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

mechanisms.mechanisms.

Outline: KineticsReaction Rates How we measure rates.

Rate LawsHow the rate depends on amounts of reactants.

Integrated Rate LawsHow to calc amount left or time to reach a given amount.

Half-lifeHow long it takes to react 50% of reactants.

Arrhenius Equation How rate constant changes with T.

MechanismsLink between rate and molecular scale processes.

Factors That Affect Reaction Rates• Concentration of Reactants

– As the concentration of reactants increases, so does the likelihood that reactant molecules will collide.

• Temperature– At higher temperatures, reactant molecules have more kinetic energy, move

faster, and collide more often and with greater energy.• Catalysts

– Speed rxn by changing mechanism.

Reaction 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

UNITS FOR RATE

• Mol/L •s

• Mol/L •hr

• Mol/L • day

• Etc…………………….

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

Products and reactants are stoichiometrically related

• 2 NO2 → 2 NO + O2

• If the rate of disappearance of NO2 is 2.0

mol/L • s what is the rate of appearance for NO and O2 ?

Page 598 – Problem 17

Rate LawsDifferential 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.”

• What is the difference between rate and time?

How to write a rate expression

• R = k [A]x [B]y

• X and Y represent the order to which the reactant is raised.

HOW DO YOU DETERMINE THE ORDER

INITIAL RATES METHOD

Writing a (differential) Rate LawWriting a (differential) Rate Law

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

• Log Rate3 = (Conc3 ) log

• Rate 4 (Conc4 )

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 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

Page 598 – problem 22

Page 598 – problem 24

Page 599 – problems 26

SHORT CUT FOR DETERMINING ORDER

• If the concentration increases and the rate stays the same – zero order

• If the concentration doubles and the rate doubles• – 1st order

• If the concentration doubles and the rate increases by 4 – 2nd order

2What if you have more than one reactant and one is not held constant?

EXPERIMENT Initial [ClO2- ]

mol/L 2

Initial [F2 ] mol/L Initial Rate of Increase of [ClO2F]

1 0.010 0.10 2.4 x 10-3

2 0.010 0.40 9.6 x 10-3

3 0.020 0.20 9.6 x 10-3

a. Write the rate law for the reaction 2 ClO2 + → 2 ClO2Fb. Calculate the numerical value of the rate constant and specify the unit.c. In experiment 2, what is the initial rate of decrease of [F2 ].d. Which of the following rate law developed in a? Justify your choice.ClO2 + F2 ↔ ClO2F2

fast

ClO2F2 → ClO2F + F slowClO2 + F → ClO2F fast

or F2 → 2 F slow2 ClO2

+ 2 F → ClO2F fast

UNITS FOR K

Determining Order withConcentration vs. Time data

(the Integrated Rate Law)

.timevs concentrationis linear

.ln( )timevs concentration is linear

1.timevs is linearconcentration

Zero Order:

First Order:

Second Order:

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

[A0 ] – [A] = kt ln[A] = kt

[A]0

1 - 1

[A] [A0 ] = kt

Plot the Plot the 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

01/ 2

[ ]

2

At

k 1/ 2

0.693t

k 1/ 2

0

1

[ ]t

k A

Solving an Integrated Rate LawTime (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

Find 2 points use∆Y/∆X = A

Find 2 more points∆Y/∆X = appx AThe slope of both should be close.Pick the plot that is closest to the same numbers

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:

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

And the winner is… Time vs. ln[H2O2]

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, kMethod #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

Determining rxn orderThe decomposition of NO2 at 300°C is described by the equation

NO2 (g) NO (g) + 1/2 O2 (g)

and yields these data:

Time (s) [NO2], M

0.0 0.01000

50.0 0.00787

100.0 0.00649

200.0 0.00481

300.0 0.00380

Graphing ln [NO2] vs. t yields:

Time (s) [NO2], M ln [NO2]

0.0 0.01000 -4.610

50.0 0.00787 -4.845

100.0 0.00649 -5.038

200.0 0.00481 -5.337

300.0 0.00380 -5.573

• The plot is not a straight line, so the process is not first-order in [A].

Determining rxn order

Does not fit:

Second-Order ProcessesA graph of 1/[NO2]

vs. t gives this plot.

Time (s) [NO2], M 1/[NO2]

0.0 0.01000 100

50.0 0.00787 127

100.0 0.00649 154

200.0 0.00481 208

300.0 0.00380 263

• This is a straight line. Therefore, the process is second-order in [NO2].

Finding the Rate Constant, kMethod #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

Reaction 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 Step

In a multi-step reaction, the slowest step is the rate-determining step. It therefore determines the rate of the 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 StepFor 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 IntermediatesFor 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 ModelKey Idea: Molecules must collide to react.However, only a small fraction of collisions produces a reaction. Why?

The Collision Model

Furthermore, molecules must collide with the correct orientation and with enough energy to cause bond breakage and formation.

Activation Energy• In other words, there is a minimum amount of energy

required for reaction: the activation energy, Ea.

• Just as a ball cannot get over a hill if it does not roll up the hill with enough energy, a reaction cannot occur unless the molecules possess sufficient energy to get over the activation energy barrier.

Collision 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 ReactionsEndothermic Reactions

Exothermic ReactionsExothermic Reactions

Maxwell–Boltzmann Distributions

• As the temperature increases, the curve flattens and broadens.

• Thus at higher temperatures, a larger population of molecules has higher energy.

The 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, 8.31451 = Gas constant, 8.31451 J/K·molJ/K·mol

The Arrhenius Equation, Rearranged1

ln( ) ln( )aEk AR 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•Catalyst: A substance that speeds up a reaction : A substance that speeds up a reaction without being consumedwithout being consumed

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

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

•Heterogeneous catalyst: Present in a different phase : Present in a different phase than the reacting moleculesthan the reacting molecules..

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

Catalysts

One way a catalyst can speed up a reaction is by holding the reactants together and helping bonds to break.

Enzymes• Enzymes are catalysts

in biological systems.• The substrate fits into

the active site of the enzyme much like a key fits into a lock.

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

Slow Initial Step

• The rate law for this reaction is found experimentally to be

Rate = k [NO2]2

• CO is necessary for this reaction to occur, but the rate of the reaction does not depend on its concentration.

• This suggests the reaction occurs in two steps.

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

Slow Initial Step• A proposed mechanism for this reaction is

Step 1: NO2 + NO2 NO3 + NO (slow)

Step 2: NO3 + CO NO2 + CO2 (fast)

• The NO3 intermediate is consumed in the second step.

• As CO is not involved in the slow, rate-determining step, it does not appear in the rate law.

Fast Initial Step• A proposed mechanism is

Step 1 is an equilibrium- it includes the forward and reverse reactions.