Question

A drug decomposes in the blood by a first-order process. A pill containing \(0.500 \mathrm{~g}\) of the active ingredient reaches its maximum concentration of \(2.5 \mathrm{mg} / 100 \mathrm{~mL}\) of blood. If the half-life of the active ingredient is \(75 \mathrm{~min},\) what is its concentration in the blood \(2.0 \mathrm{~h}\) after the maximum concentration has been reached?

Short answer

Answer: After 2 hours, the concentration of the drug in the blood will be approximately \(0.00226 \mathrm{g} / \mathrm{L}\).

Step-by-step solution

Convert the given units into standard units

First, we need to convert the given units into standard units to be used in the calculations. It is convenient to use grams for mass and liters for volume. Initial concentration: \(2.5 \mathrm{mg} / 100 \mathrm{~mL} = 0.025 \mathrm{g} / \mathrm{L}\) Half-life: \(75 \mathrm{~min}\) Duration of interest: \(2.0 \mathrm{~h} = 120 \mathrm{~min}\)

Calculate the first-order rate constant k

Using the half-life of the drug (75 minutes), we can calculate the first-order rate constant k using the standard first-order decay equation: $$k = \frac{\ln 2}{t_{1/2}}$$ Where \(k\) is the rate constant, and \(t_{1/2}\) is the half-life. Plugging in the given values: $$k = \frac{\ln 2}{75 \mathrm{~min}} \approx 0.00926 \mathrm{~min}^{-1}$$

Calculate the remaining concentration at the given time

Now that we have the rate constant, we can use the first-order decay equation to find the remaining concentration after 2 hours or 120 minutes: $$C_t = C_0 e^{-kt}$$ Where \(C_t\) is the concentration of the drug at a given time, \(C_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time duration of interest. Plugging in the given values: $$C_t = 0.025 \mathrm{~~g/L~~} e^{-(0.00926 \mathrm{~~min}^{-1})(120 \mathrm{~~min})} \approx 0.00226 \mathrm{g} / \mathrm{L}$$ So, after 2 hours following the maximum concentration, the concentration of the drug in the blood will be approximately \(0.00226 \mathrm{g} / \mathrm{L}\).