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The formula in textbooks for $t_{\frac{1}{2}}$ of a drug following first order elimination is generally given as $$t_{\frac{1}{2}}= \frac {\ln(2).V_d}{Cl}$$where $V_d$ is the volume of distribution and $Cl$ is the clearance.

Shown below is my attempt to derive the formula but I don't know where I'm going wrong.

Assuming first order kinetics of elimination, the rate of elimination ($R$) would be proportional to the plasma concentration ($Cp$)

$$ R = Cp . k$$

Where $k$ is the rate constant.

For calculation of $t_{\frac{1}{2}}$, since this is a first order elimination, $$ t_{\frac{1}{2}} = \ln(2)/k$$ Substituting, $$ t_{\frac{1}{2}} = \frac{\ln(2).Cp}{R}$$ Since $ Cl = R/Cp$, substituting would give, $$ t_{\frac{1}{2}} = \ln(2)/Cl$$

Since both the formulae don't match (this one doesn't have a $V_d$ term at all!), where have I gone wrong? Also then how is $Cl$ different from $k$?

According to the correct formula, $k=Cl/V_d$. How is that so?

EDIT

Volume of distribution is the apparent volume of blood the drug takes up and is given by $$V_d = \frac {Dose}{Pc}$$

Clearance is the volume of blood that has been cleared of the drug in unit time and is given by $ Cl = R/Cp$

Correct me if the basic definitions themselves are wrong.

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  • $\begingroup$ Should this be asked in chemistry? $\endgroup$
    – Polisetty
    Nov 5, 2016 at 7:56
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    $\begingroup$ Can you define the terms, "distribution volume" and "clearance"? The problem certainly has to do with the way you have defined these terms. $\endgroup$
    – WYSIWYG
    Nov 5, 2016 at 11:18
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    $\begingroup$ $Cl = R/Cp$ seems incorrect. Cl unit is $volume \cdot time^{-1}$, R is $weight \cdot volume^{-1} \cdot time^{-1}$ , Cp is $weight \cdot volume^{-1}$. R/Cp is $time^{-1}$ which is imcompatible with Cl. $\endgroup$
    – Eliane B.
    Nov 5, 2016 at 16:55

1 Answer 1

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Well, $Cl = R/Cp$ is correct, but $R$ is not $Cp \cdot k$.

Given $X(t)$, the actual quantity (ie weight or moles) of drug in the system, $R$ should have the dimension of $\frac{dX(t)}{dt}$, that is $weight \cdot time^{-1}$ or $mole \cdot time^{-1}$ because the base definition of clearance is given by $$- \frac{dX(t)}{dt} = Cl \cdot C(t)$$.

The volume of distribution $Vd$ comes into play when converting from $X(t)$ to $C(t)$. Following the same steps as you did :

$$ R = \frac{dX(t)}{dt} = \frac{dC(t)}{dt} * Vd = C_p \cdot k \cdot V_d$$ because (due to 1st order kinetics) $$\frac{dC(t)}{dt} = Cp \cdot k$$

Substituting, $$ t_{\frac{1}{2}} = \frac{\ln(2)}{k}$$ $$ t_{\frac{1}{2}} = \frac{\ln(2) \cdot C_p \cdot V_d}{R}$$ Since $ Cl = \frac{R}{Cp}$, substituting would give, $$ t_{\frac{1}{2}} = \ln(2) \cdot \frac{1}{Cl} \cdot V_d$$ $$ Cl = \ln(2) \cdot \frac{1}{t_{\frac{1}{2}}} \cdot V_d$$ $$ Cl = k \cdot V_d$$ $$ k = \frac{Cl}{V_d}$$

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