This document discusses accelerated stability testing and ICH guidelines. It provides definitions of stability testing and describes how exposing products to elevated temperatures can accelerate degradation reactions to predict long-term shelf life. Key points covered include common degradation reactions, advantages of accelerated testing, ICH guidance on test conditions, use of the Arrhenius equation to determine activation energy and calculate degradation rates at different temperatures, and equations for determining shelf life based on reaction order. Accelerated stability testing allows shelf life to be predicted in months rather than conducting long-term studies at room temperature.

1. Accelerated stability
testing & ICH
(Dr.) Mirza Salman Baig
Assistant Professor (Pharmaceutics)
AIKTC, School of Pharmacy,New Panvel
Affiliated to University of Mumbai (INDIA)

2. Stability
• Stability of pharmaceutical product
may be defined as the capability of a
particular formulation in a specific
container/closure system to remain
within its physical, chemical,
microbiological, therapeutic and
toxicological specification.

3. Stability testing
• A method by which a product is
exposed to elevated temperature
simulating what would happen over
longer periods on the shelf life
• Shelf life:- Time by which product
remain stable and retain 90% of its
initial potency. (10% degradation)

4. • Hydrolysis
• Oxidation- reduction
• Racemisation
• Decarboxylation
• Ring cleavage
• Photolysis
• isomerisation
Degradation reactions

5. Why?
• The stability of pharmaceutical
preparations should be evaluated by
exposing the product to normal shelf
conditions for a year or extended
periods.
• The rate of decomposition is slow at
room temperature .Such a method is
time consuming and uneconomical.

6. Objective
• To predict the shelf life of a
pharmaceutical product by
accelerating the rate of
decomposition ,preferably by
increasing the temperature.
• Stability study to predict the shelf life
of the product, by accelerating the
rate of decomposition, preferably by
increasing the temperature of reaction
conditions.

7. Advantage
• With the advancement in branch of
kinetics, shelf life of a dosage form
can be predicted within months
based on accelerated stability
reports
• Preparations are subjected to high
stresses during stability testing.
• Common high stresses include :
– Temperature
– Humidity
– Light

9. ICH

10. ICH Guidelines
• These guidelines provide definitions of key
terms & principles used in the stability
testing of drug substances& drug products.
• Q 1A (R2) :Stability testing of new drug
substances & products.
• Q 1B :Photo stability testing of new drug
substances &products.
• Q 1C :Stability testing for new dosage
forms.

11. ICH
• ICH outlined a combination of
temperature & humidities for stability
studies for most of the drug products.
These include.....
Ø -15O C ±5 C
Ø 5O C ±3 C /Ambient humidity
Ø 25O C±2 C /60%RH±5%,
Ø 30O C±2 C/60% RH±5%,
Ø 40O C±2 C/75%RH±5%.

12. Arrhenius equation
• It explains the effect of temperature
on rate of a reaction
• k = A e -Ea / RT

13. Shelf Life
• The Preparation is stored at different
elevated temperatures, to accelerate
the degradation
• Samples are withdrawn at different time
intervals
• The Order of the reaction is
determined by plotting the appropriate
function of concentration against
time and linear relationship is
determined
• Straight line in a graph permits the
estimation of k value from the slope

14. Shelf Life
• K value for each temperature are
calculated.
• By using Arrhenius relationship, Log k
values are plotted against reciprocal of
absolute temperature, energy of
activation can be calculated.
• Extrapolate the straight line to room
temperature (k 25O C ) read the log k value
on y-axis.
• Substitute the k value in the appropriate
equation to get the shelf life of the
product.

15. Undecomposed drug available
at elevated temperatures

16. Extrapolate graph to 25OC

17. Shelf life can be determined if we
know k and order of reaction
Order
of
reactio
nt
Equation t1/2 (Half life)
0
k0= x/t OR
x=k0t
t1/2= a/2k0
1
k= 2.303/t . log
a/(a-x)
t1/2=
0.693/k
2 k = 1/t . x/a(a-x) t1/2= 1/ak
t90%
t0.9= 0.9a/k
t0.9=2.303/k
t0.9= 1/ak