chapter 13 chemical kinetics. chemical kinetics the speed of a chemical reaction is called its...

Chapter 13Chemical Kinetics

Chemical Kinetics• The speed of a chemical reaction is called its

reaction rate• The rate of a reaction is a measure of how fast

the reaction makes products– or uses reactants

• The ability to control the speed of a chemical reaction is important

Defining Rate• Rate is how much a quantity changes in a

given period of time• The speed you drive your car is a rate – the

distance your car travels (miles) in a given period of time (1 hour)– so the rate of your car has units of mi/hr

Defining Reaction Rate

• The rate of a chemical reaction is generally measured in terms of how much the concentration of a reactant decreases in a given period of time– or product concentration increases

• For reactants, a negative sign is placed in front of the definition

at t = 0[A] = 8[B] = 8[C] = 0

at t = 0[X] = 8[Y] = 8[Z] = 0

at t = 16[A] = 4[B] = 4[C] = 4

at t = 16[X] = 7[Y] = 7[Z] = 1

at t = 16[A] = 4[B] = 4[C] = 4

at t = 16[X] = 7[Y] = 7[Z] = 1

at t = 32[A] = 2[B] = 2[C] = 6

at t = 32[X] = 6[Y] = 6[Z] = 2

at t = 32[A] = 2[B] = 2[C] = 6

at t = 32[X] = 6[Y] = 6[Z] = 2

at t = 48[A] = 0[B] = 0[C] = 8

at t = 48[X] = 5[Y] = 5[Z] = 3

Hypothetical ReactionRed Blue

In this reaction, one molecule of Red turnsinto one molecule of Blue

The number of moleculeswill always total 100

The rate of the reaction canbe measured as the speed of loss of Red molecules over time, or the speed of gain of Blue molecules over time

Hypothetical ReactionRed Blue

Reaction Rate Changes Over Time

• As time goes on, the rate of a reaction generally slows down – because the concentration of the reactants

decreases • At some time the reaction stops, either

because the reactants run out or because the system has reached equilibrium

Reaction Rate and Stoichiometry• In most reactions, the coefficients of the balanced

equation are not all the sameH2 (g) + I2 (g) 2 HI(g)

• For these reactions, the change in the number of molecules of one substance is a multiple of the change in the number of molecules of another– for the above reaction, for every 1 mole of H2 used, 1 mole of I2

will also be used and 2 moles of HI made– therefore the rate of change will be different

• To be consistent, the change in the concentration of each substance is multiplied by 1/coefficient

Average Rate

• The average rate is the change in measured concentrations in any particular time period– linear approximation of a curve

• The larger the time interval, the more the average rate deviates from the instantaneous rate

Hypothetical Reaction Red Blue

H2 I2HI

Stoichiometry tells us that for every 1 mole/L of H2 used, 2 moles/L of HI are made.

Assuming a 1 L container, at 10 s, we used 0.181 moles of H2. Therefore the amount of HI made is 2(0.181 moles) = 0.362 moles.

At 60 s, we used 0.699 moles of H2. Therefore the amount of HI made is 2(0.699 moles) = 1.398 moles.

The average rate is the change in the concentration in a given time period

In the first 10 s, the D[H2] is −0.181 M, so the rate is

average rate in a given time period = slope of the line connecting the [H2] points; and ½ +slope of the line for [HI]

the average rate for the first 10 s is 0.0181 M/s

Instantaneous Rate

• The instantaneous rate is the change in concentration at any one particular time– slope at one point of a curve

• Determined by taking the slope of a line tangent to the curve at that particular point– first derivative of the function

• for you calculus fans

H2 (g) + I2 (g) 2 HI (g) Using [H2], the instantaneous rate at 50 s is

Using [HI], the instantaneous rate at 50 s is

2 mol NOBr:1 mol Br2, 1 mol = 109.91g, 1 mL=0.001 L

Solve:

Conceptual Plan:

Relationships:

240.0 g NOBr, 200.0 mL, 5.0 min.

D[Br2]/Dt, M/s

Given:

Practice – If 2.4 x 102 g of NOBr decomposes in a 2.0 x 102 mL flask in 5.0 minutes, find the average rate of Br2 production

2 NOBr(g) 2 NO(g) + Br2(l)

D MD mole Br2

RateD min D s }

5.45 M Br2, 3.0 x 102 s

D[Br2]/Dt, M/s

240.0 g NOBr, 0.2000 L, 5.0 min.

D[Br2]/Dt, M/s

Measuring Reaction Rate• To measure the reaction rate you need to be able to

measure the concentration of at least one component in the mixture at many points in time

• There are two ways of approaching this problem1. for reactions that are complete in less than 1 hour, it

is best to use continuous monitoring of the concentration

2. for reactions that happen over a very long time, sampling of the mixture at various times can be used

– when sampling is used, often the reaction in the sample is stopped by a quenching technique

Continuous Monitoring• Polarimetry – measuring the change in the degree of

rotation of plane-polarized light caused by one of the components over time

• Spectrophotometry – measuring the amount of light of a particular wavelength absorbed by one component over time– the component absorbs its complementary color

• Total pressure – the total pressure of a gas mixture is stoichiometrically related to partial pressures of the gases in the reaction

Sampling the Reaction Mixture at Specific Times

• At specific times during the reaction, drawing off aliquots, (samples from the reaction mixture), and doing quantitative analysis – titration for one of the components– gravimetric analysis

• Gas chromatography can measure the concentrations of various components in a mixture– for samples that have volatile components – separates mixture by adherence to a surface

Methods for Determining Concentrations in a Mixture

Factors Affecting Reaction Rate: Nature of the Reactants

• Nature of the reactants means what kind of reactant molecules and what physical condition they are in – small molecules tend to react faster than large molecules – gases tend to react faster than liquids, which react faster than

solids – powdered solids are more reactive than “blocks”

• more surface area for contact with other reactants– certain types of chemicals are more reactive than others

• e.g. potassium metal is more reactive than sodium

– ions react faster than molecules • no bonds need to be broken

Factors Affecting Reaction Rate:Temperature

• Increasing temperature increases reaction rate– chemist’s rule of thumb—for each 10 °C rise in

temperature, the speed of the reaction doubles• for many reactions

• There is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius, which will be examined later

Factors Affecting Reaction Rate:Catalysts

• Catalysts are substances that affect the speed of a reaction without being consumed

• Most catalysts are used to speed up a reaction; these are called positive catalysts – catalysts used to slow a reaction are called negative

catalysts.• Homogeneous = present in same phase• Heterogeneous = present in different phase• How catalysts work will be examined later

Factors Affecting Reaction Rate:Reactant Concentration

• Generally, the larger the concentration of reactant molecules, the faster the reaction – increases the frequency of reactant molecule

contact– concentration of gases depends on the partial

pressure of the gas • higher pressure = higher concentration

• Concentrations of solutions depend on the solute-to-solution ratio (molarity)

The Rate Law• The rate law of a reaction is the mathematical relationship

between the rate of the reaction and the concentrations of the reactants– and homogeneous catalysts as well

• The rate law must be determined experimentally!!• The rate of a reaction is directly proportional to the

concentration of each reactant raised to a power• For the reaction aA + bB products the rate law would

have the form given below– n and m are called the orders for each reactant– k is called the rate constant

Reaction Order• The exponent on each reactant in the rate law is

called the order with respect to that reactant• The sum of the exponents on the reactants is

called the order of the reaction • The rate law for the reaction

2 NO(g) + O2(g) ® 2 NO2(g)

is Rate = k[NO]2[O2]

The reaction is

second order with respect to [NO],

first order with respect to [O2],

and third order overall 31

Sample Rate Laws

The bottom reaction is autocatalytic because a product affects the rate.Hg2+ is a negative catalyst; increasing its concentration slows the reaction.

Rate = k[acetaldehyde]2

Solve:

Conceptual Plan:

Relationships:

[acetaldehyde] = 1.75 x 10−3 M, k= 6.73 x 10−6 M−1•s−1

Rate, M/s

Given:

Practice – What is the rate of the reaction when the [acetaldehyde] = 1.75 x 10−3 M and the rate constant is 6.73 x 10−6 M−1•s−1?

k, [acetaldehyde] Rate

Finding the Rate Law:the Initial Rate Method

• The rate law must be determined experimentally• The rate law shows how the rate of a reaction

depends on the concentration of the reactants• Changing the initial concentration of a reactant

will therefore affect the initial rate of the reaction

if for the reaction A → Products

then doubling the initial concentration of A doubles the initial reaction rate

then doubling the initial concentration of A does not change the initial reaction rate

then doubling the initial concentration of A quadruples the initial reaction rate

Rate = k[A]n

• If a reaction is Zero Order, the rate of the reaction is always the same– doubling [A] will have no effect on the reaction rate

• If a reaction is First Order, the rate is directly proportional to the reactant concentration– doubling [A] will double the rate of the reaction

• If a reaction is Second Order, the rate is directly proportional to the square of the reactant concentration– doubling [A] will quadruple the rate of the reaction

Determining the Rate Law When there Are Multiple Reactants

• Changing each reactant will effect the overall rate of the reaction

• By changing the initial concentration of one reactant at a time, the effect of each reactant’s concentration on the rate can be determined

• In examining results, we compare differences in rate for reactions that only differ in the concentration of one reactant

Example 13.2: Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g)

given the data below

Comparing Expt #1 and Expt #2, the [NO2] changes but the [CO]

does not 37

n = 2, m = 0

Substitute the concentrations and rate for any experiment into the rate law and solve for k

Expt.Number

Initial [NO2], (M)

Initial [CO], (M)

Initial Rate(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

Practice – Determine the rate law and rate constant for the reaction NH4+ + NO2

− ® N2 + 2 H2O

Practice – Determine the rate law and rate constant for the reaction NH4+ + NO2

− ® N2 + 2 H2O

given the data below Rate = k[NH4

+]n[NO2]m

Finding the Rate Law:Graphical Methods

• The rate law must be determined experimentally• A graph of concentration of reactant vs. time can be

used to determine the effect of concentration on the rate of a reaction

• This involves using calculus to determine the area under a curve – which is beyond the scope of this course

• Later we will examine the results of this analysis so we can use graphs to determine rate laws for several simple cases

Reactant Concentration vs. TimeA Products

insert figure 13.6

Integrated Rate Laws

• For the reaction A Products, the rate law depends on the concentration of A

• Applying calculus to integrate the rate law gives another equation showing the relationship between the concentration of A and the time of the reaction – this is called the Integrated Rate Law

Half-Life• The half-life, t1/2, of a

reaction is the length of time it takes for the concentration of the reactant to fall to ½ its initial value

• The half-life of the reaction depends on the order of the reaction

Zero Order Reactions• Rate = k[A]0 = k

– constant rate reactions• [A] = −kt + [A]initial

• Graph of [A] vs. time is straight line with slope = −k and y–intercept = [A]initial

• t ½ = [Ainitial]/2k• When Rate = M/sec, k = M/sec

[A]initial

slope = −k

First Order Reactions• Rate = k[A]1 = k[A]• ln[A] = −kt + ln[A]initial • Graph ln[A] vs. time gives straight line with

slope = −k and y–intercept = ln[A]initial

– used to determine the rate constant

• t½ = 0.693/k• The half-life of a first order reaction is

constant• When Rate = M/sec, k = s–1

ln[A]initial

slope = −k

The Half-Life of a First-Order Reaction Is Constant

Rate Data for C4H9Cl + H2O ® C4H9OH + HCl

C4H9Cl + H2O C4H9OH + 2 HCl

slope = −2.01 x 10−3

k =2.01 x 10−3 s-1

Second Order Reactions

• Rate = k[A]2

• 1/[A] = kt + 1/[A]initial • Graph 1/[A] vs. time gives straight line with

slope = k and y–intercept = 1/[A]initial

– used to determine the rate constant

• t½ = 1/(k[A0])• When Rate = M/sec, k = M−1s−1

1/[A]initial

slope = k

Rate Data For2 NO2 ® 2 NO + O2

Rate Data Graphs for2 NO2 ® 2 NO + O2

Practice – The reaction Q 2 R is second order in Q. If the initial [Q] = 0.010 M and after 5.0 x 102 seconds the [Q] = 0.0010 M, find the rate

constant

[Q]init = 0.010 M, t = 5.0 x 102 s, [Q]t = 0.0010 M

Solution:

Conceptual Plan:

Relationships:

Given:

k[Q]init, t, [Q]t

Graphical Determination of the Rate Law for A Product

• Plots of [A] vs. time, ln[A] vs. time, and 1/[A] vs. time allow determination of whether a reaction is zero, first, or second order

• Whichever plot gives a straight line determines the order with respect to [A]– if linear is [A] vs. time, Rate = k[A]0

– if linear is ln[A] vs. time, Rate = k[A]1

– if linear is 1/[A] vs. time, Rate = k[A]2

Practice – Complete the Table and Determine the Rate Equation for the Reaction A ® 2 Product

What will the rate be when the [A] = 0.010 M?

Practice – Complete the Table and Determine the Rate Equation for the Reaction A ® 2 Product

Practice – Complete the table and determine the rate equation for the reaction A ® 2 Product

Because the graph 1/[A] vs. time is linear, the reaction is second order,

Rate k[A]2

k slope of the line 0.10 M–1 s –1

Relationship Between Order and Half-Life

• For a zero order reaction, the lower the initial concentration of the reactants, the shorter the half-life– t1/2

= [A]init/2k • For a first order reaction, the half-life is independent

of the concentration– t1/2 = ln(2)/k

• For a second order reaction, the half-life is inversely proportional to the initial concentration – increasing the initial concentration shortens the half-life– t1/2 = 1/(k[A]init)

Practice – The reaction Q 2 R is second order in Q. If the initial [Q] = 0.010 M and the rate constant is 1.8 M−1•s−1 find the length of time for

[Q] = ½[Q]init

[Q]init = 0.010 M, k = 1.8 M−1s−1

t1/2, s

Solution:

Conceptual Plan:

Relationships:

Given:

t1/2[Q]init, k

• Changing the temperature changes the rate constant of the rate law

• Svante Arrhenius investigated this relationship and showed that:

R is the gas constant in energy units, 8.314 J/(mol•K)where T is the temperature in kelvins

A is called the frequency factor, the rate the reactant energy approaches the activation energy

Ea is the activation energy, the extra energy needed to start the molecules reacting

The Effect of Temperature on Rate

Activation Energy and theActivated Complex

• There is an energy barrier to almost all reactions• The activation energy is the amount of energy

needed to convert reactants into the activated complex– aka transition state

• The activated complex is a chemical species with partially broken and partially formed bonds– always very high in energy because of its partial bonds

Isomerization of Methyl Isonitrile

methyl isonitrile rearranges to acetonitrile

for the reaction to occur, the H3C─N bond must break, and a new H3C─C bond form

As the reaction begins, the C─N bond weakens enough for the CN group to start to rotate

Energy Profile for the Isomerization of Methyl Isonitrile

the frequency is the number of molecules that begin to form the activated complex in a given period of time

the activation energy is the difference in energy between the reactants and the activated complex

the activated complex is a chemical species with partial bonds

The Arrhenius Equation:The Exponential Factor

• The exponential factor in the Arrhenius equation is a number between 0 and 1

• Tt represents the fraction of reactant molecules with sufficient energy so they can make it over the energy barrier– the higher the energy barrier (larger activation energy), the fewer

molecules that have sufficient energy to overcome it• That extra energy comes from converting the kinetic energy of

motion to potential energy in the molecule when the molecules collide – increasing the temperature increases the average kinetic energy of the

molecules– therefore, increasing the temperature will increase the number of

molecules with sufficient energy to overcome the energy barrier– therefore increasing the temperature will increase the reaction rate

Arrhenius Plots• The Arrhenius Equation can be algebraically

solved to give the following form:

this equation is in the form y = mx + bwhere y = ln(k) and x = (1/T)

a graph of ln(k) vs. (1/T) is a straight line

(−8.314 J/mol∙K)(slope of the line) = Ea, (in Joules)

ey–intercept = A (unit is the same as k)

Example 13.7: Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the following data:

slope, m = 1.12 x 104 Ky–intercept, b = 26.8

Arrhenius Equation:Two-Point Form

• If you only have two (T,k) data points, the following forms of the Arrhenius Equation can be used:

Example 13.8: The reaction NO2(g) + CO(g) CO2(g) + NO(g) has a rate constant of 2.57 M−1 s∙ −1 at 701 K and 567 M−1 s∙ −1 at 895 K. Find the activation

energy in kJ/mol

most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable

T1 = 701 K, k1 = 2.57 M−1∙s−1, T2 = 895 K, k2 = 567 M−1∙s−1

Ea, kJ/mol

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

EaT1, k1, T2, k2

Practice – It is often said that the rate of a reaction doubles for every 10 °C rise in temperature. Calculate the activation energy

for such a reaction.

Hint: make T1 = 300 K, T2 = 310 K and k2 = 2k1

Practice – Find the activation energy in kJ/mol for increasing the temperature 10 ºC doubling the rate

most activation energies are tens to hundreds of kJ/mol – so the answer is reasonable

T1 = 300 K, T1 = 310 K, k2 = 2 k1

Ea, kJ/mol

Check:

Solution:

Conceptual Plan:

Relationships:

Given:

EaT1, k1, T2, k2

Collision Theory of Kinetics• For most reactions, for a reaction to take place, the

reacting molecules must collide with each other– on average about 109 collisions per second

• Once molecules collide they may react together or they may not, depending on two factors –

1. whether the collision has enough energy to "break the bonds holding reactant molecules together"; and

2. whether the reacting molecules collide in the proper orientation for new bonds to form

Effective CollisionsKinetic Energy Factor

For a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide the activated complex can form

Effective CollisionsOrientation Effect

Effective Collisions• Collisions in which these two conditions are

met (and therefore lead to reaction) are called effective collisions

• The higher the frequency of effective collisions, the faster the reaction rate

• When two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed – the activated complex

Collision Theory and the Frequency Factor of the Arrhenius Equation

• The Arrhenius Equation includes a term, A, called the Frequency Factor

• The Frequency Factor can be broken into two terms that relate to the two factors that determine whether a collision will be effective

Collision Frequency

• The collision frequency is the number of collisions that happen per second

• The more collisions per second there are, the more collisions can be effective and lead to product formation

Orientation Factor• The orientation factor, p, is a statistical term relating

the frequency factor to the collision frequency• For most reactions, p < 1• Generally, the more complex the reactant molecules,

the smaller the value of p• For reactions involving atoms colliding, p ≈ 1 because

of the spherical nature of the atoms• Some reactions actually can have a p > 1

– generally involve electron transfer

Orientation Factor• The proper orientation results when the atoms are

aligned in such a way that the old bonds can break and the new bonds can form

• The more complex the reactant molecules, the less frequently they will collide with the proper orientation– reactions where symmetry results in multiple orientations

leading to reaction have p slightly less than 1• For most reactions, the orientation factor is less than

Molecular Interpretation of Factors Affecting Rate – Reactant Nature

• Reactions generally occur faster in solution than in pure substances– mixing gives more particle contact– particles separated, allowing more effective collisions per second– forming some solutions breaks bonds that need to be broken

• Some materials undergo similar reactions at different rates either because they have a (1) higher initial potential energy and are therefore closer in energy to the activated complex, or (2) because their reaction has a lower activation energy– CH4 + Cl2 CH3Cl + HCl is about 12x faster than CD4 + Cl2 CD3Cl + DCl

because the C─H bond is weaker and less stable than the C─D bond– CH4 + X2 CH3X + HX occurs about 100x faster with F2 than with Cl2 because

the activation energy for F2 is 5 kJ/mol but for Cl2 is 17 kJ/mol

Molecular Interpretation of Factors Affecting Rate – Temperature

• Increasing the temperature raises the average kinetic energy of the reactant molecules

• There is a minimum amount of kinetic energy needed for the collision to be converted into enough potential energy to form the activated complex

• Increasing the temperature increases the number of molecules with sufficient kinetic energy to overcome the activation energy

Molecular Interpretation of Factors Affecting Rate – Concentration

• Reaction rate generally increases as the concentration or partial pressure of reactant molecules increases– except for zero order reactions

• More molecules leads to more molecules with sufficient kinetic energy for effective collision– distribution the same, just bigger curve

Reaction Mechanisms• We generally describe chemical reactions with an

equation listing all the reactant molecules and product molecules

• But the probability of more than 3 molecules colliding at the same instant with the proper orientation and sufficient energy to overcome the energy barrier is negligible

• Most reactions occur in a series of small reactions involving 1, 2, or at most 3 molecules

• Describing the series of reactions that occurs to produce the overall observed reaction is called a reaction mechanism

• Knowing the rate law of the reaction helps us understand the sequence of reactions in the mechanism

An Example of a Reaction Mechanism• Overall reaction:

H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) • Mechanism:

1. H2(g) + ICl(g) HCl(g) + HI(g) 2. HI(g) + ICl(g) HCl(g) + I2(g)

• The reactions in this mechanism are elementary steps, meaning that they cannot be broken down into simpler steps and that the molecules actually interact directly in this manner without any other steps

H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)1) H2(g) + ICl(g) HCl(g) + HI(g) 2) HI(g) + ICl(g) HCl(g) + I2(g)

Elements of a MechanismIntermediates

• Notice that the HI is a product in Step 1, but then a reactant in Step 2

• Because HI is made but then consumed, HI does not show up in the overall reaction

• Materials that are products in an early mechanism step, but then a reactant in a later step, are called intermediates

Molecularity

• The number of reactant particles in an elementary step is called its molecularity

• A unimolecular step involves one particle• A bimolecular step involves two particles

– though they may be the same kind of particle• A termolecular step involves three particles

– though these are exceedingly rare in elementary steps

Rate Laws for Elementary Steps• Each step in the mechanism is like its own little

reaction – with its own activation energy and own rate law

• The rate law for an overall reaction must be determined experimentally

• But the rate law of an elementary step can be deduced from the equation of the step

H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)1) H2(g) + ICl(g) HCl(g) + HI(g) Rate = k1[H2][ICl] 2) HI(g) + ICl(g) HCl(g) + I2(g) Rate = k2[HI]

[ICl] 105

Rate Laws of Elementary Steps

Rate Determining Step• In most mechanisms, one step occurs slower than the other

steps• The result is that product production cannot occur any

faster than the slowest step – the step determines the rate of the overall reaction

• We call the slowest step in the mechanism the rate determining step– the slowest step has the largest activation energy

• The rate law of the rate determining step determines the rate law of the overall reaction

Another Reaction Mechanism

The first step in this mechanism is the rate determining step.

The first step is slower than the second step because its activation energy is larger.

The rate law of the first step is the same as the rate law of the overall reaction.

NO2(g) + CO(g) NO(g) + CO2(g) Rateobs = k[NO2]2

1. NO2(g) + NO2(g) NO3(g) + NO(g) Rate = k1[NO2]2 Slow2. NO3(g) + CO(g) NO2(g) + CO2(g) Rate = k2[NO3][CO] Fast

Validating a Mechanism

• To validate (not prove) a mechanism, two conditions must be met:

1. The elementary steps must sum to the overall reaction

2. The rate law predicted by the mechanism must be consistent with the experimentally observed rate law

Mechanisms with a Fast Initial Step

• When a mechanism contains a fast initial step, the rate limiting step may contain intermediates

• When a previous step is rapid and reaches equilibrium, the forward and reverse reaction rates are equal – so the concentrations of reactants and products of the step are related– and the product is an intermediate

• Substituting into the rate law of the RDS will produce a rate law in terms of just reactants

An Example

1. 2 NO(g) N2O2(g) Fast

2. H2(g) + N2O2(g) H2O(g) + N2O(g) Slow Rate = k2[H2][N2O2]

3. H2(g) + N2O(g) H2O(g) + N2(g) Fast

2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g) Rateobs = k [H2][NO]2

Example 13.9: Show that the proposed mechanism for the reaction 2 O3(g) 3 O2(g) matches the observed rate law

Rate = k[O3]2[O2]−1

1. O3(g) O2(g) + O(g) Fast

2. O3(g) + O(g) 2 O2(g) Slow Rate = k2[O3][O]

Practice – MechanismDetermine the overall reaction, the rate determining step, the rate law, and identify allintermediates of the following mechanism

1. A + B2 ® AB + B Slow2. A + B ® AB Fast

Practice – MechanismDetermine the overall reaction, the rate determining step, the rate law, and identify all intermediates of the following mechanism

1. A + B2 ® AB + B Slow2. A + B ® AB Fast

reaction 2 A + B2 ® 2 ABB is an intermediateThe first step is the rate determining stepRate = k[A][B2], same as RDS

Catalysts• Catalysts are substances that affect the rate of a

reaction without being consumed• Catalysts work by providing an alternative mechanism

for the reaction– with a lower activation energy

• Catalysts are consumed in an early mechanism step, then made in a later step

mechanism without catalyst

O3(g) + O(g) 2 O2(g) V. Slow

mechanism with catalyst

Cl(g) + O3(g) O2(g) + ClO(g) Fast

ClO(g) + O(g) O2(g) + Cl(g) Slow

Ozone Depletion over the Antarctic

Molecular Interpretation of Factors Affecting Rate – Catalysts

• Catalysts generally speed a reaction• They give the reactant molecules a different path

to follow with a lower activation energy– heterogeneous catalysts hold one reactant molecule in

proper orientation for reaction to occur when the collision takes place

• and sometimes also help to start breaking bonds

– homogeneous catalysts react with one of the reactant molecules to form a more stable activated complex with a lower activation energy

Energy Profile of a Catalyzed Reaction

polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals

Practice – MechanismDetermine the overall reaction, the rate determining step, the rate law, and identify all catalysts and intermediates of the following mechanism1. HQ2R2 + R− Û Q2R2

− + HR Fast2. Q2R2

− ® Q2R + R− Slow

Practice – Mechanism

reaction HQ2R2 ® Q2R + HRR− is a catalystQ2R2

− is an intermediateThe second step is the rate determining stepRate = k[Q2R2

−] = k[HQ2R2][R−]

Determine the overall reaction, the rate determining step, the rate law, and identify all catalysts and intermediates of the following mechanism1. HQ2R2 + R− Û Q2R2

− + HR Fast2. Q2R2

− ® Q2R + R− Slow

Catalysts• Homogeneous catalysts are in the same phase as

the reactant particles– Cl(g) in the destruction of O3(g)

• Heterogeneous catalysts are in a different phase than the reactant particles– solid catalytic converter in a car’s exhaust system

Types of Catalysts

Catalytic HydrogenationH2C=CH2 + H2 → CH3CH3

Enzymes• Because many of the molecules are large and

complex, most biological reactions require a catalyst to proceed at a reasonable rate

• Protein molecules that catalyze biological reactions are called enzymes

• Enzymes work by adsorbing the substrate reactant onto an active site that orients the substrate for reaction

1) Enzyme + Substrate Enzyme─Substrate Fast

2) Enzyme─Substrate → Enzyme + Product Slow

Enzyme–Substrate Binding:the Lock and Key Mechanism

Enzymatic Hydrolysis of Sucrose

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