Why Convert Clearance and Volume Parameters?
A two-compartment model can describe the same pathways using either clearance and volume parameters or first-order rate constants. The clearance-volume form uses \(CL\) for elimination from the central compartment, \(Q\) for exchange between compartments, \(V_c\) for central volume, and \(V_p\) for peripheral volume. The rate-constant form expresses each pathway as an amount transferred per unit time divided by the amount in its source compartment. Converting between these forms is therefore a matter of writing the amount balances and substituting the concentration-volume relationships.
The earlier material on Clearance-Volume Parameterization of the Two-Compartment Model establishes the parameterization used here. This tutorial derives the conversion explicitly and shows how to check it. The result is useful when a model is specified in one parameterization but a calculation, software implementation, or interpretation uses the other.
Deriving the Conversion From Amount Balance
Let \(A_c\) and \(A_p\) denote the amounts of drug in the central and peripheral compartments. Their concentrations are \(C_c=A_c/V_c\) and \(C_p=A_p/V_p\). The model assumes that elimination occurs from the central compartment at rate \(CL\,C_c\), and that net exchange from central to peripheral occurs at rate \(Q(C_c-C_p)\). If an input \(R_{\mathrm{in}}(t)\) enters the central compartment, the amount balances are
The corresponding microconstant description writes central elimination as \(k_{10}A_c\), central-to-peripheral transfer as \(k_{12}A_c\), and peripheral-to-central transfer as \(k_{21}A_p\). Expanding the clearance-based balances in terms of \(A_c\) and \(A_p\) gives
Matching the coefficients of \(A_c\) and \(A_p\) with the microconstant amount balances establishes the conversion. In particular, the central balance contains both elimination and outward exchange; the peripheral balance contains inward and outward exchange but no elimination term under this model.
This derivation assumes a linear model with constant clearances and volumes, exchange represented by the same \(Q\) in both directions, and elimination only from the central compartment. It also assumes the amounts and concentrations are related by the stated compartment volumes. The input term can be zero, as after an input has ended, or nonzero; it does not change the conversion of the disposition parameters.
Worked Conversions
Worked Example: Converting a Set of Clearance-Volume Values
Suppose a hypothetical model has \(CL=5.4\ \mathrm{L/h}\), \(Q=8.0\ \mathrm{L/h}\), \(V_c=18\ \mathrm{L}\), and \(V_p=32\ \mathrm{L}\). Calculate the three microconstants.
All clearances are in liters per hour and both volumes are in liters, so each quotient has units of inverse hours:
For central-to-peripheral transfer,
For peripheral-to-central transfer,
The two exchange rate constants differ because their source compartments have different volumes. The same \(Q\) appears in each direction, but \(k_{12}\) is scaled by \(V_c\) and \(k_{21}\) by \(V_p\). These values are rates, not clearances: their units are \(\mathrm{h}^{-1}\), not \(\mathrm{L/h}\).
Worked Example: Handling a Different Time Unit
Consider hypothetical values \(CL=72\ \mathrm{L/day}\), \(Q=120\ \mathrm{L/day}\), \(V_c=24\ \mathrm{L}\), and \(V_p=40\ \mathrm{L}\). First calculate the rates per day, then express them per hour.
Using the given units directly,
There are 24 hours in a day, so a rate per day is divided by 24 to obtain the numerically equivalent rate per hour:
The equality of \(k_{10}\) and \(k_{21}\) here follows from the particular hypothetical values: \(CL/V_c=Q/V_p\). It is not a general identity of the two-compartment model. A common source of a factor-of-24 error is to report rates per hour while leaving a clearance expressed per day.
Worked Example: Verifying the Rates With Amount Flows
Use hypothetical values \(CL=4\ \mathrm{L/h}\), \(Q=6\ \mathrm{L/h}\), \(V_c=20\ \mathrm{L}\), and \(V_p=30\ \mathrm{L}\). Suppose at a particular instant \(C_c=2.0\ \mathrm{mg/L}\) and \(C_p=0.5\ \mathrm{mg/L}\). Calculate the microconstants, then check the central amount rate using both parameterizations.
The conversion gives
The compartment amounts are \(A_c=C_cV_c=(2.0)(20)=40\ \mathrm{mg}\) and \(A_p=C_pV_p=(0.5)(30)=15\ \mathrm{mg}\). With no input at this instant, the microconstant balance gives
The clearance-based calculation gives the same result. Elimination is \(CL\,C_c=(4)(2.0)=8\ \mathrm{mg/h}\), and net outward exchange is \(Q(C_c-C_p)=6(2.0-0.5)=9\ \mathrm{mg/h}\). Therefore \(dA_c/dt=-8-9=-17\ \mathrm{mg/h}\). Agreement checks both the conversion and the signs of the exchange terms.
Checks, Alternatives, and Limits
A conversion is straightforward, but several checks help catch errors before the rates are used in a model. First, the clearance and volume in each quotient must use compatible units, and the resulting rate must have units of inverse time. Second, for positive finite \(CL\), \(Q\), \(V_c\), and \(V_p\), all three rate constants must be positive. Third, the amount balance must assign each rate constant to its source compartment: central loss uses \(k_{12}A_c\), while peripheral return uses \(k_{21}A_p\).
A useful algebraic check is to convert back: \(CL=k_{10}V_c\), \(Q=k_{12}V_c\), and \(Q=k_{21}V_p\). The two estimates of \(Q\) must agree, apart from rounding. If they do not, a rate may have been divided by the wrong volume, a unit conversion may be missing, or rounded values may be too coarse.
Using microconstants directly is appropriate when the model is defined in terms of first-order amount transfer rates. The clearance-volume parameterization can be more convenient when clearances and volumes are the quantities being specified or interpreted. These are alternative parameterizations of the same linear model when their values obey the conversion equations; neither parameterization alone removes limitations in the underlying structure.
The conversion is not a general rule for every compartment model. For example, if elimination also occurs from the peripheral compartment, the amount balance needs an additional peripheral elimination pathway and rate constant. If elimination or exchange is nonlinear, or if the effective clearances vary with time or concentration, fixed first-order microconstants may not represent those processes. In such cases, do not apply these formulas without first specifying the altered amount-balance model.
Check Your Understanding
Try each question, then compare your work with the solutions below.
- A hypothetical model has \(CL=3.6\ \mathrm{L/h}\), \(Q=7.2\ \mathrm{L/h}\), \(V_c=12\ \mathrm{L}\), and \(V_p=24\ \mathrm{L}\). Find \(k_{10}\), \(k_{12}\), and \(k_{21}\).
- For \(Q=96\ \mathrm{L/day}\), \(V_c=32\ \mathrm{L}\), and \(V_p=48\ \mathrm{L}\), calculate \(k_{12}\) and \(k_{21}\) in \(\mathrm{h}^{-1}\).
- Why does using the same \(Q\) in both directions not generally mean \(k_{12}=k_{21}\)?
- A proposed conversion uses \(k_{12}=Q/V_p\) and \(k_{21}=Q/V_c\). Identify the error and state the correct formulas.
Solutions. 1. \(k_{10}=3.6/12=0.300\ \mathrm{h}^{-1}\), \(k_{12}=7.2/12=0.600\ \mathrm{h}^{-1}\), and \(k_{21}=7.2/24=0.300\ \mathrm{h}^{-1}\). 2. First, \(Q/V_c=96/32=3.00\ \mathrm{day}^{-1}\) and \(Q/V_p=96/48=2.00\ \mathrm{day}^{-1}\). Dividing by 24 gives \(k_{12}=0.1250\ \mathrm{h}^{-1}\) and \(k_{21}=0.08333\ \mathrm{h}^{-1}\). 3. Each rate constant divides \(Q\) by the volume of its source compartment; if \(V_c\ne V_p\), the quotients generally differ. 4. The proposed conversion assigns each pathway the destination volume. The correct formulas are \(k_{12}=Q/V_c\) and \(k_{21}=Q/V_p\).