Chemical KineticsChemical KineticsThe area of chemistry that concerns The area of chemistry that concerns reaction reaction

rates rates andand reaction mechanisms reaction mechanisms..

Reaction RateReaction RateThe change in concentration of a The change in concentration of a reactant or product per unit of timereactant or product per unit of time

2 1

2 1

[ ] [ ]A at timet A at timetRate

t t

[ ]ARate

t

2NO2NO22(g) (g) 2NO(g) + O 2NO(g) + O22(g)(g)Reaction Rates:

2. Can measure appearance of products

1. Can measure disappearance of reactants

3. Are proportional stoichiometrically

2NO2NO22(g) (g) 2NO(g) + O 2NO(g) + O22(g)(g)Reaction Rates:4. Are equal to

the slope tangent to that point

[NO2]

t

5. Change as the reaction proceeds, if the rate is dependent upon concentration2[ ]

constantNO

t

Rate LawsRate Laws

Differential rate laws express (reveal) the relationship between the concentration of reactants and the rate of the reaction.

Integrated rate laws express (reveal) the relationship between concentration of reactants and time

The differential rate law is usually just called “the rate law.”

Writing a (differential) Rate Writing a (differential) Rate LawLaw

2 NO(g) + Cl2(g) 2 NOCl(g)

Problem - Write the rate law, determine the value of the rate constant, k, and the overall order for the following reaction:

ExperimenExperimentt

[NO][NO]

(mol/L)(mol/L)[Cl[Cl22]]

(mol/L)(mol/L)

RateRate

Mol/L·sMol/L·s

11 0.2500.250 0.2500.250 1.43 x 101.43 x 10-6-6

22 0.5000.500 0.2500.250 5.72 x 105.72 x 10-6-6

33 0.2500.250 0.5000.500 2.86 x 102.86 x 10-6-6

44 0.5000.500 0.5000.500 11.4 x 1011.4 x 10-6-6

Writing a Rate LawWriting a Rate LawPart 1 – Determine the values for the exponents in the rate law:

ExperimenExperimentt

[NO][NO]

(mol/L)(mol/L)[Cl[Cl22]]

(mol/L)(mol/L)

RateRate

Mol/L·sMol/L·s

11 0.2500.250 0.2500.250 1.43 x 101.43 x 10-6-6

22 0.5000.500 0.2500.250 5.72 x 105.72 x 10-6-6

33 0.2500.250 0.5000.500 2.86 x 102.86 x 10-6-6

44 0.5000.500 0.5000.500 1.14 x 101.14 x 10-5-5

In experiment 1 and 2, [Cl2] is constant while [NO] doubles.

R = k[NO]x[Cl2]y

The rate quadruples, so the reaction is second order with respect to [NO] R = k[NO]2[Cl2]y

Writing a Rate LawWriting a Rate LawPart 1 – Determine the values for the exponents in the rate law:

ExperimenExperimentt

[NO][NO]

(mol/L)(mol/L)[Cl[Cl22]]

(mol/L)(mol/L)

RateRate

Mol/L·sMol/L·s

11 0.2500.250 0.2500.250 1.43 x 101.43 x 10-6-6

22 0.5000.500 0.2500.250 5.72 x 105.72 x 10-6-6

33 0.2500.250 0.5000.500 2.86 x 102.86 x 10-6-6

44 0.5000.500 0.5000.500 1.14 x 101.14 x 10-5-5

R = k[NO]2[Cl2]y

In experiment 2 and 4, [NO] is constant while [Cl2] doubles. The rate doubles, so the reaction is first order with respect to [Cl2] R = k[NO]2[Cl2]

Writing a Rate LawWriting a Rate LawPart 2 – Determine the value for k, the rate constant, by using any set of experimental data:

ExperimenExperimentt

[NO][NO]

(mol/L)(mol/L)[Cl[Cl22]]

(mol/L)(mol/L)

RateRate

Mol/L·sMol/L·s

11 0.2500.250 0.2500.250 1.43 x 101.43 x 10-6-6

R = k[NO]2[Cl2]

261.43 10 0.250 0.250mol mol mol

x kL s L L

6 3 25

3 3 2

1.43 109.15 10

0.250

x mol L Lk x

L s mol mol s

Writing a Rate LawWriting a Rate LawPart 3 – Determine the overall order for the reaction.

R = k[NO]2[Cl2]

Overall order is the sum of the exponents, or orders, of the reactants

2 + 1 = 3

The reaction is 3rd order

Determining Order withDetermining Order withConcentration vs. TimeConcentration vs. Time data data

(the Integrated Rate Law)

.timevs concentrationis linear

.ln( )timevs concentration is linear

1.timevs is linearconcentration

Zero Order:

First Order:

Second Order:

Solving an Integrated Rate Solving an Integrated Rate LawLaw

Time (s)Time (s) [H[H22OO22] (mol/L)] (mol/L)

00 1.001.00

120120 0.910.91

300300 0.780.78

600600 0.590.59

12001200 0.370.37

18001800 0.220.22

24002400 0.130.13

30003000 0.0820.082

36003600 0.0500.050

Problem: Find the integrated rate law and the value for the rate constant, kA graphing calculator with linear regression analysis greatly simplifies this process!!

Time vs. [HTime vs. [H22OO22]]Time Time (s)(s)

[H[H22OO22]]

00 1.001.00

120120 0.910.91

300300 0.780.78

600600 0.590.59

12001200 0.370.37

18001800 0.220.22

24002400 0.130.13

30003000 0.0820.082

36003600 0.0500.050

y = ax + b a = -2.64 x 10-4

b = 0.841r2 = 0.8891r = -0.9429

Regression results:

Time vs. ln[HTime vs. ln[H22OO22]]

Time (s)Time (s) ln[Hln[H22OO22]]

00 00

120120 -0.0943-0.0943

300300 -0.2485-0.2485

600600 -0.5276-0.5276

12001200 -0.9943-0.9943

18001800 -1.514-1.514

24002400 -2.04-2.04

30003000 -2.501-2.501

36003600 -2.996-2.996

Regression results:

y = ax + b a = -8.35 x 10-4

b = -.005r2 = 0.99978r = -0.9999

Time vs. 1/[HTime vs. 1/[H22OO22]]

Time Time (s)(s)

1/[H1/[H22OO22]]

00 1.001.00

120120 1.09891.0989

300300 1.28211.2821

600600 1.69491.6949

12001200 2.70272.7027

18001800 4.54554.5455

24002400 7.69237.6923

30003000 12.19512.195

36003600 20.00020.000

y = ax + b a = 0.00460b = -0.847r2 = 0.8723r = 0.9340

Regression results:

And the winner is… And the winner is… Time vs. Time vs. ln[Hln[H22OO22]]

1. As a result, the reaction is 1st order

2. The (differential) rate law is:

2 2[ ]R k H O3. The integrated rate law is:

2 2 2 2 0ln[ ] ln[ ]H O kt H O

4. But…what is the rate constant, k ?

Finding the Rate Constant, Finding the Rate Constant, kk

Method #1: Calculate the slope from the Time vs. ln[H2O2] table. Time (s)Time (s) ln[Hln[H22OO22]]

00 00

120120 -0.0943-0.0943

300300 -0.2485-0.2485

600600 -0.5276-0.5276

12001200 -0.9943-0.9943

18001800 -1.514-1.514

24002400 -2.04-2.04

30003000 -2.501-2.501

36003600 -2.996-2.996

2 2 2 2 0ln[ ] ln[ ]H O kt H O

2 2ln[ ] 2.996

3600

H Oslope

t s

4 18.32 10slope x s

Now remember:

k = -slope

k = 8.32 x 10-4s-1

Finding the Rate Constant, Finding the Rate Constant, kk

Method #2: Obtain k from the linear regresssion analysis.

2 2 2 2 0ln[ ] ln[ ]H O kt H O

4 18.35 10slope a x s

Now remember:

k = -slope

k = 8.35 x 10-4s-1

Regression results:

y = ax + b a = -8.35 x 10-4

b = -.005r2 = 0.99978r = -0.9999

Rate Laws SummaryRate Laws SummaryZero OrderZero Order First OrderFirst Order Second OrderSecond Order

Rate LawRate Law Rate = k Rate = k[A] Rate = k[A]2

Integrated Integrated Rate LawRate Law

[A] = -kt + [A]0 ln[A] = -kt + ln[A]0

Plot that Plot that produces a produces a straight linestraight line

[A] versus t ln[A] versus t

Relationship Relationship of rate of rate constant to constant to slope of slope of straight linestraight line

Slope = -k Slope = -k Slope = k

Half-LifeHalf-Life

1

[ ]versus t

A

0

1 1

[ ] [ ]kt

A A

01/ 2

[ ]

2

At

k 1/ 2

0.693t

k 1/ 2

0

1

[ ]t

k A

Reaction MechanismReaction Mechanism

The reaction mechanism is the series of elementary steps by which a chemical reaction occurs.

The sum of the elementary steps must give the overall balanced equation for the reaction The mechanism must agree with the experimentally determined rate law

Rate-Determining StepRate-Determining Step

In a multi-step reaction, the In a multi-step reaction, the slowest stepslowest step is the rate-is the rate-determining step.determining step. It therefore It therefore determines the rate of the determines the rate of the reaction.reaction.

The experimental rate law must The experimental rate law must agree with the rate-determining agree with the rate-determining step step

Identifying the Rate-Determining Identifying the Rate-Determining StepStep

For the reaction:2H2(g) + 2NO(g) N2(g) +

2H2O(g)The experimental rate law is:

R = k[NO]2[H2]Which step in the reaction mechanism is the rate-determining (slowest) step?

Step #1 H2(g) + 2NO(g) N2O(g) + H2O(g)Step #2 N2O(g) + H2(g) N2(g) + H2O(g)Step #1 agrees with the experimental rate law

Identifying IntermediatesIdentifying Intermediates

For the reaction:2H2(g) + 2NO(g) N2(g) + 2H2O(g)

Which species in the reaction mechanism are intermediates (do not show up in the final, balanced equation?)

Step #1 H2(g) + 2NO(g) N2O(g) + H2O(g)Step #2 N2O(g) + H2(g) N2(g) + H2O(g) 2H2(g) + 2NO(g) N2(g) + 2H2O(g)

N2O(g) is an intermediate

Collision ModelCollision ModelKey Idea: Molecules must collide to react.However, only a small fraction of collisions produces a reaction. Why?

Collision ModelCollision ModelCollisions must have Collisions must have sufficient sufficient energyenergy to produce the reaction to produce the reaction (must equal or exceed the (must equal or exceed the activation energy).activation energy).

Colliding particles must be Colliding particles must be correctly correctly orientedoriented to one another in order to to one another in order to produce a reaction.produce a reaction.

1.1.

2.2.

Factors Affecting RateFactors Affecting RateIncreasing temperature always increases the rate of a reaction.

Particles collide more frequently Particles collide more energeticallyIncreasing surface area increases

the rate of a reaction

Increasing Concentration USUALLY increases the rate of a reaction

Presence of Catalysts, which lower the activation energy by providing alternate pathways

Endothermic Endothermic ReactionsReactions

Exothermic ReactionsExothermic Reactions

The Arrhenius EquationThe Arrhenius Equation

/aE RTk Ae kk = rate constant at = rate constant at temperature Ttemperature T AA = frequency factor = frequency factor EEaa = activation energy = activation energy RR = Gas constant, 8.314 J/K·mol = Gas constant, 8.314 J/K·mol

The Arrhenius Equation, The Arrhenius Equation, RearrangedRearranged

1ln( ) ln( )aEk A

R T

Simplifies solving for Ea

-Ea / R is the slope when (1/T) is plotted against ln(k) ln(A) is the y-intercept Linear regression analysis of a table of (1/T) vs. ln(k) can quickly yield a slope Ea = -R(slope)

CatalysisCatalysis•CatalystCatalyst: A substance that speeds up a : A substance that speeds up a reaction without being consumedreaction without being consumed

•EnzymeEnzyme: A large molecule (usually a : A large molecule (usually a protein) that catalyzes biological protein) that catalyzes biological reactions.reactions.

•Homogeneous catalystHomogeneous catalyst: Present in the : Present in the same phase as the reacting molecules.same phase as the reacting molecules.

•Heterogeneous catalystHeterogeneous catalyst: Present in a : Present in a different phase than the reacting different phase than the reacting moleculesmolecules..

Lowering of Activation Energy Lowering of Activation Energy by a Catalystby a Catalyst

Catalysts Increase the Number of Catalysts Increase the Number of Effective CollisionsEffective Collisions

Heterogeneous CatalysisHeterogeneous Catalysis

Step #1: Step #1: Adsorption and Adsorption and

activation of activation of the reactants.the reactants.

Carbon monoxide and nitrogen

monoxide adsorbed on a

platinum surface

Heterogeneous CatalysisHeterogeneous Catalysis

Step #2: Step #2:

Migration of the Migration of the adsorbed adsorbed

reactants on the reactants on the surface.surface.

Carbon monoxide and nitrogen

monoxide arranged prior to

reacting

Heterogeneous CatalysisHeterogeneous Catalysis

Step #3: Step #3:

Reaction of the Reaction of the adsorbed adsorbed

substances.substances.

Carbon dioxide and nitrogen form

from previous molecules

Heterogeneous CatalysisHeterogeneous Catalysis

Step #4: Step #4:

Escape, or Escape, or desorption, of desorption, of the productsthe products..

Carbon dioxide and nitrogen gases escape

(desorb) from the platinum surface