Kinetics

Ch 15

Kinetics

• Thermodynamics and kinetics are not directly related

• Investigate the rest of the reaction coordinate

• Rate is important!

Chemical Kinetics

• Kinetics – the study of the rates of chemical reactions

• Rate of reaction – change in concentration per unit time rate = Δ conc / Δ time

• Rate is generally not constant. It changes over the course of a reaction

• A B

A B

10 18 24 28 31 33

What is happening to the rate of the reaction as time progresses? Why?

Rate = Δ[B]/Δt = -Δ[A]/Δt

Rate = Δ[product]/Δt = -Δ[reactant]/Δt

A B

Rate of Reaction

2 N2O5 (g) 4 NO2 (g) + O2 (g)

Rate = Δ[O2]/ Δt

Stoichiometry important!

Rate = Δ[NO2]/ 4Δt = - Δ[N2O5]/ 2Δt

2 N2O5 (g) 4 NO2 (g) + O2 (g)

Defining Rate• Rate is defined arbitrarily by one pdt or rxt• To be self consistent,• Example:

Another Example

Data

Calculated Rates

• Collect concentration data for reactants and products, then graph

• Effect of stoichiometry

• Average rate• Instantaneous

rate

Rate Law

• Study rates to understand mechanism of reaction

• True rate depends on forward and reverse reactions (remember equilibrium?)

• But we can write rate law based on reactants– Many reactions functionally irreversible– Use initial rates (reverse rate is negligible)

Rate Laws

• Two forms of rate law• Differential Rate Law (Rate Law)

– How rate depends on concentration of reactants– Experiment: Initial Rates of multiple trials

• Integrated Rate Law– How concentrations of species depend on time– Experiment: One trial sampled at multiple times

Relationship Between Rate and Concentration

• 2 NO2 (g) + F2 (g) 2 NO2F (g)

• Rate = Δ[NO2F]/ 2Δt = -Δ[F2]/ Δt

= -Δ[NO2]/ 2Δt

• Rate α [NO2] and [F2]

• Rate = k [NO2]x [F2]y

• k = rate constant• x and y are the orders of reaction, these are determined

experimentally – not from stoichiometry!

• 2 NO2 (g) + F2 (g) 2 NO2F (g)• Rate = k [NO2]x [F2]y

• From experiment, x = 1 , y = 1• Rate = k [NO2] [F2] = Rate Law• 1st order in NO2 , 1st order in F2, 2nd

order overall• One way to determine the rate law is from

initial rates.

H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) I3- (aq) + 2 H2O (l)

Expt #

[H2O2] [I-] [H+] Initial rateM/s

1 0.010 0.010 0.0005 1.15 x 10-6

2 0.020 0.010 0.0005 2.30 x 10-6

3 0.010 0.020 0.0005 2.30 x 10-6

4 0.010 0.010 0.001 1.15 x 10-6

Rate = k [H2O2]x [I-]y [H+]z

H2O2 (aq) + 3 I- (aq) + 2 H+ (aq) I3- (aq) + 2 H2O (l)

Expt #

[H2O2] Rel [I-] Rel [H+] Rel Initial rateM/s

Rel

1 0.010 1 0.010 1 0.0005 1 1.15 x 10-6 1

2 0.020 2 0.010 1 0.0005 1 2.30 x 10-6 2

3 0.010 1 0.020 2 0.0005 1 2.30 x 10-6 2

4 0.010 1 0.010 1 0.001 2 1.15 x 10-6 1

Rate = k [H2O2]x [I-]y [H+]z

Rate = k [H2O2] [I-]

Example Problem

• 2 NO2 (g) 2 NO (g) + O2 (g)

Expt # [NO2] Rate, M/s

1 0.010 7.1 x 10-5

2 0.020 2.8 x 10-4

Relationship Between Concentration and Time

• We want to use a single experiment to determine the rate law.

• We will do this by plotting concentration versus time.

• We will deal with simplest cases initially—only one reactant, generally “A”

Zero Order Reaction

• How can a reaction rate be concentration independent?

Zero Order Reactions

• Zero order reaction A B– Rate = k[A]0 = k – Rate = -d[A]/dt– k = -d[A]/dt– Rearrange and integrate from time = 0 to time = t– [A]t – [A]o = -kt– [A]t = -kt + [A]o

Graphing Zero Order

– [A]t = -kt + [A]o – y = mx + b– Plot of conc. vs. time

gives straight line with slope of -k

– Units of k are M/s

First Order Reactions• Plot of conc vs time does not give straight

line (not 0 order)• Rate changes over time: Doubling

concentration of A doubles the rate• A products• Rate = k[A] Rate = -d[A]/ dt• k[A] = -d[A]/ dt • UTMOC ln[A]/[A]0 = - kt• ln[A] = -kt + ln[A]0

Example Problem

• The decomposition of N2O5 to NO2 and O2 is first order with k = 4.80 x 10-4 s-1 at 45 oC.– If the initial concentration is 1.65 x 10-2 M,

what is the concentration after 825 sec?– How long would it take for the concentration of

N2O5 to decrease to 1.00 x 10-2 M?

Graphing First Order

• Plot of ln[A] vs t gives straight line with a slope of -k and a y-intercept of ln[A]0

• Units of k = s-1

2 N2O5 4 NO2 + O2

First Order Reactions

Determine Order of Reaction by plotting data!

T (sec) [N2O5] (M)

0 2.15 x 10-3

4000 1.88 x 10-3

8000 1.64 x 10-3

12000 1.43 x 10-3

16000 1.25 x 10-3

2 N2O5 (g) 4 NO2 (g) + O2 (g)

0.27

0.24

0.18

0.001

0.0012

0.0014

0.0016

0.0018

0.002

0.0022

0 5000 10000 15000

time (sec)

[N2O

5] (

M)

NOT Zero order!

T (sec) [N2O5] (M) ln [N2O5]

0 0.00215 -6.14

4000 0.00188 -6.28

8000 0.00164 -6.41

12000 0.00143 -6.55

16000 0.00125 -6.68

0.14

0.13

0.14

0.13

-6.8

-6.7

-6.6

-6.5

-6.4

-6.3

-6.2

-6.1

time (sec)

ln [

N2O

5]

This is First Order

Second Order Reactions• Plot of [conc] vs t and plot of ln[conc] vs t

do not give straight lines. (not 0 or 1st )• A products• Rate = k[A]2 Rate = -Δ[A]/ Δt• k[A]2 = -Δ[A]/ Δt • UTMOC 1/[A] = kt + 1/[A]0

Graphing Second Order

• Plot of 1/[A] vs t gives straight line, with a slope of k and a y-intercept of 1/[A]0

Time, sec [HI], M

0 1.000

1000 0.112

2000 0.061

3000 0.041

4000 0.031

2 HI (g) H2 (g) + I2 (g) @ 580K

0

0.2

0.4

0.6

0.8

1

1.2

0 1000 2000 3000 4000 5000

time (sec)

[HI]

(M

)

Not Zero Order

Time, sec [HI], M ln [HI]

0 1.000 0

1000 0.112 -2.19

2000 0.061 -2.80

3000 0.041 -3.19

4000 0.031 -3.47

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

time (sec)

ln [

HI]

(M

)

Not First Order

Time, sec [HI], M ln [HI] 1/[HI]

0 1.000 0 1

1000 0.112 -2.19 8.93

2000 0.061 -2.80 16.4

3000 0.041 -3.19 24.4

4000 0.031 -3.47 32.3

0

5

10

15

20

25

30

35

0 1000 2000 3000 4000 5000

time (sec)

1/[H

I]

Second Order Rate = k[HI]2

Reactions Involving Gases

• A (g) products• PV = nART [A] = nA / V = P/RT

• ln[A]/[A]0 = -kt

• ln(P/RT) /(P0/RT) = -kt

• ln P/P0 = -kt

• Can use the pressures of gases for the concentrations.

Half Life• k = describes speed of the reaction

– Large k = fast reaction• Another way to describe speed is to use t½,

the half life.• This is the time needed to decrease to ½

[A]0.• For a first order reaction,• t = (1/k)ln[A]0/[A]• t½ = (1/k) ln[A]0/([A]0/2)• t½ = (1/k) ln2 = 0.693/k

Comparison of Half-Lives

• Use same procedure to derive each half-life

• For zero order, each half life is half as long as previous one• For first order, each half-life is the same• For second order, each half life is twice as long as the

previous one

Order Zero First Second

t1/2 [A]o / 2k 0.693 / k 1/ k[A]o

Application Question

• Kinetic data were plotted for A 2B + C

• What can these data tell you about this reaction?

More Complicated System

• So far we have assumed one reactant• How do we study A + 2B C + 3D• Run experiment with B in huge excess

– Rate = k [A]x[B]y but [B] remains constant– Called “psuedo” kinetics—can be used to

determine order of A

• Repeat with excess A to get pseudo-order of B• Combine experiments to get real rate law

Reaction Mechanisms

• A reaction may be more complex that 1 simple collision – may form intermediates.

• It is unlikely that 3 or more molecules will collide simultaneously.

• Elementary steps – describe a molecular event.

NO2 + CO NO + CO2

2 elementary steps

NO2 + NO2 + NO3 + CO NO3 + NO + NO2 + CO2

NO2 + CO NO + CO2

NO3 is an intermediate

Cl2 2 Cl

Cl + CHCl3 HCl + CCl3

Cl + CCl3 CCl4

Cl2 + CHCl3 HCl + CCl4 (overall)

Rate laws for elementary steps can be written from stoichiometry. (unlike overall)

1. Rate = k1[Cl2] unimolecular

2. Rate = k2[Cl][CHCl3] bimolecular

3. Rate = k3[Cl][CCl3] bimolecular

Rate Determining Step

• Rate Determining Step (rate limiting step) – is the slowest step leading to the formation of the products (slow step).

• The rates of any steps after the slow step are not important.

O3 + 2 NO2 O2 + N2O5

1. O3 + NO2 NO3 + O2 slow

2. NO3 + NO2 N2O5 fast

Rate = k[O3][NO2]

This explains why stoichiometry and rate law are independent

2 NO2 + O3 N2O5 + O2

Rate = k[NO2][O3]

2 NO + Cl2 2 NOCl

Rate = k[NO]2[Cl2]

Temperature Changes Rate

• 2NO (g) + Cl2 (g) 2NOCl (g)

• Rate = k[NO]2 [Cl2]

• k @ 25 oC = 4.9 x 10-6 M-1s-1

• k @ 35 oC = 1.5 x 10-5 M-1s-1

• This is more than 3x increase!• Why is there a temperature dependence on k?• Can k and temperature be related

theoretically and quantitatively?

Dependence of Rate Constant on Temperature

• Exponential of rate constant on absolute temperature

• Every curve different• This one represents

double rate every 10 K

Collision Theory• Collision Theory – molecules must collide

in order to react.• Rate α # collisions / sec α [reactants]• From KMT

• Increase temp, increase speed• Accounts for higher rate• Kinetic energy made into

potential energy to break

bonds

Activation Energy• Arrhenius – expanded collision theory (1888)• Molecules must collide with enough energy

to rearrange bonds.• If not, they just bounce off.• Activation Energy = Ea = minimum amount

of energy required to initiate a chemical reaction.

• Activated Complex (transition state) – temporary species in reaction sequence, least stable, highest energy, often undetectable.

Activation Energy and Transition States

Molecules with Enough KE to Overcome Activation Energy

• Boltzmann distribution

• Doubling temperature more than doubles fraction of molecules with enough energy

• Fraction =

Steric Factor• Rate constant k depends on three things:

– 1. Need to collide (z = collision rate)– 2. Need E > Ea (fraction of collisions is

– 3. Need to be oriented in the right way to react!

• Steric factor (orientation factor) = p 0>p>1• k = zp • k = A A is pre-exponential factor, or

frequency factor, for the Arrhenius equation

Orientation of Molecules

Arrhenius Equation• k = Ae-Ea/RT • ln k = ln Ae-Ea/RT

• y = m x + b• If you experimentally determine

___ and ____, then you can graphically determine ____ and ____.

• Can also be used to determine k at any ____.

ATR

Ek a ln

1ln

m = -Ea/R

b = ln A

ln k vs. 1/T

lnk1

k2

Ea

R

1 1

T1T2-=

Useful Form of Arrhenius Equation

• In principle, best to do many experiments, graph line, and determine Ea

• In practice, can get decent value from only two experiments

• Rearrange Arrhenius equation to get

Example Problem

H2 + I2 2 HI

k = 2.7 x 10-4 M-1s-1 @ 600 K

k = 3.5 x 10-3 M-1s-1 @ 650 K

a. Find Ea

b. Calculate k @ 700 K

Catalysis• To increase the rate of a reaction

– 1. Increase temperature– 2. Add a catalyst

• Catalyst – a substance which increases the rate of a reaction but is not consumed.

• Catalysts are involved in the course of a reaction.• Usually by lowering Ea

• Homogeneous – catalyst the same phase as the reactants

• Heterogeneous – catalyst in different phase (usually a solid)

Ammonia Formation with

a Catalyst

N2 + 3 H2 2 NH3

Catalytic Hydrogenation

• Food processing• Trans fatty acids