Acid Base Catalysis : Kinetic Studies - PowerPoint PPT Presentation

1 / 40
About This Presentation
Title:

Acid Base Catalysis : Kinetic Studies

Description:

Reaction stops in certain organic solvents (chloroform, pyridine, m-cresol) ... 4. Neither pyridine (basic) nor cresol or phenol (acidic) alone was very effective ... – PowerPoint PPT presentation

Number of Views:3526
Avg rating:3.0/5.0
Slides: 41
Provided by: tkvarad3
Category:

less

Transcript and Presenter's Notes

Title: Acid Base Catalysis : Kinetic Studies


1
Acid Base Catalysis Kinetic Studies
2
CHEMICAL KINETICS
  • What is Chemical Kinetics ?
  • In what way it is different from
    Chemical
  • Thermodynamics ?
  • Consider the reaction 2 H2 O2 2
    H2O Keq 10 41
  • (i) at room temperature t1/2
    gt 10 25 y or 10 32 s
  • (ii) in the presence of spark
    t1/2 lt 10 - 6 s
  • Reason ?

3
Do You Know ?
  • What is a catalyst ?
  • Who coined the word catalysis ?
  • How old is this concept ?
  • What is specific acid catalysis ?
  • What is specific base catalysis ?
  • What is general acid catalysis ?
  • What is general base catalysis ?
  • What are electrophilic and nucleophilic catalysis
    ?
  • What is the simple mechanism of acid base
    catalysis ?
  • What are protolytic and prototropic mechanisms ?
  • What are Arrhenius and vant Hoff intermediates ?
  • When do you apply steady state approximations ?
  • How to evaluate the various rate constants
    involved in a reaction ?
  • What is the importance of salt effect in acid
    base catalysis ?
  • What is the limitation of pH scale ?
  • How to express the acid strength of concentrated
    solutions ?

4
History of catalyst
  • Catalysts have been used by mankind for over 2000
    years
  • The term "catalysis" was coined in 1835 by the
    chemist J.J. Berzelius (1779-1848)
  • The first Catalysis Nobel Prize in Chemistry 1909
    - Wilhelm Ostwald
  • To day,catalysts play a major economic role in
    the world market

5
What is a Catalyst ?
A catalyst is a substance that accelerates the
rate of a chemical reaction without being part of
its final products A catalyst is a substance
that increases the rate of a reaction
without affecting the position of equilibrium A
catalyst is a substance that increases the
reaction other than by the medium effect A
catalyst is a substance that makes available a
reaction path with a lower free energy of
activation A catalyst is a substance that in
some manner increases the probability of
reaction A catalyst is a substance that in some
manner assist in achieving the necessary
orientation for a reaction
6
Catalysed reactions where E is constant and A
varies
7
Relative Velocity
E kcal
Catalyst
Decomposition of formic acid vapour at 200 0C
Glass surface
0.05
24.5
Gold surface
23.5
2
Platinum surface
100
22.0
Rhodium surface
500
25.0
Hydrogenation of ethylene
Tungsten surface
10.4
1
10
10.4
Iron surface
60
10.4
Nickel surface
200
10.4
Platinum surface
1,600
10.4
Palladium surface
10,000
10.4
Rhodium surface
8
Relative conductivities and Catalytic effects of
different acids
9
Acid Hydrolysis of an ester Inversion of
sucrose
  • Rate k HsubstrateH2O
  • k' substrate
  • k' k H
  • log10k' log10k log10H
  • log10k ? pH
  • How to follow the kinetics ?
  • How to evaluate Ea ?

10
Relation between rate constant and H ion
concentration
Skrabal diagram
11
k k0 kH H kOH OH- pseudo rate
constant
k k0 KH H KOH OH- k k0 kH H
KOH Kw /H dk/dH kH kOH Kw /H2 0
at minimum k Hmin ( kOH Kw / kH )1/2 pHmin
½ pKw log ( kH / kOH)
12
Types of Catalysis
  • Specific Acid Catalysis (H3O)
  • Specific Base Catalysis (OH-)
  • General Acid Catalysis
  • uncharged acids (CH3COOH)
  • cationic acids (NH4)
  • 4. General Base Catalysis
  • uncharged bases (NH3)
  • anionic bases (CH3COO-)
  • 5. Electrophilic Catalysis (Lewis acids)
  • Nucleophilic Catalysis (Lewis bases)

13
CONJUGATE PAIRS
H3O A-
HA H2O
H3O CH3COO-
CH3COOH H2O
BH OH-
B H2O
NH4 OH-
NH3 H2O
14
Mechanism
1
First stage BH S
B SH
2
3
Second stage SH H2O
P H3O (Protolytic)
4
P BH
(Prototropic)
(OR) SH B
  • Protolytic Mechanism
  • dP/dt k3SH
  • k3k1 BHS
  • --------------------------
  • k2B k3
  • If k3 gtgt k2B v k1 SBH general acid
    catalysis vant Hoff

  • k1k3BHS
  • (b) If k3 ltlt k2 B v ---------------------
    Arrhenius complex
  • k2B

15
Since for the equilibrium
B H, BH/B H / K BH
BH
k1k3 rate --------- SH -
specific acid catalysis
k2KBH Protoropic Mechanism
dP/dt k4 BSH
k1k4 ----------
SBH - General Catalysis
k2 k4
16
Summary of conclusions for various mechanisms of
Acid-Base catalysis
SH B
Proton transfer to solute
SH B
BH S
SH B
P BH
S- BH
P OH-
P B
17
Activation Energies for Catalysed Reactions
  • Arrhenius intermediate (k-1 gtgt k2), the rate
    controlling step is the crossing of
  • the second barrier
  • vant Hoff intermediate (k2 gtgt k-1), the crossing
    of the first barrier is the rate
  • controlling step

18
Examples
  • Specific Acid Catalysis
  • Inversion of sucrose
  • Hydrolysis of acetals
  • Decomposition of diazoacetic ester
  • Specific Base Catalysis
  • Claisen condensation
  • Michael condensation
  • Aldol condensation
  • Specific Acid and Base Catalysis
  • 1. Hydrolysis of ?-Lactones
  • 2. Hydrolysis of amides
  • 3. Hydrolysis of esters

19
  • General Acid Catalysis
  • Decomposition of acetaldehyde hydrate
  • Hydrolysis of ortho esters
  • Formation of nitro compounds
  • General Base Catalysis
  • Decomposition of nitramide
  • Bromination of nitromethane
  • General Acid and Base Catalysis
  • Mutarotation of glucose
  • Halognation of ketones

20
IODINATION OF ACETONE
CH3COCH3 I2
CH3COCH2I HI
Mechanism CH3COCH3 BH
OH CH3-C-CH3 B
1
-1
2
H3O
CH3-C(OH)
CH2
AH H2O
-2
3
CH3COCH2I
HI
E I2
dP/dt k3 E I2
k2 AH E --------------------
------ k-2H3O k3I2
k1 A BH k-2 E
H3O AH -----------------------------------
--------- k-1 B k2
21
k1 k2 A
BH dP/dt k3 I2 --------------------------
-------------------------------
k-1 k-2 B H3O k3 (k2 k-1 B)
I2
At high I2 k1k2 A
BH rate ------------------------
k2 k-1 B
  • When k2 gtgt k-1B rate k1 A BH ---- Gen.
    acid
  • (vant Hoff intermediate)
  • b) When k2 ltlt k-1 B k1k2
    A BH
  • (Arrhenius intermediate) rate
    -------------------- k A H

  • k-1 B

22


Observed Ea
1. Consider the rate law
rate k1 k2 / k -1 S
k obs k1 k2 / k -1

Ea obs Ea 1 Ea 2 - Ea
-1
Ea obs Ea2 ?H 2.
Consider the rate law
rate ( k 1 k 2 k 3 / k 4 ) ½ S

E obs ½ ( E1 E2 E 3 - E 4 )

23
Parallel Reactions
  • B ( E1 )
  • A
  • C ( E2 )
  • k1 E1 k2 E2
  • Eobs -----------------------
  • k1 k2

24
MUTAROTATION OF GLUCOSE
(a D Glucose ß D Glucose )
The a and ß forms interconvert over a timescale
of hours in aqueous solution, to a final stable
ratio of aß 3664, in a process called
mutarotation.
Mutarotation of Glucose
?-D-glucose open-chain form
?-D-glucose m.p. 146?C, ? 112? m.p. 150?C,
? 19?
25
  • Reaction proceeds both in water and in certain
    organic solvents
  • Reaction is accelerated by the addition of both
    acids and bases
  • Reaction stops in certain organic solvents
    (chloroform, pyridine, m-cresol)
  • in the absence of water
  • 4. Neither pyridine (basic) nor cresol or phenol
    (acidic) alone was very effective
  • 5. In a mixture of 1 part of pyridine and 2
    parts of m-cresol mutarotation proceeds
  • 20 times faster than in water

26
6. Thus, the simultaneous presence of an acid and
base is needed 7. Since water acts both as an
acceptor and donor of a proton, reaction
proceeds in water 8. 2-hydroxy pyridine is
a better catalyst than an equivalent mixture of
pyridine and phenol due to the simultaneous
presence of both acidic and basic group 9.
2-hydroxy pyridine is a weaker base than pyridine
and weaker acid than phenol 10. Mutarotation is
an example of general acid-general base catalysis
27
-------OH---------


H..O--------
Base
Acid
Base
Acid
28
Determination of rate constants
k k0 kH H kOH OH- kHA HA kA
A- k k' kHA HA kA A- at a constant
pH k k' kHA kA q HA, where q A- /
HA
Slope kHA kAq Intercept k
Primary Plot
29
Secondary Plot
Determination of kH and kOH The intercept, k
of the primary plot depends on the buffer ratio,
q k k0 kH H kOH OH- k0 kH (K
/ q) kOH (Kw q / K) Where, K H A- /
HA H q and OH- Kw/H By solving
the two equations kH and kOH can be obtained
30
Hydrolysis of Acetal ( HCOOH HCOONa ) 3.0
  • The first five results show that the rate
    constant increases with the increase in the
  • concentration of HCOOH suggesting the reaction
    is General Acid Catalysis
  • The last three results show that the rate
    constant is independent of HCOOH Con.
  • In the last three cases, the total ionic strength
    is kept constant
  • The above reaction is Specific Acid Catalysed
    reaction

31
PRIMARY SALT EFFECT
The reactants are A Co(NH3)5Br2 Hg2
(zAzB 4) B S2O82- I-
(zAzB 2) C CO(OC2H5)NNO2- OH-
(zAzB 1) D Cr(urea)6 3 H2O (open
circles) (zAzB 0) CH3COOC2H5 OH- (closed
circles) (zAzB 0) E H Br- H2O2
(zAzB -1) F Co(NH3)5Br2 OH-
(zAzB -2) G Fe 2 Co(C2O4)3-
(zAzB -6)
Plots of log10(k/k0) versus the square root of
the ionic strength for ionic reactions of
various types. The lines are drawn with slopes
equal to zAzB.
32
Secondary Salt Effect
Consider a
reaction catalysed by a weak acid, HA
HA
H A-
H A- ? H ? A- Ka
----------------------------
HA ? HA
HA ?HA
K H Ka ---------------------
--------------- A- ?H ?A-
?H ?A- log10H log10K
log10?H - log10?A- log10K
(-Avµ) (-Avµ)
  • But, -log10?i AZi2 vµ ------ Debye-Huckel
    limiting law
  • log10H log10K 2A vµ
  • H increases as a result of an increase in
    ionic strength, µ
  • Rate k H S increases.

33
Consider the buffer system, BH H2O
B H3O
(or) simlpy, BH
B H
B H ?B ?H K BH
---------------------------
BH ? BH
K BH BH ?BH ?BH
H ------------------------ K
------------ B ?B ?H
?B ?H As the salt concentration
increases-ionic strength increases, activity
coeff. decreases to the same extent in both the
numerator and denominator. Consider the system,
B H2O BH OH-
BH OH- ?BH ?OH-
KB -----------------------------
B ?B
KB B ?B K
OH- ---------------------------
-------------- (?B 1)
BH ?BH ?OH- ?BH ?OH- As
ionic strength increases OH- increases and
hence base catalysed reaction increases
34
HAMMETT ACIDITY FUNCTION
pH scale is valid only for dilute aqueous
solutions Hammett Acidity Function is used in
concentrated solutions Since most of the
organic bases are weak, high acid concentrations
are needed to catalyse the reaction Consider an
uncharged base, B in concentrated acid medium

B H
BH
B H ?B ?H aH ?B B
Ka --------------------
-------------------- BH ? BH
?BH BH B
a H ?B h0
------------- where, h0 -------------
BH ?BH
35
log10 KBH log10h0 log10 B / BH pK
BH H0 log10 B / BH H0 pK BH
log10 B / BH
  • H0 Property of the solvent independent of the
    base used if it is of similar
  • type
  • 2. (i) measurement of the ratio of B/BH
  • (ii) determination of pK BH
  • (i) use the base B in a dil. Solution measure
    pH ? H0 (dilute)
  • (ii) use in conc. medium measure B/BH
    and get pK BH


36
H function
H In OH - ? H2O
In - H In ? H
In - a H ? In
In- In -
K ---------------------- h ?
---------- ? H
In H In H In

a H ?In where h -
--------------------
? H In and
H - log 10 h ?
37
Numerical Problems
  • For a base B in 0.02M HClO4 ,the ratio of BH /
    B is 0.01. Calculate the value
  • of pKBH
  • 2. A base B is 8 protonated in a medium of 90
    H2SO4. If the pK BH of the
  • protonated base is -10 , calculate the H0
    of the medium.
  • A second order reaction was investigated using
    acetic acid sodium acetate
  • buffer ( KHA 1.80 x 10 -5 ) at a constant
    temp, and ionic strength. From the
  • available data, evaluate as many rate
    constants as possible. (k0 0 )
  • A - / HA
    1.0 5.0
  • Slope ( M-2 min-1 )
    25 x 10 4 38.89 x 10 4
  • Intercept ( M-1min-1 )
    0.92 x 10 4 2.06 x 10 - 4

38
SOLUTION
  • 1. H0 pH 1.70
  • H0 pK BH log10
    B / BH
  • 1.70 pK BH log10
    100
  • pK BH - 0.30
  • 2. H0 -10.00 log10 (
    92 / 8 )

  • -10.00 1.07 - 8.94
  • 3. k k0 ( kHA
    kA q )
  • slope kHA kA q
  • substitute the
    corresponding q and slope values and solve

39
Steady - state Approximation
Consider a reaction A
1
2
B
C
-1
  • dA / dt k1 A k-1 B
  • dB / dt k1 A k-1 B k2 B
  • dC /dt k2 B
  • If dB / dt is assumed to be zero, then
  • k1 A k-1 B k2 B
  • rate -dA /dt k2 B dC / dt
  • Also, A0 A B C
  • dA0 / dt dA /dt dB /dt dC / dt 0
  • If dB / dt 0, the dA / dt dC / dt

40
Actually , dB / dt ? 0 B
k1A / ( k-1 k2 ) dB / dt k1
/ ( k-1 k2 ) dA / dt If dB / dt
0 then dA / dt 0 Therefore no
reaction would occur The necessary
condition is B ltlt A C
A0 A B C A C
dA0 / dt 0 dA / dt dB / dt
dC / dt dA / dt
dC / dt Therefore dB / dt
0
Write a Comment
User Comments (0)
About PowerShow.com