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Absorption Models · Tutorial 67 of 797

Sequential Zero-Order and First-Order Absorption

Learn to formulate and calculate a sequential zero-order-then-first-order absorption profile, including the equal-rate limit and checks on dose mass and model assumptions.

Intermediate 11 min read

What You'll Learn

  • Define the depot and central-compartment equations for sequential zero-order and first-order absorption.
  • Calculate concentrations during the zero-order stage and after the depot switches to first-order absorption.
  • Use the continuous limiting formula when the absorption and elimination rate constants are equal.
  • Check the dose constraint, units, transition continuity, and total exposure implied by the model.
  • Distinguish sequential absorption from a constant input that continues beyond the modeled zero-order stage.

Why Model Absorption in Two Stages?

Some absorption models represent a dose entering the systemic circulation at a constant rate for a finite period. A different assumption is that the dose first leaves its depot at a constant rate, then the remaining depot is absorbed more gradually according to first-order kinetics. This tutorial develops that sequential model and shows how each stage contributes to the concentration profile.

The key modeling distinction is that the two input mechanisms do not operate simultaneously. The zero-order stage ends at a specified time \(T\); only the depot amount left at that time supplies first-order input afterward. The model therefore needs to keep track of both the drug already absorbed and the drug still in the depot.

Definition: Let \(D\) be the dose initially in the absorption depot, \(R_0\) the constant depot-depletion rate during the first stage, and \(T\) the duration of that stage. Let \(A_g(t)\) be the amount remaining in the depot, \(k_a\) the first-order absorption rate constant after time \(T\), \(k_e\) the elimination rate constant, \(V\) the apparent volume of distribution, and \(F\) the fraction of the dose reaching the systemic circulation. Assume \(D>0\), \(T>0\), \(0<R_0T\leq D\), \(k_a>0\), \(k_e>0\), \(V>0\), and \(0<F\leq1\). The amount left at the switch is \(A_T=D-R_0T\).

The model assumes one-compartment linear disposition, instantaneous mixing in the central compartment, and the same bioavailability fraction \(F\) for drug absorbed in either stage. There is no additional absorption or elimination from the depot beyond the stated processes. The condition \(R_0T\leq D\) prevents the constant-rate stage from absorbing more than the available dose.

Model Equations and Concentration Profile

During the first stage, the depot loses drug at rate \(R_0\). Once that stage ends, its remaining amount declines exponentially. The systemic input rate is \(F R_0\) before \(T\), and \(F k_a A_g(t)\) after \(T\).

$$ A_g(t)= \begin{cases} D-R_0t, & 0\leq t\leq T,\\ A_T e^{-k_a(t-T)}, & t>T, \end{cases} \qquad A_T=D-R_0T. $$

Let \(A_c(t)\) denote the central-compartment amount and \(C(t)=A_c(t)/V\) its concentration. The central-compartment amount follows \(dA_c/dt=F(\text{depot depletion rate})-k_eA_c\), with \(A_c(0)=0\). Solving this equation gives the profile below. The post-switch time \(u=t-T\) is used to make the contributions from the two stages explicit.

$$ C(t)= \begin{cases} \displaystyle \frac{F R_0}{V k_e}\left(1-e^{-k_e t}\right), &0\leq t\leq T,\\[1.2em] \displaystyle \frac{F}{V}\left[ \frac{R_0}{k_e}\left(1-e^{-k_eT}\right)e^{-k_eu} + \frac{k_aA_T}{k_a-k_e} \left(e^{-k_eu}-e^{-k_au}\right) \right], &t>T,\quad u=t-T,\quad k_a\ne k_e. \end{cases} $$

After the switch, the first term is drug absorbed during the zero-order stage and subsequently eliminated. The second term is drug absorbed from the residual depot after the switch. This decomposition also helps check units: each bracketed term has units of amount, and multiplication by \(F/V\) gives concentration.

Equal-rate limit: If \(k_a=k_e=k\), the expression with denominator \(k_a-k_e\) is replaced by its continuous limit. For \(t>T\), the residual-depot contribution to concentration is \(\frac{F}{V}kA_Tu e^{-ku}\). The first-stage contribution remains \(\frac{F}{V}\frac{R_0}{k}(1-e^{-kT})e^{-ku}\).

Derivation and Continuity at the Switch

During zero-order absorption, the central amount satisfies \(dA_c/dt=FR_0-k_eA_c\). Multiplying by the integrating factor \(e^{k_et}\) and integrating from 0 to \(t\) yields the first part of the concentration profile:

$$ \frac{d}{dt}\left(e^{k_et}A_c(t)\right)=FR_0e^{k_et} \quad\Longrightarrow\quad A_c(t)=\frac{FR_0}{k_e}\left(1-e^{-k_et}\right). $$

After the switch, put \(u=t-T\). The central amount at the switch is \(A_c(T)=\frac{FR_0}{k_e}(1-e^{-k_eT})\), and the subsequent input is \(Fk_aA_Te^{-k_au}\). Solving the central-compartment equation from this initial condition gives

$$ A_c(T+u)=A_c(T)e^{-k_eu} + Fk_aA_T\int_0^u e^{-k_a s}e^{-k_e(u-s)}\,ds. $$

For \(k_a\ne k_e\), evaluating the integral gives \(\frac{e^{-k_eu}-e^{-k_au}}{k_a-k_e}\). For \(k_a=k_e=k\), the integrand simplifies to \(e^{-ku}\) for every \(s\), so the integral is \(ue^{-ku}\). Dividing the resulting amount by \(V\) gives the stated profile, including the equal-rate limit.

The concentration is continuous at \(T\): setting \(u=0\) makes the residual-depot contribution zero and leaves exactly \(A_c(T)/V\), the concentration from the first-stage solution. The input rate itself need not be continuous; it changes from \(FR_0\) to \(Fk_aA_T\) at the specified switch. That abrupt change is an assumption of this model, not an algebraic error.

Worked Example: Concentrations Before and After the Switch

Consider hypothetical values \(D=100\) mg, \(R_0=20\) mg/h, \(T=3\) h, \(k_a=0.5\) h\(^{-1}\), \(k_e=0.2\) h\(^{-1}\), \(V=20\) L, and \(F=0.8\). The dose constraint holds because \(R_0T=60\) mg is less than \(D\); therefore \(A_T=100-60=40\) mg.

At \(t=2\) h, absorption is still in its zero-order stage. Substitution gives

$$ C(2)=\frac{0.8(20)}{20(0.2)}(1-e^{-0.2(2)}) =4(1-e^{-0.4}) =4(0.329680) =1.31872\ \text{mg/L}. $$

At \(t=5\) h, \(u=5-3=2\) h. The contribution from the first stage is \(4(1-e^{-0.6})e^{-0.4}=1.209762\) mg/L. The residual-depot contribution is \[ \frac{0.8}{20}\frac{0.5(40)}{0.5-0.2}(e^{-0.4}-e^{-1}) =2.666667(0.302441) =0.806508\ \text{mg/L}. \] Thus \(C(5)=2.016270699\ldots\) mg/L, or \(2.016271\) mg/L to six decimal places and \(2.016\) mg/L to three decimal places. Both terms are positive, as expected for a positive dose and positive rate constants.

Worked Example: When Absorption and Elimination Rates Are Equal

Suppose \(D=80\) mg, \(R_0=10\) mg/h, \(T=4\) h, \(F=1\), \(V=20\) L, and \(k_a=k_e=0.25\) h\(^{-1}\). Then \(A_T=80-10(4)=40\) mg. Find the concentration at \(t=6\) h, so \(u=2\) h.

Because the rates are equal, use the limiting expression rather than dividing by \(k_a-k_e=0\). The first-stage contribution is \[ \frac{1}{20}\frac{10}{0.25}(1-e^{-1})e^{-0.5} =2(0.632121)(0.606531) =0.766801\ \text{mg/L}. \] The residual-depot contribution is \[ \frac{1}{20}(0.25)(40)(2)e^{-0.5} =1(0.606531) =0.606531\ \text{mg/L}. \] Therefore \(C(6)=0.766801+0.606531=1.373332\) mg/L, approximately \(1.373\) mg/L. This finite result is what the unequal-rate expression approaches as \(k_a\) approaches \(k_e\).

Worked Example: Checking Dose Mass and Total Exposure

For a third hypothetical case, let \(D=60\) mg, \(R_0=12\) mg/h, \(T=2\) h, \(k_a=0.4\) h\(^{-1}\), \(k_e=0.1\) h\(^{-1}\), \(V=15\) L, and \(F=0.75\). The zero-order stage absorbs \(R_0T=24\) mg; the residual depot is \(A_T=36\) mg. Their sum is \(24+36=60\) mg, so the depot mass balance is satisfied.

At \(t=6\) h, \(u=4\) h. The first-stage contribution is \[ \frac{0.75}{15}\frac{12}{0.1}(1-e^{-0.2})e^{-0.4} =0.729050\ \text{mg/L}. \] The residual-depot contribution is \[ \frac{0.75}{15}\frac{0.4(36)}{0.4-0.1}(e^{-0.4}-e^{-1.6}) =1.124216\ \text{mg/L}. \] Thus \(C(6)=1.853266927\ldots\) mg/L, or \(1.853267\) mg/L to six decimal places and \(1.853\) mg/L.

With complete eventual absorption from the depot, the total systemic input is \(FD=0.75(60)=45\) mg. Under linear one-compartment disposition, the AUC over time is therefore \[ \mathrm{AUC}_{0-\infty}=\frac{FD}{Vk_e} =\frac{45}{15(0.1)} =30\ \text{mg}\cdot\text{h/L}. \] This exposure check depends on \(k_e>0\); if elimination were absent, the concentration would not approach zero and this finite AUC formula would not apply.

Assumptions, Diagnostics, and Alternatives

The formula is appropriate only when the change in absorption mechanism is a plausible structural description. Check the following before interpreting a fitted profile:

  • Check the depot constraint. \(R_0T\) cannot exceed \(D\). A negative \(A_T\) is physically impossible under this model and signals incompatible parameter values or units.
  • Check units and time origins. \(R_0\) must be dose per time; \(k_a\) and \(k_e\) must be inverse time; and \(T\) and each observation time must use the same time unit.
  • Check the transition. The predicted concentration must be continuous at \(T\). A jump in concentration indicates a coding or implementation error; a change in input rate at \(T\) is allowed.
  • Check limiting behavior. The profile should start at zero for an initially empty central compartment. With \(k_e>0\), it should eventually decline toward zero after the depot input becomes negligible.
  • Check the model against the sampling times. Sparse observations near or after \(T\) may not distinguish the assumed switch from other absorption patterns. A plausible-looking curve alone does not establish that the sequence is supported by the data.

A common alternative is pure first-order absorption, in which the entire dose declines exponentially from the beginning, or pure zero-order absorption for a fixed duration, in which the input stops when that stage ends. Those models impose different input histories. The sequential model instead has a constant early depot-depletion rate and then a first-order decline of the remaining depot. It should not be interpreted as a constant-rate input that continues after \(T\), nor as two absorption mechanisms acting at once.

A practical pitfall is to use the post-switch expression with \(A_T=D\), forgetting that \(R_0T\) has already left the depot. That overcounts the dose and can inflate predicted concentrations and AUC. Another is to use the unequal-rate formula when \(k_a=k_e\); the denominator is then zero, although the underlying concentration has a well-defined limit. Use the equal-rate expression or a numerically stable limit near equality. Finally, this model assumes a sharp transition at \(T\). If data suggest a gradual change, a different structural model may be needed; the sequential formulation alone cannot describe every absorption profile.

Key takeaway: Track the depot amount through both stages. The first stage absorbs \(R_0T\), the second begins with \(A_T=D-R_0T\), and the central concentration after the switch is the sum of the surviving first-stage contribution and the contribution from the residual depot.

Check Your Understanding

Use the model assumptions and equations to answer these questions. Solutions follow the problems.

  1. If \(D=90\) mg, \(R_0=15\) mg/h, and \(T=4\) h, what amount remains in the depot at the switch? Is the dose constraint satisfied?
  2. Which concentration formula applies at \(t=T\), and why must the profile be continuous there?
  3. For \(k_a=k_e=k\), what replaces the residual-depot term after the switch?
  4. For \(D=50\) mg, \(F=0.8\), \(V=10\) L, and \(k_e=0.2\) h\(^{-1}\), what is the total AUC if the entire dose is eventually absorbed?
  5. Why is \(R_0T>D\) not an admissible parameter combination?

Solutions. (1) \(A_T=D-R_0T=90-15(4)=30\) mg; the constraint holds because \(60\leq90\). (2) Either piece gives the same concentration at the switch: the post-switch time is \(u=0\), so its residual-depot term is zero and its first-stage term equals the concentration accumulated by time \(T\). (3) The residual contribution is \(\frac{F}{V}kA_Tu e^{-ku}\). (4) \(\mathrm{AUC}_{0-\infty}=FD/(Vk_e)=0.8(50)/(10(0.2))=20\) mg·h/L. (5) It would imply that the zero-order stage absorbs more drug than the initial depot contains, producing a negative residual amount and violating mass balance.