Zero order kinetics pharm: concentration–time calculation
Zero order kinetics pharm refers to a saturated (capacity-limited) elimination regime in which the rate of concentration decrease is approximately constant, matching the zero-order rate law from chemical kinetics.
Problem data
- Zero-order elimination constant: \(k_0 = 1.5\ \text{mg/L·h}\)
- Initial concentration: \(C_0 = 20.0\ \text{mg/L}\)
- Time of interest: \(t = 6.0\ \text{h}\)
- Target concentration: \(C = 5.0\ \text{mg/L}\)
Step 1: Write the zero-order differential rate law
For a zero-order decrease in concentration:
\[ -\frac{dC}{dt} = k_0 \]
The negative sign indicates that \(C\) decreases as time increases.
Step 2: Integrate to obtain the zero-order integrated rate law
Integrate both sides from \(t=0\) (where \(C=C_0\)) to time \(t\) (where \(C=C(t)\)):
\[ \int_{C_0}^{C(t)} dC = -\int_{0}^{t} k_0\, dt \] \[ C(t) - C_0 = -k_0 t \] \[ C(t) = C_0 - k_0 t \]
Step 3: Compute the concentration after 6.0 h
Substitute \(C_0 = 20.0\ \text{mg/L}\), \(k_0 = 1.5\ \text{mg/L·h}\), and \(t = 6.0\ \text{h}\):
\[ C(6.0) = 20.0 - \left(1.5 \times 6.0\right) \] \[ C(6.0) = 20.0 - 9.0 = 11.0\ \text{mg/L} \]
Step 4: Find the time to reach 5.0 mg/L
Set \(C(t)=5.0\ \text{mg/L}\) and solve for \(t\):
\[ 5.0 = 20.0 - 1.5t \] \[ 1.5t = 20.0 - 5.0 = 15.0 \] \[ t = \frac{15.0}{1.5} = 10.0\ \text{h} \]
Quick validity check
Zero-order decay predicts \(C(t)\ge 0\) up to \(t_{\text{zero}}=\dfrac{C_0}{k_0}\):
\[ t_{\text{zero}}=\frac{20.0}{1.5}=13.333\ldots\ \text{h} \]
Since \(10.0\ \text{h} < 13.333\ldots\ \text{h}\), the computed time is within the physically meaningful interval.
Concentration values over time (helpful check table)
| Time \(t\) (h) | \(C(t)=C_0-k_0 t\) (mg/L) |
|---|---|
| 0 | \(20.0\) |
| 2 | \(20.0 - (1.5 \times 2)=17.0\) |
| 4 | \(20.0 - (1.5 \times 4)=14.0\) |
| 6 | \(20.0 - (1.5 \times 6)=11.0\) |
| 8 | \(20.0 - (1.5 \times 8)=8.0\) |
| 10 | \(20.0 - (1.5 \times 10)=5.0\) |
Visualization: zero-order (linear) concentration decrease
Final results
- \(C(6.0\ \text{h}) = 11.0\ \text{mg/L}\)
- Time to reach \(5.0\ \text{mg/L}\): \(t = 10.0\ \text{h}\)