1

Unit 3: Reaction RatesIcons are used to prioritize notes in this section.

Make some notes: There are SOME important items on this page that should be copied into your notebook or highlighted in your printed notes.

Copy as we go: There are sample problems on this page. I EXPECT YOU to copy the solutions into a notebook— Even if you have downloaded or printed the notes!!!

Look at This: There are diagrams or charts on this page. Look at them and make sure you understand them. (but you don’t need to copy them)

Extra Information: This page contains background information that you should read, but you don’t need to copy it.

R Review: There is review material on this page. It is up to you to decide if you want to make notes or highlight it, depending on how well you remember it.

2

Chapter 8

Measuring Reaction Rate

3

Reaction Rates

Overview:Chemical reactions don’t all occur at the same

rate. Some are fast, some are slow. In chapters 8 to 10 we will examine reaction rates: How we measure them, how rates relate to kinetic theory and collisions, and what factors affect reaction rates. We will start with what rates are.

p.313

4

Expressing Reaction Rate

• A rate is a change in some quantity or amount over a certain amount of time. In its simplest form:

Unfortunately, both amount and time can be measured in many different units…

Amount in grams (g), millilitres (mL), litres(L), or moles (mol) Time in hours (h), minutes (m) or seconds (s).

5

Rates in a Chemical Reaction

• During a chemical reaction, reactants change into products:• Reactants Products

• As the reaction proceeds, the amount of reactant decreases, and the amount of product increases

• The change is always negative from the point of view of the reactants, and the change is positive from the point of view of the products.

Reactant Product

6

Graph of Reaction Progress

• From the sample graph above, you can see how, as the reaction progresses, the amount of product increases and the amount of reactant decreases.

• You will also notice that the rate changes. The reaction starts off quickly and then tends to slow down.

Reactant

Product

Reaction Progress (time)

Amou

nt o

f sub

stan

ce

7

Positive Reaction RatesThe Reaction rate is a positive value that expresses the change in the quantity of a reactant or product as function of time during a chemical reaction.• Note: It is traditional to express the reaction rate as a

positive number, even for the reactants where the actual change is negative.

For reactants:

or

For products:Where: r = reaction rate

Δt = change in time (tf -ti)Δq = change in quantity of material (qf -qi)

8

General vs. Individual Reaction Rates

• The formula on the previous slide can give you the reaction rate with respect to any individual reactant or product, but in all but the simplest of reactions, the individual reaction rates will be different for each substance involved.

• Sometimes we want to have a single “general” reaction rate that describes the rate of the entire reaction. In order to do this, however, we must consider the stoichiometry of the reaction.

9

Graph of Two ProductsLet’s consider a reaction in which water vapour (steam) is decomposed by a

hypothetical catalyst to create hydrogen and oxygen gas inside a sealed container

Sample Reaction2H2O 2H2 + O2

At any given point on the graph, the amount of hydrogen that has been produced is double the amount of oxygen. This means that the rate of hydrogen production is double the rate of oxygen production. This corresponds to the molar ratios!

H2

O2

(mol

es)

H2O

10

Reaction Rates and Stoichiometry

• The individual reaction rates are related to the stoichiometric ratios… the mole ratios… of the reaction’s equation.

• In the example on the previous slide, the H2 rate was twice the O2 rate, or conversely the O2 rate was half the H2 rate.

• In the reaction 2N2O5 4NO2 + O2, the production of NO2 will be four times the production of O2. The NO2 will also appear at a rate that is twice as fast as the N2O5 is disappearing!

11

General Reaction RateFor a generalized reaction, such as:

aA + bB cC + dDWhere: a, b, c, and d are coefficients

A, B, C and D are chemical formulasThe general reaction rate is given by:

Where: r = the general rate of the reactiona, b, c, and d are coefficients from the reaction’s equation.Δ[A], Δ[B], Δ[C], and Δ[D] are the changes in concentration.Δt is the change in time between concentration measurements.

Special note: You only need to know any one of the four data sets to solve for r

t

D

dt

C

ct

B

bt

A

ar general

1111

)(

12

Example

• For the equation C3H8 + 5O23CO2+4H2O, you could find the general rate of reaction using:

• Or simplified a bit: t

OH

t

CO

t

O

t

HCr general

22283)( 4

1

3

1

5

1

1

1

t

OH

t

CO

t

O

t

HCr

43522283

13

New symbol [ ]

• The square brackets around a quantity indicates “the concentration of”, normally expressed in mol/L

[NaCl] means “the concentration of salt solution.”

[CaBr2] means the concentration of Calcium bromide

[A] means the concentration of substance A

14

• Since concentration is often used to measure changes, the equation given two slides ago is often written as shown below:

• Later in the course, we will use this notation when determining forward and reverse rates in an equilibrium situation.

t

D

dt

C

ct

B

bt

A

ar general

][1][1][1][1

)(

15

Example

• Write the mathematical expression for the reaction rates of the following equation.

• Na2S(aq) 2Na+(aq) +S2-

(aq)

Answer:

t

S

t

Na

t

SNar

][

2

][][ 22

16

rA=-ar

\rB=-brrC=crrD=dr

Corresponding Individual Rates• Once you know the general rate, you can use it to

find the corresponding individual rates–eg. for aA + bB cC + dD:

Where:a, b, c, and d are coefficientsr is the general rate of the reaction.rA is the individual rate for substance ArB is the individual rate for substance BrC is the individual rate for substance CrD is the individual rate for substance D

17

• Read pages 213 to 217• Study the example problems on page 218

• Page 219• Do the seven questions

18

Ways to Measure Reaction Rate

Overview:There are many ways to measure the rate of a change… • A reactant may disappear, the product dispersing and

changing the apparent mass.• A gas may be produced and its volume measured. • The concentration of a solution may change• A solution may change its pH

All of these things can be used to measure a rate of change. In an experiment, you should choose the easiest one to measure.

8.3

19

Parameters of Measurement

There are many parameters of reactants and products that may be measured during a reaction. The four most common are:• Mass• Volume• Concentration• Number of moles

20

What Parameter to Use?

• When planning an experiment, use the parameter that is easiest to measure

• Whatever parameter you choose, remember that a rate is always a unit of the parameter, divided by a unit of time:

• Litres / second• Kilograms / hour• Moles / minute• grams/(Litre·minute)• Moles/(Litre·second)

time

Amountrate

These are all examples of rates: Parameters units divided by time units.

(some are not used in chemistry)

21

Summary of Common Parameters to Calculate the Rate of Reaction

Parameter measured

Preferred rate units

Best used with these physical states

Sample Rate Equations

Mass g/s Solid, liquid r = Δm/ΔtVolume mL/s or L/s Liquid, gas r = ΔV/ΔtConcentration mol/(L s)∙ Aqueous solution r = Δ[A]/ΔtParticles mol/s Solid, liquid, gas

(calculated) r = Δn/Δt

Other units of time can be used, eg. g/min,

mL/h, mol/min, etc., but seconds are preferred

22

• Read pages 220 to 223• Do questions on page 225, #1-6

23

Average vs. Instantaneous Rate

OverviewA reaction rate is based on time, and since the rate may change with time you have two choices for calculating the reaction rate.• Average rate between two specified times, or• Instantaneous rate at a specific time.Of the two ways, the average rate is the one we use

most often, simply because it is easier to calculate. Arguably, the instantaneous rate is a more accurate reflection of what actually happens during the reaction

8.4

24

Comparison of Average and Instantaneous Rates

• Average rate is found by picking two points in time and calculating the slope of the secant between them.

• Instantaneous rate is found by picking one point, and drawing a tangent. The slope of the tangent is then calculated.

Time

Amou

nt

Δt

Δt

ΔA

ΔA

secant

tangent

2 4 7

Inst

anta

neou

s

Average

25

The Difficulty of Instantaneous Rates • It is not possible to calculate an exact tangent

slope from data without using calculus, a branch of mathematics usually taught in the first year of university.

• There are methods of estimating the slope of a tangent that are fairly simple:• You can draw the graph and sketch a tangent line,

then calculate its slope (a rather unreliable method shown previously), or

• You can use the method given in your text book (shown on the next slide).

26

Estimating Tangent Slope• Let’s say you have been asked to find

the instantaneous rate at 10 seconds.• It’s impossible to get the exact slope of

the tangent without calculus, but let’s estimate.

• Draw a secant (a “fake” tangent) spanning equal times on each side of the time we are interested in, say from 5 seconds to 15 seconds (with our 10 second time exactly in the middle)

• Calculate the slope of this “fake” tangent. It will be very close to the same as the slope of a real tangent at the same point.

27

Does the rate always slow down?Hypothetical situation: Jennifer does an experiment in which she puts a small piece of magnesium wire into concentrated hydrochloric acid. She measures the amount of hydrogen gas produced, records the data on the left and draws a graph below:

Time mL of H2

0 sec 0 mL

5 sec 6 mL

10 sec 12 mL

15 sec 18 mL

20 sec 24 mL

25 sec 29 mL

30 sec 35 mL

35 sec 40 mL

40 sec 40 mL

45 sec 40 mL

50 sec 40 mL

55 sec 40 mL 0 5 10 15 20 25 30 35 40 45 50 55 Time (s)

mL 40 35 30 25 20 15 10 05 00

Why does her graph not show the normal curve representing a gradual slowing of the reaction rate as the reactants are used up?(Discuss. Answer is on the next slide)

Why not this?

28

Answer to previous slide

• Jennifer used a fairly large amount of concentrated acid, and a small amount of magnesium. Because the magnesium is all used up before the concentration of the acid changes much, the reaction ends before the normal “curve” can be established.

Magnesium used up

29

Rate of Reaction Lab Activity

• Lab Activity: Determine the rate of the reaction of magnesium and hydrochloric acid under at least three different concentration conditions.

• Mg(s) + 2HCl(aq) MgCl2(aq) + H2(g)

30

• Exercises on page 228, #1 to 5• Question 3 requires a pencil and ruler to draw an

accurate graph.

31

Chapter 9

Collision Theory

32

Types of Collision

Overview:• Collision theory is an extension of the kinetic theory. It

states that in order to react, particles must collide with each other.

• Not all collisions are effective in starting reactions. The particles must collide with just the right angle and enough energy in order to react.• Ineffective collisions do not result in a chemical reaction

(your text book calls these elastic collisions)• Effective collisions do result in a chemical reaction (your

textbook calls these inelastic collisions)

9.1

33

Ineffective (elastic) collisions

In a small (1mL) sample of gas at room temperature there can be as many as 1028 collisions every second.• (That’s 1 000 000 000 000 000 000 000 000 000 000 collisions!!!!)

• The vast majority of these collisions are “ineffective”. They do not result in a chemical reaction. The particles simply rebound in the same shape as they hit. The collision is “elastic”• That’s a good thing. If every collision between

molecules caused a reaction, ALL chemical reactions would occur at explosive speeds!

34

Effective (Inelastic) Collisions

• In order for a collision to be effective there are two conditions that must be met:1) Orientation. The colliding particles must hit

each other in the correct orientation to break apart and reassemble.

2) Activation energy. The particles must hit with enough kinetic energy to break apart and reassemble in a new pattern.

35

Pink area represents the number

of ineffective elastic collisions

Activation Energy

• Recall the Maxwell-Boltzmann curve discussed back in module 2. It shows the number of particles with various amounts of kinetic energy available during collisions.

Kinetic Energy available during collisions

Num

ber o

f Par

ticle

s

The minimum kinetic energy needed for the collisions to be effective can be shown by a vertical line. This represents the Activation Energy (Ea)

Activ

ation

Ene

rgy

(EA)

Not enough energy Enough Energy

Green area represents thenumber of effective collisions

36

• In the example on the previous slide, the number of effective collisions was much smaller than the number of elastic collisions.

• This corresponds to a slow reaction—one with a low reaction rate.• By changing the parameters of the reaction, the

temperature, the concentration, the catalyst, etc. we could change the size of the green area, and therefore change the rate of reaction.

Effective collisionsElastic collisions

37

Reaction Mechanisms & Collisions

Overview:Collision theory explains the interactions between particles in a reaction mechanism. As particles collide, kinetic energy changes into potential energy. If enough kinetic energy (the activation energy) is absorbed, then a chemical change occurs. If not, the potential energy changes back into kinetic, and the particles bounce apart unchanged. If a reaction mechanism has several steps, the step with the highest activation energy determines the rate.

9.2

38

Simple Reactions

• In a simple reaction, the rate is controlled by the number of effective collisions.

• This in turn depends on the activation energy of the reaction

• The higher the activation energy, the slower the reaction is

Reactants activated products complex

Ea

39

Complex Reactions

• In a complex reaction mechanism there may be several steps, each with its own activation energy

• The step with the highest point on the energy graph is the “rate determining step”… the slowest step.

• The reaction cannot proceed faster than the slowest step

• In this example, step 2 is the rate determining step.

Step 1 Step 2 Step 3

Ea1

Ea2

Ea3

40

Chapter 10

Factors that Affect Reaction Rates

41

Factors That Affect Reaction Rates

Overview:• The rate of a chemical reaction can be affected,

and even controlled, by several factors. We keep food in a refrigerator to slow down the reactions that cause spoilage. We grind up reactants into fine powders to make reactions happen faster. We can heat materials to speed up reactions, we can control the concentration of reactants and sometimes we add catalysts to hasten reactions.

10.1

42

The Five Factors

The five most important factors affecting the rate of a reaction:1. Nature of the Reactants2. Surface Area of the Reactants3. Concentration of the Reactants4. Temperature of the Reaction Environment5. Catalysts

We will examine these five factors in more detail in the slides that follow.

43

Factor 1:The Nature of the Reactants

Generalizations:• Reactions involving gases are faster than those

involving liquids or solids• Homogeneous reactions (all reactants the same

phase) tend to be faster than heterogeneous reactions that must occur at an “interface”

• The more bonds which must be broken, the slower a reaction will be

• The stronger the bonds that need to be broken, the slower a reaction will be.

Collision Theory:• Effects of collisions are on the next slides

44

Summary

Slower FasterSolids Are slower than... Gases

Heterogeneous(phase boundaries)

Are slower than... Homogeneous(all same phase)

More Bonds Are slower than... Fewer BondsCovalent Bonds Are slower than... Ionic BondsStrong Bonds

Are slower than...Weak Bonds

These are generalizations, not absolute rule. However, when faced with an exam question that asks you to compare the rate of two reactants, you can use these to help you out.

45

• Read pages to 246 to 249• Do exercises 1 to 7 on page 250

46

Factor 2: Surface Area

Generalizations:• The more surface area of solid reactants, the

faster the reaction will be.• Finely ground or powdered reactants will react

faster than larger chunks of solid reactant.

Collision Theory Justification:• The more surface area a solid has, the more places

particles can collide with it.

47

Effect of Surface Area

• Grinding the reactants into smaller pieces increases their effective surface areas.

• Reactions involving powdered reactants will be faster than reactions with solid chunks.

One large cubeSurface area = 24cm2

Eight small cubesSurface area = 48cm2

48

Summary

Slower FasterSolid Chunks of

Reactant Are slower than...Powdered reactants

Tight packed piles of reactant Are slower than... Mist or Dust

Small Surfaces Are slower than... Large Surfaces

49

Factor 3:Concentration of Reactants

Generalization:• The more concentrated the reactants in a liquid

solution or gaseous mixture are, the faster the reaction will be.

• This only affects reactants in gaseous mixtures and reactants in aqueous solutions. Solids and pure substances do not have easily calculated concentrations.

Collision Theory Justification:• The closer together particles are, the more

frequently they will collide.

50

SummarySlower Faster

Dilute reactants slower than... concentratedLow mol/L slower than... High mol/L

Concentration is such an important factor in controlling the rate of a reaction that we have developed a mathematical model to correspond to it called “the Rate Law” which is discussed before we go on to the remaining factors.

51

The Rate Lawrelating concentration and rate

• Concentration change is one of the most widely used methods of finding rates when using aqueous solutions.

• Since concentration is so directly linked to rates, an alternate method of calculating reaction rates exists, based on concentration.

• There are limitations to this method– It only works for elementary reactions in solutions at fixed temperatures– but it is quite useful when a reaction satisfies those conditions.

52

• In general, reactions proceed more quickly when the reactants are more concentrated• Remember: The concentration of a reactant is

represented by putting square brackets around the formula of the reactant.

• Eg. [NaOH] means “the concentration of sodium hydroxide in moles per litre”

• In a solution at unchanging temperature, the rate of a simple reaction is directly related to its concentration.

Rates of Reaction and Concentration

R

53

Rate Constant (k)

• What the previous slide means is that for a simple (first order) reaction with just one reactant , the rate of the reaction can be found by multiplying a certain, special number by the concentration of the reactant.

• This special number is called the rate constant of the reaction. In rate law equations it is symbolized by the letter k.

54

The Simple Rate Law(for 1st order elementary reactions)

• The rate of a simple reaction is therefore given by the formula:

• Where: r is the general rate of the reaction, k is the rate constant, and [A] is the concentration of the reactant.

][Akr

55

Example for Simple Reaction• A chemist has developed a process for

changing dissolved carbon dioxide into carbon and oxygen. Under a certain set of conditions, the k value for the reaction CO2(aq) C(s)+O2(g) is 0.955 L/s

• Find the rate of reaction when the concentration of carbon dioxide is:

a) [CO2]=2 mol/Lb) [CO2]=0.5 mol/L

Answer:1.91 mol/sec

Formula: rate = k [CO2]

Answer:0.478 mol/sec

Work 1: rate = 0.955 (2)

Work 2: rate = 0.955 (0.5)

56

Finding k for a simple 1st order reaction

• For a simple first order reaction

• Where:• k is the rate constant• r is the rate (given or determined by experiment)• [A] is the concentration of the one reactant (given or

determined by experiment)

][A

rk

r

r

k [A]

57

Higher Order Reactions

• In some reactions, the rate depends on the product* of the concentration of two reactants.

• In other reactions, the rate may depend on the square of the concentration of a certain reactant.

• These are higher order reactions (2nd order, 3rd order etc.)*I mean the mathematical product, the result of

multiplication, not the chemical product.

58

The Coefficients of the Equation usually give the “order”

• The coefficient of a reactant is its order.• Note: there are exceptions to this rule!

• The sum of the coefficients is normally the overall order of the reaction.

• In a rate law calculation, the concentration of a reactant is raised to the power of its coefficient (or order)

• PURE solids(s) and liquids(l) are not included in rate law calculations.

59

• For reactions of the type: a A + b B +… p P...

• Where: • a,b are coefficients. (numbers in front)• A, B are reactants. (chemical formulas)• k is the rate constant of reaction.• [A], [B] are reactant concentrations.

Note: The concentration of the product has no direct effect on the rate!

The General Rate Law(for higher order elementary reactions)

...][][ ba BAkr

60

Example for 2nd Order Reaction

• Under a certain set of conditions, the k value for the reaction 2H2O2 2H2O+O2 is 0.84 L2/mol-s

• Find the rate of reaction when the concentration of hydrogen peroxide is:

a) [H2O2]=2 mol/L

b) [H2O2]=0.5 mol/L

Answer:3.36 L/sec

Formula: rate = k [H2O2]2

Answer:0.21 L/sec

Work a): rate= 0.84(2)2

Work b): rate= 0.84(0.5)2

61

Finding k for higher order reactions

• For sample reaction aA + bB +... cC +...

• Where:• r is the rate• [A] and [B] are the concentrations of reactants and a

and b are coefficients • The values of [A], [B] and r may be given in the

problem, or may be determined by experiment

...][][ ba BA

rk

62

Restrictions on the Rate Law

• Reactants which do not have a concentration (ie. Pure solids(s), pure liquids(l)) are not used in rate law calculations. Only include gases(g) and aqueous(aq) reactants!

• Rate law cannot be used with complex reactions, that is reactions with multi-step mechanisms.

• Actually, the rate law can apply if you know the exact order of the slowest step of the reaction, but since this information is not usually available to you, avoid using the rate law for complex reactions.

63

Example of High Order Reaction

• 2 H2 + O2 2 H20 the coefficients are in yellow • 2 H2 + 1 O2 2 H20 remember there is one not shown

• So the rate formula for this reaction would be:• Rate= k [H2]2 [O2]1 Note: the 1 is not usually written!

• It is a 3rd order reaction overall (2+1=3)• It is 2nd order with respect to hydrogen.[H2]2

• It is 1st order with respect to oxygen.[O2]1

64

Example for 3rd Order Reaction

• Under a certain set of conditions, the k value for the reaction 2H2+O2 2H2O is 1.84 L/mol-s

• Find the rate of reaction when the concentrations are:

a) [H2]=3 mol/L,[O2]=2 mol/L

Formula: rate = k [H2]2 [O2]

Answer:33.1 mol/sec

work: rate = 1.84 [3]2 [2]

65

Example

• The Haber process turns nitrogen into ammonia, N2(g) + 3H2(g) 2NH3 .

• Under a hypothetical temperature and pressure the reaction has a general rate of 0.5 mol/s when the concentration of N

66

The Value of k

• The value of the rate constant (k) is different for each reaction,

• However once it has been found for a reaction, it can be used whenever that reaction occurs under the same conditions.

• The value of k is set at specific conditions (eg. temperature=25°C, pressure = 101 kPa, etc.)

67

The Units of k

• The units of k vary depending on the problem. Units are chosen to make the units of the rate work out as desired usually to mol/sec (mols-1) or mol/(Lsec) (molL-1s-1)

• Since the units are variable, they are frequently not given. Sometimes they are “reverse engineered” from the units of the answer.

• Common units are:• s-1 (or per sec) 1st order, answer in mol/Lsec• Ls-1 (or L/sec) 1st order, answer in mol/sec• L2mol-1s-1 (or L2/molsec) 2nd order, answer in mol/sec• L3mol-2s-1 (or L3/mol2sec) 3nd order, answer in mol/sec

• My advice: Don’t worry too much about the units of k, think about what the units of the rate should be.

68

• Read pages 251 to 257• On page 258 do: • Questions 1 to 7, plus • Questions 9, 11, 13, and 15

69

Factor 4:Temperature

Generalization:• Most reactions are faster at higher temperatures.• In fact, for every increase of 10°C, many reactions will

approximately double their rate.

Collision Theory Justification:• The higher the temperature, the faster particles

move. The faster particles move, the more often they will collide.• The overall number of collisions increase at higher

temperatures. The activation energy does not change.

70

Maxwell-Boltzmann Energy Distribution Curves

• Not all molecules move at the same speed (ie. With the same kinetic energy)

• Some are faster, some are slower, most are average. If we could graph their kinetic energies, we would get something like this:

71

Effect of Temperature on distribution curve

Kinetic Energy in kJ/mol

Num

ber o

f mol

ecul

es

Cold: no molecules reach activation energy

Warmer: some molecules reach activation energy

Hot: many molecules reach activation energy

Increasing temp.Moves peak

Forward.

72

Generalization

Slower FasterCold Reactants slower than... Hot Reactants

Cold Environment slower than... Hot Environment

Which is why we often heat up reactants in order to encourage a reaction.

73

Factor 5:Catalysts

Definition: A catalyst is a substance that increases the rate of a reaction without being used up by the reaction.

Generalizations:• A catalyst increases the rate of a reaction.• Catalysts do not change the outcome of a reaction.• Catalysts do not change the ΔH of a reaction.

74

How Do Catalysts Work?• All catalysts work by lowering the activation

energy (Ea) of a reaction.

• By lowering the Ea barrier they cause more of the collisions to become effective (inelastic).

uncatalyzed

catalyzed

Reaction Progress

Ener

gy

Ea

Ea

Kinetic Energy of Particles

Num

ber o

f Par

ticle

s

E a Bar

rier (

unca

taly

zed)

E a Bar

rier (

cata

lyze

d)Ea

Ea

Maxwell-Boltzmann CurveEnergy Curve

SLOW

FAST

75

Graphic Representation

None of the molecules have enough energy to climb over the activation energy barrier.NO REACTION!

Now let’s heat up the molecules so they move faster!

Some of the molecules now have enough energy to make it over the “activation energy barrier”These molecules REACT and form new products

The more molecules make it over the barrier, the faster the reaction is.

76

Graphic Representation

Cool molecules again, but this time with a catalyst.

The Catalyst makes the

“barrier” lower so it is easier to get

over

Some of the molecules now have enough energy to make it over the “activation energy barrier”These molecules REACT and form new products

The more molecules make it over the barrier, the faster the reaction is.

77

Collision Theory Justification

• The more effective collisions there are, the faster a reaction will occur.

• By lowering the activation energy of a reaction, a catalyst creates more effective collisions.• Note: A catalyst does not increase the total

number of collisions. It increases the percentage of collisions that are effective in producing a chemical change!

78

How Do Catalysts Work, Really?There are three ways catalysts work:• Homogeneous Catalysts are the same phase as the

reactants (gaseous or aqueous). They work by providing a reaction mechanism with a lower activation energy.

• Biological Catalysts (enzymes) are protein molecules in living systems that use a “lock and key” mechanism to speed up a reaction.

• Heterogeneous Catalysts are solids (often powdered) added to a solution or gaseous mixture that provide a surface on which reactions can take place. The surface may change the orientation of collisions.

See pages 264 to 266 for details.

79

Inhibitors

An inhibitor is a substance added to a reaction to slow it down. Inhibitors work in several ways:• Oil on the surface of iron slows rusting by

acting as a barrier to prevent oxygen or water from reacting with the iron.

• Preservatives and antioxidants absorb and remove reactants that would spoil food.

• Some inhibitors deactivate or destroy catalysts that are affecting reactions.

80

Generalization

Slower FasterNo Catalyst slower than... Catalyst

Inhibitor slower than... No Inhibitor

If you can find a catalyst for your reaction, it will speed the reaction up. Not all reactions have catalysts.

81

• Read pages to 266• Do Questions 1 to 7 on page 267• Do Questions page 272 #1 to 7(standard)

plus 9, 11 and 13 (advanced)