William L Masterton

Cecile N. Hurley http://academic.cengage.com/chemistry/masterton

Edward J. Neth University of Connecticut

PHYSICAL CHEMISTRY

Nor Azira Irma Muhammad

UiTM Perlis

azira_irma@perlis.uitm.edu.my

3.1 Rates of reaction

3.2 Factors affecting rates of reaction

3.3 Rate Law and order of reaction

3.4 Methods to determine order of reactions

2

Chemical kinetics- a branch of chemistry that deals with the rate of chemical reaction

Many chemical reactions are fast reactions, especially those involved reactions between oppositely charged ions in aqueous solution (e.g, neutralisation, precipitation reaction)

Reaction involving organic compounds are usually slow reactions because they involved breaking and making covalent bonds

Reaction rate: Rate of reaction the change in concentration of a reactant with

time (M/s) Consider the following reaction: C4H9Cl + H2O C4H9OH + HCl The above reaction was conducted using 0.1 M butyl chloride

aqueous solution and its concentration was determined as a function of time

3

Graph of concentration C4H9Cl vs time was plotted

4

[C4

H9

Cl]

Time (s)

Rate of reaction = concentration time. = decrease in concentration of butyl chloride time taken = - d [C4H9Cl] dt The average rate = concentration (of the first 50 s) 50 s (value taken from table / expt)

The instantaneous rate is a rate of change at particular instant in time

The instantaneous rate at time t = gradient (slope) of the tangent to the curve at time, t (specific time) The instantaneous rate at the start of the reaction (t = o) is called

the initial rate Initial rate = gradient (slope) of the tangent to the curve at time = 0

5

concentration

time t

Instantaneous

rate

initial rate

time

concentration

example;

the instantaneous rate at t

= y2 - y1 mol dm-3s-1

x2 - x1 x2

x1

y1

y2

example;

the initial rate

= y2 - y1 mol dm-3s-1

x2 - 0 y1

y2

0 x2

(1) Collision theory (2) Transition state theory 1) Collision theory Three ideas;

i. Molecules must collide to react ii. Molecules must posses a certain minimum kinetic energy,

called the activation energy, Ea to initiate the chemical reaction. (otherwise they are unable to react)

iii. Molecules must collide in the right orientation

The basis concept of collision is that, for the reaction between two particles to occur, an effective collision must take place

Not all collision are effective in producing reaction, but only small proportion of the molecules react after collision

Collision frequency of reaction = total number of collision between molecules per unit time at specific T 6

The effect of concentration on the rate of reaction the reaction rate will increase if the concentration of one or

more reactant increase [ ] high, frequency of collision high, because more particles

present in the same volume and more likely to collide - probability of collision with sufficient energy for reaction to occur. So, rate of reaction increase.

The effect of temperature Rate of chemical reaction increase with T increase At higher T, Arhenius suggested colliding molecules can react only

if they have total kinetic energy equal or greater than Ea (activation energy)

Ea (activation energy) is a minimum amount of energy required to initiate a chemical reaction, which means that Ea is a minimum kinetic energy that molecules must possess in order for a chemical reaction to occur

Ea is shown on diagram called reaction profiles /energy profile - Energy versus progress of reaction

7

8

H reactant

H product

Ea

energy energy

Progress of reaction Progress of reaction

Ea

reactants

reactants poducts

poducts

H = - ve H = + ve

a) Reaction profile of an exothermic reaction b) Reaction profile of an endothermic reaction

Ea = activation energy

H = difference in energies between reactant and product/ heat of reactions = H product H reactants

2) The transition state theory This theory focus on what happens to the reactant molecules as

they change into products The presence of a transitory intermediate stage (that lies

between the reactants and the products) which is called the transition state, which involved hypothetical species is activated complex

e.g : H-H + Cl [H----H---Cl] H. + HCl reactants activated complex products The complex does not always changed to product. (if it does, it

means that the reaction is occuring / proceed to give the product)

The reaction progress on x- axis for diagram of reaction profile represents the extent of the reaction.

The reaction starts with the reactants on the left, progress through an activated complex (transition state), and ends with the products on the right.

The activated complex is very unstable. The energy gap between energies of the reactants and the activated complex is called the Ea. 9

The rate law is an equation that relates the rate of reaction to the concentration of the reactants. This rate law can only be determined by experiment

Example: aA + bB cC + dD

If the rate is to the concentration of A only and not B, then

Rate of reaction [A] [B]0

Rate of reaction = k [A][B]0

= k [A]

The reaction is said to be first order with respect to A and zero order with respect to B

The overall order is 1 + 0 = 1, that is first order

10

Rate = k [A]m [B]n

(the expression is called rate law / rate equation) m = order of reaction with respect to A

n = order of reaction with respect to B

Overall order of reaction = m + n

The proportionality constant, k is called the rate constant.

k value depends on the specific reaction, the temperature of the reaction and the presence of a catalyst (if any).

The larger value of k, the faster a reaction proceeds. Once the k and order of reaction are known, the reaction rate can be predicted for any concentration of A and B

11

The order of reaction (m + n) and the rate constant (k) can be determined by using:

I. The reaction rate method

II. The linear plots

III.The half-life method, t1/2

IV. Initial rate method

12

The order of reaction can be found by plotting a reaction rate against the concentration

13

k

rate rate rate

[A] [A] [A]

First order Zero order Second order

y= kx y= kx2 y= k

Example: Bromine reacts with methanoic acid according to the equation: The table below shows the rates of reactions at specific concentrations

)(2)(2)()(2 aqHaqBraqHCOOHaqBr

14

[Br2] mol dm-3 Rate (10-5 mol dm-3s-1)

8.0 2.75

7.0 2.40

5.0 1.70

4.0 1.35

2.0 0.70

Show that the reaction between bromine and methanoic acid is a first order by a) calculation b) A graphical method

Solution:

a) Calculation (Refer to the table)

when the concentration of bromine is doubled, the rate of reaction is doubled.

Thus, the reaction is first order with respect bromine. so, rate [Br2].

b) By graph

0

0.5

1

1.5

2

2.5

3

0 2 4 6 8 10

[Br2]

Rate

of

reacti

on

15

A sloping straight line graph is obtained, when the graph of rate against concentration is plotted. This shows that the reaction is first order with respect to bromine Rate [Br2]

The order of reaction can be found by plotting a linear plots with respect to a given reactant

16

ln [A]t

time

y

x

time time

x

y x

y

[A]t 1/ [A]t

Slope = y / x = - k /2.303

First order Zero order second order

Slope = y / x = k

Slope = y / x = - k

A first order reaction with respect to a reactant A, is a reaction in which the rate of reaction is directly proportional to the concentration of A.

If the concentration of a reactant is doubled, the rate of reaction is also doubled.

The rate equation for a first order reaction is, rate reactions = k [A] where, k = rate constant [A] = concentration in mol dm-3

the unit of k is time-1 (that is s-1, min-1, h-1) Examples of first order reactions:

i. Radioactive decay

ii. Catalytic decomposition of hydrogen peroxide

iii. Thermal decomposition of dinitrogen pentoxide

UkeradioactivofrateHeThU

238

92

4

2

234

90

238

92

)()(2)(2 222 gasOliqOHaqOH

17

24252 22 OONON

For the reaction:

A products

Having the rate law :

where [A] t = concentration of A at time t

Rearrange the expression:

Ak

dt

Adreactionofrate

18

ot

ot

t

o

A

A

AnlktAnl

ktAnlAnl

dtkA

Ad

tegratein

dtkA

Ad

t

O

:

ln [A]t

t

A straight line with -ve slope

19

Rate = - d [A] = k[A]1

dt

d[A] = - k[A]1

dt Integrate:

1 d[A] = - k dt [A] ln [A]t = - kt + c , t = 0 c = ln [A]0 ln [A]t = - kt + ln [A]0

ln [A]t

t

A straight line with -ve slope

A B (reactant) (product)

y -mx c

CH3CO-C2H5 + H2O CH3COOH + C2H5OH

For example, the hydrolysis of ethyl ethanoate is first order

respect to the ester and first order with respect to the water and acid.

In the presence of excess H2O and H+, only a small fraction of acid

or water used in reaction The constant of acid and water hardly alter during the course of

the reaction So, the reaction appear to be zero order with respect to the acid

and water and the rate of hydrolysis is depend only on the concentration of ester

Rate of hydrolysis = k [ ester] Thus, acid hydrolysis of ester is first order The order of reaction which sometimes altered by the conditions

20

O

[H+]

A second order reaction is a reaction in which the reaction rate is proportional to the product of the concentrations two reactants

The rate equation for a second order reaction is, rate = k [A] [B] or rate= k [A]2

where, k = rate constant [A] = concentration in mol dm-3

the unit of k is dm3 mol-1 time-1 Examples of second order reactions:

i. The hydrolysis of iodomethane is second order

i. The thermal decomposition of hydrogen iodide is second

order

][][

)()()()(

3

33

NaOHICHkhydrolysisofrate

aqNaIliqOHCHaqNaOHliqICH

21

2

22

][

)()()(2

HIkiondecompositofrate

gasIgasHgasHI

Rate of reaction = - (rate of disappearance of A) = k [A]2

ot

ot

o

t

Akt

A

Akt

A

ACt

cktA

dtkA

Ad

kdtA

Ad

Akdt

Ad

11

11

1,0

][

1

][

2

2

2

22

A straight line graph with a + ve slope

1/[A]

t

y mx c

The rate of zero order reaction does not depend on the concentration of the reactants

For zero order reaction, the rate law is rate = k [A]0 = k where, k = rate constant [A] = concentration in mol dm-3

the unit of k is mol dm-3 time-1

Examples of zero order reactions:

i. Reaction between iodine and propanone

The reaction is slow, even at high temperature. However,

addition of dilute acid provides H+ ion which catalyze the reaction. The graph of the reaction rate against the iodine concentration is a straight line parallel to the horizontal axis. This means reaction rate remains constant even though the reactant, iodine is being used. Thus, the reaction is zero order with respect to iodine

)()()()( 2323 aqHIaqICOCHCHaqIaqCOOHCH

23

o

o

o

o

AktA

AC

AAotat

CktA

dtkAd

kdtAd

kdtAd

kAkdt

Ad

,

][

24

[A]

t

A straight line graph with -ve slope

y c mx

t

t

The half-life, t1/2, of a reaction is the time required for the concentration of a reactant to decrease to half-life of its initial concentration

i. For the first order reaction the half life is independent of the initial concentration

Thus first half-life (t 1/2) = Second half-life (t1/2)

2

1

2

1

693.0

2

tk

ort

nlk

25

[A], concentration

Times (minutes)

12

10

8

6

4

2

t1/2 t1/2 t1/2

0 2 4 6

First order reaction

3

1.5

ii. For the second order reaction, the half life is inversely proportional to the initial concentration. Thus, second half life (t1/2) =2t1/2

2

1

10 tA

k

26

t1/2

Second order reaction

[A]

times 2t1/2

5

2.5

10

iii. For the zero order reaction, a sloping straight line is obtained

2

0

12

][

t

Ak

27

Zero order reaction

[A]

times

nA

A

r

r

2

1

2

1

28

Where: r1 and r2 = initial rates of experiment 1 and 2 respectively A1 ,A2 , B1 & B2 = initial concentration of A and B for experiment 1 and 2 respectively m = order of reaction with respect to A n = order of reaction with respect to B

r1 = k [A]

1 [B]

1

r2 = k [A]

2 [B]

2

m n

m n

m

Example;

Reaction A + 2B C have been studying at 25C and the result are shown below.

Determine the a) rate law or rate of equation b) order of reaction c) rate constant, k

d) write the rate equation for the reaction

Exp [A]

(mol dm-3)

[B]

(mol dm-3)

Initial reaction rate

(mol dm-3s-1)

1 0.1 0.1 5.5 x 10-6

2 0.2 0.1 2.2 x 10-5

3 0.4 0.1 8.8 x 10-5

4 0.1 0.3 1.65 x 10-5

5 0.1 0.6 3.3 x 10-5

29

30

To find order of reaction for A consider exp. 1 and 2, where [B] is keep constant that is [B] = 0.1

So, we have r1=5.5 x 10-6, r2 = 2.2 x 10-5 A1=0.1 A2=0.2

To find order of reaction for B consider exp. 4 and 5, where [A] is keep constant that is [A] = 0.1

So, we have r1=1.65 x 10-5, r2 = 3.3 x 10-5 B1=0.3 B2=0.6

A respect to with

order second isreaction theThus,

2

5.025.0

5.025.0

1.0

1.0

2.0

1.0

102.2

105.55

6

2

1

2

1

2

1

m

glmgl

x

x

B

B

A

A

r

r

m

nm

nm

a) So, rate of reaction or rate law = k [A]m[B]n= k [A]2[B]1 b) And order of reaction = m+ n = 2 +1 =3

B respect to with

orderfirst isreaction theThus,

1

5.05.0

5.05.0

6.0

3.0

1.0

1.0

103.3

1065.15

5

2

1

2

1

2

1

n

glngl

x

x

B

B

A

A

r

r

n

nm

nm

1623

323

136

2

105.5

1.01.0

105.5

][][

][][

2

sdmmolx

moldmmoldm

smoldmx

BA

reactionofratek

BAkreactionofrate

31

c) Rate constant, k - can use exp 1, 2, 3,4 & 5 for calculation because the value of k is same for all experiments

d) Write the rate equation Rate = k [A]2[B] = 5.5 x 10-3[A]2[B]

1. Titration method 2. Colour changes by visual inspection or by using colorimeter 3. Gas syringe for measuring the volume of gas released 4. Pressure changes 5. Conductivity changes for reactions in which the number of

ions in solution changes 6. Monitoring the rotation of plane polarised light

Example: Titration method

H2O2 + 2H+ + 2I- I2 + 2H2O Measuring the concentration of I2 using standard S2O3-

(thiosulphate). At measured time interval, samples of the reaction mixture are removed using pipette and titrated with S2O3-

32

Is a series of simple steps that lead from the initial reactants to the final products of a reaction

An elementary reaction represent, at the molecular level, a single stage in the progress of the overall reaction

One of the main reasons for determining the order of a reaction is to see whether the experimentally determined overall order sheds any light on the detailed mechanism by which the reaction occurs

By knowing the reaction mechanism, a chemist can more effectively control the reaction or predict new reaction

Two requirements must be met: 1) The mechanism must account for the experimentally determined

rate 2) The mechanism must be consistent with the stoichiometry of the

overall/net reaction

33

The exponent in the rate law for an elementary reaction are the same as the stoichiometric coefficients in the chemical equation for the reaction (this is not usually the case with the rate law of the overall reaction)

Reversible reaction One elementary reaction may be much slower than all the others

the rate determining step (1) A mechanism with a slow step followed by a fast step. Example: decomposition of hydrogen peroxide, H2O2

2H2O2 (aq) 2H20 (liq) + O2 (gas) In the decomposition of H202, the facts (determined by the

experiment) are: i. The rate of decomposition of H2O2 is first order in both H2O2

and I-(second order overall) ii. The reactant I- (iodide) is unchanged during the reaction (acts as

a catalyst) and therefore does not appear in the equation for the net reaction.(overall becomes first order)

34

I-

The mechanism suggested, Slow step: H2O2 + I

- H20 + OI-

Fast step: H2O2 + OI- H2O + O2 + I

-

overall/net: 2H2O2 2H20 + O2 I- = is a catalyst (recovered at the end of the reaction) OI- = is an intermediate The slow step determines the rate of the overall rate So, the slow step is the rate determining step We set the reaction rate to that of the slow step; rate = rate of the slow step = k [H2O2][I-] rate = k [H2O2] where k = k [I-] = constant (the [I-] remains constant through out the reaction)

35

(2) A mechanism with a fast reversible step followed by a slow step The smog forming reaction: 2NO (gas) + O2 (gas) 2NO2 (gas) The rate law investigated experimentally: rate = k [NO]2 [O2] Suggested mechanism: fast step : slow step : N2O2 + O2 2NO2 overall : 2NO + O2 2NO2 The rate equation based on the rate determining step (slow step) rate = k2 [N2O2] [O2] We can state the experimentally determined rate law only in terms of

substances in the net equation. So, we must cancell out/ eliminate N2O2

2NOk1 N2O2k -1

36

k2

Fast step:

Forward rate = reverse rate equilibrium Rate formation of N2O2 = rate of disappearance of N2O2 k1 [NO] 2 = k-1[N2O2] [N2O2] = k1 [NO]2 k-1 Substitute [N2O2] in the rate- law for the rate-determining

step rate = k2 [N2O2] [O2] = k2 ( k1 [NO]2 ) [O2] k-1 = k2k1 [NO]2 [O2] k-1 rate = k [NO]2[O2] So, the proposed mechanism is consistent with the observed

rate law

2NOk1 N2O2k -1

37

Example of Question; For the reaction: H2 (gas) + I2 (gas) 2HI (gas) A proposed mechanism is fast step : slow step : 2I- + H2 2HI What is a) The net equation based on this mechanism b) The order of reaction according to this mechanism Answer: a) Net equation: I2 + H2 2HI b) Rate law for the rate determining step: rate = k2 [I-]2 [H2] 1 to eliminate I-, we assume rapid equilibrium in the first step rate forward = rate of the reverse reaction k1 [I2] = k-1[I-]2 [I-]2 = k1 [I2] k-1 substitute in equation 1 rate = k2k1 [I2] [H2] = k [I2] [H2] k-1

I2k1

2I-

k-1

38

k2

Arrhenius conducted research on the effects of temperature on reaction rates

For most reactions, the increase in rate with increasing temperature is not linear but exponential

Arrhenius equation is useful in analysing the effect of temperature on the rate constant and

the reaction rate it allow us to determine the activation energy, Ea, if the value of

the rate constants are known at different temperatures

ntconstaratek

(K)etemperaturT

)Jmol(8.31ntconstagasR

energyactivationEa

collision)of(frequencyntconstaArrheniusA:where

Aek

11

RT

Ea

K

39

2.303RT

EaAglkgl

RT

EaAlnknl

Aek RTEa

40

k

T (k)

lg k

1/ T

Gradient = -Ea / 2.303 R

At the temperature high, the rate constant k also high and therefore the reaction rate also increases

Example: in the presence of platinum as a catalyst, hydrogen iodide

decomposes to form hydrogen and iodine. The activation energy for this reaction is 58 KJmol-1. Calculate the ratio of the rate constant at 30C and 20C. Comment on your answer.

Answer: Let the rate constants at 30C and 20C be represented by k1

and k2, respectively

2.20.1

2.2

1052.4

109.9

11

11

2

1

2027331.8

100058

3027331.8

100058

2

1

11

1

11

1

k

k

k

k

KKJmol

Jmol

KKJmol

Jmol

e

e

41

The calculation shows that for an increase of 10oC, the initial reaction rate increases by approximately twofold

42

Exercise:

Calculate activation energy, Ea for the reaction:

2HI H2 + I2 from the k at (i) 250C = 2.5 x 10-5s-1

(ii) 550C = 1.0 x 10-3 s-1

R = 8.314 Jmol-1K-1

Answer: 9.99 x 104 J/mol

43

ln k1 = ln A1 - Ea 1 - 1

k2 A2 8.314 T1 T2

ln 2.5 x 10-5 s-1 = ln A1 - Ea 1 - 1

1.0 x 10-3 s-1 A2 8.314 298 328

-3.688 = -Ea (3.069 x 10-4)

8.314

Ea = - 3.688 x 8.314

- 3.069 x 10-4

= 9.99 x104 J/mol

44

Answer: 9.99 x 104 J/mol

45

A catalyst is a a substance that alters the rate of chemical reaction without itself being chemically changed at the end of the reaction

Catalyst (positive catalyst)- speed up the reaction Inhibitor (negative catalyst)- slow down the reaction Only a small amount of catalyst is needed to achieve a large/big

increase in reaction rate Catalyst lower the activation energy of the reaction

46

Ea catalysed uncatalysed

energy

Progress of reaction

Catalyst are often highly specific a catalyst for one reaction is not necessarily a catalyst for another reaction

Catalyst may be poisoned/ becomes ineffective by small amounts of substances such as lead, As or CN-

Homogenous catalyst is a catalyst that exists in the same phase as the reactants. e.g acid catalyst of ester

CH3COOCH3 + H2O+ CH3COOH + CH3OH

Heterogenous catalysis occurs when the catalyst and the reactants are in different phases. e.g. the transition metals, Al2O3, SiO3, V2O5

Autocatalysis occurs when one of the products of the reaction is a catalyst for the reaction

47

H+