chapter 13 chemical kinetics chemistry ii. kinetics is the study of the factors that affect the...

Chapter 13Chemical Kinetics

Chemistry II

• kinetics is the study of the factors that affect the speed of a reaction and the mechanism by which a reaction proceeds

• Factors that influence the speed of a reaction: physical state of reactants, temperature, catalysts, concentration

The Rate of a Chemical Reaction

The Rate of a Chemical ReactionDefining Rate

rate is how much a quantity changes in a given period of time, e.g.

time

distance Speed

The Rate of a Chemical Reaction

• Chemical reaction - concentration change with time

• for reactants, a negative sign is used to show a decrease in concentration

• as time goes on, the rate of a reaction generally slows down and stops because the concentration of the reactants decreases

time

ionconcentrat Rate

time

[reactant]

time

[product] Rate

at t = 0[A] = 8[B] = 8[C] = 0

at t = 0[X] = 8[Y] = 8[Z] = 0

at t = 16[A] = 4[B] = 4[C] = 4

at t = 16[X] = 7[Y] = 7[Z] = 1

25.0

061

04Rate

tt

CC

t

CRate

12

12

25.0

061

84Rate

tt

AA

t

ARate

12

12

0625.0

061

01Rate

tt

ZZ

t

ZRate

12

12

0625.0

061

87Rate

tt

XX

t

XRate

12

12

125.0

061

46Rate

tt

CC

t

CRate

12

12

0625.0

061

76Rate

tt

XX

t

XRate

12

12

125.0

061

42Rate

tt

AA

t

ARate

12

12

at t = 16[A] = 4[B] = 4[C] = 4

at t = 16[X] = 7[Y] = 7[Z] = 1

at t = 32[A] = 2[B] = 2[C] = 6

at t = 32[X] = 6[Y] = 6[Z] = 2

0625.0

061

12Rate

tt

ZZ

t

ZRate

12

12

0625.0

061

65Rate

tt

XX

t

XRate

12

12

125.0

061

20Rate

tt

AA

t

ARate

12

12

at t = 32[A] = 2[B] = 2[C] = 6

at t = 32[X] = 6[Y] = 6[Z] = 2

at t = 48[A] = 0[B] = 0[C] = 8

at t = 48[X] = 5[Y] = 5[Z] = 3

125.0

061

68Rate

tt

CC

t

CRate

12

12

0625.0

061

23Rate

tt

ZZ

t

ZRate

12

12

8

174.0

046

08overall Rate

tt

CC

t

Coverall Rate

12

12

174.0

046

80overall Rate

tt

AA

t

Aoverall Rate

12

12

9

065.0

046

03overall Rate

tt

ZZ

t

Zoverall Rate

12

12

065.0

046

85overall Rate

tt

XX

t

Xoverall Rate

12

12

rate of change reactants = rate of change products

The Rate of a Chemical ReactionStoichiometry

• e.g.

H2 (g) + I2 (g) 2 HI(g)

• for the above reaction, for every 1 mole of H2 used, 1 mole of I2 will also be used and 2 moles of HI made therefore the rate of change will be different

• in order to be consistent, the change in the concentration of each substance is multiplied by 1/coefficient

t

HI][

2

1

t

]I[

t

]H[ Rate 22

The Rate of a Chemical ReactionStoichiometry

• e.g.

H2 (g) + I2 (g) 2 HI(g)

The Rate of a Chemical ReactionAverage Rate

• the average rate is the change in measured concentrations in any particular time period linear approximation of a curve

• the larger the time interval, the more the average rate deviates from the instantaneous rate

H2 I2

HI

Stoichiometry tells us that for every 1 mole/L of H2 used, 2 moles/L of HI are made.

Assuming a 1 L container, at 10 s, we used 0.181 moles of H2. Therefore the amount of HI made is 2(0.181 moles) = 0.362 moles

At 60 s, we used 0.699 moles of H2. Therefore the amount of HI made is 2(0.699 moles) = 1.398 moles

The average rate is the change in the concentration in a given time period.

In the first 10 s, the Δ[H2] is -0.181 M, so the rate is

s

M0181.0

s 10.000

M 181.0

Avg. Rate, M/s Avg. Rate, M/s

Time (s) [H2], M [HI], M -[H2]/t 1/2 [HI]/t

0.000 1.000

10.000 0.819

20.000 0.670

30.000 0.549

40.000 0.449

50.000 0.368

60.000 0.301

70.000 0.247

80.000 0.202

90.000 0.165

100.000 0.135



0.000 1.000 0.000

10.000 0.819 0.362

20.000 0.670 0.660

30.000 0.549 0.902

40.000 0.449 1.102

50.000 0.368 1.264

60.000 0.301 1.398

70.000 0.247 1.506

80.000 0.202 1.596

90.000 0.165 1.670

100.000 0.135 1.730

Avg. Rate, M/s

Time (s) [H2], M [HI], M -[H2]/t

0.000 1.000 0.000

10.000 0.819 0.362 0.0181

20.000 0.670 0.660 0.0149

30.000 0.549 0.902 0.0121

40.000 0.449 1.102 0.0100

50.000 0.368 1.264 0.0081

60.000 0.301 1.398 0.0067

70.000 0.247 1.506 0.0054

80.000 0.202 1.596 0.0045

90.000 0.165 1.670 0.0037

100.000 0.135 1.730 0.0030



0.000 1.000 0.000

10.000 0.819 0.362 0.0181 0.0181

20.000 0.670 0.660 0.0149 0.0149

30.000 0.549 0.902 0.0121 0.0121

40.000 0.449 1.102 0.0100 0.0100

50.000 0.368 1.264 0.0081 0.0081

60.000 0.301 1.398 0.0067 0.0067

70.000 0.247 1.506 0.0054 0.0054

80.000 0.202 1.596 0.0045 0.0045

90.000 0.165 1.670 0.0037 0.0037

100.000 0.135 1.730 0.0030 0.0030

Ave. rate slows down as reaction proceeds

Rate of loss reactant = Rate gain product

Concentration vs. Time for H2 + I2 --> 2HI

0.000

0.200

0.400

0.600

0.800

1.000

1.200

1.400

1.600

1.800

2.000

0.000 10.000 20.000 30.000 40.000 50.000 60.000 70.000 80.000 90.000 100.000

time, (s)

con

cen

trat

ion

, (M

)

[H2], M

[HI], M

average rate in a given time period = slope of the line connecting the [H2] points; and ½ +slope of the line for [HI]

the average rate for the first 10 s is 0.0181 M/sthe average rate for the first 40 s is 0.0150 M/sthe average rate for the first 80 s is 0.0108 M/s

Ave. rate slows down as reaction proceeds

The Rate of a Chemical Reaction Instantaneous Rate

• average rate becomes less accurate over longer time spans

• the instantaneous rate is the change in concentration at any one particular time slope at one point of a curve

• determined by taking the slope of a line tangent to the curve at that particular point first derivative of the function

for you calculus fans

H2 (g) + I2 (g) 2 HI (g) Using [H2], the instantaneous rate at 50 s is:

s

M 0.0070 Rate

s 40

M 28.0 Rate

Using [HI], the instantaneous rate at 50 s is:

s

M 0.0070 Rate

s 40

M 56.0

2

1 Rate

rate reactants = rate products

Generalized rate law:

aA + bB → cC + dD

t

D][1

t

C][1

t

B][1

t

A][1 Rate

dcba

Ex 13.1 - For the reaction given, the [I] changes from 1.000 M to 0.868 M in the first 10 s. Calculate the average rate in the first 10 s and the Δ[H+].

H2O2 (aq) + 3 I(aq) + 2 H+(aq) I3

(aq) + 2 H2O(l)

Solve the equation for the Rate (in terms of the change in concentration of the Given quantity)

Solve the equation of the Rate (in terms of the change in the concentration for the quantity to find) for the unknown value

s

M3-104.40 Rate

s 10

M 000.1M 868.0

3

1

t

]I[

3

1 Rate

s

M3-s

M3- 108.80 104.402t

]H[

Rate2t

]H[

t

]H[

2

1 Rate

Factors Affecting Reaction Rate- nature of reactants

• nature of the reactants means what kind of reactant molecules and what physical condition they are in. small molecules tend to react faster than large molecules;

gases tend to react faster than liquids which react faster than solids;

powdered solids are more reactive than “blocks” more surface area for contact with other reactants

certain types of chemicals are more reactive than others e.g., the activity series of metals

ions react faster than molecules no bonds need to be broken

• increasing temperature increases reaction rate chemist’s rule of thumb - for each 10°C rise in temperature, the

speed of the reaction doubles

• there is a mathematical relationship between the absolute temperature and the speed of a reaction discovered by Svante Arrhenius which will be examined later

Factors Affecting Reaction Rate- Temperature

• catalysts are substances which affect the speed of a reaction without being consumed

• most catalysts are used to speed up a reaction, these are called positive catalysts catalysts used to slow a reaction are called negative catalysts

• homogeneous = present in same phase

• heterogeneous = present in different phase

• how catalysts work will be examined later

Factors Affecting Reaction Rate- Catalysts

• generally, the larger the concentration of reactant molecules, the faster the reaction increases the frequency of reactant molecule contact concentration of gases depends on the partial pressure of the

gas higher pressure = higher concentration

• concentration of solutions depends on the solute to solution ratio (molarity)

Factors Affecting Reaction Rate- Reactant Concentration

The Rate Law: Effect of Concentration On Reaction Rate

• Mathematical relationship between the rate of the reaction and the concentrations of the reactants

• for the reaction aA + bB products the rate law would have the form given below n and m are called the orders for each reactant k is called the rate constant

mnk [B][A] Rate


• sum of the exponents is called the order of the reaction • The rate law for the reaction:

2 NO(g) + O2(g) ⇌ 2 NO2(g)

Rate = k[NO]2[O2]

The reaction is second order with respect to [NO], first order with respect to [O2],

and third order overall

Reaction CH3CN CH3NC

CH3CHO CH4 + CO

2 N2O5 4 NO2 + O2

H2 + I2 2 HI


Rate Law Rate = k[CH3CN]

Rate = k[CH3CHO]

Rate = k[N2O5]2

Rate = k[H2][I2]

Sample Rate Laws


Example:

A → Products

Rate = k[A]1 k = 0.015 / 0.10 = 0.15 s-1

If concentration of A doubles, the new rate, Rate2 = k[2A]1 = 2 k[A]1 = 2 x Rate

[A] (M) Initial Rate (M/s)

0.10 0.015

0.20 0.030

0.30 0.060


Example:

Zero order: Second order:

Concentration of A doubles, the rate is constant Conc. A doubles, new rate

Rate1 = Rate2 = k Rate2 = k[2A]2 = 4 x k[A]2

Rate2 = 4 x Rate


0.10 0.015

0.20 0.015

0.30 0.015


0.10 0.015

0.20 0.060

0.30 0.240

Reactant Concentration vs. TimeA Products

Rate = k[A]2

Rate = k[A]Rate = k

0: Concentration dec. linearly with time. Rate Is constant, reaction does not slow down as [A] dec.

1 and 2: Rate slows as reaction proceeds since [A] dec.

The Integrated Rate Law

• can only be determined experimentally• graphically

rate = slope of curve [A] vs. time

if graph [A] vs time is straight line, then exponent on A in rate law is 0, rate constant = -slope

if graph ln[A] vs time is straight line, then exponent on A in rate law is 1, rate constant = -slope

if graph 1/[A] vs time is straight line, exponent on A in rate law is 2, rate constant = slope

The Integrated Rate Law

• the half-life, t1/2, of a reaction is the length of time it takes for the concentration of the reactants to fall to ½ its initial value

• the half-life of the reaction depends on the order of the reaction

Zero Order Reactions (n = 0)

Rate = -d[A] = k[A]0 = k dt

constant rate reactions

Solution: [A] = -kt + [A]0 (= integrated rate law) y = mx + b

graph of [A] vs. time is straight line with slope = -k and y-intercept = [A]0

[A] = [A0]/2, t ½ = [A0]/2k

Units: when Rate = M/sec, k = M/sec

[A]0

[A]

time

slope = - k

∫ -d[A]/dt = ∫ k∫ d[A] = -∫ k dt[A] = -kt + C, where C = [A]0

First Order Reactions (n = 1)

Rate = -d[A] = k[A] dt

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

graph of ln[A] vs. time gives straight line with

slope = -k and y-intercept = ln[A]0

used to determine the rate constant

[A] = [A0]/2, t½ = ln 2 [lna-lnb = ln(a/b)] k

the half-life of a first order reaction is constant

Units: when Rate = M/sec, k = sec-1

(dim. Analysis: M/s = k.M)

Rate slows as reaction proceeds since [A] dec.

∫ -d[A]/dt = ∫ k[A]∫ 1 d[A] = -∫ k dt [A]ln[A] = -kt + C, where C = ln[A]0

ln[A]0

ln[A]

time

slope = −k

Rate Data for hydrolysis of C4H9Cl

Time (sec) [C4H9Cl], M

0.0 0.1000

50.0 0.0905

100.0 0.0820

150.0 0.0741

200.0 0.0671

300.0 0.0549

400.0 0.0448

500.0 0.0368

800.0 0.0200

10000.0 0.0000

Show reaction is first-order and find k

C4H9Cl + H2O C4H9OH + 2 HCl Concentration vs. Time for the Hydrolysis of C4H9Cl

0

0.02

0.04

0.06

0.08

0.1

0.12

0 200 400 600 800 1000

time, (s)

conc

entr

atio

n, (

M)

C4H9Cl + H2O C4H9OH + 2 HClRate vs. Time for Hydrolysis of C4H9Cl

0.0E+00

5.0E-05

1.0E-04

1.5E-04

2.0E-04

2.5E-04

0 100 200 300 400 500 600 700 800

time, (s)

Rat

e, (

M/s

)

C4H9Cl + H2O C4H9OH + 2 HCl

LN([C4H9Cl]) vs. Time for Hydrolysis of C4H9Cl

y = -2.01E-03x - 2.30E+00

-4.5

-4

-3.5

-3

-2.5

-2

-1.5

-1

-0.5

0

0 100 200 300 400 500 600 700 800

time, (s)

LN(c

once

ntra

tion)

slope = -2.01 x 10-3

k =2.01 x 10-3 s-1

s 345s 1001.2

693.0

693.0t

1-3

21

k

Second Order Reactions

Rate = -d[A] = k[A] = k[A]2

dt

Solution: 1/[A] = kt + 1/[A]0

y = mx + b

graph 1/[A] vs. time gives straight line with

slope = k and y-intercept = 1/[A]0

used to determine the rate constant

t½ = 1 k[A0]

when Rate = M/sec, k = M-1∙sec-1

l/[A]0

1/[A]

time

slope = k

Example Second-Order Reaction

Time (hrs.)PNO2,

(mmHg) ln(PNO2) 1/(PNO2)

0 100.0 4.605 0.01000

30 62.5 4.135 0.01600

60 45.5 3.817 0.02200

90 35.7 3.576 0.02800

120 29.4 3.381 0.03400

150 25.0 3.219 0.04000

180 21.7 3.079 0.04600

210 19.2 2.957 0.05200

240 17.2 2.847 0.05800

Show that the reaction: NO2 → NO + O is incorrect according to the data below

Rate Data Graphs ForNO2 → NO + O

Partial Pressure NO2, mmHg vs. Time

0.0

10.0

20.0

30.0

40.0

50.0

60.0

70.0

80.0

90.0

100.0

0 50 100 150 200 250

Time, (hr)

Pres

sure

, (m

mH

g)

Non-linear so not zero order

Rate Data Graphs ForNO2 ® NO + O

ln(PNO2) vs. Time

2.4

2.6

2.8

3

3.2

3.4

3.6

3.8

4

4.2

4.4

4.6

4.8

0 50 100 150 200 250

Time (hr)

ln(p

ress

ure)

Non-linear so not first order

Rate Data Graphs ForNO2 → NO + O

1/(PNO2) vs Time

1/PNO2 = 0.0002(time) + 0.01

0.00000

0.01000

0.02000

0.03000

0.04000

0.05000

0.06000

0.07000

0 50 100 150 200 250

Time, (hr)

Inve

rse

Pres

sure

, (m

mH

g-1

)

We can deduce actual reaction should be:2NO2 → 2NO + O2

k = 2 x 10-4 M-1 s-1

Tro, Chemistry: A Molecular Approach 47

Ex. 13.4 – The reaction SO2Cl2(g) SO2(g) + Cl2(g) is first order with a rate constant of 2.90 x 10-4 s-1 at a given set of conditions. Find the [SO2Cl2] at 865 s when [SO2Cl2]0 = 0.0225 M

the new concentration is less than the original, as expected

[SO2Cl2]0 = 0.0225 M, t = 865, k = 2.90 x 10-4 s-1

[SO2Cl2]

Check:

Solution:

Concept Plan:

Relationships:

Given:

Find:

[SO2Cl2][SO2Cl2]0, t, k

0ln[A]tln[A] :processorder 1st afor k

M 0.0175 ]Cl[SO

4.043.790.251]Clln[SO0.0225lns 865s 102.90]Clln[SO

]Clln[SOt]Clln[SO

(-4.04)22

22

1-4-22

02222

e

k

49

Ex 13.2 – Determine the rate law and rate constant for the reaction NO2(g) + CO(g) NO(g) + CO2(g)

given the data below.

Expt.

Number

Initial [NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

Write a general rate law including all reactants

Examine the data and find two experiments in which the concentration of one reactant changes, but the other concentrations are the same

Expt.

Number

Initial [NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

Comparing Expt #1 and Expt #2, the [NO2] changes but the [CO] does not

mnk [CO]][NO Rate 2



Determine by what factor the concentrations and rates change in these two experiments.

Expt.

Number

Initial [NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

2M 10.0

M 20.0

][NO

][NO

1expt 2

2expt 2 4 0021.0

0082.0

Rate

Rate

sM

sM

1expt

2expt



Determine to what power the concentration factor must be raised to equal the rate factor.

Expt.

Number

Initial [NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

2M 10.0

M 20.0

][NO

][NO

1expt 2

2expt 2 4 0021.0

0082.0

Rate

Rate

sM

sM

1expt

2expt

242

Rate

Rate

][NO

][NO

1expt

2expt

1expt 2

2expt 2

n

n

n



Repeat for the other reactants

Expt.

Number

Initial [NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

2M 10.0

M 20.0

[CO]

[CO]

2expt

3expt 1 0082.0

0083.0

Rate

Rate

sM

sM

2expt

3expt

012

Rate

Rate

[CO]

[CO]

2expt

3expt

2expt

3expt

m

m

m

Expt.

Number

Initial [NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033



Substitute the exponents into the general rate law to get the rate law for the reaction

Expt.

Number

Initial [NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

mnk [CO]][NO Rate 2n = 2, m = 0

22

022

][NO Rate

[CO]][NO Rate

k

k



Substitute the concentrations and rate for any experiment into the rate law and solve for k

Expt.

Number

Initial [NO2], (M)

Initial

[CO], (M)

Initial Rate

(M/s)

1. 0.10 0.10 0.0021

2. 0.20 0.10 0.0082

3. 0.20 0.20 0.0083

4. 0.40 0.10 0.033

1-1-

2s

M

2s

M

22

sM 21.0M 01.0

0.0021M 10.0 0.0021

1expt for ][NO Rate

k

k

k

The Effect of Temperature on Reaction Rate

• Rate constant k is temperature dependent

• Arrhenius investigated this relationship and showed that:

RT

Ea

eAk

R is the gas constant in energy units, 8.314 J/(mol∙K)

where T is the temperature in kelvin

A is a constant called the frequency factor

Ea is the activation energy, the extra energy needed to start the molecules reacting

As x (temperature) increases e-1/x will increase up to a maximum value of 1, k increases

As Ea increases k will decrease (follows e-x graph)

The Effect of Temperature on Reaction RateActivation Energy and the Activated Complex

• Ea is an energy barrier to the reaction

• amount of energy needed to convert reactants into the activated complex aka transition state

• the activated complex is a chemical species with partially broken and partially formed bonds always very high in energy

because partial bonds

The Effect of Temperature on Reaction RateThe Exponential Factor

• e-Ea/RT is a number between 0 and 1• it represents the fraction of reactant molecules with sufficient energy

to make it over the energy barrier• that extra energy comes from converting the KE of motion to PE in the

molecule when the molecules collide

• e-Ea/RT decreases as Ea increases

Reaction rate inc.:-Increasing T increases the Ave. KE of the molecules-Increases no. of molecules with sufficient energy to overcome the energy barrier

The Effect of Temperature on Reaction RateArrhenius Plots

• the Arrhenius Equation can be algebraically solved:

AR

Ek a ln

T

1)ln(

y = mx + bwhere y = ln(k) and x = (1/T)

a graph of ln(k) vs. (1/T) is a straight line

slope of the line = -Ea/R so Ea = -mR

ey-intercept = A, (unit is the same as k …why?)

k = Ae-Ea/RT

lnk = ln(Ae-Ea/RT)lnk = lnA + lne-Ea/RT

lnk = lnA – Ea/RT

Ex. 13.7 Determine the activation energy and frequency factor for the reaction O3(g) O2(g) + O(g) given the following data:

Temp, K k, M-1∙s-1 Temp, K k, M-1∙s-1

600 3.37 x 103 1300 7.83 x 107

700 4.83 x 104 1400 1.45 x 108

800 3.58 x 105 1500 2.46 x 108

900 1.70 x 106 1600 3.93 x 108

1000 5.90 x 106 1700 5.93 x 108

1100 1.63 x 107 1800 8.55 x 108

1200 3.81 x 107 1900 1.19 x 109


use a spreadsheet to graph ln(k) vs. (1/T)


Ea = m∙(-R)

solve for Ea

molkJ

molJ4

KmolJ4

1.93

1031.9314.8K 1012.1

a

a

E

E

A = ey-intercept

solve for A11-11

118.26

sM 1036.4

1036.4

A

eA

The Effect of Temperature on Reaction RateThe Collision Model

• for most reactions, in order for a reaction to take place, the reacting molecules must collide into each other.

• once molecules collide they may react together or they may not, depending on two factors –

1. whether the collision has enough energy to "break the bonds holding reactant molecules together";

2. whether the reacting molecules collide in the proper orientation for new bonds to form.


Effective Collisions

• collisions in which these two conditions are met (and therefore lead to reaction) are called effective collisions

• the higher the A value (frequency of effective collisions), the higher k value and the faster the reaction rate

• when two molecules have an effective collision, a temporary, high energy (unstable) chemical species is formed - called an activated complex or transition state


Orientation Effect


• A is the factor called the frequency factor and is the number of molecules that can approach overcoming the energy barrier

• there are two factors that make up the frequency factor – the orientation factor (p) and the collision frequency factor (z)

RT

E

RT

E aa

pzeeAk


Orientation Factor

• proper orientation is when the atoms are aligned so that old bonds can break and the new bonds can form

• the more complex the reactants, the less frequently they will collide with the proper orientation reactions between atoms generally have p = 1 reactions where symmetry results in multiple orientations leading to reaction have p

slightly less than 1

• for most reactions, the orientation factor is less than 1 For many, p << 1

e.g. H(g) + I(g) → HI(g)

H2(g) + I2(g) → 2HI(g)

HCl(g) + HCl(g) → H2(g) + Cl2(g) Smallest p

Reaction Mechanisms

• we generally describe chemical reactions with an equation listing all the reactant molecules and product molecules

• but the probability of more than 3 molecules colliding at the same instant with the proper orientation and sufficient energy to overcome the energy barrier is negligible

• most reactions occur in a series of small reactions involving 1, 2, or at most 3 molecules

• describing the series of steps that occur to produce the overall observed reaction is called a reaction mechanism

• knowing the rate law of the reaction helps us understand the sequence of steps in the mechanism

Reaction Mechanisms

• Overall reaction:

H2(g) + 2 ICl(g) 2 HCl(g) + I2(g) • Mechanism:

1) H2(g) + ICl(g) HCl(g) + HI(g)

2) HI(g) + ICl(g) HCl(g) + I2(g)

• the steps in this mechanism are elementary steps, meaning that they cannot be broken down into simpler steps and that the molecules actually interact directly in this manner without any other steps

1) H2(g) + ICl(g) HCl(g) + HI(g) 2) HI(g) + ICl(g) HCl(g) + I2(g)

H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)

Reaction MechanismsIntermediates

• notice that the HI is a product in Step 1, but then a reactant in Step 2

• since HI is made but then consumed, HI does not show up in the overall reaction

• materials that are products in an early step, but then a reactant in a later step are called intermediates

Reaction MechanismsMolecularity

• the number of reactant particles in an elementary step is called its molecularity

• a unimolecular step involves 1 reactant particle

• a bimolecular step involves 2 reactant particles

though they may be the same kind of particle

• a termolecular step involves 3 reactant particles though these are exceedingly rare in elementary steps

Reaction MechanismsRate Laws for Elementary Steps

• each step in the mechanism is like its own little reaction – with its own activation energy and own rate law

• the rate law for an overall reaction must be determined experimentally

• but the rate law of an elementary step can be deduced from the equation of the step

H2(g) + 2 ICl(g) 2 HCl(g) + I2(g)

1) H2(g) + ICl(g) HCl(g) + HI(g) Rate = k1[H2][ICl] 2) HI(g) + ICl(g) HCl(g) + I2(g) Rate = k2[HI][ICl]

Reaction MechanismsRate Laws for Elementary Steps

Reaction MechanismsRate Determining Step

• in most mechanisms, one step occurs slower than the other steps

• the result is that product production cannot occur any faster than the slowest step – the step determines the rate of the overall reaction

• we call the slowest step in the mechanism the rate determining step the slowest step has the largest activation energy

• the rate law of the rate determining step determines the rate law of the overall reaction

Another Reaction MechanismNO2(g) + NO2(g) NO3(g) + NO(g) Rate = k1[NO2]2 slow

NO3(g) + CO(g) NO2(g) + CO2(g) Rate = k2[NO3][CO] fast

NO2(g) + CO(g) NO(g) + CO2(g) Rateobs = k[NO2]2

The first step in this mechanism is the rate determining step.

The first step is slower than the second step because its activation energy is larger.

The rate law of the first step is the same as the rate law of the overall reaction.

Reaction MechanismsValidating a Mechanism

in order to validate (not prove) a mechanism, two conditions must be met:

1. the elementary steps must sum to the overall reaction

2. the rate law predicted by the mechanism must be consistent with the experimentally observed rate law

Reaction Mechanisms Mechanisms with a Fast Initial Step

• when a mechanism contains a slow initial step, the rate law will not contain intermediates

• when a mechanism contains a fast initial step, the rate limiting step, and hence the rate law may contain intermediates

• We can express[intermediate] in terms of [reactant]• If first step is fast, intermediate products build up (limited by slower step down

the line) as they build up they react to re-form reactants reaches equilibrium, the forward and reverse reaction rates are equal – so

the concentrations of reactants and products of the step are related• substituting into the rate law of the RDS will produce a rate law in terms of just

reactants

An Example

2 NO(g) ⇌ N2O2(g) Fast

H2(g) + N2O2(g) H2O(g) + N2O(g) Slow (rate limiting)

H2(g) + N2O(g) H2O(g) + N2(g) Fast

k1

k-1

2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g)

Experimentally observed Rateobs = k [H2][NO]2

Proposed mechanism:

2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g)

Reaction Mechanisms Mechanisms with a Fast Initial Step

• Is the mechanism valid?

1. steps must sum to the over all reaction

2. rate law predicted by mechanism must be consistent with exp. observation

Since 2nd step is rate limiting, Rate = k2[H2][N2O2]

BUT! Contains intermediate [N2O2] , not consistent with observation

Since 1st step is in equilibrium we can express [intermediate] in terms of reactants

An Example

2 NO(g) ⇌ N2O2(g) Fast

H2(g) + N2O2(g) H2O(g) + N2O(g) Slow Rate = k2[H2][N2O2]

H2(g) + N2O(g) H2O(g) + N2(g) Fast

k1

k-1

2 H2(g) + 2 NO(g) 2 H2O(g) + N2(g) Rateobs = k [H2][NO]2

for Step 1 Rateforward = Ratereverse

2

1

122

2212

1

[NO]]O[N

]O[N [NO]

k

k

kk

22

1

12

2

1

122

2222

][NO][HRate

[NO]][HRate

]O][N[HRate

k

kk

k

kk

k

Let k2k1/k-1 = k, we have obs rate law

Ex 13.9 Show that the proposed mechanism for the reaction 2 O3(g) 3 O2(g) matches the observed rate law

Rate = k[O3]2[O2]-1

O3(g) O⇌ 2(g) + O(g) Fast

O3(g) + O(g) 2 O2(g) Slow Rate = k2[O3][O]

k1

k-1

for Step 1 Rateforward = Ratereverse

123

1

1

2131

]][O[O[O]

][O][O ][O

k

k

kk

1-2

23

1

12

1-23

1

132

32

][O][ORate

]][O[O][ORate

][O][ORate

k

kk

k

kk

k

(Slow rate has intermediate)

Catalysts

• catalysts are substances that affect the rate of a reaction without being consumed

• catalysts work by providing an alternative mechanism for the reaction with a lower activation energy

• catalysts are consumed in an early mechanism step, then made in a later step

mechanism without catalyst

O3(g) + O(g) 2 O2(g) V. Slow

mechanism with catalyst

Cl(g) + O3(g) O⇌ 2(g) + ClO(g) Fast

ClO(g) + O(g) O2(g) + Cl(g) Slow

O3(g) + O(g) 2 O2(g)

CatalystsDemo

http://academics.rmu.edu/~short/chem1215/1215-demos/oscillating-methanol-7mins.mov

2CH3OH(l) + 3O2(g) → 2CO2(g) + 4H2O(g)



Ozone Depletion over the Antarctic

Catalysts

polar stratospheric clouds contain ice crystals that catalyze reactions that release Cl from atmospheric chemicals

CatalystsHomorgeneous and Heterogeneous Catalysts

• homogeneous catalysts are in the same phase as the reactant particles Cl(g) in the destruction of O3(g)

• heterogeneous catalysts are in a different phase than the reactant particles solid catalytic converter in a car’s exhaust system

Catalysts Enzymes

• because many of the molecules are large and complex, most biological reactions require a catalyst to proceed at a reasonable rate

• protein molecules that catalyze biological reactions are called enzymes

• enzymes work by adsorbing the substrate reactant onto an active site that orients it for reaction

Enzymatic Hydrolysis of Sucrose

chapter 13 chemical kinetics chemistry ii. kinetics is the study of the factors that affect the...

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