2. Basicsβ¦.
1) Reaction rate How we measure rate
2) Rate laws How the rate depend on amount of factors
3) Half life (π‘1/2) How long it takes to react 50% of its reactants.
4) Arrhenius equation How rate constants changes with T.
5) Mechanisms Link between rate and molecular scale
processes
3. Basicsβ¦..
β’ Molarity refers to the concentration of a compound or ion in a
solution, normality refers to the molar concentration only of the
acid component or only of the base component of the solution.
M= Moles/Lit of solvent used in solution
N= Eq/V
Eq- No. of gm equivalents of solute
V- Volume of solvent in Lit.
β’ The relation between normality and molarity is N = M x n where
N refers to normality, M is molarity, and n denotes the number of
equivalents.
4. Need of Chemical Kinetics
β’ To study, understand and interpret conditions of instability of
pβceutical product as well as to be able to offer
β’ Recognize alterations in stability may occur when a drug is combined
with other ingredients.
β’ Knowing the rate at which a drug deteriorates at various hydrogen ion
concentrations allows one to choose a vehicle that will retard or
prevent the degradation.
β’ The pharmacist is able to assist the physician & patient regarding
storage & use of medicinal agents.
β’ Molecularity of reactions
Unimolecular, Bimolecular & Termolecular reactions
5. Rates, Order and Molecularity of Reactions
Law of Mass Action:
The rate of chemical reaction is proportional to the product of the
molar concentration of the of the reactants each raised to a power
usually equal to the number of molecules, a and b, respectively,
a A + b B+β¦β¦= Products (1)
Rate/velocity/Speed of reaction
dc/dt
6. APPLICATIONS
1) Drug stability: help to predict shelf life of drug.
2) Dissolution: Drug is expected to release from solid dosage form and
immediately get into molecular solutions.
3) Drug release: prodrugs (agents who donβt have therapeutic activity
but converted back in vivo to their parent compounds)
Eg. Levodopa is prodrug of Dopamine.
4) Pharmacokinetics: ADME mechanism
5) Drug action: Interaction of drugs with bio membranes or receptors
7. Order of reaction
β’ Zero order reaction
β
ππ
ππ‘
= ππ
Ko= Specific rate constant
m+n=0
Eg.: 1) Colour loss of liquid multisulphonamide preparation.
2) Oxidation of vitamin A in an oily solution.
3)Photochemical degradation of chlorpromazine in aqueous solution.
8. Derivation
β
ππ΄
ππ‘
= ππ
A=absorbance of preparation
- ve sign= colour is fading
Integrating above equation between initial absorbance ,Ao at t=0 time
& absorbance, π΄π‘ at t=t gives
π΄π
π΄π‘
ππ΄ = βππ
0
π‘
ππ‘ β ππ
0
π‘
ππ‘
π΄π‘βπ΄0= -k0π‘
Or π0= π΄0- π΄π‘/t
An integral equation for zero order reaction. Also permits us to
calculate the con. Of drug remain undecomposed after time t.
9. Type equation here. Initial con. = βaβ & con. at any time t, is βcβ. Then
equation becomes
ππ = (π β π)/π‘
It may be written as π = π β πππ‘
Half life: π =
π
2
, π‘ = π‘1/2
10. Substituting values in rearranged equation
π‘1
2
=
π β π
ππ
= π β
π
2
ππ
=
1
2
ππ
=
π
2ππ
Unit is time scale, i.e., sec/con, min/con, h/con
According to equation half life period od zero order is directly
proportional to the initial concentration of reactant.
SHELF LIFE : π =
90π
100
t=π‘90
Substituting values in rearranged equation
π‘90 =
π β 0.9
ππ
=
0.1π
ππ
11. APPARENT ZERO ORDER - SUSPENSIONS
ο It may be first order but
behaves like a zero order,
depending on the
experimental conditions.
ο Degradation is possible only
when drug is available in
solution form.
ο As the drug in solutions starts
to degrade suspended
particles act as reservoir &
continuously release drug in
solution. Thus con. Remains
constant throughout the
process.
ο In case of no reservoir; β
π π΄
ππ‘
=
π1[π΄]
]
12. FIRST ORDER REACTION
β’ β
ππ
ππ‘
β π or β
ππ
ππ‘
= π1π
Derivation
A Products
According to definition of 1st order,
β
ππ
ππ‘
= π1π
π0
π1
ππ
π
= βπ1
π
π‘
ππ‘
[πΌπ π]ππ
ππ‘
=-k1[π‘]0
π‘
13. πΌπ π1 β πΌπ π0 = βπ1(π‘ β π)
πΌπ ππ‘ = πΌπ ππ β π1π‘
Converting equation to the base 10
log π1 = log π0 β
π1π‘
2.303
Rearranging gives,
π1 =
2.303
π‘
log
ππ
ππ‘
K1 value explains the fraction of the reactant consumed per unit time
15. Substituting in rearranged equation
π‘90 =
2.303
π1
log
ππ
0.9π0
π‘90 =
2.303
π1
log
10
9
= 2.303 β
0.04575
π1
=
0.105
π1
Pseudo first Order Reaction
β
ππ
ππ‘
= π2[π΄][π΅]
A& B= reactants, K2= 2nd order rate constant
Reaction conditions are maintained such that B present in excess
amount compared to A. Thatβs why equation changes to
β
ππ
ππ‘
= π2 π΄ ππππ π‘πππ‘ = π1[π΄]
16. SECOND ORDER REACTION
A+B=Products
Rate equation can be written as:
β
ππ΄
ππ‘
= β
ππ΅
ππ‘
= π2[π΄]1[π΅]1
[A] & [B]= concentration of A & B respectively
K2= Specific rate constant for 2nd order.
i.e. Rate of reaction is 1st order W.R.T. A & again 1st order W.R.T B.
Therefore, m+n=2
For identification/evidence purpose; Plot graph of log con. Vs. time.
Straight line indicating 1st order w.r.t A & again 1st order w.r.t B.
17. DERIVATION
β
ππ΄
ππ‘
= β
ππ΅
ππ‘
= π2[π΄][π΅]
Let βaβ & βbβ be the initial con. Of A&B, and βxβ be the con. Of each
species reacting in time t. On substituting these gives,
ππ₯
ππ‘
= π2 π β π₯ (π β π₯)
By considering case in which a=b i.e. (both having same con.) then the
equation will be
ππ₯
ππ‘
= π2(π β π₯)2
Integrating above equation on employing the conditions x=0 at t=0 &
x=x at t=t, gives
0
π₯
ππ₯/(π β π₯)2 = π2
0
π‘
ππ‘
20. Determination Of Order
1) Graphic method :
οMore reliable coz deviation from best fit line can be easily observed.
οA straight line that gives better fit is identified, & reaction is
considered to be that order.
2) Substitution method:
οFrom kinetic expt. Data are collected on time course of change in con.
Of reactants.
οData are substituted in the integral equations of zero, first, & second
order reactions to get k values.
οZero order: ππ =
π΄πβπ΄π‘
π‘
οFirst order: π1 =
2.303
π‘
log
π0
ππ‘
22. 3) Half life method
Average k value is calculated using the data for zero, first and second
orders as given in substitution method or graphic method. Followed by
it half life are calculated for each time period in kinetic study.
Zero order:
π‘1
2
=
π
2π0
First order:
π‘1
2
=
0.693
π1
Second order:
π‘1
2
=
1
ππ2
(where a=b)
From above equations we may consider that