Zero Order Absorption
*
Types of chemical administration or exposure that provide (or approximate) zero-order chemical input include:
- Intravenous (IV) infusion
- Prolonged breathing of chemical vapors
- Prolonged skin exposure to constant chemical concentration
- Implantable controlled/sustained release system
*
*
Model Equations
D = k0 * T
k0 = D/T
*
During Chemical Exposure (t≤T)
- dA/dt = (rate in) – (rate out)
= (absorption rate) – (elimination rate)
dA/dt = k0 – (k*A)
dA/dt + (k*A) = k0
*
A = c1 + (c2 * e-kt)
- The unknown constant terms are then evaluated by employing the initial condition (A=0 at t=0) and the differential equation to get:
c1 = -c2 = k0/k
or
A = (k0/k)(1-e-kt)
*
-
- As before, it is more useful to write the final equation in terms of plasma concentration, which can be done through the transformation C = A/V, giving:
C = [k0/V*k] [1-e-kt] for t≤T
*
After Chemical Exposure Stopped (t>T)
- After chemical exposure has stopped, the differential mass balance can be written:
dA/dt = -k*A
A = AT at t = T
*
A = AT * e-k(t-T) = (k0/k)(1- e-kT)e-k(t-T)
C = CT * e-k(t-T) = (k0/V*k)(1- e-kT)e-k(t-T)
CT = (k0/V*k)(1- e-kT)
*
*
31.bin
Steady-State Concentration (CSS)
- Css occurs when the rate of chemical elimination exactly equals the rate of chemical input
- Chemical Elimination Rate = Chemical Input Rate
CL * Css = k0
V * k * Css = k0
Css = k0/V*k
Note: that the equations for C at t≤T can be rewritten in terms of Css:
C = (k0/V*k)(1-e-kt) = Css(1-e-kt) for t≤T
*
- Concentration drops exponentially immediately after exposure stops, so after infusion (t>T):
C = CT * e-k(t-T)
- taking the natural logarithm of both sides gives:
lnC = lnCT -k(t-T)
or
lnC = {lnCT + kT} – kt
*
*
Effect of Model Parameters on Plasma Concentrations
Dosing Rate (k0): Concentration proportional to k0 at all times
*
Effect of Model Parameters on Plasma Concentrations
Distribution Volume (V): Concentration proportional to 1/V at all times
*
Effect of Model Parameters on Plasma Concentrations
Elimination Rate Constant (k) and Half-life (t1/2): Affect approach to plateau and rate of drop after stopping input, k and t1/2 have opposite effects
*
-
- The approach to steady state (plateau) conditions is dependent on half-life in a manner similar to elimination:
at t1/2 → 50% of way to equilibrium → C=0.5Css
at 2t1/2 → 75% of way to equilibrium → C=0.75Css
at 5t1/2 → ~97% of way to equilibrium → C~0.97Css
at 7t1/2 → ~99% of way to equilibrium → C~0.99Css
at 10t1/2 → ~99.9% of way to equilibrium → C~0.999Css
*
*
Determination of Chemical Input Rate (k0)
-
- IV Infusion
- Known dose (D) infused over known time period (T) → k0 = D/T
- k0 = D/T → 100mg/2hr = 50 mg/hr
- Skin Exposure Absorption
- k0 = Absorption Rate = Pskin*Sskin [Cskin – Cblood]
*
Determination of Chemical Input Rate (k0)
-
- Lung Exposure Absorption
-
- Low Solubility Gas (Kblood/air <<1, perfusion limited)
- K0 = Absorption Rate = Qlungblood [Cblood out – Cblood in] ~ Qlung blood [Kblood/air*Cair in-Cvein]
- High Solubility Gas (Kblood/air >>1, ventilation limited)
- K0 = Absorption Rate = Qlung air [Cair in – Cair out] ~ Qlungair *Cair in
*
Example: Estimate the toluene absorption rate for rats exposed to 100ppm toluene via lung absorption
-
- kblood/air = 18; MW = 92.15; Qlung air = 3.75 L/hr; Qlung blood = 3.75 L/hr
-
- remember: C(mg/L) = C(ppm)*MW/24450
- Cair in = 100ppm * [92.15/24450] = 0.377 mg/L
- kblood/air = 18, therefore kblood/air >> 1 → High solubility gas
- k0 = Qlung air * Cair in ~ 3.75 L/hr * 0.377mg/L = 1.4mg/hr
*
-
- Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1
- Estimate C for t<T: calculate C at 2hr after exposure (t = 2hr)
- Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)
- Estimate theoretical Css: Calculate Css if exposure lasted for a very long time (e.g., T>10 t1/2,elim)
- Estimate C for t>T: calculate C at 6hr after exposure starts (t=6hr), same as C at 2hr after exposure ends (t-T = 6hr – 4hr = 2hr)
- Estimate t for given C if t<T: calculate t when C first rises to 0.5mg/L
- Estimate t-T for given C if t>T: calculate t-T when C falls back down to 0.5mg/L
-
- Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1
-
- Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)
CT = [k0/Vk][1-e-kt]
CT = [1.4mg/hr/(1.0L * 1.07hr-1)][1-e-1.07hr-1*4hr]
CT = 1.31mg/L * 1-0.0138
CT = 1.29 mg/L
-
- Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1
-
- Estimate C for t<T: calculate C at 2hr after exposure (t = 2hr)
C = [k0/Vk][1-e-kt]
C = [1.4mg/hr/(1.0L * 1.07hr-1)][1-e-1.07hr-1*2hr]
C = 1.31mg/L * 1-0.1177
C = 1.16mg/L
-
- Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1
-
- Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)
CT = [k0/Vk][1-e-kt]
CT = [1.4mg/hr/(1.0L * 1.07hr-1)][1-e-1.07hr-1*4hr]
CT = 1.31mg/L * 1-0.0138
CT = 1.29 mg/L
-
- Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1
-
- Estimate theoretical Css: Calculate Css if exposure lasted for a very long time (e.g., T>10 t1/2,elim)
Css = [k0/Vk]
Css = [1.4mg/hr/(1.0L * 1.07hr-1)]
Css = 1.31mg/L
-
- Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1
-
- Estimate C for t>T: calculate C at 6hr after exposure starts (t=6hr), [same as C at 2hr after exposure ends (t-T = 6hr – 4hr = 2hr)]
C = CT * e-k(t-T)
C = 1.29mg/L * e-1.07hr-1(2hr)
C = 1.29mg/L * 0.1177
C = 0.152 mg/L
*
-
- Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1
-
- Estimate t for given C if t<T: calculate t when C first rises to 0.5mg/L
C = [k0/Vk][1-e-kt]
0.5mg/L = 1.31mg/L * e-1.07hr-1(t)
(0.5mg/L)/(1.31.mg/L) = 0.3817 = 1 – e-1.07hr-1(t)
e-1.07hr-1(t) = 1 – 0.3817 = 0.6183
ln e-1.07hr-1(t) = ln0.6813
-1.07hr-1 * t = -0.4807 → -0.4807/-1.07hr-1 = 0.45 hr
or 0.45 hr after start of exposure
*
-
- Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1
-
- Estimate C at t=T: calculate CT at 4hr after exposure starts (t = T = 4hr)
CT = [k0/Vk][1-e-kt]
CT = [1.4mg/hr/(1.0L * 1.07hr-1)][1-e-1.07hr-1*4hr]
CT = 1.31mg/L * 1-0.0138
CT = 1.29 mg/L
-
- Rats are exposed to 100 ppm toluene for 4 hrs, after which the exposure to toluene stops. k0 = 1.4 mg/hr (from previous slide); T = 4hrs; V = 1.0L; t1/2,elim = 0.65 hr; k=ln2/t1/2,elim = 0.693/0.65hr = 1.07 hr-1
-
- Estimate t-T for given C if t>T: calculate t-T when C falls back down to 0.5mg/L
C = CT * e-k(t-T)
0.5 mg/L = (1.29 mg/L) * e-(1.07/hr)(t-T)
(0.5mg/L)/((1.29 mg/L) = 0.3876 = e-(1.07/hr)(t-T)
ln0.3876 = lne-(1.07/hr)(t-T)
-1.07hr-1 * (t-T) = -0.9478
t-T = (-0.9478/-1.07hr-1) = 0.86hr or 0.86hr after exposure ends
*
Analysis of plasma concentration data (C vs t)
- Parameters from Post-Exposure Measurements (t≥T):
- for t>T: C = CT * e–k(t-T)
- taking the natural logarithm of both sides gives:
lnC = ln CT – k(t-T)
- Plot post-exposure data in the form:
y = lnC vs x = t-T
- To yield: slope = m = -k and y = intercept = lnCT
*
*
-
- CT = [k0/Vk][1-e-kT]
-
- V = [k0/CT*k ][1-e-kT]
- CL = V * k
*
*
Example: Estimate toluene k, t1/2, CT, V and CL from rat concentrations following a 4hr exposure to 100ppm toluene (from earlier graph, data provided below). Previous example gave k0 ~ 1.4mg/hr
| Time, t (hr) | Concentration, C (mg/L) |
| 0.00 | 0.000 |
| 0.25 | 0.278 |
| 0.50 | 0.537 |
| 1.00 | 0.830 |
| 2.00 | 1.114 |
| 3.00 | 1.243 |
| 4.00 | 1.277 |
| 4.08 | 1.201 |
| 4.50 | 0.722 |
| 5.00 | 0.392 |
| 5.50 | 0.261 |
| 6.00 | 0.153 |
*
How do we go about solving this problem?
- First, calculate (t-T) and lnC for time points after exposure stopped (t≥4hrs)
| Time, t (hr) | t – T (hr) | C (mg/L) | lnC |
| 4.00 | 0 | 1.277 | 0.245 |
| 4.08 | 0.08 | 1.201 | 0.183 |
| 4.50 | 0.50 | 0.722 | -0.326 |
| 5.00 | 1.00 | 0.392 | -0.936 |
| 5.50 | 1.50 | 0.261 | -1.343 |
| 6.00 | 2.00 | 0.153 | -1.877 |
- Next, plot lnC vs (t-T)
- Determine useable points (points falling on linear portion of line)
*
- Perform linear regression analysis
m = -1.07hr-1 → k = -m = -(-1.07hr-1) → k = 1.07 hr-1
t1/2 = 0.693/k → 0.693/1.07hr-1 → t1/2 = 0.65hr
b = 0.228 → CT = eb = e0.228 → CT = 1.26mg/L
V = [k0/(CT*k)][1-e-kT] = [(1.4mg/hr)/(1.26mg/L * 1.07hr-1)][1 – e-1.07hr-1 * 4hr] → V = 1.02L
CL = V*k = 1.02L * 1.07hr-1 = 1.10 L/hr (intercept method)
*
Parameters from Two-Post Exposure Plasma Samples (monitoring after accidental exposure)
Ct1 at t1, Ct2 at t2 with t1 and t2 ≥ T
Slope = (ln Ct1 – ln Ct2) / (t1 – t2)
= [ln(Ct1 / Ct2)]/(t1 – t2)
k = -slope = [ln(Ct1 / Ct2)]/(t2 – t1)
Ct1 = [k0/V*k] [1-e-kT] e-k(t1-T)
V = [k0/(Ct1*k)][1-e-kT]e-k(t1-T)
Example: Use 4ht and 6hr measurements from toluene example.
Ct1 = 1.277 mg/L at t1 = 4.0hr
Ct2 = 0.153 mg/L at t2 = 6.0hr
k = [ln(1.277mg/L / 0.153mg/L)]/(6.0 – 4.0 hr) = 1.06 hr-1
t1/2 = 0.693/1.06hr-1 = 0.65hr
V = [(1.4mg/hr)/(1.277mg/L * 1.07hr-1)][1-e-1.07hr-1*4hr)][e -1.06hr-1(4hr – 4hr)]
= 1.01 L
CL = V*k = 1.01 L * 1.06hr-1 = 1.08 L/hr
*
Parameters from measurements during exposure period (t≤T)
- In order to perform this analysis, it is necessary to first estimate Css
- Analysis only works for data from long chemical exposure period that clearly reaches a steady-state plateau
- Css estimate is more accurate when 2 – 3 plateau points can be averaged
- For a chemical fitting a one-compartment model, concentrations during the exposure period are given by:
C = Css * (1-e-kt)
- This equation can be rearranged as follows:
C = Css – Css*e–kt → Css – C = Css * e–kt
- Then taking the natural logarithm of each side gives:
ln(Css – C) = lnCss – kt
- Which is now in a linear form with
Slope = m = -k
Intercept = b = lnCss
*
ln(CSS – C)
t
lnCSS
slope = -k
- then if one performs a linear regression with
y = ln(Css – C) and x = t
- the values of k and V can be determined by
k = -m
Css = k0/(V*k) → V = k0/(Css * k)
- one should always plot the data to determine if the one-compartment model is appropriate and to identify unusable points
*
ln(CSS – C)
t
Late points that
jump around are
too close to C
ss
for
use in the analysis.
If early points do not
fall on line, may need
different model.
Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)
Estimate the Css
Calculate ln(Css-C) for each measurement
Plot ln(Css-C) vs t
Determine useable points
Perform linear regression to calculate necessary variables
| Time, t (hr) | Concentration, C (mg/L) |
| 0.00 | 0.000 |
| 0.25 | 0.278 |
| 0.50 | 0.537 |
| 1.00 | 0.830 |
| 2.00 | 1.114 |
| 3.00 | 1.243 |
| 4.00 | 1.277 |
Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)
Estimate the Css
Since concentration is still rising between last two timepoints can only use the last timepoint to estimate Css
Css ~ 1.277 mg/L
| Time, t (hr) | Concentration, C (mg/L) |
| 0.00 | 0.000 |
| 0.25 | 0.278 |
| 0.50 | 0.537 |
| 1.00 | 0.830 |
| 2.00 | 1.114 |
| 3.00 | 1.243 |
| 4.00 | 1.277 |
*
Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)
Estimate the Css: Css~ 1.277 mg/L
Calculate ln(Css-C) for each measurement
| Time, t (hr) | Concentration, C (mg/L) | ln(Css – C) |
| 0.00 | 0.000 | 0.245 |
| 0.25 | 0.278 | -0.001 |
| 0.50 | 0.537 | -0.301 |
| 1.00 | 0.830 | -0.805 |
| 2.00 | 1.114 | -1.814 |
| 3.00 | 1.243 | -3.381 |
| 4.00 | 1.277 |
Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)
Estimate the Css
Calculate ln(Css-C) for each measurement
Plot ln(Css-C) vs t
Determine useable points
*all points are useable
*can use linear regression to describe data
*
Example: Estimate toluene k, t1/2, V and CL from rat concentrations during 4hr exposure to 100ppm toluene (as previously estimated, k0 ~ 1.4 mg/hr)
Perform linear regression to calculate necessary variables
m = -1.18hr-1 → k = -m = -(-1.18hr-1) → k = 1.18hr-1
t1/2 = 0.693/k → 0.693/1.18hr-1 →t1/2 = 0.59hr
b = 0.315 → Css,intercept = eb = e0.315 → Css = 1.37 mg/L
V = k0/(Css * k) = (1.4 mg/hr)/[1.28mg/L * 1.18hr-1) → V = 0.93L
CL = V*k = 0.93L * 1.18hr-1 → CL = 1.10L/hr
AUC Calculations from C vs t Data
-
- Calculate AUC(t1 →tf) using the trapezoidal rule for consecutive points
- AUC(t1 →tf) = ½[Ct1 + Ct2][t1-t2]
- Use C0 = 0 at t=0, even if C0 is not actually measured
- If CT is not measured, estimate it from post-exposure regression
- Evaluate the AUC of the tail region just as for bolus IV injection
- AUCtail = AUC(tf →∞) = Ctf/k
*
AUC Calculations from C vs t Data
- As a short-cut method, a reasonable estimate of AUC during post-exposure period is given by
- AUC(T →∞) ~ CT/k
- There is unfortunately no general short-cut method for the exposure period (t<T)
- If Css can be estimated, then total AUC ~ Css * T
*
Evaluation of V and CL from AUC (AUC Method)
-
- AUC = D/(V*k) = D/CL (always true for linear kinetics)
-
- Remember: D = k0 * T yields:
V = [k0 * T]/[AUC * k]
CL = [k0 * T]/AUC
Example: Determine AUC, V, and CL (using AUC Method) for 100ppm rat toluene exposure data examined earlier (use k = 1.07hr-1 and k0 = 1.4mg/hr as determined previously)
- Make table showing t1, t2, Ct1, Ct2 (include t=0 and t=∞)
- Calculate AUC for each interval and for tail region (after tf)
- Sum AUC values to get running sum of AUC and total AUC
- Calculate V and CL by the AUC Method (compare to intercept method used previously)
Example: Determine AUC, V, and CL (using AUC Method) for 100ppm rat toluene exposure data examined earlier (use k = 1.07hr-1 and k0 = 1.4mg/hr as determined previously)
- Make table showing t1, t2, Ct1, Ct2 (include t=0 and t=∞)
| t1 (hr) | t2 (hr) | Ct1 (mg/L) | Ct2 (mg/L) |
| 0.00 | 0.25 | 0.000 | 0.278 |
| 0.25 | 0.50 | 0.278 | 0.537 |
| 0.50 | 1.00 | 0.537 | 0.830 |
| 1.00 | 2.00 | 0.830 | 1.114 |
| 2.00 | 3.00 | 1.114 | 1.243 |
| 3.00 | 4.00 | 1.243 | 1.277 |
| 4.00 | 4.08 | 1.277 | 1.201 |
| 4.08 | 4.50 | 1.201 | 0.722 |
| 4.50 | 5.00 | 0.722 | 0.392 |
| 5.00 | 5.50 | 0.392 | 0.261 |
| 5.50 | 6.00 | 0.261 | 0.153 |
| 6.00 | ∞ | 0.153 |
Example: Determine AUC, V, and CL (using AUC Method) for 100ppm rat toluene exposure data examined earlier (use k = 1.07hr-1 and k0 = 1.4mg/hr as determined previously)
-
- Make table showing t1, t2, Ct1, Ct2 (include t=0 and t=∞)
- Calculate AUC for each interval and for tail region (after tf)
AUC(0→0.25hr) = ½[0 + 0.278mg/L][0.25 – 0 hr] = 0.035 mg hr/L
AUCtail = (0.153mg/L)/(1.07hr-1) = 0.143 mg hr/L
3. Sum AUC values to get running sum of AUC and Total AUC
| t1 (hr) | t2 (hr) | Ct1 (mg/L) | Ct2 (mg/L) | AUC(t1→t2) (mg hr/L) | AUC run SUM (mg hr/L) |
| 0.00 | 0.25 | 0.000 | 0.278 | 0.035 | 0.035 |
| 0.25 | 0.50 | 0.278 | 0.537 | 0.102 | 0.137 |
| 0.50 | 1.00 | 0.537 | 0.830 | 0.342 | 0.478 |
| 1.00 | 2.00 | 0.830 | 1.114 | 0.972 | 1.450 |
| 2.00 | 3.00 | 1.114 | 1.243 | 1.178 | 2.629 |
| 3.00 | 4.00 | 1.243 | 1.277 | 1.260 | 3.889 |
| 4.00 | 4.08 | 1.277 | 1.201 | 0.099 | 3.988 |
| 4.08 | 4.50 | 1.201 | 0.722 | 0.404 | 4.392 |
| 4.50 | 5.00 | 0.722 | 0.392 | 0.278 | 4.670 |
| 5.00 | 5.50 | 0.392 | 0.261 | 0.163 | 4.834 |
| 5.50 | 6.00 | 0.261 | 0.153 | 0.104 | 4.937 |
| 6.00 | ∞ | 0.153 | 0.143 | 5.080 |
As a comparison, using the short-cut method for the post-exposure period gives:
AUC(T→∞) ~ CT/k = (1.277mg/L)/(1.07hr-1) = 1.193 mg hr/L
AUC = AUC(0→T) + AUC(T→∞)
= 3.889 mg hr/L + 1.193 mg hr/L
= 5.082 mg hr/L
4. Calculate V and CL by the AUC Method (compare with intercept method)
V = [k0 * T] / [AUC * k]
= [1.4mg/hr * 4hr] / [5.080mg hr/L * 1.07hr-1]
= 1.03 L (compared to 1.02 L)
CL = [k0 * T] / AUC
= [1.4 mg/hr * 4hr] / 5.080 mg hr/L
= 1.10 L (compared to 1.10 L)
*
Chemical
Absorption
Rate
Time
0
Instantaneous
Absorption
Zero-Order
Absorption
T
C
k
V
k
0
Short Exposure
T
C
Time
Prolonged Exposure
T
C
Time
C
T
C
T
Short Infusion
T
lnC
Time
Prolonged Infusion
T
lnC
Time
lnC
T
lnC
T
slope = -k
slope = -k
T
C
Time
Large
k
0
Small
k
0
T
lnC
Time
Large
k
0
Small
k
0
Slope not
affected
by k
0
T
C
Time
Small
V
Large
V
T
lnC
Time
Small
V
Large
V
Slope not
affected
by V
T
C
Time
Small t
1/2
Large k
Large t
1/2
Small k
T
lnC
Time
Slope, C
�
,
and approach
to plateau all
affected
Large t
1/2
Small k
Small t
1/2
Large k
0.0
1.0
2.0
3.0
4.0
5.0
6.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Toluene Concentration in Plasma (mg/L)
Time (hr)
100 ppm Toluene
200 ppm Toluene
0.1
1.0
10.0
0.0
1.0
2.0
3.0
4.0
5.0
6.0
7.0
Toluene Concentration in Plasma (mg/L)
Time (hr)
100 ppm
Toluene
200 ppm
Toluene
lnC
Time after Chemical Input Stopped (t – T)
b = lnC
T
m = -k
Late points may jump
around near limit of
drug assay, not reliable
=> do not use
lnC
t – T
Early high points suggest need
for more than one compartment
=> we will consider later
Useable
Points
-2.0
-1.5
-1.0
-0.5
0.0
0.5
0.0
0.5
1.0
1.5
2.0
2.5
y = 0.22844 + -1.0674x R= -0.99808
lnC
Time after Exposure Ended, t – T (hr)
C
t
If possible, average multiple
plateau points to estimate C
ss
in order to improve accuracy.
l
n
(
C
SS
–
C)
t
l
nC
SS
s
lope
=
–
k
l
n
(
C
SS
–
C)
t
L
ate
poi
n
t
s t
ha
t
j
u
m
p
a
r
ou
n
d a
r
e
t
oo
c
lo
s
e to
C
ss
fo
r
u
s
e
i
n
t
he
ana
l
y
s
i
s
.
I
f ea
r
l
y
poi
n
t
s d
o no
t
f
a
ll o
n
li
n
e,
m
ay
n
eed
d
i
ffe
r
e
n
t
m
ode
l
.
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
-0.5
0.0
0.5
0.0
0.5
1.0
1.5
2.0
2.5
3.0
3.5
y = 0.31522 + -1.1777x R= -0.99534
ln(C
ss
-C)
Time (hr)
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