2O(g) s lows t he reverse react ionwit hout init ially affect ing t he rat e of t he forward react ion. The equilibrium is

disrupt ed, and t he syst em shifts toward more products because t he


c. The temperature is increased from 25 C to 190 C. (In the industrial production ofurea, ammonia and carbon dioxide are heated to 190 C.)

Using Le Cht elier's Principle, we predict t hat t he syst em shift s in t heendot hermic direct ion t o part ially count eract t he increase in t emperat ure.

Because t he forward react ion is exot hermic, t he reverse react ion must be

endot hermic. As t he syst em shifts toward more reactants, energy is

absorbed, and t he t emperat ure decreases.

The increased t emperat ure increases t he rat es of bot h t he forward andreverse react ions, but it has a great er effect on t he endot hermic react ion.

Thus t he syst em shifts toward more reactants because t he reverse rat e

becomes great er t han t he forward r at e.

68. Hydriodic acid, which is used to make pharmaceuticals, is made from hydrogen iodide. The

hydrogen iodide is made from hydrogen gas and iodine gas in the following exothermicreaction.

H2(g) + I2(g) 2HI(g) + 9.4 kJ

What changes could you make for this reaction at equilibrium to shift the reaction to the right

and maximize the concentration of hydrogen iodide in the final product mixture?

The addition of either H2 or I2 (or both) would increase t he concent rat ions of

react ant s, increasing t he rat e of collision bet ween t hem, increasing t he forward rat e,

and shift ing t he syst em t oward more product .

Lower t emperat ure favors t he exot hermic direct ion of t he react ion, so lower

temperature would shift t his react ion t o a higher percent age of product s at equilibrium.


20/20


70. Phosgene gas, COCl2, which is a very toxic substance used to make pesticides and

herbicides, is made by passing carbon monoxide gas and chlorine gas over solid carbon,

which acts as a catalyst.

CO(g) + Cl2(g) COCl2(g)

If the carbon monoxide concentration is increased by adding CO to an equilibrium system ofthis reaction, what effect, if any, does it have on the following? (Assume constant

temperature.)

a. The concentration of COCl2 after the system has shifted to come to a newequilibrium.

The syst em will shift t oward product s, which leads t o increasedCOCl2.

b. The concentration of Cl2 after the system has shifted to come to a new equilibrium.The syst em will shift t oward product s, which leads t o decreasedCl2.

c. The equilibrium constant for the reaction.Equilibrium const ant s are unaffect ed by react ant and product concent rat ions,

so t he equilibrium constant remains the same.

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Unit 7 Chemical Processes Study Guide

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