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 of Reaction

A



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

Defining Rate

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

Stoichiometry

important!

• Example: 2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g)

2

N 2 O 5

Rate

= Δ[O 2 ]/ Δt (g) 

4

NO 2 (g) + O 2 (g)

Rate

= Δ[NO 2 ]/

4

Δt = - Δ[N 2 O 5 ]/

2

Δt

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 NO 2 (g) + F 2 (g)  2 NO 2 F (g) • Rate = Δ[NO 2 F]/ 2Δt = -Δ[F 2 ]/ Δt = -Δ[NO 2 ]/ 2Δt • Rate α [NO 2 ] and [F 2 ] • Rate = k [NO 2 ] x • k = rate constant [F 2 ] y

x

and

y

are the

orders of reaction

, these are determined experimentally –

not

from stoichiometry!

• 2 NO 2 (g) + F 2 (g)  2 NO 2 F (g) • Rate = k [NO 2 ] x [F 2 ] y • From experiment, x = 1 , y = 1 • Rate = k [NO 2 ] [F 2 ]

=

• 1 st , 1 st Rate Law order in F 2 , 2 nd order in NO 2 order overall • One way to determine the rate law is from initial rates.

H 2 O 2 (aq) + 3 I (aq) + 2 H + (aq)  I 3 (aq) + 2 H 2 O (l) Rate = k [H 2 O 2 ] x [I ] y [H + ] z [I ] [H + ] Expt # [H 2 O 2 ] 1 0.010

2 0.020

3 0.010

4 0.010

0.010

0.010

0.020

0.010

0.0005

0.0005

0.0005

0.001

Initial rate M/s 1.15 x 10 -6 2.30 x 10 -6 2.30 x 10 -6 1.15 x 10 -6

H 2 O 2 (aq) + 3 I (aq) + 2 H + (aq)  I 3 (aq) + 2 H 2 O (l) Rate = k [H 2 O 2 ] x [I ] y [H + ] z Expt # [H 2 O 2 ] Rel 1 0.010

1 [I ] 0.010

Rel 1 [H + ] 0.0005

Rel 1 Initial rate M/s 1.15 x 10 -6 Rel 1 2 0.020

2 0.010

1 0.0005

1 2.30 x 10 -6 2 3 4 0.010

0.010

1 0.020

2 0.0005

1 2.30 x 10 -6 2 1 0.010

1 0.001

2 1.15 x 10 -6 1 Rate = k [H 2 O 2 ] [I ]

Example Problem

• 2 NO 2 (g)  2 NO (g) + O 2 (g) Expt # 1 2 [NO 2 ] 0.010

0.020

Rate, M/s 7.1 x 10 -5 2.8 x 10 -4

Relationship Between Concentration and Time

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

experiment to • 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 – – Rate = -d[A]/dt k = -d[A]/dt – Rearrange and integrate from time = 0 to time = t [A] t – [A] o = k = -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

line ( not 0 order ) give straight • 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 N 2 O 5 to NO 2 is first order with k = 4.80 x 10 -4 s -1 and O at 45 2 o C.

– 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 N 2 O 5 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 N 2 O 5  4 NO 2 + O 2

First Order Reactions

Determine Order of Reaction by plotting data!

2 N 2 O 5 (g)  4 NO 2 (g) + O 2 (g) T (sec) 0 4000 8000 12000 16000 [N 2 O 5 ] (M) 2.15 x 10 -3 1.88 x 10 -3 1.64 x 10 -3 1.43 x 10 -3 1.25 x 10 -3

0.27

0.24

0.18

0.0022

0.002

0.0018

0.0016

0.0014

0.0012

0.001

0 5000 10000 time (sec)

NOT Zero order!

15000

T (sec) 0 4000 8000 12000 16000 [N 2 O 5 ] (M) ln [N2O 5 ] 0.00215

-6.14

0.00188

0.00164

0.00143

0.00125

-6.28

-6.41

-6.55

-6.68

0.14

0.13

0.14

0.13

-6.1

-6.2

-6.3

-6.4

-6.5

-6.6

-6.7

-6.8

time (sec) 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 1 st ) • A  products • Rate = k[A] 2 • k[A] 2 = -Δ[A]/ Δt Rate = -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

2 HI (g)  H 2 (g) + I 2 (g) @ 580K Time, sec 0 1000 2000 3000 4000 [HI], M 1.000

0.112

0.061

0.041

0.031

1.2

1 0.8

0.6

0.4

0.2

0 0 1000 2000 time (sec) 3000 4000 5000

Not Zero Order

Time, sec 0 1000 2000 3000 4000 [HI], M 1.000

0.112

0.061

0.041

0.031

ln [HI] 0 -2.19

-2.80

-3.19

-3.47

0 -0.5

-1 -1.5

-2 -2.5

-3 -3.5

-4 time (sec)

Not First Order

Time, sec 0 1000 2000 3000 4000 [HI], M 1.000

0.112

0.061

0.041

0.031

ln [HI] 0 -2.19

-2.80

-3.19

-3.47

1/[HI] 1 8.93

16.4

24.4

32.3

15 10 5 0 35 30 25 20 0 1000 2000 time (sec) 3000 4000 5000

Second Order Rate = k[HI] 2

Reactions Involving Gases

• A (g)  products • PV = n A RT [A] = n A • ln[A]/[A] 0 = -kt • ln(P/RT) /(P 0 /RT) = -kt • ln P/P 0 = -kt • Can use the pressures / V = P/RT 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 t ½ , 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] • t ½ = (1/k) ln[A] 0 /([A] 0 /2) • t ½ = (1/k) ln2 = 0.693/k

Comparison of Half-Lives

• Use same procedure to derive each half-life

Order

t 1/2

Zero

[A] o / 2k

First

0.693 / k

Second

1/ k[A] o • 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

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.

NO 2 + CO  NO + CO 2 2 elementary steps NO 2 + NO 2 + NO 3 + CO  NO 3 + NO + NO 2 + CO 2 NO 2 + CO  NO + CO 2 NO 3 is an intermediate

Cl 2  2 Cl Cl + CHCl 3  HCl + CCl 3 Cl + CCl 3  CCl 4 Cl 2 + CHCl 3  HCl + CCl 4 (overall) Rate laws for elementary steps

can

be written from stoichiometry. (unlike overall) 1. Rate = k 1 [Cl 2 ] unimolecular 2. Rate = k 2 [Cl][CHCl 3. Rate = k 3 [Cl][CCl 3 ] 3 ] bimolecular 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.

O 3 + 2 NO 2  O 2 + N 2 O 5 1. O 3 + NO 2 2. NO 3 + NO 2  NO 3 + O 2  N 2 O 5 slow fast Rate = k[O 3 ][NO 2 ] This explains why stoichiometry and rate law are independent

2 NO

2

+ O

3 

N

2

O

5

+ O

2 Rate = k[NO 2 ][O 3 ]

2 NO + Cl

2 

2 NOCl

Rate = k[NO] 2 [Cl 2 ]

Temperature Changes Rate

• 2NO (g) + Cl 2 • Rate = k[NO] 2 (g) [Cl 2 ]  • k @ 25 o C = 4.9 x 10 -6 2NOCl (g) M -1 s -1 • k @ 35 o C = 1.5 x 10 -5 M -1 s -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 = E a = 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 = 𝑒 −𝐸𝑎 𝑅𝑇

Steric Factor

• Rate constant

k

depends on three things: – 1. Need to collide (z = collision rate) – 2. Need E > E a (fraction of collisions is 𝑒 −𝐸𝑎 𝑅𝑇 – 3. Need to be oriented in the right way to react!

• Steric factor (orientation factor) = p 0>p>1 • k = zp 𝑒 −𝐸𝑎 𝑅𝑇 • k = A 𝑒 −𝐸𝑎 𝑅𝑇 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 ln

k

    

E R a

      1

T

    ln

A

• y = m x + b • If you experimentally determine ___ and ____, then you can graphically determine ____ and ____. • Can also be used to determine k at any ____.

ln k vs. 1/T

m = -E a /R b = ln A

Useful Form of Arrhenius Equation

• In principle, best to do many experiments, graph line, and determine E a • In practice, can get decent value from only two experiments • Rearrange Arrhenius equation to get

ln k 1 k 2 = E a R 1 T 2 1 T 1

Example Problem

H 2 + I 2  2 HI k = 2.7 x 10 -4 M -1 s -1 @ 600 K k = 3.5 x 10 -3 M -1 s -1 @ 650 K a. Find E a 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 E a • Homogeneous – catalyst the same phase as the reactants • Heterogeneous solid) – catalyst in different phase (usually a

Ammonia Formation with a Catalyst

N 2 + 3 H 2



2 NH 3

Catalytic Hydrogenation

• Food processing • Trans fatty acids