Dr. Williamson’s Kinetics Notes Chemical Kinetics

transcript

Dr. Williamson’s Kinetics Notes

Chemical Kinetics or How fast do we go!

Student versionDr. Vickie M. WilliamsonDepartment of Chemistry

Texas A&M University

Thermodynamics-vs.-Kinetics2HCl(aq) + Mg(OH)2(s) --> MgCl2(aq) + 2H2O(l)

DGo = ________C(diamond) + O2(g) --> CO2(g)

DGo = _______

• Thermodynamics: Is a reaction spontaneous (favorable)?

• Kinetics: Will it actually occur within a given time period?

Thermodynamics-vs.-KineticsIs a reaction

thermodynamicallyfavored?

Is a reaction kinetically favored?

Reaction does not

occur without

continual energyinput

Reaction is very

Reaction occurs

Chemical Kinetics• Study of rates of chemical reactions: how

fast reactants are __________ and products are _________ in a reaction

Reaction:

Reaction rate: Change of ______________of reactant or product per unit time

Chemical Kinetics• Study of reaction mechanisms:

sequences of elementary _______ that lead to formation of product(s)

O3 + NO O2 + NO2

Reaction mechanism: Shows how reactions occur at the ________________

Wide Range of Reaction Rates

Measurement of Reaction Rates

Dye 1 ---> Dye 2

[dye1]

Obj: Give the rate of rxn in terms of conc. variations

• For a reaction such as2N2O5(g) ---> 4NO2(g) + O2(g)

• the rate of the reaction may be written as:

Rate of _______of concentration

of reactants

Rate of _______of concentration

of products} } }

2H2O2(l) ---> 2H2O (l) + O2(g)• Rate of rxn =

• Rate of disappearance of H2O2 _______ equal to the rate of rxn

• Rate of disap of H2O2 =

• This is _________ the rate of rxn

2HI ---> H2 + I2• Rate of rxn =

• Rate of disap of HI =

• If given rate of disappearance of HI(say 4.2 M/s) and asked for rate of rxn

or rate of appearance of I2

• Rate of rxn =

Obj: Give the rate in terms of conc. variations

For the reaction________(g) --> ______(g) + ____ (g)∆[Br2]/∆t =

(a) (1/2) ∆[BrNO]/∆t (b) 2 ∆[BrNO]/∆t (c) - (1/2) ∆[BrNO]/∆t (d) - 2 ∆[BrNO]/∆t

Obj: Give the rate in terms of conc. variations

Which of the following is _____for the reaction:____ (g) + _____ (g) --> _____(g) +

_____(g)

(a) a ∆[A]/∆t = b ∆[B]/∆t = c ∆[C]/∆t(b) a ∆[A]/∆t = b ∆[B]/∆t = -c ∆[C]/∆t(c) (1/a) ∆[A]/∆t = (1/b) ∆[B]/∆t = (1/c) ∆[C]/∆t(d) (1/a) ∆[A]/∆t = (1/b) ∆[B]/∆t = -(1/c) ∆[C]/∆t

Obj: Calculate average rate• Average rate: The _________ of the slope of

the reactant concentration-vs.-time curve for some ____________ D t for rxn Dye1--> Dye2

7.04 e-6 M/min

Obj: Calculate average rateFor the rxn, find the aver. rate between ___ and

_____ seconds:2 BrNO (g) --> 2 NO (g) + Br2 (g)

Time (sec) [BrNO]0 0.10 M50 0.082 M100 0.070 M

(A)_________M/s (B)__________M/

(C)_________ M/s (D) _________ M/s

(E)_________ M/s

Obj: Calculate Instantaneous rate• Instantaneous rate: The ________ of the slope of

the _______to the reacant concentration-vs.-time curve at a _____________ t for rxn Dye1--> Dye2

4.16 e-6 M/min at t = 3 min

Obj: Calculate Initial rate• Initial rate: The negative of the slope of the

_______ to the reactant concentration-vs.-time curve at ______ for rxn Dye1--> Dye2

1.18 e-5 M/min = initial rate

C4H9Cl(aq) + H2O(l) ® C4H9OH(aq) + HCl(aq)

Ave Rate 0-800=

Ave Rate =

Inst Rate 600s=

Obj: Explain the factors that affect rxnrates

An ________ in surface area ________ the rate of a reaction

An _______________ in reactant concentrationgenerally ____________ the rate of a reaction

Obj: Explain the factors that affect rxn rates

Obj: Explain the factors that affect rxnrates

An ________ in temperature increasesthe rate of any reaction

_________ are substances added to ________the rates of reactions

Aggie chemists have found a reaction for flubber! But the reaction is ____________. One reactant is NaCl. In order to _______ the rxn for mass production, they should:

(a) Use rock salt instead of table salt as a reactant

(b) Decrease the amount of salt (c) Conduct the experiment at high

temperatures (d) Add the appropriate catalysis

Reaction Rates:Effect of Energy and Orientation

Collision Theory of Reaction Rates

• For a chemical reaction to occur:1. Reactants (molecules, atoms, or ions) _____________ with each other

2. Collisions must be effective:✶Reactants must possess _________

_____________✶Reactants must have _________

______________at time of collision

Effect of Temperature on Ea

Rate directly related to • absolute temperature– Number of ____________________ increases–More molecules _____________________

• reactant concentration–More particles = more collisions

• Surface area

• Use of a catalyst– Different mechanism with ________________

_________________

Effect of Concentration on Reaction Rate: The Rate Law

• For a general reaction:aA + bB + .... --> cC + dD + ....

the rate law has the formRate = k[A]x[B]y.....

k = _______________________x and y = reaction orders for A & Bx + y = _________ reaction order

[A] = _______________________ of A in ________• The reaction orders x and y are not necessarily

related to a, b, c, d, ...

Reaction Orders and Rate Laws• Determined from experiment, not from the net

____________ ____________• Reaction orders are usually positive or zero,

but may be fractional or negative• e.g.: For the reaction in water:

H2O2 + 3I– + 2H+ ----> 2H2O + I3–

Rate = k[H2O2][I–] (given)________order in H2O2, ______ order in I–; _______ order in H+; _______order overall

Other Rate Laws• Experimentally observed rate laws for some

reactions are given below. What is the overall reaction orders for each?

3NO(g) --> N2O(g) + NO2(g) (given)Rate = k______

2NO2(g) + F2(g) --> 2NO2F(g) (given) Rate = k______________

CH3COCH3 + I2 --> CH3COCH2I + I-

(given)Rate = k____________________H+

For the Rate LawMeasuring the rate of a reaction

Determine the rate law

Propose a reasonable mechanism

• For multiple reactions with different beginning conc. (Experimental data )

• For each find instantaneous rate (t = 0) • Rate Law = k [A]m[B]n

• Rate Law relates the beginning conc. of reactants to speed.

• m and n are the orders

• What is the order with respect to Ba) 0.5 b) 1.5c) -0.5 d) 2.5 e) -2

rate = k [A]___

[B]____

What is the overall order of a reaction?a) -2.5 b) 2.5c) -1.5 d) 1.5

Obj: Determine the order

For the Rate LawMeasuring the rate of a reaction

Determine the rate law

Propose a reasonable mechanism

• For multiple reactions with different beginning conc. (Experimental data )

• For each find instantaneous rate (t = 0) • Rate Law = • Rate Law relates the ______________ of

reactants to speed.• m and n are the orders

Determination of the Rate Law• Run multiple experiments: Measure _____rate as a

function of different reactant concentrations– Concentrations of species may be monitored by

pressure, spectroscopic, conductivity, other measurements

• Find the order of each reactant by comparing _____________.

• Specific rate constant (k): Calculate from reaction-rate data for _____________ of reactant concentrations

Determination of a Rate LawReaction: A + 2B --> AB2

Expt Initial Initial Initial Rate [A, M] [B, M] (AB2, M/s)

1 0.010 0.010 1.5 x 10–4

2 0.010 0.020 1.5 x 10–4

3 0.020 0.030 6.0 x 10–4

Rate Law: Rate = k[A]x[B]yFind x, y and k

Determination of a Rate Law• Examine how the rate of reaction varies

with beginning concentration

• Expts ___ and ___: [A] fixed, [B] is ______, reaction rate is same ->rate is independent of B -> ____ order in B (y = ____)

• The rate law reduces toRate = k = k

Determination of a Rate Law• Write the rate laws for any two sets of

experiments where the concentration of only one species is varied

rate = k [A]xExpt 1: 1.5 x 10–4 = k(.01)x

Expt 3: 6.0 x 10–4 = k(.02)x

• Divide result from 1 into 3 to obtain:3 rate = k [A]x

1 rate = k [A]x

4 = _______ so x = _____• The rate law is: Rate = k[A]___

• To find the value of k, use data from any experimentExpt 3: 6.0 x 10–4 Ms–1 = k (.02 M)2

1.5 _________ = k

• The rate law for the reaction is then:Rate = 1.5 ________[A]____

Determination of a Rate Law

Determination of a Rate LawReaction:2A + B2 + C à A2B + BC

Expt [A]o [B2]o [C]o Rate[BC]o(M) (M) (M) (M/min)

1 0.20 0.20 0.20 2.4 x 10–6

2 0.40 0.30 0.20 9.6 x 10–6

3 0.20 0.30 0.20 2.4 x 10–6

4 0.20 0.40 0.60 7.2 x 10–6

Rate Law: Rate = k[A]x[B2]y[C]z

Find x, y, z and k

Determination of a Rate LawReaction: 3A + 2B --> 2C + D

Expt. [A]o [B]o Rate [D]o(M) (M) (M/min)

1 0.010 0.010 6.0 x 10–3

2 0.020 0.030 0.144 3 0.010 0.020 0.012

Rate Law: Rate = k[A]x[B]y

Evaluate x, y and k

Effect of Exponent When Comparing 2 Initial Rates

exponent Effect on initial rates0 No change in rate with

changing conc.1 Rate up by amount of

change in conc.___ Conc. up by 2; rate up by _____ Conc. up by 2; rate up by _____ Conc. up by 2; rate up by __

Rate order• For A + B --> C + D

The rate of the reaction increased ___-fold when A is ______ and ___-fold when B is ________. The order of the reaction is:a) 9 in A and 2 in B b) 3 in A and 2 in B c) 2 in A and 2 in B d) 2 in A and 1 in Be) 3 in A and 1 in B

The Rate Constant k• for a ______________________• from ___________________________ • does NOT change with changes in initial

___________________• does NOT change with _________• changes with ___________• changes with whether or not a _______ is

present• units depend on the ____________ of the

reaction

The Collision Theory says:Temperature increases, the rate ____________

Typical K’s = 10-2 to 10-8

But for:H+ + OH- -> H2Ok is _______

The Rate Constant k= a measure of rxn speed

The Rate Constant k• Its units depend on the ________________ of

the reaction– If reaction is 1st order:

rate = k [A]1M/s = units on k =

– If reaction is 2nd order:rate = k [A]2M/s = units on k =

– If reaction is zero order:rate = k [A]0 or rate = units on k =

Summary: Kinetics to Date• From equation: can get rate in terms of changes

in reactants or products• From 1 exp. (time and [ ] data) and equation:– Can get average rate between 2 times– Can get instanteous rate for a specfic time

from tangent line• From multiple experiments using different initial

concentrations: use conc. and initial rates (initial rate from time and [ ] data)– Can get rate law (rate based on initial conc.),

order of rxn, value for rate const.

Concentration-vs.-Time:The Integrated Rate Equation

• The integrated rate equation relates ________________as a function of time

• It allows determination of _________ and __________ at some specified time

• It allows for the evaluation of half-life (t1/2),the time it takes for half of reactants to be converted to products

• The integrated rate equation and t1/2 depend on the overall ______________

Half-Life

Zero-Order Reactions• The rate law for a zero-order reaction is:

Rate = k[A]0 = k(1) or Rate = _____• The integrated rate law for this reaction

y = m x + b

See math

Zero-Order Reaction:[A]-vs.-Time Plot

D tSlope = D[A] = –ak

Zero Order Half-Life[A] =

MT only uses integrated rate equations where a =1. Some books, as yours, ignore �a�, but only use chemical equations where the coefficient is 1. Other books stating that it is wrapped into k. I will use �a�.

First-Order Reactions• The rate law for a first-order reaction

(aA-->B) is:Rate = k[A]1 = k______

• The integrated rate law for this reaction is:

y = m x + b

See Math

First-Order Reaction:[A]-vs.-Time Plot

ln [A]0

ln [A]

D ln [A]

Slope = D ln[A] = –akDt

Second-Order Reactions• The rate law for a second-order reaction

is:Rate = k____

• The integrated rate law for this reaction is:

y = m x + b

See math

Second-Order Reaction:[A]-vs.-Time Plot

1/[A]0

Slope = D1/[A] = akDt

1 [A]D

Summary of Relationships for Various Reaction Orders

for aA--> products (a=coeficient)

M/s or M s-1

s-1 or 1/s

s-1 M-1or 1/(s•M)

Use graphing to determine order!

[A] = -akt + [A]o

a Reactant à products

RateEqu

Straight Line with

Integrated Equation

Slope k units

[A] vs t

ln[A] vs t

1 vs t[A]

2 1 = akt + 1[A] [A]o

ln[A] = -akt + ln[A]o

Intercept?

Equations Given1 1 ln[A] = -akt + ln [A]0

[A] = akt + [A]o

[A]o = t1/2 [A] = -akt + [A]0

0.693 = t1/2 1 = t1/2

ak ak[A]o

Order from Graphs

• Which is 2nd order?• Which is zero order?• Which is 1st order?

Using the Integrated Rate Equation• The gas-phase reaction:

2N2O5 --> 2N2O4 + O2obeys first-order kinetics with k = 0.00840s–1 at a certain temperature.

If 2.50 moles of N2O5 are placed in a 5.00-L container, we can calculate the number of moles N2O5 left after one minute

Using the Integrated Rate Equation• Use the first-order integrated rate equation in the

form:ln [N2O5] = – akt + ln [N2O5]o

• where: [N2O5]o =______________________k = 0.00840s–1

t = ________(the units on t and k must be =)a = 2

• ln [N2O5] = – akt + ln [N2O5]o= – 2(0.00840s–1)(60s) + ln 0.500

=• [N2O5] = ____________M• mol N2O5 = 5.00 L x _________mol/L = _____ mol

• MT only uses integrated rate equations where a =1. Some books, as yours, ignore �a�, but only use chemical equations where the coefficient is 1. Other books stating that it is wrapped into k. I will use �a�.

• For practice when a = 2, 3, …. See our homepage under �frequently asked questions� and �kinetics�. There are two sets of problems with solutions for your practice.

• Read about the integrated rate laws using ‘a’ from the pdf on the protected page.

• http://chemed.tamu.edu/chem102

Transition State Theory• Chemical reactions involve the breaking

and forming of bonds; energy is needed to break bonds

• Activation energy, Ea: Minimum energy reactants must have to form the activated complex

• There is an energy of activation for both ____________ and ______________ reactions

• Reaction rates ______________ due to differences in activation energies

ActivatedComplex

Transition State TheoryEn

Progress of a Reaction

DEreaction

ReactantsA + B2

ProductsAB + BDEreaction = Ea,f – Ea,r

A– –B– –B

Obj. Read Energy Profile

Find _____ a) 80 J b) -100 Jc) -75 J d) -80 J

Find of reverse rxna) -155 J b) -75 Jc) 75 J d) 80 Je) -80J

Find Ea ________________a) 155 J b) 100 Jc) 75 J d) 80 J

Find ________________a) 155 J b) 100 Jc) 75 J d) 80 J

Obj. Read Energy Profile

A + BEner

Find_________a) 150 Jb) 135 Jc) 85 Jd) 50 J

Find __________________a) 135 J b) 100 Jc) 50 J d) 85 J

Transition State Theory

Transition State Theory• Magnitude of Ea cannot be predicted from

the enthalpy of reaction_________

• Increasing the ___________ increases the rate of reaction by increasing the fraction of molecules that can overcome the activation barrier

• Catalyst increases reaction rate by _______________the activation barrier

Effect of a Catalyst on a Reaction

Effect of a Catalyst

Decreases Activation Energy2H2O2(aq) --> 2 H2O(l) + O2(g)

Catalyzed and Uncatalyzed Rxn

Find_______ of the __________ ________ rxna) b) c) d) e)

Find _______of ____________________rxna) b) c) d) e)

The Arrhenius Equation• The Arrhenius equation is:

k = Ae–Ea/RT

wherek = specific rate constantA = frequency factor based on geometryEa = activation energyR = 8.314 J/mol.KT = absolute temperaturee–Ea/RT = fraction of molecules having min. energy to react

Evaluation of Ea• Take an ln of both sides to get:

ln k = ln A + ln e–Ea/RT

ln k = ln A -Ea/RTFinally for the equation of a line:ln k = - Ea/R 1/T + ln A

y = Determine rate constants at two different temperaturesSolve for Ea:

Graphical Determination of Ea

1/T (K–1)

Slope = – Ea/R

ln k = - Ea/R 1/T + ln A

Obj: Determine Ea• Find Ea for the rxn V--> W if

k1 = ______ e-3 s-1T1= 700. Kk2 = ______ e-3 s-1 T2 = 900. K

a) 36.3 kJb) 5.29 e-5 Jc) 0.106 Jd) 105 kJ R = 8.314 J/mol*K

R = 0.0821 L*atm/mol*K

Reaction Mechanism• Reaction mechanism: sequence of elementary steps that leads to formation of products

• Sum of elementary steps = ____________• Steps show bonds breaking and making to

form products• Most chemical reactions occur in a series of

elementary steps, but can be _________• If a series, one step will be slower =

______________

Elementary Steps• Elementary reaction: single event at the

molecular level• For an elementary step, reaction order equals

the number of reactant molecules (the coefficient)

• Unimolecular: one-molecule reactionBr2 --> 2Br

• Bimolecular: two-molecule reactionO3 + NO --> O2 + NO2

• Termolecular: three-molecule reaction A + B + C --> PA + A + C --> P

Guidelines in Constructing Reaction Mechanisms

• Overall equation is:2NO2Cl -->2NO2 + Cl2

• Elementary steps must add up to give overall equation:

NO2Cl --> NO2 + ClNO2Cl + Cl --> NO2 + Cl2

Rules for Rate Laws• Rate law for an overall equation does not

include _____________, but can include products if mechanism dictates.

• Rate law for an overall equation comes from _________________or from the __________________ NOT from _______________

• Rate Law for a step in a mechanism _____ be read from the _____________

Reaction Mechanism• Mechanism: (1) A + B --> 2C

(2) 2C + A --> D(3) D --> E + F

Net reaction:• Rate Law for a step in a mechanism CAN

be read from stoichiometry!!– So rate of step 1 = – So rate of step 2 =– So rate of step 3 =

• Substances formed, then consumed are _______________

Reaction Mechanisms

From experimentNO2(g)+ CO(g)--> NO(g)+ CO2(g)

Exp rate law = k [NO2]2Is this one-step mechanism for the reaction possible? What rate law would it give?

Evidence for a Mechanism:Isotope Labeling

NO2(g)+ CO(g)--> NO(g)+ CO2(g)

Two-step mechanism for the reaction:

(slow)

(fast)

The Rate-Determining Step(RDS)• The slowest step in a mechanism• Reaction: NO2(g) + CO(g) --> NO(g) + CO2(g)

From experiment: Rate = k[NO2]2• Proposed two-step mechanism:

NO2(g) + NO2(g) --> NO3(g) + NO(g) (slow)NO3(g) + CO(g) --> NO2(g) + CO2(g) (fast)

would give this overall equation and rate:

Rate = k________• If first step in mechanism is rds (is slow), then

observed rate law is rate law for rds

• Mechanism must be consistent with the experimental rate lawE.g. 2NO2 + F2 --> 2NO2F

Exp. Rate law: Rate =Is mechanismpossible?

NO2 + F2 --> NO2F + F (slow)NO2 + F --> NO2F (fast)

The rate law from the mechanism is:Rate =

• Elementary steps must be physically reasonable; unless evidence exists, select bimolecular over termolecular steps

• E.g.: 2NO2 + F2 --> 2NO2FExp. Rate law: Rate =k[NO][F2]

• Is a 1 step mechanism possibleNO2 + NO2 + F2 --> 2NO2F

not likely since termolecular reactions are rare. • The 1-step mechanism gives Rate = k____ _____ • SO is it possible?

Selecting a Reaction Mechanism2NO + H2 --> N2O + H2O

Expt. [NO] [H2] Initial Rate(M/s)

1 0.1 M 0.1 M ______2 0.5 M 0.5 M ______3 0.5 M 0.1 M _______

Rate law : Rate = k[NO]x[H2]y

A. Rate = k[NO]2[H2]0 B. Rate = k[NO]2[H2]1

C. Rate = k[NO]0[H2]2 D. Rate = k[NO]5[H2]1

Selecting a Reaction Mechanism2NO + H2 --> N2O + H2O obs. Rate =_____ A. NO + H2 --> H2NO slow

H2NO + NO --> N2O + H2O fastB.2NO --> N2O2 slow

N2O2 + 2H2 --> 2N2O + H2O + 2H+ fast C. 2NO --> N2O + O slow

O + H2 --> H2O fastD.2NO + H2 --> N2(OH)2 slow

N2(OH)2 --> N2 + H2O2 fast E.2NO + H2 --> N2O + OH2 slow

OH2 --> H2O fast

Mechanism with Fast Initial Step• Reaction: 2NO + Br2 --> 2NOBr

From experiment: Rate = k[NO]2[Br]

• Proposed two-step mechanism:

NO + Br2 <--> NOBr2 (fast, eq)NOBr2 + NO --> 2NOBr (slow, rds)

• Is this mechanism consistent with observed rate law?

Mechanism: Consistency CheckNO + Br2 <--> NOBr2 (fast, eq)NOBr2 + NO --> 2NOBr (slow, rds)

• Rate law from slow step:Rate2 = k2[NOBr2][NO] BUT it contains an intermediate, which can�t be in the overall rate

• For equilibrium step(initial fast step): ratef = rater

kf[NO][Br2] =[NOBr2] =

• Now sub into slow step for [NOBr2] Rate2 = k2 [NO]

Rateo = k

Sample Problem• The proposed mechanism for the

reaction between H2 and CO is:H2 <--> 2H (fast, eq)H + CO --> HCO (slow)H + HCO --> HCHO (fast)

• Write the balanced overall equation• The observed rate-dependence is

_____order in H2 and ___ order in CO. Is this mechanism plausible?A. Yes B. No

Catalyst• Substance added to __________ the

reaction rate without being _____________in reaction (does not appear in the balanced equation)

• ________________ both forward and reverse reactions

• Is part of the ___________• Lowers the __________________ by

providing a different reaction ________________

Homogeneous Catalysis• Catalyst exists in same phase as reactants

(solid, liquid, gas, aqueous)• Oxidation of Ce4+ catalyzed by Mn2+ -all

substances are aqueous :Net: 2 Ce4+ + Tl+ --> 2Ce3+ + Tl3+

Steps: Ce4+ + Mn2+ --> Ce3+ + Mn3+

Ce4+ + Mn3+ --> Ce3+ + Mn4+

Tl+ + Mn4+ --> Tl3+ + Mn2+

A catalysts is first a ________, then a ___________in the mechanism steps

Recognizing Various Speciesin a Reaction Mechanism

• ___________: More consumed than formed• ____________: More formed than consumed• _________: Formed in an early step and entirely

consumed in later step(s)• _____________: Consumed in an early step

(reactant )and entirely regenerated as a ____________ in later step(s)

Ce4+ + Mn2+ --> Ce3+ + Mn3+Ce4+ + Mn3+ --> Ce3+ + Mn4+

Tl+ + Mn4+ --> Tl3+ + Mn2+

Heterogeneous Catalysis• Catalyst present in _____________ from

reactants

• Usually _______; provide surfaces on which reactions occur

• The larger the surface area, the more _____________the catalyst

• When solid catalyst employed, reactions usually follow ____-order kinetics

Heterogeneous Catalysts at WorkCatalytic Converters

Heterogeneous Catalysis at Work• The catalytic converter in automobile exhaust

systems uses noble metals and transition metal oxides (Pt/NiO) to catalyze the following reactions:2C8H18(g) + 25O2(g) --> 16CO2(g) + 18H2O(g)2CO(g) + O2(g) --> 2CO2(g)2NO(g) --> N2(g) + O2(g)

Catalytic NO à N2 + O2 Conversion

• Proteins in living systems that catalyze veryspecific biochemical reactions

• Function as both homogeneous and heterogeneous catalyst

• Increase rates by 108 to 1020 times• Mechanism of enzyme catalysis:

E + S --> ES (fast, reversible)ES --> E + P (slow, rds)

E = enzyme (catalysis); S = substrate (reactant)

Enzymes: Biological Catalysts

Mechanism of Enzyme Action

• In the presence of Enzyme, the reaction:Substrate --> Product

occurs by 3 stepsE + S --> ES (fast, reversible)

ES --> EP (slow)EP --> E + P (fast)

The ______________for this reaction is:(a) (b) (c) (d)

Enzymes: Biological Catalysts

Reaction Mechanisms:A Closer Look

What is the Rate Law?Elementary Steps:

1. O3 <---> O2 + O ________2. O + O3 ---> 2O2 ________

Rate = ? RXN 2O3 ---> 3O2

(A) k (B) k (C) k (D) k (E) k

Kinetics Equations1 1 ln[A] = -akt + ln [A]0[A] = akt + [A]o

[A]o = t1/2 [A] = -akt + [A]0 0.693 = t1/22ak ak1 = t1/2

ak[A]o lnk = -Ea/R (1/T) + ln A

ln k1 = Ea [ 1 - 1 ]k2 R [ T2 T1 ]

Dr. Williamson’s Kinetics Notes Chemical Kinetics

Documents