ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) ERT 108/3 : PHYSICAL CHEMISTRY Chemical Kinetics By; Mrs Haf iza Bint i Shu kor

ERT 108/3 PHYSICAL CHEMISTRYSEM 2 (2010/2011)

ERT 108/3 : PHYSICAL CHEMISTRY

Chemical KineticsBy; Mrs Hafiza Binti Shukor


TOPIC TOPIC COVERED…; COVERED…;

Experimental Chemical and Kinetics Reactions

First Order Reactions

Second Order Reactions

Reaction Rates and Reaction Mechanisms

Light Spectroscopy and Adsorption Chemistry (Experimental methods for fast reactions).


CHEMICAL KINETICS??CHEMICAL KINETICS??

Also called reaction kinetics

Study of the rates & mechanisms of chemical reactions

2 types of reaction;

a)homogeneous – reaction occurs

in 1 phase (gas @liquid phase)

b)heterogeneous – reaction occurs

in 2 @ > phase


Experimental Chemical and Kinetics Reactions

Rates of chemical Reactions: the rate of speed with which

a reactant disappears or a product appears.

the rate at which the concentration of one of the reactants decreases or of one of the products increases with time.

mol L-1 s-1. 4


Example 1

• The decomposition of dinitrogen pentoxide (N2O5) in an inert solvent (carbon tetrachloride) at 450C:

• The data of the formation of O2(g) and the disappearance of N2O5 is shown in Table 1.

• The initial concentration [N2O5] = 1.40M.

What is the concentration, [N2O5] at time, t=423s?

5


Rate of Reaction: A variable quantity

• Rate of reaction is expressed as either:

or

6

t

treacrateaction

tan

Re

t

productrateaction

Re

[ Negative value ]

[ Positive value ]

Disappearance of reactant

Formation of products


Answer (Example 1)• At t=0, Initial [N2O5] = 1.40M

• At t = ∞, Final [N2O5] = 0M [decomposes completely]

• 5.93cm3 O2(g) is obtained at STP.

• After 423s, the volume of O2 (g) collected is 1.32cm3 of a possible 5.93cm3.

• The fraction of the N2O5 decomposed is 1.32/5.93.

• The decrease in concentration of N2O5 at this point

= (1.32/5.93) x 1.40M = 0.312 M.• After 423s, [N2O5] remaining undecomposed

= 1.40-0.31 = 1.09M.7


• From the figure , determine the rate of decomposition of N2O5 at 1900s.

• What is the initial rate of reaction?

8

Example 2:

Note: the rate of reaction can be expressed as the slope of a tangent line.


Answer (Example 2)

9

• Based on the graph of concentration of reactant vs time,

the slope of a tangent line at t=1900s,

114

52

106.2

800

/21.0

tan

Re

SLmolx

s

Lmolt

ON

gentofslope

rateaction


10

• The initial rate =

= 8.0 x 10-4 mol N2O5 L-1 s-1

s

LONmol

200

/40.124.1 52


The Rate Law for Chemical Reactions

• The rate law or rate equation – mathematical equation.

Reaction rate, r = k[A]m[B]n ….. The rate, r at time t is experimentally found

to be related to the concentrations of species present at that time, t .

The exponents in the rate reaction are called the order of the reaction.

The term k in the equation is called the rate constant.

it is a proportionality constant that is characteristic of the particular reaction & is significantly dependent only on temperature.

11

............ hHgGbBaA


Method of Initial Rates

• This simple method of establishing the exponents in a rate equation involves measuring the initial rate of reaction for different sets of initial concentration.

12


Example 3

• The data of three reactions involving S2O8

2- and I- were given in the below table.

(i) Use the data to establish the order of reaction with respect to S2O8

2-, the order with respect to I- & the overall order.

13

aqIaqSOaqIaqOS 324

282 )(2)(3)(


Cont…Example 3

(ii) Determine the value of k for the above reaction.

(iii) What is the initial rate of disappearance of S2O8

2- reaction in which the initial concentrations are [S2O8

2- ] =0.050M & [I-]=0.025M?

(iv) What is the rate of formation of SO4

2- in Experiment 1?

14


Answer (Example 3)

15

(i) In the experiments 1 & 2, [I-] is held constant & [S2O82-] is increased

by a factor of 2, from 0.038 to 0.076M. The reaction rate, R increased by a factor of 2 also. R2 = k (0.076)m(0.060)n = k (2x0.038)m(0.060)n

= k (2)m (0.038)m (0.060)n = 2.8 x 10-5 mol L-1 s-1

R1 = k (0.038)m(0.060)n =1.4 x 10-5 mol L-1 s-1

If 2m =2, then m =1. The reaction is first order in S2O8

2-.

2104.1

108.22

060.0038.0

060.0038.0)2(5

5

1

2

x

x

k

k

R

R mnm

nmm

nm IOSkR 82


16

R2 = k (0.076)m(0.060)n = k (0.076)m(2x0.030)n

= k (0.076)m (2)n(0.030)n = 2.8 x 10-5 mol L-1 s-1

R3 = k (0.076)m(0.030)n =1.4 x 10-5 mol L-1 s-1

If 2n =2, then n =1. The reaction is first order in I-.

The overall order of the reaction is m + n = 1+1 = 2 (second order).

2104.1

108.22

030.0076.0

030.02076.05

5

3

2

x

x

k

k

R

R nnm

nnm


TRT401 Physical Chemistry BBLee@UniMAP 17

(ii) Use any one of the three experiments:

k = 6.1 x 10-3 L mol-1 s-1.

(iii) Once the k value is determined, the rate law can be used to predict the rate of reaction.

Reaction rate, R = 6.1 x 10-3 L mol-1 s-1 x 0.050 mol L-1 x 0.025 mol L-1 = 7.6 x 10-6 mol L-1 s-1.

11

115

282

1

060.0038.0

104.1

LmolxLmol

sLmolx

IOS

Rk

IOSkR 82


18

• Based on the stoichiometry, 2 moles of SO42- are produced

for every mole of S2O82- consumed.

= 2.8 x 10-5 mol SO42-(L-1 s-1).

282

24112

825112

4 1

2104.1)(.

OmolS

SOmolxsLOSmolxsLSOmolNo


Zero-order, First-order, Second-order Reactions

19


Cont………

20


Zero-order, First-order,

Second-order

Reactions

21

Zero order

First order

Second order


Example 4

(a) When [N2O5] =0.44M, the rate of decomposition of N2O5 is 2.6 x 10-4 mol L-1 s-1.

what is the value of k for this first-order reaction?

(b) N2O5 initially at a concentration of 1.0 mol/L in CCl4, is allowed to decompose at 450C. At what time will [N2O5] be reduced to 0.50M?

22


Answer (Example 4)

23

(a) Rate of disappearance of N2O5 (R) = k [N2O5]

= 5.9 x 10-4 s-1

(b) For 1st order of reaction, to determine t, we can use:

log [A]0 = log [N2O5]0 = log 1.0 = 0.

log [A]t = log [N2O5]t = log 0.50 = -0.30.

use, k = 5.9 x 10-4 s-1.

Lmol

sLmolx

ON

Rk

/44.0

106.2 114

52

0log303.2

]log[ Atk

A t

0303.2

109.530.0

14

tsx

sxsx

xt 3

14101.1

109.5

30.0303.2

0ln]ln[ AktA


Example 5

• The data of the above table were obtained for the decomposition reaction: A → 2B + C.

(a)Establish the order of the reaction.

(b)What is the rate constant, k?

24

Time, min [A], M log [A] 1/[A]

0 1.00 0.00 1.00

5 0.63 -0.20 1.59

10 0.46 -0.34 2.17

15 0.36 -0.44 2.78

25 0.25 -0.60 4.00


Answer (Example 5)

25

(a) Plot graph based on the data given in the Table.

(b) The slope of the 3rd graph:

Not Straight line – Not Zero order

Not Straight line – Not First order

Straight line – 2nd order

11 min12.0min25

/00.100.4

molLmolL

k


Reaction rates: Effect of temperature

• Chemical reactions tend to go faster at higher temperature.

slow down some reactions by lowering the temperature.

• Increasing the temperature increases the fraction of the molecules that have energies in excess of the activation energy.

this factor is so important that for many chemical reactions it can lead to a doubling or tripling of the reaction rate for a temperature increase of only 100C. 26


Cont….• In 1889, Arrhenius noted that the k data for many

reactions fit the equation:

where, A & Ea are constants characteristics of the reaction

R = the gas constant.Ea = the Arrhenius activation energy (kJ/mol or

kcal/mol)A = the pre-exponential factor (Arrhenius factor). ( the unit of A is the same as those of k.)• Taking log of the above equation:

27

RTEaAek

ART

Ek a lnln A

RT

Ek a

1010 log303.2

log


• If the Arrhenius equation is obeyed: a plot of log10 k versus 1/T is a straight

line with slope: -Ea/2.303 R and intercept log10 A.

This enables Ea and A to be found.

• Another useful equation:

(eliminate the constant A). T2 and T1 - two kelvin temperatures.

k2 and k1 - the rate constants at these temperatures.

Ea – the activation energy (J/mol) R – the gas constant (8.314 Jmol-1 K-1). 28

12

12

303.2log

1

2

TT

TT

R

Eakk

Cont….


Example 6

(a) Use the figure given to find A and Ea for:

(b) Calculate Ea for a reaction where rate constant at room temperature is doubled by a 10Kelvin increase in T.

29

2252 42 ONOON

Figure: Rate constant versus temperature for the gas-phase first order decomposition reaction


Answer (Example 6a)

30

• Tabulate the data as follows.

• Construct the Arrhenius plot of log10k versus 1/T for the reaction.

Intercept (log10A)=13.5

A = 3x1013s-1

Slope=-5500K,

Ea=25kcal/mol

=105 kJ/mol

Temp, 0C Temp, K 1/Temp, 1/K k, s-1 log10 k

25 298 0.0034 0.001 -3

R

Ea

303.25500

Figure: Arrhenius plot of log10 k versus 1/T for this reaction. Note: the long extrapolation needed to find A.


Answer (Example 6b)

31

• Based on the given info: k2 = 2k1 , T1 = room temperature (298K), T2=298+10 = 308K,• The Arrhenius equation:

• Substitute:

Ea = 53 kJ/mol

12

12

303.2log

1

2

TT

TT

R

Eakk

)298(308

298)308(

303.2log

1

12

R

Eakk


Reaction Mechanisms• Each molecular event that

significantly alters a molecule’s energy or geometry is called an elementary process (reaction).

• The mechanism of a reaction:the sequence of elementary

reactions that add up to give the overall reaction.

A mechanism is a hypothesis about the elementary steps through which chemical change occurs.

32


Reaction Mechanisms• Elementary processes in which a single molecule

dissociates (unimolecular) or two molecules collide (bimolecular) much more probable than a process requiring the simultaneous collision of three bodies (termolecular).

• All elementary processes are reversible and may reach a steady-state condition. In the steady state the rates of the forward & reverse processes become equal.

• One elementary process may occur much more slower than all the others. In this case, it determines the rate at which the overall reaction proceeds & is called the rate-determining/ limiting step.

33


The Hydrogen-Iodine Reaction H2 (g) + I2 (g) → 2HI (g)• Rate of formation of HI = k [H2][I2]• The hydrogen-iodine reaction is

proposed to be a two-step mechanism [Sullivan J. (1967). J.Chem.Phys.46:73].

1st step: iodine molecules are believed to dissociate into iodine atoms.

2nd step: simultaneous collision of two iodine atoms and a hydrogen molecule.

(this termolecular step is expected to occur much more slowly – the rate-determining step).

34


The Hydrogen-Iodine Reaction1st step: [Fast]

2nd step: [Slow]

Net:

• If the reversible step reaches a steady state condition: rate of disappearance of I2 = rate of formation of I2

35

)(22

1

2 gIgIk

k

)(2)(2 32 gHIgHgI k

gHIgHgI 2)()( 22

2221 ][][ IkIk

22

12 Ik

kI


The Hydrogen-Iodine Reaction• For the rate-determining

step: Rate of formation of HI = k3 [I]2[H2] where

= K[H2][I2] where (K=k1k3/k2)

36

2232

1 IHkk

k

)(2)(2 32 gHIgHgI k

2221 ][][ IkIk


Example 7

• The thermal decomposition of ozone to oxygen: 2O3 (g) → 3O2 (g)

• The observed rate law: Rate of disappearance of O3 =

• Show that the following mechanism is consistent with this experiment rate law.

1st:

2nd:

37

2

23

O

Ok

OOOk

k

23

2

1

2k

3 2O O O 3


Answer (Example 7)

38

• Assume the 1st step reaches the steady state condition: Rate of formation of O = Rate of disappearance of O k1 [O3] = k2 [O2] [O]

• Assume the 2nd step is the rate-determining step: Rate of disappearance of O3 = k3 [O][O3]

(where k = k1k3/k2)

22

31

Ok

OkO

2

23

22

3331

O

Ok

Ok

OOkk


Experimental methods for fast reactions• Many reactions are too fast to follow by the classical

methods.• Several ways to study fast reactions :

39

1. Rapid flow methods: (i) Continuous flow (ii) Stopped flow

2. Relaxation methods: (i) Temperature jump (T-jump) method

(ii) Pressure jump method(iii) Electric field jump method

3. Flash photolysis

4. Shock tube

5. Nuclear-magnetic-resonance (NMR) spectroscopy


ASSIGNMENT 1

Write a short note for the following fast reaction:a) Rapid flow methodsb) Relaxation methodsc) Flash photolysisd) Shock tubee) Nuclear-magnetic-resonance (NMR)

spectroscopy

40


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Documents

ERT 108/3 PHYSICAL CHEMISTRY SEM 2 (2010/2011) ERT 108/3 : PHYSICAL CHEMISTRY Chemical Kinetics By; Mrs Haf iza Bint i Shu kor