Pharmaceutical Calculations Module

Tutorial 12: Drug Stability

About This Tutorial This tutorial was developed by the Victorian College of Pharmacy, Monash University in conjunction with the Australian Pharmacy Examining Council Incorporated (APEC Inc) (now known as the Faculty of Pharmacy and Pharmaceutical Sciences, Monash University and the Australian Pharmacy Examining Committee respectively).It remains the property of Monash University. The tutorial cannot be used or copied without prior authorisation.

This tutorial is intended as an aid for students enrolled in Monash University’s Bachelor of Pharmacy & Pharmacy Internship programs.

Original tutorial prepared by in1996:

Arthur Pappas Lecturer in Pharmacy Practice Dr Louis Roller Head, Department of Pharmacy Practice

with the assistance of : May Admans Senior Assistant-Lecturer in Pharmaceutics Dr Denis Morgan Reader in Pharmaceutics Dr Kay Stewart Lecturer in Pharmacy Practice Prof Peter Stewart Professor of Pharmaceutics

Original Tutorial Manual produced by Arthur Pappas - December 1996

Adaptation & updating of the material for online delivery to students of Monash University was undertaken by Phil Bergen, Assistant Lecturer in Pharmacy Practice -February 2005

© Copyright 2005 This publication is copyright. Apart from any fair dealing for the purpose of private study, research, criticism or review as permitted under the Copyright Act, no part may be reproduced by any process or placed in computer memory without written permission. Any assembled extracts each attract individual copyright and enquiries about reproduction of this material should be addressed to the appropriate publishers. Enquiries about all other material should be made to the Faculty named above. Last amended by EM- November 2008

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Pharmaceutical Calculations 12. Drug Stability

12. Drug Stability

Topic Index

Page 12.1 Introduction 114 12.2 Objectives 114 12.3 Rate Laws 114

12.3.1 Zero Order Reactions 114 12.3.2 First Order Reactions 115

12.4 Half-life and Shelf-life 117 12.4.1 Zero-order Reactions 117 12.4.2 First-order Reactions 118

12.5 Relationship between reaction rate and temperature (Arrhenius Equation) 122

12.5.1 Using Arrhenius Equation to Extrapolate k2 at T2 122

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Pharmaceutical Calculations 12. Drug Stability

12.1 Introduction

The concentration of active ingredients in a formulation can vary with time depending on the chemical stability of the drug. A drug can therefore degrade to form various byproducts which maybe inactive therapeutically or even toxic. It is important that we know the likely degradation pathway of a drug in a particular formulation vehicle and the rate at which it degrades under specified conditions. Having this information will enable us to ascertain the storage period and conditions (eg: temperature, pH, presence of light, catalysts) for which we are confident that the concentration of the drug is within acceptable limits.

12.2 Objectives

The aim of this tutorial is to give you an appreciation of the main issues concerning drug stability and how a knowledge of degradation rate laws enables us to predict shelf-life of products.

At the end of this tutorial you will be able to:

- define the term 'order of a reaction’ - state the formula for a zero order and first order reaction - given appropriate information (eg: reaction order, rate constant (at a specific

temperature), initial concentration, activation energy) - be able to calculate:

- the half-life of a drug - the shelf-life of a drug - the amount of drug remaining at a particular time - the rate constant at another temperature - shelf-life at another temperature

12.3 Rate Laws

Drug degradation can be described by the following fundamental rate relationship.

rate C p

The rate of degradation is proportional to the concentration of drug (C) raised to some power (P), P identifies the order of this equation. eg: P =0 (zero order)

P = 1 (first order)

P = 2 (second order)

The two commonly used rate laws are zero and first order.

12.3.1 Zero Order

rate C 0

rate = k0C°

where ko = zero order rate constant

Since any number raised to zero power has a value of unity, this equation can be written as:

rate = k0

'rate' can be mathematically expressed as - dC/dt

Therefore: -dC/dt = k0

This expression represents the change in drug concentration with time. 114

Pharmaceutical Calculations 12. Drug Stability

The negative sign indicates that there is a decrease in concentration with time. Integrating the above expression yields:

C = Co - k0 x t

where: C = concentration at a particular time (units: g/mL, mg/mL, mol/L) Co = initial concentration (units: g/mL, mg/mL, mol/L) t = time (units: seconds, minutes, hours) k0 = zero order rate constant (units: concentration units divided by time units)

When you plot C versus t, there is a straight line with a gradient of -ko

Worked Example 12.1

A 0.50% w/v solution of Drug X is prepared. It is known to follow a zero order degradation with a rate constant of 0.05mg/mL/year. What concentration of Drug X will remain after 3 years of storage at 25°C ?

Working: Using:

C = C0 - k0 . t

since k0 is in units of mg/mL, then concentration will have to be in those units too. ie: 0.50% w/v = 0.50g/100mL = 500mg/100mL = 5mg/mL

C = 5mg/mL - (0.05mg/mL/year x 3 years) = 5 mg/mL - 0.15mg/mL

= 4.85mg/mL

12.3.2 First Order

rate C1

rate = k1C1

where k1 = first order rate constant (units: time'1)

Mathematically,

-dC/dt = k1C

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Pharmaceutical Calculations 12. Drug Stability

Integrating the above expression yields:

C = Co. e-k1t

or using natural logarithms (to the base e)

In C = In Co – k1t

or using logarithms to the base 10 k1

log C10 = log Co - ..................... t 2.303

(Note: the units of k1 are sec-1, hour-1, day-1 etc.)

When you plot log C versus time, there is a straight line with a gradient of -ki/2.303

Note: If a drug (eg: Drug X) follows first-order degradation, it will have the same rate constant k1, irrespective of the initial concentration.

Worked Example 12.2

The degradation of Drug Y follows first-order kinetics. The initial concentration was 2.0 mg/mL. It was found that 10% of its activity was lost after 3 months.

How much drug will remain at the end of 6 months ?

Working:

C = Co. e-k1t

Co = 2.0 mg/mL

At the end of 3 months, C = (90% of 2.0mg/mL = 1.8mg/mL)

1.8 = 2. e-k1

3

1.8/2 = 0.9 = e-k1

3

-0.105 = -k1

3

Therefore k1 = 0.035 month-1

Pharmaceutical Calculations 12. Drug Stability

Now, to determine the concentration after 6 months:

C = Co. e-k1

t

C = 2.e-0.035 x 6

C = 1.62 mg/mL (answer)

Quiz Question 12.1

Drug K follows first-order kinetics when in solution and loses 20% potency in 25 days at 26°C. What concentration will remain if 0.18M is stored for 80 days at 26°C?

Answer:

C = Co. e-k1

t

Co =100%

At the end of 25 days, C = (80% of initial)

80 = 100. e-k1

25 =

80/100 = 0.8 = e-k1

25

-0.223 = -k1

25

Therefore k1 = 0.00893 day-1

Now, to determine the concentration after 80 days. Using Co = 0.18M

C = Co. e-k1

t

C = 0.18.e-000893 x 80

C = 0.088M (answer)

12.4 Half-life and Shelf-life

'Half-life' is the time required to reduce the concentration of a drug to 50% of its initial value, ie: C = 0.5 Co

The following symbols are used: t1/2, t50%, t0.5

'Shelf-life' is the time required to reduce the concentration of a drug to 90% of its initial value, ie: C = 0.9 Co

The following symbols are used; t90%, t0.9

12.4.1 Zero-order Reactions

C = Co - ko . t

For half-life C = 0.5C0

Therefore: 0.5C0 = Co - k0. t1/2

-0.5C0 = - ko . t1/2

0.5C0

ie: t1/2 = ----------- (Half-life Zero-Order Reaction)

ko 117

Pharmaceutical Calculations 12, Drug Stability

For shelf-life C = 0.9C0

Therefore: 0.9C0 = Co - k0. t90%

-0.1 Co = -ko. t90%

0.1 Co

ie: t90% = ---------- (Shelf-life Zero-Order Reaction)

ko

12.4.2 First-order Reactions

k1

log C = log Co - .................. t

2.303

For half-life

C = 0.5 Co

Therefore:

log 0.5C0 = log Co- k1 t1/2

2.303

log 0.5C0 = 0 - k1 t1/2

2.303

(Note: taking Co as having the value 1, then log10 (1) = 0)

-0.301 - - k1 . t1/2

2.303

t1/2 = -0.301 x 2.303 /-k1

0.693 t1/2

= --------- (Half-life First-Order Reaction)

k1

For shelf-life

C = 0.9C0

Therefore:

log 0.9C0 = log Co -k1 . t90% 2.303

log 0.9C0 = 0 - k1 . t90%

2.303 (Note: taking Co as having the value 1, then log10 (1) = 0)

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Pharmaceutical Calculations 12. Drug Stability

t90% = log 0.9C0 x 2.303/-k1

t90% = - 0.0458 x 2.303/-k1

0.105 t90% = --------- (Shelf-life First Order Reaction)

k1

Worked Example 12.3

A stability test conducted on a batch of tablets under simulated "patient use" conditions showed that half of the potency was lost after 27 months. On the basis of this information and assuming the degradation process is first order what should the shelf-life of the product be?

Working:

Using t1/2 = 0.693/k,

t1/2 = 27 months = 0.693/k1

k1 = 0.02567 month-1

Now to determine shelf-life t90% = 0.105/ k1

t90% = 0.105/0.02567 month-1 = 4.09 months

Note: The dispensed label should therefore read "Discard 4 months after first opening".

Quiz Question 12.2

The degradation constant of a drug in water is 0.69 hours-1 at 25°C. Degradation is known to follow first order kinetics. What is the half-life of this drug ?

Answer:

Using t1/2 = 0.693/k1

t1/2 = 0.693/0.69hr-1 = 1.0 hour

Worked Example 12.4

A solution of Cephalothin Sodium 1g/100 mL retained 90.6% activity in Dextrose Injection when stored at 5°C for 14 days. What is the shelf-life of the drug under these conditions (assuming it follows first-order kinetics) ?

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Pharmaceutical Calculations I2- Dm& Stability

Working:

Using:

k1 log C = log Co - .......... t

2.303

Determine the value of k1:

log 90.6 = log 100 - k1 / 2.303 . 14 days

1.9571 = 2- k1 / 2.303 .14 days

-0.0429 = -k1 . 6.079

k, = 0.00706 day-1

Determine the value of t90%:

Using t90% = 0.105/ k1

t1/2 = 0.105/0.00706day-1

t1/2 = 14.87 days

Quiz Question 12.3

A solution formulation of the diuretic Frusemide (10 mg/10 mL) was found to retain 86.2% potency after 90 days stored at 24°C. Degradation follows first-order kinetics. What would be the amount of Frusemide remaining in 30 mL of this mixture when stored under these conditions for one year?

Answer:

Using:

k1

log C = log Co ............. t

2.303

Determine the value of k1

log 86.2 = log 100 - k1 / 2.303 . 90 days

1.9355 = 2 - k1 / 2.303 .90 days

-0.0645 = - k1. 39.079 days

k1 = 0.001651 day-1

Determine the % remaining after 365 days:

log C = log C0 - k1_____ . t 2.303 120

Pharmaceutical Calculations 12. Drug Stability

0.001651day-1 log C = log 100 - ........ . 365 days

2.303

log C = 2 - 0.2617 = 1.7383

C = 54.74%

In 30mL of the original mixture (10mg/10mL) there is 30mg drug.

Therefore 54.75% of 30mg = 16.42mg (answer)

Quiz Question 12.4

A solution containing 0.75g/100mL of Drug P degrades by a first-order process with a rate constant of 0.0035 day-1 at 20°C. Calculate the shelf-life of this solution.

Answer:

Using: t90% = 0.105/k1

= 0.105/0.0035day-1 = 30.0 days at 20°C.

Quiz Question 12.5

A new eye drop formulation has an initial drug concentration of 1.6mg/mL It is known that the half-life of the drug is 850 days. Degradation follows first-order kinetics. What is its shelf-life ?

Answer:

Firstly, determine the rate constant using:

t1/2= 0.693/k1

850 days = 0.693/ k1

k1 = 0.000815day-1

Then, to determine the shelf-life:

t90% = 0.105/k1

= 0.105/0.000815day-1

= 128.8 days (answer)

Quiz Question 12.6

A drug degrades via a zero-order process. It has a k0 value of 2mg/mL/day. If a solution of concentration 25mg/mL is prepared, what is the shelf-life of this formulation?

121

Pharmaceutical Calculations 12. Drug Stability Answer: Using t90% = C0

10k0

= 25mg/ml 10x2mg/mL/day = 1.25 days

12.5 Relationship between reaction rate and temperature (Arrhenius Equation)

The rate at which a chemical reaction proceeds can depend on several factors, the most important being pH and temperature.

The Arrhenius Equation shows the relationship between rate constant and temperature.

k = A.e-(Ea/RT)

k = rate constant A = constant Ea = activation energy (calories/mol or joules/mol) R = gas constant (R = 1.987 cal deg-1 mol-1 or 8.314 J deg-1 mol-1) T = absolute temperature (ie: °C + 273)

Logarithm form of the Arrhenius Equation:

Ea log10 k = log10 A - ....................

2.303 R.T

12.5.1 Using Arrhenius Equation to Extrapolate k2at T2

If k1 and Ea are known at a particular temperature (T1), the rate constant (k2) at a second temperature (T2) can be determined using:

Ea (T2 – T1) logk2 = logk1 + -------------

2.3 R.T1. T2

Worked Example 12.5

Drug Y has a k, value of 0.085 hour1 (35°C and pH6) What is the k value at 50°C (and pH6) ? What is the half-life ?

[The activation energy Ea = 18kcal/mol]

Working: Using:

logk2 = logk1 + Ea(T2 -T1)

2.303. R.T1. T2

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Pharmaceutical Calculations 12. Drug Stability

Ea = 18kcal.mol-1

T1= 35°C = 308K

T2 = 50°C = 323K

R = 1.987 cal deg-1 mol-1

18,000 cal.mol-1(323 - 308) logk2 = log (0.085hour-1) + -----------------------------------------------------------

2.303 x 1.987 cal deg-1 mol-1 x 308 x 323

= -1.071 + 0.593 = -0.478

Therefore k2 = 0.333 hour-1

(Notice that k values are greater at higher temperatures) (As a practical rule: an increase of 10"C leads to an approximate doubling of the reaction

rate)

Quiz Question 12.7

Drug Y has a k value of 0.085 hour-1 (35°C and pH6) What is the half-life at refrigeration temperature (4°C) and pH6 ? Degradation follows first-order kinetics. [The activation energy Ea = 18kcal.mol-1]

Working: Using:

Ea (T2 – T1) logk2 = logk1 + —

2.303. R.T1T2

Ea = 18kcal.mol-1 T1 = 35°C = 308K T2 = 4°C = 277K R = 1.987 cal deg-1 mol-1

18,000 cal.mol-1 (277-308) logk2 = log (0.085hour-1) +

2.3 x 1.987 cal.deg-1.mol-1 x 308 x 277

= -1.071 +-1.429 =-2.50

Therefore k2 = 0.0032 hour-1

To determine the half-life t1/2 at 4°C:

t 1/2 = 0.693/k

= 0.693/0.0032 = 216.6 hours

Quiz Question 12.8

The shelf-life of a new drug is 36 minutes at 36 °C. Degradation follows first-order kinetics. What is the shelf-life (in hours) of the drug at refrigeration temperature (4 °C) ? The activation energy is 15.7 kcal/mol.

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Pharmaceutical Calculations 12. Drug Stability

Answer:

Firstly determine the k1 value.

At 36°C:

t 90% = 0.105/k1

36 min = 0.105/ki Therefore k1 = 0.00292 min-1

Use the Arrhenius Equation to find k2:

Ea (T2 – T1) logk2 = logk1 + -----------------

2.303 . R.T1T2

Ea= 15.7kcal/mol

T1 = 36°C = 309K

T2 = 4°C = 277K

R= 1.987 cal deg-1 mol-1

15,700 cal/mol (277-309) logk2 = log (0.00292min-1) + ---------------------------------------------------------

2.303 x 1.987 cal deg-1 mol-1 x 309 x 277

logk2 = -2.535 + -1.283 = -3.818

Therefore k2 = 0.000152 min-1

To determine the shelf- life at 4°C:

t90% = 0.105/k2

= 0.105/0.000152 = 690.8 minutes = 11.5 hours

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