) An 81 kg cancer patient receives a 6.5 mg/kg IV dose of a chemotherapy drug. Drugplasma levels were monitored over a period of 7.5 hours. At 4.0 hr after doseadministration, the plasma concentration was 69 mg/L, while after 7.2 hr it was 52 mg/L.Note that the drug exhibits first order elimination kinetics.(i) Estimate the elimination rate constant, the volume of distribution, and the half-life of the drug

To solve this problem, we need to use the following formulas:

1. Elimination rate constant (k) = (ln(C1) - ln(C2)) / (t2 - t1)
2. Volume of distribution (Vd) = Dose / C0
3. Half-life (t1/2) = 0.693 / k

Let's start with the elimination rate constant:

(i) Elimination rate constant (k)

First, we need to convert the plasma concentrations from mg/L to mg/kg. We do this by dividing the concentration by the patient's weight (81 kg).

C1 = 69 mg/L / 81 kg = 0.85185 mg/kg C2 = 52 mg/L / 81 kg = 0.64198 mg/kg

Then we can plug these values into the formula:

k = (ln(0.85185) - ln(0.64198)) / (7.2 hr - 4.0 hr) k = -0.286 hr^-1

(ii) Volume of distribution (Vd)

The initial dose (D) is 6.5 mg/kg * 81 kg = 526.5 mg.

The initial concentration (C0) can be estimated using the concentration at 4.0 hr (C1), since this is close to the time of administration.

C0 = 0.85185 mg/kg

Then we can plug these values into the formula:

Vd = 526.5 mg / 0.85185 mg/kg Vd = 618.1 L

(iii) Half-life (t1/2)

Finally, we can calculate the half-life using the elimination rate constant:

t1/2 = 0.693 / -0.286 hr^-1 t1/2 = 2.42 hr

So, the elimination rate constant is -0.286 hr^-1, the volume of distribution is 618.1 L, and the half-life is 2.42 hr.