reaction kinetics and drug stability the rate expression the rate of a chemical reaction is...

Reaction rate is the change in the concentration of a

reactant or a product with time.

A B

The Rate Expression The rate, velocity, or speed of a reaction

Reaction Kinetics and Drug Stability

rate = dB

dt

dA = decrease in concentration of A over

time period dt

dB = increasee in concentration of B over

time period dt

Because [A] decreases with time, dA is negative.

rate = - dA

dt

dt

Bd

dt

dAR f

This expression gives the increase (+) or

decrease (-) of concentration, C, within a given

time interval, dt.

A + B C + D

dt

dC

The Rate Expression




time interval, dt.

A + B C + D

dt

dC

The Rate Expression


dt

Bd

dt

Ad )(

2

1)(



time interval, dt.

A + 2B C + D

dt

dC

The Rate Expression


Reaction Rate and Stoichiometry



time interval, dt.

A + 2B C + D

dt

dC

The Rate Expression


The rate of a chemical reaction is proportional to

the product of the molar concentration of the

reactants each raised to a power equal to the

number of molecules of the substance

undergoing reaction:

The rate law


Effect of concentration on reaction rate

In which k is known as the reaction rate constant.

)()( BAkdt

Bd

dt

dAR f

A + B C + D

)()( BAdt

Bd

dt

dAR f

The rate law

The rate equation can be written as:


a A + b B → g G + h H

The overall order of a chemical reaction is the

sum of the exponents of the concentration terms

in a reaction kinetics equation that after

integration gives a linear plot.


nm BAKdt

Hd

hdt

Gd

gdt

Bd

bdt

Ad

a)()(

)(1)(1)(1)(1

The order of reaction

reaction is mth order in A

reaction is nth order in B

reaction is (m +n)th order overall


a A + b B → g G + h H

The overall order of a chemical reaction is the

sum of the exponents of the concentration terms

in a reaction kinetics equation that after

integration gives a linear plot.


nm BAKdt

Hd

hdt

Gd

gdt

Bd

bdt

Ad

a)()(

)(1)(1)(1)(1

The order of reaction


Concentration term exponents (m and n) are usually small whole

numbers but may be fractional, negative or zero.

They are unlikely to be the stoichiometric factors for the overall

rate law.

Reaction Mechanisms

Simple elementary

step

Sequence of elementary

steps A B

Elementary processes are reversible

The reactions are unimolecular or bimolecular

• Exponents for concentration terms are the same as

the stoichiometric factors for the elementary

process


Reaction Mechanisms

Simple elementary

step

Sequence of elementary

steps

• One elementary step is usually slower than all the

others and is known as the rate determining step

• Intermediates are produced in one elementary process

and consumed in another. They do not appear in the

overall chemical equation or the rate law.


F2 (g) + 2ClO2 (g) → 2FClO2 (g)

rate = k [F2]x[ClO2]

y

Double [F2] with [ClO2] constant

Rate doubles x = 1

Quadruple [ClO2] with [F2] constant

Rate quadruples y = 1

Experimental determination of rate law

F2 (g) + 2ClO2 (g) → 2FClO2 (g)

rate = k [F2][ClO2] 1

(method of initial rates)


I. Zero Order Reaction:

Are reactions, in which the rate of reaction

is independent of the concentration of the

reactants.

The velocity of the reaction is constant.




Zero-order reactions are typically found

when a material that is required for the

reaction to proceed, such as a surface or a

catalyst, is saturated by the reactants.

1) occurs in a closed system

2) there is no net build-up of intermediates

3) there are no other reactions occurring

https://www.boundless.com/definition/zero-order-reaction/





https://www.boundless.com/definition/catalyst/



Haber Process

Zero-order reactions are typically found

when a material that is required for the

reaction to proceed, such as a surface or a

catalyst, is saturated by the reactants.

Decomposition of multi-sulfa drugs

https://www.boundless.com/chemistry/textbooks/boundless-chemistry-textbook/chemical-kinetics-13/the-rate-law-concentration-and-time-99/zero-order-reactions-419-1010/images/haber-process/






https://www.boundless.com/definition/catalyst/

rate = - d(C)

dt rate = k (C)0 = k


= K d(C)

dt


The rate equation is a differential

equation. It can be integrated to obtain an

integrated rate equation that links

concentrations of reactants or products

with time.

https://www.boundless.com/definition/reactant/

https://www.boundless.com/definition/product/

[Ct] is the concentration of C at any time t

[C]0 is the concentration of C at time t=0 C

Time (t)

Slope = - K

Intercept = Co

On Integration Ct = Co - Kt

= K d(C)

dt - dC = K dt

rate = - d(C)

dt rate = k (C)0 = k



We can Also calculate the constant of reaction by

rearranging the linear equation:


0

)(K

dt

Cd

11../ Slitermole

Second

litermole

110 ../

Slitermloe

Second

litermole

t

CCk t

time

Conc.

Ct = Co - Kt

UNITS ?


t½ = t when Ct= 1/2 C0

The half-life (t½) the time required for a quantity

to fall to half its value as measured at the

beginning of the time period.



Ct = Co - k t

t = Co - Ct

k

t1/2 = Co - 1⁄2Co

k

t1/2 = 1⁄2Co

k

Putting Ct = 1⁄2 Co

in the linear equation:



The half-life (t½)

t90% = Co – 0.9Co k

t90% = 0.1Co

k

The shelf life (t90%) is the time required for 10%

of the drug to decompose. In other word 90% of

the drug remains.

Putting Ct = 0.9 Co

in the linear

equation:



tlife = Co – 0 k

tlife = Co

k

Time life (tlife) is the time required for 100% of

the drug to decompose. Time for a drug to

decompose completely.

Putting Ct = zero

in the linear

equation:




The decomposition of a multi-sulfa

compound, was found to follow zero

order kinetics. If its concentration was

0.47 mole/liter, when freshly prepared,

and after 473 days its concentration

reached 0.225 mole/liter, calculate its

degradation rate constant and t1/2.

Example

k = Co - Ct k = 0.470 – 0.225

t 473

= 0.00052 mole/liter . days-1

t ½ = Co = 0.47 = 452 days = 15 months

2K 2x 0.00052

Solution


II. 1st Order Reaction:

The speed of the reaction depends on the

concentration of only one reactant, raised

to the first power.


There can be other reactants present in the reaction, but

their concentrations do not affect the rate.

First-order equations are often seen in decomposition

reactions. Absorption, metabolism, distribution and

elimination (not linked with a carrier or a carrier with

unsaturable state).



The speed of the reaction depends on the

concentration of only one reactant, raised

to the first power.

CKdt

Cd1

)(

Integrating the previous equation between C0 at

time t = 0 and concentration C at some later time t

gives:

tkCC 0lnln

Converting to common logarithms yields:

303.2loglog 0

tkCC


CKdt

Cd1

)(




tkCC 0lnln303.2

loglog 0

tkCC

t

ln c

Intercept = ln C0

Slope = - k

t

log c

Intercept = log C0

Slope = - k

2.303



tkCC 0lnln303.2

loglog 0

tkCC

The reaction rate constant

C

C

tk olog

303.2

t

CCk o lnln

)()(

1 CKdt

Cd 1

)/(

/ SlitermoleSec

litermole

time

1

UNITS ?

kteCt

C 0

303.2/

0 10 ktCt

C

The relation between concentration and time is

exponential:


t1/2

Concentration at t∞

Initial concentration

1/2 C0

C0

1/4 C0

1/8 C0

t1/2 t1/2

C

t


half life when C = ½ Co

= 2.303 log Co = 2.303 . log 2 = k ½ Co k

C

C

tk olog

303.2

The Half Life "t½”

C

C

kt olog

303.2


T1/2

693.0

k


= 2.303 log Co = 2.303 . (0.0457) = k 0.9 Co k

0.105

k t90%

The Shelf Life (t90%)


C

C

tk olog

303.2

C

C

kt olog

303.2

the shelf life when C = 0.9 Co


Time life (tlife) is the time required for 100% of

the drug to decompose.

Putting Ct = zero in the linear equation:


?


A solution of a drug contained 500 units per ml when

prepared. It was analyzed after a period of 40 days and

was found to contain 300 units per ml. Assuming that

the decomposition is first order, at what time will the

drug have decomposed to one–half of its original

concentration?

Solution:

K = 2.303 log Co = 2.303 log 500 = 0.0128 day -1

t Ct 40 300

t½ = 0.693 = 0.693 = 54 days

K 0.0128



What is the half-life of N2O5 if it decomposes

with a rate constant of 5.7 x 10-4 s-1?

t½ 0.693

k =

0.693

5.7 x 10-4 s-1 = = 1200 s = 20 minutes



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