Answer:
See explanation below
Step-by-step explanation:
In this case, let's answer this by parts:
<u>a) Concentration at time t when Co is concentration at t = 0:</u>
In this case, we will use the initial expression but without the 2, so:
C(t) = -kC
In this case, we want to know the concentration at time t so:
C'(t) = -kC(t) derivating:
dC/dt = -kC
dC/C = -kdt From here, we can do integrals so:
lnC = -kt + C₁ (1)
Now, it's time to replace t = 0 and C = C₀:
lnC₀ = -k(0) + C₁
lnC₀ = C₁ (2)
Replacing (2) in (1) we have:
lnC = -kt + lnC₀
lnC - lnC₀ = -kt
ln(C/C₀) = -kt
C/C₀ = e^(-kt)
C(t) = C₀ e^(-kt) (3)
This is the expression for C at given time t.
<u>2. time to eliminate 80% of the drug:</u>
With the first data, we need to calculate the value of k, which will be constant at any given time so:
C(t) = C₀ e^(-kt)
0.5C₀ = C₀ e^(-30k)
0.5 = e^(-30k)
ln(0.5) = -30k
k = 0.02310
Now with this value we can calculate the time to eliminate 80% of the drug or simply in other words, that we just have a remaining of 0.2C₀:
0.2C₀ = C₀ e^(-0.0231t)
ln(0.2) = -0.0231t
-ln(0.2) / 0.0231 = t
t = 69.67 h
This is the time to eliminate 80% of the drug
