ln (60/100) = -k . (3h)
k = 0,17 h-1
C0 = 100%

t = 3 hs

C=?

C = C0 . e –kt
C = 100 . e -0,17 1/h 9 h = 21,65 %
Rta: 21,65%

Problema 14
Qiny = 15 mmol
C =

Q iny
Vd

Vd = 15 L
15 mmol

=

15 L

C = C0 = ?

t1/2 = 3 h

= 1 mmol/L

t1/2 = ln 2 / k
k = ln 2 / t1/2
k = 0,693 / 3 h = 0,231 h-1
C = C0 . e –kt
C = 1 mmol/L . e -0,231 1/h . 12 h = 0,063 mmol/L = 0,063 mM
Rta: 0,063 mM

Problema 15
Peso = 70 Kg

Qiny = 1 g (ACT)

100 %

70 Kg

60 %

x = 42 Kg = 42 L

C =

Q iny
Vd

=

1000 mg
42 L

k = 0,34 h-1

= 23,80 mg/L

C = C0 . e –kt
C = 23,80 mg/L . e -0,340 1/h . 2 h = 12,06 mg/L
Rta: 12,06 mg/L

ACT = 60% de 70 Kg

t=2h