SME
Biotransport · Problem type
An oral dose: peak time and peak level
Use this for “A 500 mg tablet is absorbed at kₐ = 1 h⁻¹ into 50 L and cleared at k = 0.1 h⁻¹. When does the level peak, and how high?”
Chart
Peak
at
tₘₐₓ
=
ln(kₐ ÷ k)
kₐ − k
=
2.56,
Cₘₐₓ
=
7.74
This is an example. Type your own values to start a new problem.
tₘₐₓ (h)
=
ln
kₐ (h⁻¹)
k (h⁻¹)
kₐ (h⁻¹)
−
k (h⁻¹)
F
Fraction absorbed
example
D
Dose
example
V
Volume of distribution
example · L
Cₘₐₓ
Peak concentration
worked out (not typed)
Formulas
tₘₐₓ
=
ln(kₐ ÷
k
)
kₐ −
k
2.56
=
ln(1 ÷ 0.1)
1 − 0.1
Cₘₐₓ
=
F
×
D
×
kₐ
V
× (kₐ −
k
)
×
(
e
−
k
× tₘₐₓ
−
e
−kₐ × tₘₐₓ
)
7.74
=
1
×
500
×
1
50 × (1 − 0.1)
×
(
e
−0.1 × 2.56
−
e
−1 × 2.56
)
Assumptions
•
One well-mixed compartment, filled from the gut at first order (kₐ) and cleared at first order (k).
•
A fraction F of the dose reaches the blood; F = 1 means all of it.
•
At the peak, absorption in equals elimination out: kₐ e−kₐ t = k e−kt.
•
kₐ ≠ k: with equal rates the formula divides 0 by 0.
Step-by-step
WE KNOW
D = 500 mg, F = 1, V = 50 L, kₐ = 1 h⁻¹, k = 0.1 h⁻¹
FIND
time of the peak (tₘₐₓ), peak concentration (Cₘₐₓ)
Step 1 · Find time of the peak (tₘₐₓ)
Use tₘₐₓ = ln(kₐ ÷ k) ÷ (kₐ − k)
Solve
kₐ
e
−kₐ t
=
k
e
−kt
for
t:
the
log
of
the
rate
ratio
over
the
rate
gap.
tₘₐₓ
=
ln(1 ÷ 0.1)
1 − 0.1
tₘₐₓ
=
ln(10)
0.9
tₘₐₓ
=
2.3026
0.9
tₘₐₓ = 2.56 h
Step 2 · Find peak concentration (Cₘₐₓ)
Use Cₘₐₓ = F × D × kₐ ÷ (V × (kₐ − k)) × (e−k × tₘₐₓ − e−kₐ × tₘₐₓ)
Put tₘₐₓ into the Bateman curve: the level climbs while absorption leads, then falls.
Cₘₐₓ
=
1
×
500
×
1
50 × (1 − 0.1)
×
(
e
−0.1 × 2.56
−
e
−1 × 2.56
)
Cₘₐₓ
=
500
50 × 0.9
×
(
e
−0.256
−
e
−2.56
)
Cₘₐₓ
=
500
45
×
(0.77414
−
0.077305)
Cₘₐₓ
=
500
45
×
0.69684
Cₘₐₓ = 11.111 × 0.69684
Cₘₐₓ = 7.74 mg/L
Check
Put the numbers back in. Both sides should match.
2.56
=
ln(1 ÷ 0.1)
1 − 0.1
✓
7.74
=
1
×
500
×
1
50 × (1 − 0.1)
×
(
e
−0.1 × 2.56
−
e
−1 × 2.56
)
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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