NAME____________________________________ PER____________ DATE DUE____________

ACTIVE LEARNING IN CHEMISTRY EDUCATION

“ALICE”

CHAPTER 18

RATES OFCHEMICAL

REACTIONS"Kinetics"

18-1 ©1997, A.J. Girondi

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© 1997 A.J. Girondi, Ph.D.505 Latshmere DriveHarrisburg, PA 17109

alicechem@geocities.com

Website: www.geocities.com/Athens/Oracle/2041

18-2 ©1997, A.J. Girondi

ACTIVITY 18.1 Comparing Rates of Reactions Between Metals and Acids

You have, thus far, completed various activities in which different types of chemical reactionsoccurred. These chemical reactions involved color changes, temperature changes, precipitate formation,and the evolution of gases. These reactions occurred quickly. Many chemical reactions, however, canproceed rather slowly. In this chapter, you will study the factors that determine the rates of reactions andthe extent to which chemical reactions occur. These two factors are generally called reaction kinetics.

Let's begin by looking at two simple chemical reactions that occur at different rates. Get thematerials labeled 18.1 from the materials shelf. Place about 4 mL of 3 M HCl into two 150 mm Pyrex testtubes. Into one test tube place a small piece of zinc metal. Put a small piece (≈2 cm) of magnesium metalin the other test tube. Answer the following questions about what you just did.

1. How can you tell that a chemical reaction has occurred in the HCl–zinc combination?

{1}________________________________________In the magnesium–HCl combination? {2]_______

______________________________________________________________________________

2. How does the rate of the zinc–HCl reaction compare to the rate of the magnesium–HCl reaction?_____

______________________________________________________________________________

3. Is your answer to the previous question a qualitative or a quantitative statement?{3}____________

Explain: {4}______________________________________________________________________

4. After you have compared the reaction rates, you should test each test tube for the presence of aflammable gas. (Wear safety glasses.) Allow the gas bubbles to accumulate for about 20 to 30 secondswhile you keep your thumb loosely over the mouth of the tube. If needed, put more metal in the tubes.Hold one tube at a 45 degree angle, remove your thumb and very quickly place a flaming wood splint atthe top of the tube. Repeat with the second tube. Identify the gas given off in this reaction: {5}________

SECTION 18.2 Solving Problems Involving Reaction Rates

It is a fairly simple task to quantitatively compare reaction rates. But, is there a way that we mightquantitatively express the rates of the reactions of magnesium and zinc with HCl which you witnessed inactivity 18.1.? Well, actually, there are several ways in which this could be done.

rate =

moles reactant used

time required for change

1. We could weigh the magnesium and record the time requiredfor it to react completely. Rate would then be defined by how fasta reactant is consumed.

rate =

moles produce formed

time required for change

2. Or, we could time the reaction and measure the volume of gasproduced during the experiment. Rate would then be defined byhow fast a product is formed. From this data, the rate of thereaction could then be calculated using the formula at right:

Often it is more convenient for scientists to calculate rate in units of grams/second, orgrams/week, or moles/day, etc. When this is the case, the situation will specify the units you are to usewhen solving a problem.

18-3 ©1997, A.J. Girondi

rate =

moles reactant used

second

If specific units are not requested, you should calculate the rate in "moles reactant used / second."

Sample problem: If 0.048 grams of magnesium, Mg, completely reacted with acid in 20. seconds, whatwould be the rate of this reaction in units of "moles Mg / sec?"

Note that in this problem you are given a quantitative relationship between Mg and time: 0.048 gMg / 20 sec. To change this ratio to "moles Mg / sec" will require a short fencepost. Complete thefencepost below, and calculate an answer. Fill in the blanks.

0.048 g Mg

20. sec X = {6} ___________ mole Mg /sec

If you got a small answer, don't be suspicious. Remember, a mole is a very large number of molecules!Now, let's try another problem.

Sample Problem: In an experiment 0.0070 g of zinc metal completely reacts with acid in 30.0 seconds.Calculate the average rate of this reaction. Complete the fencepost below and do the calculations to see ifyou get the answer given.

0.0070 g Zn X X = 3.6 X 10-6 mol Zn /sec

Next, let's try a problem with a different twist to it.

Sample Problem: It is known that the rate at which magnesium reacts with HCl is 5.0 X 10-2 moles/secand that the reaction takes 5.0 seconds to occur. Calculate the number of moles of magnesium that havereacted.

Complete the fencepost below, and see if you can get the answer that is given.

5.0 X 10-2 mole Mg

1 sec X

1 = 0.25 mole Mg

Problem 1. Calculate the number of grams of Mg that reacted in the problem above. Show your work.

__________ g Mg

Before moving on, it is important that you feel certain about your ability to solve rate problems. Foradditional practice, solve the following problems.

18-4 ©1997, A.J. Girondi

Problem 2. After 2.0 minutes, 0.50 grams of zinc react completely in HCl. Calculate the rate of reactionin grams of zinc per second.

__________ g Zn/sec

Problem 3. Referring to the last problem, calculate the rate of the reaction in moles Zn used persecond.

__________ mole Zn/sec

Problem 4. Given that magnesium reacts with air at a rate of 4.00 X 10-3 moles per 1.00 second, howmany minutes would it take for 4.00 moles of Mg to react?

__________ minutes

Problem 5. Referring to problem 4 above, how many grams of Mg would react after 140. seconds?

__________ g Mg

Problem 6. In 3.0 minutes 0.10 g of CaCO3 will decompose in HCl. Calculate the rate of the reaction inmoles CaCO3 per second.

__________ moles CaCO3/sec

18-5 ©1997, A.J. Girondi

SECTION 18.4 Collision Theory and Activation Energy

Now that you know how to express reaction rates quantitatively, you will do several experimentsthat will enable you to determine the effects of certain factors on the rates of chemical reactions. Theyinclude: (1) nature of the reactants; (2) concentrations; (3) temperature; (4) surface area; and, (5) catalysts.It is usually quite simple to determine the effects of these factors on reaction rates once the basicprinciples involved in chemical reactions are known. An explanation based on the collision theory plays amajor role in this.

The collision theory is based on the assumption that for a chemical reaction to occur, particlesmust first collide with each other. In these collisions, atoms and electrons are rearranged by a reshufflingof chemical bonds that results in the formation of products. A reaction does not necessarily occur everytime there is a collision. First, the reactant molecules must collide in just the right way. According to thecollision theory, the rate of a reaction depends on two factors: (1) the number of collisions per secondbetween the reacting particles, and (2) the fraction of these collisions that are effective. The collidingparticles must collide with sufficient energy to break the reactant's bonds. If sufficient force is not exerted,no reaction will occur. This minimum amount of energy required to rearrange the reactants into products iscalled their activation energy.

Activation energy is easiest to visualize using graphs. Figure 18.1, shows that the products of aparticular reaction possess more energy than the reactants do. There is only a 5 kJ (kilojoule) differencebetween the energy of the reactants and the products. Logically, one would think that adding 5 kJ ofenergy to the reactants would be enough to convert them into products. This is exactly what Figure 18.1shows. Now look at Figure 18.2. This plot is of the same reaction shown in Figure 18.1, but the activationenergy has been added. Recall that the activation energy is the minimum energy needed to rearrangereactants into products. This means that before the reaction in Figure 18.2 can occur, about 12 kJ (not 5kJ) of energy must be added to the reactants.

When reactant molecules collide they form what is called an unstable activated complex whichquickly decomposes to form the products of the reaction. When this decomposition occurs, some of theactivation energy is given off. So while about 12 kJ of activation energy were needed to get the reactionin Figure 18.2 to go, the actual difference in the energy content of the reactants and products is only 5 kJ.

kJ

5

10

15

20

reaction progress

reactants

products

Figure 18.1

kJ

5

10

15

20

reaction progress

reactants

products

Figure 18.2

Activation.Energy

As you will learn in chapter 19, exothermic reactions give off heat energy, while endothermicreactions absorb heat energy. Therefore, in exothermic reactions the products have less energy than thereactants. In endothermic reactions the products have more energy than the reactants. Since theproducts possess 5 kJ more energy than the reactants in Figure 18.2, would this plot represent an

exothermic or an endothermic reaction? {7}_______________________________

18-6 ©1997, A.J. Girondi

Study Figures 18.3 and 18.4. Label the reactants and products, and draw an arrow showing the

activation energy on each curve. How much energy would need to be added to the reactant in Figure

18.3 to get a reaction to occur? {8}________________. How much energy would need to be added to

the reactant in Figure 18.4 to get a reaction to occur?{9}_______________ Is the reaction represented in

Figures 18.3 exothermic or endothermic {10}________________________ Is the reaction represented

in Figures 18.4 exothermic or endothermic? {11}_________________________

kJ

5

10

15

20

reaction progress

Figure 18.3

kJ

5

10

15

20

reaction progress

Figure 18.4

In any reaction system, the molecules do NOT all have equal amounts of energy. That's why wedefine the temperature of a system as a measure of the average kinetic energy of the molecules. As aresult, some molecules may not collide with sufficient energy to produce a reaction. If the activationenergy required for a particular reaction is high, fewer molecules are likely to have enough energy to reactwhen they collide. If a particular reaction has a very high activation energy, would you expect the reactionrate to be fast or slow?{12}___________. Explain why. {13}___________________________________

______________________________________________________________________________

___________________________________________________. Would this same reaction have a

large or a small value for its reaction rate? {14}______________________

The "nature of the reactants" refers to the identity and properties of the reactants. Activationenergy is just one of these properties. Some combinations of reactants have high activation energies,while that for other combinations is low.

Without specific knowledge about the reactants, you can't be certain whether a particular reactionwill occur rapidly or slowly. You may recall that when you were studying equations of single and doublereplacement reactions, you had to predict whether or not a reaction would occur. However, it's possiblethat some of those reactions may occur instantly, while others which are predicted to occur may occurvery, very slowly – perhaps even taking years until they are completed! Educated guesses about reactionrates are possible if some characteristics of the reactants are known. Chemists have developed two rulesconcerning relative rates of reactions:

1. Many reactions that do not involve the breaking of bonds occur rapidly at room temperature.

2. Many reactions in which bonds are broken tend to occur slowly at room temperature.

18-7 ©1997, A.J. Girondi

Use these two general rules to determine which of the reactions below would probably be faster. Writethe word faster next to that equation. Explain why you made your choice.

Reaction 1: H + H -----> H2

Reaction 2: 2 CH4 + 2 Br2 -----> 2 CH3Br + 2 HBr

Explanation: {15}_________________________________________________________________

______________________________________________________________________________

______________________________________________________________________________

The two reactions shown above involve either atoms or molecules of substances. It is alsopossible for ions to react with each other. Ions are charged particles consisting of one or more atomswhich can exist as dissolved (aqueous) particles in water solutions or as gases at high temperatures.Reactions between ions tend to occur very rapidly. Two examples of ionic reactions are shown below.Note that the ions have the charges shown:

Ag1+(aq) + Cl1-(aq) ----> AgCl(s)

5 Fe2+(aq) + MnO41-(aq) + 8 H1+(aq) ---> Mn2+(aq) + 4 H2O(l) + 5 Fe3+(aq)

ACTIVITY 18.5 Comparing the Rates of Two Reactions

In this activity you are going to compare the rate at which two compounds react with a solution.The first compound is ferrous sulfate, FeSO4, which is ionicly bonded. The second compound is oxalicacid, H2C2O4, which is covalently bonded. The solution they are going to react with is potassiumpermanganate, KMnO4, which is ionicly bonded. Wear your safety glasses and an apron. Obtain adropper bottle containing a solution of 0.01 M potassium permanganate (KMnO4) which is purple in color.Handle this solution with care since it can stain your skin or clothes. (Be sure to follow proper lab safetyprocedures.) Obtain two 100 or 150 mL beakers and fill each about one-third full of distilled water. Labelthem as beaker 1 and beaker 2. Add a small amount of solid FeSO4 to beaker 1 and a small amount of solidoxalic acid to beaker 2. (Handle oxalic acid carefully. Wash your hands as needed.) Stir each solution for acouple of minutes, rinsing the stirring rod before you move from one solution to the other. Don't beconcerned if some of the solid in either beaker remains undissolved.

Add a drop of the KMnO4 solution to the FeSO4 in beaker 1, stir the solution, and make aqualitative estimate (such as rapid or slow) of the time required for the color to disappear. Repeat thisprocedure by adding another drop of KMnO4 to beaker 1.

Repeat the procedure in the paragraph above, but add the KMnO4 to the oxalic acid solution inbeaker 2. Again, make a qualitative estimate of the time required for the color to disappear compared tothe result for beaker 1. There should be an obvious difference in rate.

Compare the rates of reaction for FeSO4 in beaker 1 and for H2C2O4 in beaker 2? __________________

______________________________________________________________________________

Suggest possible reasons for the difference in rates. (Refer to section 18.4 if you need help.) ________

______________________________________________________________________________

18-8 ©1997, A.J. Girondi

SECTION 18.6 Reaction Mechanisms

There is one additional factor related to the nature of the reactants that must be discussed here.Look at the reactants and products in this reaction:

4 HBr + O2 ----> 2 H2O + 2 Br2

Remember that there must be collisions of sufficient energy between the HBr and O2 molecules beforethe products can be formed. Is it possible that the O2 molecule could collide with the HBr with enoughenergy to become bonded to the HBr, but not with enough energy to actually break the H–Br bond? Ifthis happened, you would have one larger molecule containing all of the reactants:

HBr + O2 ----> HBrO2

Let's take a closer look at what this means. Many chemical reactions involve several intermediate steps.The pathway of intermediate steps through which the reactants pass as they are converted to products iscalled the reaction mechanism. The reaction mechanism for the HBr + O2 reaction involves the threesteps shown below:

Step 1: HBr + O2 ----> HBrO2 <--------- (slow)

Step 2: HBrO2 + HBr ----> 2 HBrO <--------- (fast)

Step 3: 2 HBrO + 2 HBr ----> 2 H2O + 2 Br2 <--------- (fast)

Sum: 4 HBr + O2 + HBrO2 + 2 HBrO ----> HBrO2 + 2 HBrO + 2 H2O + 2 Br2

Net: 4 HBr + O2 ----> 2 H2O + 2 Br2

In step 1, a molecule of HBr combines with an O2 molecule to form a larger molecule of HBrO2. In step 2,the HBrO2 molecule collides with another HBr molecule to form two molecules of HBrO. In step 3, twomore molecules of HBr collide with two HBrO molecules to form the final products which are two moleculesof H2O and two molecules of Br2. Note that the HBrO2 and HBrO molecules are both formed andconsumed during the reaction; therefore, they do not appear as either reactants or products in the overallor net equation. Such substances are sometimes called "intermediates." If you "add up" the threeequations in the reaction mechanism, the result will be the net equation.

Two steps in this reaction mechanism occur extremely fast; however, one step occurs much moreslowly. For that reason, we say that the rate of the reaction depends on the slowest step in themechanism. Just like a relay team of runners, if even one runner is slow, the entire team may beconsidered slow.

No chemical reaction can occur faster than the slowest step in its mechanism. This slow step iscalled the rate-determining step. In the 3-step reaction shown above, which step is the rate-determiningstep? {16}__________ Would the overall reaction be fast or slow?{17}___________________.

Explain: {18}_____________________________________________________________________

______________________________________________________________________________

(Note: whether a reaction has a one-step mechanism or a multi-step mechanism depends on the nature ofthe particular reactants involved.)

18-9 ©1997, A.J. Girondi

The rate of a chemical reaction is also affected by the concentration of the chemicals used.Collision theory proposes that molecules must collide with each other before they can react. Using thisprinciple from collision theory, make a prediction about how increasing the concentration of reactantswould affect the rate at which substances react. Explain both how and why: {19}_______________________________________________________________________________________________________________________________________________________________________________

ACTIVITY 18.7 Solution Concentration and Reaction Rate

Now let's investigate the relationship between concentration and reaction time. For this activityyou will need two solutions from the materials shelf labeled solution 1 and solution 2. You will be timingreactions in this series of experiments. It is important that you read through the entire activity beforebeginning. In this way, you will know what is expected of you. It is always a good idea to do this beforeperforming an experiment.

Procedure:

1. Measure 5 mL of solution 1 and pour it into a 50 mL flask or beaker.

2. Add 5 mL of water (at room temperature) to the same flask. Swirl the container to mix the solution.

3. Carefully measure 5 mL of solution 2.

4. Get a watch ready (use a stopwatch, if available) and then mix solutions 1 and 2. Immediately stir thesolution by swirling. Begin timing the reaction at the moment solution 2 is added to solution 1. Record thetime required for a noticeable reaction to occur in Table 18.1. If you place the flask or beaker on a sheet ofwhite paper, the reaction will be more visible.

What evidence do you see that a chemical reaction has occurred?_____________________________

You have already carried out reaction 1 in Table 18.1. Now prepare the correct combinations shown inTable 18.1 for each of the remaining three reactions. Be sure to use water which is at room temperature.Time each reaction and record your results in the Table.

The last column of Table 18.1 asks that you calculate the concentration of HSO31- ion present ineach of the reaction mixtures. The bisulfite ion, HSO31-, is present in solution 1. To calculate thisconcentration you need to know that the concentration of HSO31- in the original stock solution is 0.072 M.When you add solution 2 and water to the stock HSO31- solution (solution 1), you reduce theconcentration of the HSO31- ion. The total volume in each flask at the end of the reactions is always 15mL.

Study the calculations shown below for determining the concentration of HSO31- in reaction 1.The formula you will use is:

McVc = MdVd Mc = molarity of concentrated sol'n

Vc = volume of concentrated sol'n Md = molarity of dilute sol'n Vd = volume of dilute sol'n

18-10 ©1997, A.J. Girondi

Mc will have a value of 0.072 M HSO31- in each of your calculations. Vc will vary in the four reactions: (5.0mL, 4.0 mL, 3.0 mL,and 2.0 mL). Md is the concentration of the diluted HSO31- solution (the unknown). Vd

will be 15 mL in each reaction.

For reaction 1 the calculation is as follows:

Mc Vc = MdVd so, M d =

McVcVd

and, M d = (0.072) (5.0 mL)

(15 mL) = 0.024 M

The equation above can be used whenever you wish to calculate the new concentration of a solution afteryou have diluted it. Perform the calculations for the remaining three reactions. Show your work below.Enter the results in Table 18.1.

Reaction 2:

__________ M

Reaction 3:

__________ M

Reaction 4:

__________ M

Table 18.1Concentration vs. Reaction Time

Rx No. Volume Sol'n 1 Volume Sol'n 2 Rx Time Molarity of HSO31-

+ water (sec) (mol/L)

1 5 mL + 5 mL 5 mL ______ _________

2 4 mL + 6 mL 5 mL ______ _________

3 3 mL + 7 mL 5 mL ______ _________

4 2 mL + 8 mL 5 mL ______ _________

On the grid which follows, prepare a graph of reaction time versus the concentration of HSO31-.Label each axis, placing the independent variable on the horizontal (x) axis and the dependent variable onthe vertical (y) axis. The independent variable in this example is concentration. The value of thedependent variable depends on the value of the independent variable.

Which variable is dependent here?{20}_______________________

18-11 ©1997, A.J. Girondi

State a general rule that relates the concentration of reactants to the rate of a chemical reaction: {21}_____

______________________________________________________________________________

What effect would diluting solutions with water have on the rate at which they react? {22}____________________________________________________________________________________________

ACTIVITY 18.8 The Effect of Temperature on Reaction Rate

(Note: This activity is best done by all lab groups on the same day. Ask your instructor.)

Another factor that affects the rate of chemical reactions is temperature. Before actuallyconducting this activity, give some thought to what is going to be happening to the kinetic energy(motion) of the particles of the reactants as they are heated.

Scientists use a hypothesis as a means of predicting what is going to happen before they actuallyperform an experiment. This gives them a chance to generate an "educated guess" about the results ofthe experiment.

What effect will temperature have on the rate of a chemical reaction? (State a hypothesis in terms of the

collision theory.) :{23}_______________________________________________________________

______________________________________________________________________________

To test your hypothesis, we will once again be using solutions 1 and 2. To be effective, both ofthese solutions should be at the experimental temperature. This can be done quite easily by preparing awater bath with a known temperature. Your class will be measuring the rate of the reaction at four differenttemperatures. These temperatures should be close to the following: 5oC, 25oC, 40oC, and 50oC. Eachof the time trials will require 3 mL of solution 1, 7 mL of water, and 5 mL of solution 2.

A hot or cold water bath is prepared by filling a 600 mL beaker Figure 18.5 about half-full of water.If there is a need to raise the temperature, use your burner to supply heat. Ice cubes can be used to lower

18-12 ©1997, A.J. Girondi

the temperature. The general setup is shown in Figure 18.5. Be sure that the level of water in the bath ishigher than that of the solutions in the test tubes that will be placed in the bath.

Your lab group will be assigned one temperature at which to time the reaction. Other labs groupswill be assigned different temperatures. Each group should repeat its time trials several times. The timesshould be fairly close. Discard any times which are not close to the others which you obtained, thenaverage the results. If other lab groups in your class are assigned the same temperature as your group,include their times when you calculate your average time.

Procedure:

1. To prepare your solutions, place 5 mL of solution 2 into a large (25 X 200 mm) test tube. Then,measure 3 mL of solution 1 and 7 mL of water into a second large test tube. Do not mix these twosolutions, yet! They can be stoppered and saved. Be sure to label them.

2. Bring your water bath to the assigned temperature for your lab group.When the proper temperature has been achieved, place each unstopperedtest tube in the water bath and give each enough time to reach thetemperature of the water bath. (This will be about 3 to 5 minutes.) Be sure tomaintain a constant and even temperature in the bath while the test tubes arein it. Small pieces of ice or small amounts of heat with lots of stirring will do this.The test tubes, themselves, can be used as stirring rods.

3. After time has elapsed, quickly mix the two solutions by pouring thecontents of one tube into the other tube. To provide good mixing, you shouldimmediately pour the mixture back into the empty tube and quickly place thetube back into the water bath. Record the reaction time. Repeat the reactionat the same temperature until you get two or three values which are close.Average the values and record them in Table 18.2. (As noted above, if othergroups worked at the same temperature as yours, include their values in youraverage.)

thermometer

sol'n 2

sol'n 1 +water

Figure 18.5Heating Solutions

4. Obtain the average time values for the other three temperatures from other lab groups in your class.Enter those values in Table 18.2.

Table 18.2Temperature vs. Reaction Time

Rx No. Relative Experimental Rx Time Temperature Temperature (sec)

1 Cold __________ ______(about 5oC)

2 Room Temp __________ ______(about 25oC)

3 Warm __________ ______(about 40oC)

4 Hot __________ ______(about 50oC)

18-13 ©1997, A.J. Girondi

5. Prepare a graph of temperature vs. reaction time on the grid which follows. Be certain to label each axison your graph. Put the dependent variable on the vertical (y) axis.

What is the dependent variable?{24}_________________________ Does the graph support your

hypothesis concerning temperature and reaction rate?__________ If not, explain: _______________

______________________________________________________________________________

Was the prediction you made in your hypothesis in agreement with the results that you obtained for this

activity? ___________ Experimental error tells us that some of the data we collect during experiments

is not always accurate. A lot of experimental error is really human error. We are often not demanding

enough of ourselves with our laboratory techniques. The end result is data that is not wrong, but not very

accurate, either. List a few sources of experimental error in this activity: _________________________

______________________________________________________________________________

ACTIVITY 18.9 The Effect of Surface Area on Reaction Rate

The next variable which you will study that influences chemical reaction rates is surface area.Surface area has a large effect on rates of reactions in which solids are involved. Chemical reactionsinvolving solids take place only on the surface of the solid. Justify this statement using the collision theoryas your basis. Now, hypothesize about what effect increased surface area has on the rate of a reaction

involving a solid: {25}_______________________________________________________________

______________________________________________________________________________

18-14 ©1997, A.J. Girondi

Part A. Steel Wool

Obtain a sample of steel wool. Tear off two small and approximately equal pieces about the size ofa quarter. Form one piece into a small, marble–sized ball. Separate and spread the fibers of the othersample. Wear your safety glasses! Light a burner and use crucible tongs to hold the small ball of steelwool directly over the flame, allowing the flame to touch the wool. Note the rate at which the oxygen in theair reacts with the iron. Next, do the same thing with the other sample of steel wool. Was the rate ofreaction the same for both samples?__________. If not, reveal which reacted faster and why: ________

______________________________________________________________________________

______________________________________________________________________________

Part B. Iron Filings or Powdered Iron

Next, obtain a bottle of iron filings from the materials shelf. Obtain a "pinch" of the iron filingsbetween your thumb and forefinger, and while holding your hand above the burner flame, sprinkle thefilings into the flame. Watch what happens. Iron filings are very small particles which provide a greatamount of surface area between the iron and the air. When the iron meets the flame it reacts very quicklywith the oxygen in the air. You see "sparkles" in the air!

Many people who own backyard barbecue grills use charcoal briquettes which are pieces ofcarbon, that yield heat as they react with oxygen in the air. Gunpowder also contains carbon that reactswith other chemicals to give off heat. Why must the carbon in the gunpowder be in the form of a powder?______________________________________________________________________________

Part C. A Dust Explosion Using Lycopodium Powder - (Teacher Demonstration)

<--- blow air in

candle

Lid

Paint Can

powder

Figure 18.6Dust Explosion Apparatus

As you witnesses in the demonstration withiron filings or powder, substances that normally burnonly very poorly or slowly burn much more quicklywhen in powdered form. Many dangerous explosionshave occurred at flour mills. We don't normally think ofwheat as being explosive! However, when in thepowdered form it can be very hazardous. Todemonstrate the power of a dust explosion, yourinstructor will use a fine powder which is actually apollen from the Lycopodium plant.

The powder will be placed in a small pile insidean empty paint can. A candle is lit and placed inside thecan. The lid is tightly sealed to the can and a puff of airis blown into the can to suspend the dust. If all goeswell, the lid will be launched toward the ceiling as theexothermic dust explosion causes the gas inside thecan to rapidly expand!

ACTIVITY 18.10 The Effect of a Catalyst on Reaction Rate

The final factor that we will study which affects the rates of chemical reactions is the use ofcatalysts. You experimented briefly with a catalyst called manganese dioxide, MnO2, in an earlier chapter.A catalyst is a substance that can alter the rate of a chemical reaction. Most catalysts are used to speed upthe rate of reactions. You will be performing a "catalyzed" reaction which you saw in a previous chapter.Procedure:

18-15 ©1997, A.J. Girondi

1. Place roughly 10 mL of hydrogen peroxide, H2O2, in each of four clean test tubes (they don't have tobe dry). Even though you cannot see it because it is so slow, there is a chemical reaction occurring inwhich H2O2 decomposes. The equation for this reaction is: 2 H2O2(l) ---> 2 H2O(l) + O2(g) Certain factorscan speed up the rate of this decomposition.

Propose a reason that might explain why H2O2 is normally stored in a dark glass or dark plastic bottle.

{26}_______________________________________ Why is it a good idea to store H2O2 in the

refrigerator?{27}___________________________________________________________________

2. Add a little of the catalyst, MnO2, to one of the tubes of H2O2. Rapid bubbling indicates that the H2O2 israpidly decomposing into water and oxygen gas, and that the catalyst is working. Do you observe rapidbubbling? ______________

3. Try adding a little Fe2O3 (iron (III) oxide) to the second tube and a little common dirt to the third. To the

fourth tube add a little bit of granular aluminum. Which of these substances appear to be good catalysts

for this reaction?_________________________________________________ Which (if any) do not

seem to catalyze this reaction?_______________________________________________________

A catalyst is a substance that can alter a reaction rate, but interestingly enough, it is not consumedin the reaction. The catalyst is not a reactant or a product. For that reason, it does not appear in theequation for the reaction (except that it may be written over or under the arrow (the yield sign) to indicatethat it is being used:

Uncatalyzed reaction: 2 H2O2 ----------------------------------> 2 H2O + O2 (slow)

Catalyzed reaction: 2 H2O2 ----------------------------------> 2 H2O + O2 (fast)catalyst

ACTIVITY 18.11 Catalysts in Matches and in Cigarette Tobacco

--> --> --> A Teacher Demonstration <-- <-- <--

Hopefully, you are getting a bit curious about how a catalyst actually works. It is neither a reactantnor a product, but it must take some part in the reaction, or else it could not affect the reaction rate.

O

O H

H

O O

H

H

Surface Catalyst

Bonds weaken whenH2O2 contacts the catalyst

Figure 18.7Action of Surface Catalysts

A catalyst enters into a reaction by alteringthe reaction mechanism. You will recall thatthe reaction mechanism is the series ofreactions that together comprise the stepsin the overall changes observed. A catalystcan alter the reaction mechanism in a varietyof ways. Some surface catalysts, such asnickel, provide a surface on which one ormore of the reactants can be adsorbed.Once adsorbed onto the surface of thenickel, it is believed that reactant bonds maybe stretched and weakened so much that aless energetic collision is sufficient to causethe reaction.

18-16 ©1997, A.J. Girondi

kJ

5

10

15

20

reaction progressFigure 18.8

Effect of Catalyst onActivation Energy

reactants

products

AB

A = activation energy without catalystB = activation energy with catalyst

Other catalysts - sometimes called cyclic catalysts -may actually combine chemically with a reactant, making itmore reactive. Later on in the reaction, the catalyst ischanged back to its original state, leaving the same mass ofcatalyst at the end of the experiment as there was at thebeginning of it.

In spite of the specific mechanism involved,catalysts perform their function by allowing reactions tooccur without adding as much energy to the reactants.Recall that the activation energy is defined as the minimumamount of energy required to rearrange the reactants intothe products. A catalyst lowers the activation energy (seeFigure 18.8).

Your teacher will follow this procedure:

1. Obtain a book of safety matches and a small piece of very fine sandpaper from the materials shelf. Tearone match from the book and attempt to light it using the sandpaper. What's the result?

The activation energy required to start the reaction is greater than the small amount of heat generated bythe friction between the sandpaper and the match head.

2. Next, try to light another match by rubbing it on the striking pad on the book. Result?

______________________________________________________________________________

The striking pad on a pack of "safety" matches contains a catalyst that mixes with the chemicals on thehead of the match when you strike it. The small amount of heat produced is sufficient to cause a reactionwhen the catalyst is present. This is because the activation energy is lower when the catalyst is present.

Exactly how catalysts do this is an active area of research in science. Why do you think they call these

"safety" matches? ________________________________________________________________

3. Next, obtain two sugar cubes and a container of cigarette ash from the materials shelf. Rub a few sidesof one of the cubes in the ash so that some of the ash adheres to the sides. Place the cubes on a Pyrexwatch glass supported on a ring stand and try to ignite them by flaming them directly with a burner flamewhile holding the burner in your hand.

Describe what happens: ___________________________________________________________

______________________________________________________________________________

Tobacco ash, itself, does not serve as a catalyst. However, the temperature at which cigarettes burn isoften not high enough to insure that cigarettes will continue to burn slowly. To aid the burning process, acatalyst is added to the tobacco when it is processed into cigarettes. The catalytic action prevents thecigarette from going out by lowering the activation energy needed to keep the cigarette burning. Thissame catalyst also aids the burning process with the sugar cube.

18-17 ©1997, A.J. Girondi

A catalyst lowers the activation energy for a reaction by changing the pathway by which the reactionoccurs. (Pathway refers to the intermediate series of reactions which cause reactants to becomeproducts.) Altering the pathway is kind of like finding another way to walk home from school - perhaps aroute that will not require you to walk up a hill! Remember that the rate of a reaction depends on the rate atwhich molecules collide and also the number of collisions which are effective. Raising the temperature will{28}_____________ the rate at which collisions occur, thus speeding up the reaction. Lowering theactivation energy required can increase percentage of collisions which are effective. This is the job of a{29}__________________.

There are also catalysts that can raise the activation energy of a reaction, which serves to decrease thereaction rate. Such substances are called inhibitors .

SECTION 18.12 Some Uses of Catalysts

A catalytic converter is a device that helps to reduce the pollutants emitted by an automobile'sengine. The converter contains a wire screen which is plated with a catalytic metal, probably platinum orpalladium. Engines are not 100% efficient and do not burn gasoline completely into CO2 and H2O.Instead, some unburned hydrocarbons escape through the exhaust system. These are compounds thatcontain both carbon and hydrogen, and they are considered to be pollutants. When the hot butunreacted hydrocarbons pass over the catalyst they react, reducing the hydrocarbon pollution producedby the vehicle. The catalyst in this case is a metal that does not take part in and is not consumed by thereaction. The reaction merely occurs on the surface of the catalyst. How do you think the catalyst affectsthe activation energy needed to allow the burning of the hydrocarbons? {30}_____________________

Only "unleaded fuels" are supposed to be used in cars that have catalytic converters. If fuelscontaining lead are used, the lead can form a coating on the surface of the metal in the converter which willdestroy its catalytic properties.

Catalysts are widely used by industry to increase the rates of many chemical reactions thatotherwise would take place too slowly to be practical. Catalysts are used to increase the yield of high-octane gasoline from petroleum. Catalysts are widely used to increase the rate of formation of ammonia, amajor constituent of fertilizer. Catalysts also occur naturally.

The metal in a catalytic converter functions as a surface catalyst. Nitric oxide (also known asnitrogen monoxide) is an example of a cyclic catalyst. A typical example of catalysis involving NO is itseffect on the decomposition of ozone, O3, in the upper atmosphere of the Earth. Without the presenceof NO, ozone decomposes slowly. In the presence of NO, which serves as a catalyst, the reaction occursrapidly:

2 O3 -----------------> 3 O2 <---- slow

2 O3 -----------------> 3 O2 <---- fast[NO]

The steps involved in the reaction mechanism or pathway for the catalytic decomposition of ozone are asfollows.

Step 1: NO + O3 -----> NO2 + O2 <---- (fast)Step 2: NO2 -----> NO + O <---- (fast)Step 3: O + O3 -----> 2 O2 <---- (fast)

Overall: 2 O3 ---------------> 3 O2 <---- (fast)

NO is consumed in this step

But, NO is reproduced in this step

Overall, the NO is neitherproduced nor consumed. Itis a cyclic catalyst.

[NO]

18-18 ©1997, A.J. Girondi

In the space below, list all of the reactants found in the three steps of this mechanism. (Do not considerthe overall reaction.)

Next, in the space below, list all of the products found in the three steps of this mechanism. (Do notconsider the overall reaction.)

Finally, draw a slash through any particles which appear in both lists, and write the formulas of theremaining particles in the space below.

Remaining reactants:{31}____________________; Remaining products:{32}_____________________

Can you now use the remaining reactants and products to write the overall reaction?_________. Thesum of the steps in any reaction mechanism should "add up to" the overall reaction.

One specific type of catalyst is called an enzyme . Enzymes are complex substances present inbiological systems which function as catalysts for biochemical processes. Pepsin in gastric juice andptyalin in saliva are two examples of enzymes. Ptyalin is the catalyst that accelerates the conversion ofstarch to sugar. Starch will react with water to produce sugar without the presence of the enzyme ptyalin,but it takes weeks for the conversion to occur. We could not survive if our digestive processes occurredthat slowly. We would literally starve to death even as we eat!

SECTION 18.13 Review Questions and Problems

Problem 7. The equation: 2 H2 + O2 ----> 2 H2O describes a reaction in which 0.042 moles of H2O formafter 1.3 hours. Calculate the rate of this reaction in moles of O2 used per minute.

Problem 8. The equation: 2 Al + 3 H2SO4 ----> Al2(SO4)2 + 3 H2 describes a reaction in which 30.0 gramsof Al react with sulfuric acid in 10.0 minutes. Calculate the rate in moles of Al used per hour.

Problem 9. In the reaction: Cu + 2 AgNO3 ----> Cu(NO3)2 + 2 Ag it is found that if 0.082 grams of Cu isconsumed per 1.00 minute, how many moles of Ag will be produced in 1.00 hour?

18-19 ©1997, A.J. Girondi

Problem 10. In the reaction 2 H2 + O2 ----> 2 H2O, it is found that hydrogen gas reacts at a rate of 10.0grams H2 per 1.00 minute. How many minutes will it take for 267 grams of oxygen gas, O2, to react?

Problem 11. On the axes below, sketch an "energy profile curve" (like those found in Figures 18.1 -18.4) for a reaction which has an activation energy of 42 kJ and in which the total energy of the reactants is103 kJ more than the total energy of the products which is 20 kJ. Put a kJ scale on the "y" axis.

kJ

Reaction Progress ---->

Problem 12. Assume that you have 1.25 L of a solution of NaCl which is 0.36 M. If you add enoughwater to this solution to increase the total volume to 2.10 L, what will the molarity (M) of the diluted solutionbe? (See Activity 18.7 if you need help.)

Problem 13. If you have 1.00 L of a 0.65 M solution of HCl which you want to dilute until its molarity isreduced to 0.45 M, how much water will you have to add?

18-20 ©1997, A.J. Girondi

SECTION 18.14 Learning Outcomes

Place a check mark to the left of each learning outcome on the next page after you believe thatyou have mastered it. Arrange to take any chapter exams or quizzes, and then move on to Chapter 19.

_____1. Explain the effects of the following on the overall rate of a chemical reaction: the nature of the reactants, the concentration of the reactants, the temperature of the reactants, and catalysts.

_____2. Explain how the rate-determining step effects the overall rate of a chemical reaction.

_____3. Calculate rates of chemical reactions given the needed information.

_____4. Given the rate of a chemical reaction, calculate the number of grams or moles of product produced after a specified amount of time.

_____5. Given the rate of a reaction and the number of moles produced, calculate the reaction time.

_____6. Define activation energy and explain how a catalyst and an inhibitor affect it.

_____7. Given several chemical equations, hypothesize about which reaction occurs fastest.

_____8. Explain the relationship between the collision theory and rates of chemical reactions.

18-21 ©1997, A.J. Girondi

SECTION 18.15 Answers to Questions and Problems

Questions:

{1} A gas is evolved (given off); {2} A gas is evolved (given off); {3} Qualitative; {4} It is based onobservation rather than measurement; {5} H2; {6} 9.9 X 10-5 mole Mg/sec; {7} endothermic;{8} about 6 kJ needed; {9} about 10 kJ; {10} exothermic; {11} endothermic; {12} slow:{13} not many of the collisions would involved particles with enough energy to react; {14} small;{15} H + H ---> H2 would be faster since no bonds need to be broken; {16} Step 1; {17} slow;{18} If even only one step in a mechanism is slow, the overall reaction will be slow; {19} If substances aremore concentrated there will be more collisions between the reacting particles per unit of time, andtherefore a more rapid reaction; {20} reaction time; {21} More concentrated reactants will react faster thanless concentrated ones; {22} It would reduce the rate of the reaction; {23} At higher temperature there willbe more particle collisions per unit time and a faster rate of reaction; {24} reaction time; {25} Increasedsurface area between reactants will increase reaction rate; {26} Light may act as a catalyst; {27} At coldtemperatures the decomposition of H2O2 would be slower than at room temperature; {28} increase;{29} catalyst; {30} The catalyst lowers the activation energy; {31} 2 O3; {32} 3 O2

Problems:

1. 6.1 g Mg2. 4.2 X 10-3 g Zn/sec3. 6.4 X 10-5 mole Zn/sec4. 16.7 min5. 13.6 g Mg6. 5.6 X 10-6 mole CaCO3/sec7. 0.0027 mole O2/min8. 6.67 mole Al/hr9. 0.155 mole Ag/hr10. 3.37 min11.

kJ

Reaction Progress ---->

20

123

165

12. 0.21 M

13. Must add 0.40 L of water to obtain a final volume of 1.4 L of diluted solution

18-22 ©1997, A.J. Girondi