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Choosing the Structural Model · Tutorial 113 of 599

First-Order Elimination

You will derive first-order elimination relationships and use them to calculate concentration decline, half-life, clearance, and exposure while recognizing when the model does not apply.

Intermediate 10 min read

What You'll Learn

  • Define first-order elimination as an elimination rate proportional to the amount of drug present.
  • Derive exponential concentration decline from the one-compartment differential equation.
  • Convert between the elimination rate constant, volume of distribution, and clearance.
  • Calculate half-life and the fraction of drug remaining at a specified time.
  • Relate dose and bioavailability to AUC when systemic clearance is linear and constant.
  • Recognize when ongoing input, distribution phases, or concentration-dependent disposition complicate interpretation.

What First-Order Elimination Means

Elimination describes the removal of drug from the body. Under first-order elimination, the amount eliminated per unit time is proportional to the amount of drug present. The proportionality constant is the elimination rate constant, \(k\), with units of inverse time. This is a concentration-independent process in the sense that \(k\) does not change with concentration: doubling the amount doubles the elimination rate, while the fraction eliminated per unit time remains constant.

Definition: In a one-compartment model with amount \(A(t)\), first-order elimination is represented by an elimination rate \(kA(t)\), where \(k>0\) is constant. If the apparent volume of distribution \(V\) is constant and concentration is \(C(t)=A(t)/V\), then the elimination rate is also \(CL\,C(t)\), where \(CL=kV\) is clearance.

The distinction between \(k\) and \(CL\) matters. The rate constant \(k\) describes the fraction of the amount eliminated per unit time. Clearance \(CL\) describes the volume of plasma or blood from which drug is cleared per unit time. For example, \(k\) may be measured in \(\mathrm{h}^{-1}\), whereas \(CL\) may be measured in \(\mathrm{L/h}\). In a one-compartment model, the two are related through \(CL=kV\).

The simple exponential concentration curve requires more than first-order elimination alone. It also assumes a one-compartment disposition model, constant \(k\), and no ongoing input during the interval being modeled. If drug is still entering the body, concentration reflects both input and elimination. In a multicompartment model, elimination from the central compartment can be first order even though the observed concentration curve is not a single exponential.

Deriving Exponential Decline

After input has stopped, the amount in the one-compartment model changes only through elimination. Its rate of change is therefore the negative of the elimination rate:

$$ \frac{dA(t)}{dt}=-kA(t). $$

Separating variables and integrating from time zero, when the amount is \(A_0\), to time \(t\), gives:

$$ \int_{A_0}^{A(t)}\frac{dA}{A} = -k\int_0^t dt, \qquad \ln\!\left(\frac{A(t)}{A_0}\right)=-kt. $$

Exponentiating both sides yields \(A(t)=A_0e^{-kt}\). With constant \(V\), concentration is proportional to amount, so it follows that:

$$ C(t)=C_0e^{-kt}, \qquad \ln C(t)=\ln C_0-kt. $$

Thus, while the model’s assumptions hold, concentration declines exponentially and its natural logarithm declines linearly with time, with slope \(-k\). The fraction remaining after time \(t\) is \(e^{-kt}\); the fraction eliminated is \(1-e^{-kt}\). Neither fraction depends on the starting concentration.

Key result: For a one-compartment model after input has stopped, \(C(t)=C_0e^{-kt}\). The elimination half-life, the time required for concentration to fall to one-half its current value, is \(t_{1/2}=\ln(2)/k\). The result applies at any starting time during the exponential phase, not only immediately after dosing.

The half-life expression follows by setting \(C(t_{1/2})=C_0/2\). Substitution into the exponential model gives \(1/2=e^{-kt_{1/2}}\). Taking natural logarithms gives \(-\ln(2)=-kt_{1/2}\), and hence \(t_{1/2}=\ln(2)/k\). A larger \(k\) therefore means a shorter half-life. In the one-compartment model, substituting \(k=CL/V\) gives the equivalent relationship \(t_{1/2}=\ln(2)V/CL\). This formula should not be applied as though it necessarily describes every observed phase in a multicompartment profile.

Worked Example: Concentration Decline After an IV Bolus

Suppose a hypothetical one-compartment patient receives a \(500\)-mg IV bolus. Assume \(V=25\) L and constant \(CL=5\) L/h. Calculate \(k\), the initial concentration, the concentration after 6 hours, the half-life, and the AUC to infinity.

Check the model conditions. This calculation assumes linear, time-invariant clearance, a constant apparent volume, and one-compartment disposition. After an IV bolus there is no continuing dose input. Under these stated assumptions, the exponential model applies.

First, \(k=CL/V=(5\ \mathrm{L/h})/(25\ \mathrm{L})=0.200\ \mathrm{h}^{-1}\). The initial concentration is \(C_0=D/V=(500\ \mathrm{mg})/(25\ \mathrm{L})=20.0\ \mathrm{mg/L}\).

At 6 hours, \(C(6)=20.0e^{-(0.200)(6)}=20.0e^{-1.20}=6.024\ \mathrm{mg/L}\).

The half-life is \(t_{1/2}=0.693147/0.200=3.466\ \mathrm{h}\). The AUC to infinity for this IV dose is \(D/CL\), so \(\mathrm{AUC}_{0-\infty}=(500\ \mathrm{mg})/(5\ \mathrm{L/h})=100\ \mathrm{mg\cdot h/L}\). The units follow because dividing amount by volume per time gives amount-time per volume.

Clearance and Exposure

For first-order elimination in the one-compartment setting, the elimination rate can be written either as \(kA\) or as \(CL\,C\). Since \(A=VC\), equating these expressions gives \(kVC=CL\,C\), and therefore \(CL=kV\). This relationship connects the exponential rate of decline to the clearance and apparent volume that describe the model.

A useful exposure result follows from conservation of drug. Let \(R_{\mathrm{in}}(t)\) denote the rate at which drug enters the systemic circulation. With first-order elimination, the amount balance is \(dA/dt=R_{\mathrm{in}}(t)-CL\,C(t)\). For a single dose, integrate this balance over the full period from before dosing until all drug from that dose has been eliminated. If the initial and final amounts attributable to the dose are zero, the integrated change in amount is zero. The total systemic input is \(FD\), where \(D\) is the administered dose and \(F\) is the fraction reaching systemic circulation. Consequently:

$$ 0=FD-CL\int_0^\infty C(t)\,dt, \qquad \mathrm{AUC}_{0-\infty}=\frac{FD}{CL}. $$

For an IV dose, \(F=1\). For an extravascular dose, the expression assumes that \(F\) appropriately represents the fraction reaching systemic circulation and that the full concentration-time profile is captured. The relationship does not require instantaneous absorption: under linear, time-invariant clearance, changing the rate of input can change the shape and peak of the profile without changing this total exposure, provided the total systemic input remains \(FD\).

Worked Example: Half-Life and Fraction Remaining

A hypothetical one-compartment drug has \(k=0.120\ \mathrm{h}^{-1}\). At the start of the modeled elimination phase, its concentration is \(12.0\ \mathrm{mg/L}\). Find the half-life and the concentration 8.00 hours later.

Check the model conditions. The calculation assumes that input is negligible over the 8-hour interval, \(k\) remains constant, and the concentration is in a monoexponential one-compartment phase. With these assumptions, the exponential decline equation is appropriate.

The half-life is \(t_{1/2}=0.693147/0.120=5.776\ \mathrm{h}\). The fraction remaining after 8.00 hours is \(e^{-kt}=e^{-(0.120)(8.00)}=e^{-0.960}=0.3829\). Therefore, \(C(8.00)=12.0(0.3829)=4.595\ \mathrm{mg/L}\).

About \(38.29\%\) of the starting concentration remains, and about \(61.71\%\) has declined over this interval. The 8-hour interval is longer than one half-life, so the remaining fraction is less than one-half, consistent with the calculated result.

Worked Example: Oral Dose and AUC

Suppose a hypothetical oral dose of \(300\) mg has systemic bioavailability \(F=0.800\), and the drug has constant systemic clearance \(CL=4.00\) L/h. Assuming linear disposition and complete characterization of the dose-related concentration profile, calculate \(\mathrm{AUC}_{0-\infty}\).

Check the model conditions. The exposure relationship requires that clearance be linear and time-invariant over the relevant concentration range, and that \(F\) and \(CL\) refer to this dose and system. The absorption process may take time, but the full systemic input must be represented by \(FD\).

The systemic amount is \(FD=(0.800)(300\ \mathrm{mg})=240\ \mathrm{mg}\). Thus, \(\mathrm{AUC}_{0-\infty}=FD/CL=(240\ \mathrm{mg})/(4.00\ \mathrm{L/h})=60.0\ \mathrm{mg\cdot h/L}\).

For comparison, if the same \(300\)-mg dose were delivered IV with the same clearance, the predicted AUC would be \((300\ \mathrm{mg})/(4.00\ \mathrm{L/h})=75.0\ \mathrm{mg\cdot h/L}\). The lower oral AUC follows from \(F=0.800\), not from a different elimination rate in this calculation.

Ongoing Input and Observed Profiles

When drug is entering the body during the period of interest, the one-compartment amount balance is \(dA/dt=R_{\mathrm{in}}(t)-kA(t)\). The observed concentration can rise, remain nearly stable, or decline depending on the balance between input and elimination. The exponential decline equation describes the elimination-only interval, not every part of a concentration-time profile.

For example, after an oral dose, absorption may still contribute materially to concentration while elimination is occurring. Once input becomes negligible, the terminal slope reflects disposition. If absorption is slow and remains an important contributor to concentrations, the observed terminal slope can instead be absorption-limited, as in flip-flop kinetics. A straight line on a log-concentration plot is therefore not, by itself, proof that the observed slope represents the disposition elimination rate constant.

The one-compartment model is also distinct from first-order elimination in a multicompartment model. In the latter, elimination from the central compartment can be proportional to the amount there, but exchange between compartments also shapes plasma concentrations. The overall profile can contain multiple exponential phases. A terminal phase may be useful for describing late concentrations, but its slope is not automatically interchangeable with a one-compartment \(k=CL/V\) for the whole profile.

Diagnostics, Alternatives, and Limitations

A model should be checked against the design and the observed concentration-time data. A semilogarithmic plot of concentration against time can help assess whether a post-input interval is approximately linear, as expected under a single exponential. Curvature, systematic residual patterns, or different slopes across time may indicate that input is ongoing, distribution phases are present, the one-compartment structure is inadequate, or clearance changes with concentration or time. Sparse sampling can conceal these patterns, so apparent linearity is not conclusive evidence.

The first-order model is a natural choice when a constant fraction of the amount is removed per unit time and dose-normalized exposure is consistent with the assumed linear disposition. It contrasts with elimination models in which the removal rate is not proportional to amount across the relevant concentration range. Those alternatives may be needed when data show concentration-dependent or capacity-limited behavior. They should not be inferred from a single noisy curve or from a half-life calculated without confirming that the relevant phase is disposition-driven.

The most useful alternatives depend on what the data require. A multicompartment structural model can represent distribution and elimination phases when a single exponential fails. A model that allows nonlinearity may be appropriate when the elimination rate no longer scales proportionally with concentration. A time-varying clearance model may be considered when clearance changes over time. Each adds assumptions and parameters; the simplest model that adequately represents the data is preferable, but simplicity must not override clear evidence of misspecification.

Common pitfall: “The concentration fell by half, so the half-life is constant for this drug at every concentration.” This conclusion is justified only if the relevant decline follows the assumed first-order, time-invariant model. A changing log-concentration slope, ongoing absorption, multicompartment distribution, or concentration-dependent clearance can make an estimated half-life phase-specific or otherwise misleading.

First-order elimination is valuable because it yields transparent links among rate, half-life, clearance, volume, and exposure. These links support dose and concentration calculations, but they are conditional results, not universal properties of a drug. The assumptions should be stated whenever the equations are used.

Check Your Understanding

Answer each question, then compare your reasoning with the solutions below.

  1. A one-compartment model has \(CL=3.00\ \mathrm{L/h}\) and \(V=15.0\ \mathrm{L}\). What is \(k\), and what is the elimination half-life?
  2. Under first-order elimination, what fraction of the amount remains after two half-lives? State why this result does not depend on the starting amount.
  3. For a hypothetical IV dose of \(240\) mg and constant clearance \(6.00\) L/h, calculate \(\mathrm{AUC}_{0-\infty}\).
  4. Why does a straight terminal line after an oral dose not necessarily identify the disposition elimination rate constant?
  5. In a multicompartment model, can first-order elimination from the central compartment coexist with a plasma concentration curve that is not a single exponential? Explain.

Solutions. (1) \(k=CL/V=(3.00\ \mathrm{L/h})/(15.0\ \mathrm{L})=0.200\ \mathrm{h}^{-1}\). Thus \(t_{1/2}=0.693147/0.200=3.466\ \mathrm{h}\). (2) Two half-lives leave \((1/2)^2=0.250\), or \(25.0\%\), of the starting amount. The fractional decline is determined by \(k\) and elapsed time, not by the initial amount. (3) For IV dosing, \(F=1\), so \(\mathrm{AUC}_{0-\infty}=D/CL=(240\ \mathrm{mg})/(6.00\ \mathrm{L/h})=40.0\ \mathrm{mg\cdot h/L}\). (4) Absorption may still contribute materially to the observed concentrations, so the terminal slope can be absorption-limited rather than disposition-driven. Once input is negligible, the terminal slope reflects disposition. (5) Yes. First-order elimination describes removal from the central compartment, while drug exchange among compartments also shapes plasma concentration. The resulting profile can have multiple exponential phases rather than a single exponential.

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