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Please post to pkin website:
I am occasionally asked to explain why 0.693 is used to convert the
elimination rate constant to half-life. I have been offering the following
explanation. Please tell me if I am correct in making the following
assumption:
The factor 0.693 "unlogs" the elimination rate constant to yield half-life
by the following relationship: 1/e^0.693 = 50%.
If I am incorrect, please someone set me straight.
Thanks.
R. Timothy Gendron
Pharmacy Department
Naval Medical Center
27 Effingham Street
Portsmouth, VA 23708
Phone (757) 953-0252
Fax (757) 953-3328
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[A few replies - I'm rather partial to the second but then..:-) - db]
=46rom: "JVIS (Jennifer Visich)"
Date: Mon, 3 Dec 2001 13:51:22 -0800
To: david.-at-.boomer.org
Subject: RE: PharmPK Defining 0.693 correction factor for converting
Ke to t1/2
The following message was posted to: PharmPK
Richard,
converting by 0.693 is a result of the following
the first order equation:
C(t)=3D C(0)e^-kt
rearranged you get: C(t)/C(0) =3D e^-kt
when C(t) is one half of C(0)you get:
0.5 =3D e^-kt
take the natural log of both sides to get:
-.693 =3D -kt
rearranged
t1/2=3D.693/k
Hope that helps,
Jenn Visich, Ph.D.
Sr Scientist
ZymoGenetics, Inc.
1201 Eastlake Ave E
Seattle, WA 98102
(206) 442-6728
jvis.-a-.zgi.com
---
=46rom: Stephen Day
Date: Mon, 3 Dec 2001 17:29:08 -0500 (EST)
To: david.-a-.boomer.org
Subject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t=
1/2
The following message was posted to: PharmPK
Richard,
David Bourne has a nice explanation here:
http://www.boomer.org/c/p1/
All you need to know is in section 4 "One Compartment
I.V. Bolus"
0.693 is the natural logarithm of 2, almost (i.e.,
ln(2)=3D 0.693147....)
Steve
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=46rom: "Thomas A Torda"
Date: Tue, 4 Dec 2001 10:19:09 +1100
To: david.-a-.boomer.org
Subject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t=
1/2
Status: R
The following message was posted to: PharmPK
Perhaps, simpler migt be:
1 Assuming a linear fall in log(concentration) v time, otherwise 1/2
lives do not exist:
Using the Naperian (natural) logs, the equation for the ln(conc), y, is y =
=3D
C(0 time) + b.time (b will be negative and is the rate constant of
elimination or whatever, the slope of the concentration line.)
2 1/2 life =3D time for halving of concentration, that is, time for conc(=
2)
=3D conc(1)/2 or ln(conc(1)) - ln(conc(2)) =3D ln 2 =3D 0.693
0.693 =3D (Co + b.time1) - (Co + b.time2)
=3D b(time1 - time2)
3 1/2 life or 1/2 time =3D 0.693/b
Tom Torda
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=46rom: "samia ezzine"
Date: Mon, 03 Dec 2001 23:03:04
To: david.aaa.boomer.org
Subject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t=
1/2
Status: R
The following message was posted to: PharmPK
Hi Timothy,
As Cp =3D Coe-ket
At t1/2, Cp =3D C0/2 and t=3D t1/2 yields to C0/2 =3D C0 e-ket1/2
by simplifying e-ket1/2 =3D 1/2 , Ket1/2 =3D Ln 2 so
t1/2 =3D Ln2/ke and Ln 2=3D 0.693 so t1/2 =3D 0.693/ke
I hope that this can help you.
Samia Ezzine
ph.D student
=46acult=E9 de pharmacie
Universit=E9 de Montr=E9al
Qc, Canada
---
=46rom: "Bruce CHARLES"
Date: Tue, 4 Dec 2001 10:23:56 10
To: david.at.boomer.org
Subject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t=
1/2
Status: R
The following message was posted to: PharmPK
Send reply to: PharmPK.at.boomer.org
Date sent: Mon, 3 Dec 2001 14:26:05 -0600
=46rom: "Gendron, Richard T. (GS)"(by way of David Bourne)
To: Multiple recipients of PharmPK - Sent by
Subject: PharmPK Defining 0.693 correction factor for
converting Ke to t1/2
Going back a step further, define half-life (t1/2) as time to go from
concentration 2C (at time, t2) to C (at time, t)
k (=3Dslope) =3D Ln (2C) - Ln(C) / (t2-t)
=3D Ln (2C/C) / t1/2
=3D Ln 2 / t1/2
k =3D 0.693 / t1/2
q.e.d.
Cheers,
BC
---
From: Michael.D.Karol.aaa.abbott.com
Date: Tue, 4 Dec 2001 10:08:09 -0600
MIME-Version: 1.0
Richard:
In pharmacokinetics, we refer to half-life, the time for half of the
elimination process to be completed. The value of 0.693 is ln(2), reciproca=
l
of 1/2.
Thus, we have K =3Dln(2)/ "half life" =3D 0.693/ "half life".
In physics or engineering, the term "time constant" is used. This is the ti=
me
for 1/e to be completed (where e represents the base of natural log).
To convert from a time constant to rate constant the following would apply:
K =3D ln(e)/ "time constant". Note that ln(e) =3D 1.
Other constant times could be used to describe first order processes. For
example, you could use the time for one third of the elimination to be
completed. This might be called a third life (as opposed to a half life). =
In
this case the equation would be:
K =3D ln(3)/ "third life"
Hope this helps.
MK
---
=46rom: Glenn Whelan
Date: Tue, 4 Dec 2001 11:27:05 -0500
To: david.-a-.boomer.org
Subject: RE: PharmPK Defining 0.693 correction factor for converting
Ke to t1/2
The following message was posted to: PharmPK
Dr. Gendron:
0.693 comes from the natural log of 2 thus to find time, t, for half of a
concentration, Xo:
0.5Xo =3D Xo e-kt1/2
(0.5Xo/Xo)=3De-kt1/2 Xo cancels out
Ln(0.5)=3D -kt1/2
0.693=3D -kt1/2
0.693/-k =3D t1/2
Glenn Whelan, PharmD, Research Fellow
Nemours Children's Clinic
807 Children's Way
Jacksonville, FL 32250
Phone 904.390.3406
=46ax 904.390.3425
email gwhelan.-a-.nemours.org
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[Three more replies - db]
From: David Jaworowicz
Date: Tue, 04 Dec 2001 12:23:57 -0500
To: david.-a-.boomer.org
Subject: Re: PharmPK Defining 0.693 correction factor for converting Ke to t1/2
A more straightforward way of explaining where 0.693 comes from and why it's
used:
For a drug exhibiting monoexponential disposition: Cp=Co*e^(-k*t)
where Co is initial drug concentration, Cp is drug at time t, and k is the
elimination rate constant.
Rewritten: Cp/Co=e^(-k*t)
At t=t1/2 (i.e half-life): Cp/Co=0.5 and thus 0.5=e^(-k*t1/2)
Solving for t1/2: ln 0.5 = -0.693 = -k * t1/2
and thus t1/2= 0.693/k
Sincerely,
David Jaworowicz
---
From: "Daniel Sitar"
Organization: University of Manitoba
To: david.at.boomer.org
Date: Tue, 4 Dec 2001 11:44:00 -0600
Subject: Re: PharmPK Re: Defining 0.693 correction factor for
converting Ke to t1/2
Reply-to: sitar.aaa.Ms.UManitoba.CA
X-Confirm-Reading-To: sitar.at.ms.UManitoba.CA
X-pmrqc: 1
Priority: normal
Many students don't follow the math presented in these
solutions. It is better show
getting rid of the "-" sign by taking the reciprocal of e^-kt. Then
1/e^kt = 0.5
e^ kt = 2
Ln of both sides yields kt = 0.693
Dan Sitar, Pharmacology & Therapeutics
University of Manitoba
---
From: "Ronald A. Herman"
Date: Tue, 4 Dec 2001 12:54:31 -0600
To: david.-a-.boomer.org
Subject: RE: PharmPK Re: Defining 0.693 correction factor for
converting Ke to t1/2
The following message was posted to: PharmPK
As is accurately pointed out, Ln 2 = 0.693147...... so let's all stop using
0.693 and start using Ln 2. Why add rounding error? Ln 2 is on my
calculator and only takes me two keystrokes whereas, 0.693 is 5 keystrokes!
Ron Herman
Ronald A. Herman, Ph.D.
Asst. Professor (Clinical), Director
Iowa Drug Information Network
University of Iowa College of Pharmacy
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Copyright 1995-2010 David W. A. Bourne (david@boomer.org)