reaction kinetics and drug stability reaction...

11
Reaction Mechanisms Simple complex Reaction Kinetics and Drug Stability Parallel Reversible Consecutive K f A B K r K 1 K 2 A N P A B C K 1 K 2

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Reaction Mechanisms

Simple complex

Reaction Kinetics and Drug Stability

Parallel Reversible Consecutive

Kf

A B Kr

K1

K2

A

N P

A B C

K1 K2

Reaction Kinetics and Drug Stability

reactions Kf

A B

Kr

Reversible

Time

C (

mg/

ml)

Reactant

Product

Reversible

Reaction Kinetics and Drug Stability

reactions Kf

A B

Kr

BkAkdt

dArf Integration tkk

AkAkk

Akrf

rrf

f)(

)(ln

0

0

This equation may be simplified by introducing the equilibrium condition: Kf Aeq = kr Beq in which Ao – Aeq = Beq

tkkAA

AArf

eq

eq)(ln

0

t

kk

AA

AA rf

eq

eq

303.2log

0

OR

Reversible

Reversible

Reaction Kinetics and Drug Stability

reactions This equation may be simplified by introducing the equilibrium condition: Kf Aeq = kr Beq in which Ao – Aeq = Beq

tkkAA

AArf

eq

eq)(ln

0

t

kk

AA

AA rf

eq

eq

303.2log

0

OR

eq

eq

AA

AA

0log

t

303.2

rf kk Slope =

kf and kr can be calculated

Slope x 2.303 = kf + kr

equilibrium constant of the reaction:

eq

eq

r

f

A

B

k

kK

Tetracycline undergoes epimerization. At the equilibrium

point a distribution consisted of 32% of tetracycline and

68% of epi-tetracycline was obtained. Measuring the

concentration of tetracycline at different time interval and

plotting the data according to the following equation lead

to a straight line the slope of which equals 0.01 min-1,

calculate both Kf and kr

Reversible

Reaction Kinetics and Drug Stability

reactions

Reversible

Reaction Kinetics and Drug Stability

reactions

kf = 2.1 kr

Slope = 0.01

0.01 x 2.303 = kf + kr

0.01 x 2.303 = 2.1 kr + kr = 3.1 kr

kr = 0.02303 = 0.00767 min -1

3.1

kf = 2.1 x 0.00767 = 0.06 min -1

1.232

68

eq

eq

r

f

A

B

k

kKt

kk

AA

AA rf

eq

eq

303.2log

0

Parallel

Reaction Kinetics and Drug Stability

reactions K1

K2

A

N P

Co

nc

Time

N

P A

K1

K2

A

N

P

The corresponding rate equation is:

PkPkPkdt

dp 21

k = k1+ k2

P = Po e-kt

Rate of formation of the acidic

product

Pkdt

dA1

kt

o ePkdt

dA 1

)1(10

kt

o ePk

kAA

Neutral product equation )1(2 kt

o ePk

kN

`

ln P = ln Po – kt

)1(1 kt

o ePk

kA

Parallel

Reaction Kinetics and Drug Stability

reactions

Parallel

Reaction Kinetics and Drug Stability

reactions

t

ln p

Slope =- k

(1 - e-kt)

A

oPk

kSlope 1

oP

kxSlopek 1

k2 = K – K1

ln P = ln Po – kt )1(1 kt

o ePk

kA

Consecutive

Reaction Kinetics and Drug Stability

reactions

Rate of decomposition of A

Akdt

dA1

Rate of change in conc. of B

BkAkdt

dB21

Rate of formation of C

Bkdt

dC2

tkeAA 1

0

)( 21

12

10 tktkee

kk

kAB

)(

11 21

12

21

0

tktkekek

kkAC

A B C

K2 K1

Consecutive

Reaction Kinetics and Drug Stability

reactions

First Step is Rate-Limiting Second Step is Rate-Limiting

A B C

K1 K2