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Chapter 11: Problem 104

In a first-order reaction, suppose that a quantity \(X\) of a reactant is added at regular intervals of time, \(\Delta t\). At first the amount of reactant in the system builds up; eventually, however, it levels off at a saturation value given by the expression $$\text { saturation value }=\frac{X}{1-10^{-a}} \quad \text { where } a=0.30 \frac{\Delta t}{t_{1 / 2}}$$ This analysis applies to prescription drugs, of which you take a certain amount each day. Suppose that you take \(0.100 \mathrm{~g}\) of a drug three times a day and that the half-life for elimination is \(2.0\) days. Using this equation, calculate the mass of the drug in the body at saturation. Suppose further that side effects show up when \(0.500 \mathrm{~g}\) of the drug accumulates in the body. As a pharmacist, what is the maximum dosage you could assign to a patient for an 8 -h period without causing side effects?

Short Answer

Expert verified
Answer: The maximum dosage is approximately 0.403 g.

Step by step solution

01

Calculate the time interval at which the drug is taken

The drug is taken three times a day. So, the time interval between two drug intakes is: $$\Delta t = \frac{24\,\text{h}}{3} = 8\,\text{h}$$
02

Calculate the value of a

Using the given formula for a, $$a = 0.30 \frac{\Delta t}{t_{1/2}}$$ where \(t_{1/2} = 2.0\,\text{days} = 48\,\text{hours}\). Plugging in the values, $$a = 0.30 \frac{8}{48} = 0.05$$
03

Calculate the saturation value

Saturation value is given by $$\text{saturation value} = \frac{X}{1-10^{-a}}$$ where \(X = 0.100\,\mathrm{g}\). Plugging in the values, $$\text{saturation value} = \frac{0.100}{1-10^{-0.05}} \approx 0.10074\,\mathrm{g}$$ This is the mass of the drug in the body at saturation.
04

Determine the maximum allowed dosage for 8 hours

From the problem, side effects show up when a total of \(0.500\,\mathrm{g}\) of drug accumulates in the body. So, we need to find the dosage that will not exceed the side effects threshold: First, we need to find the number of saturation value equivalent to reach \(0.500\,\mathrm{g}\): $$\frac{0.500\,\mathrm{g}}{0.10074\,\mathrm{g}} \approx 4.97$$ We can round \(4.97\) to \(4\) times the accepted saturation value to be on the safer side and avoid the side effects. In this case, the maximum dosage within 8 hours should be: $$4 \times 0.10074\,\mathrm{g} \approx 0.403\,\mathrm{g}$$ Therefore, the maximum dosage that can be assigned to a patient for an 8-hour period without causing side effects is approximately \(0.403\,\mathrm{g}\).

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Most popular questions from this chapter

For the reaction $$\mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}$$ the rate expression is rate \(=k[\mathrm{~A}][\mathrm{B}]\) (a) Given three test tubes, with different concentrations of \(\mathrm{A}\) and \(\mathrm{B}\), which test tube has the smallest rate? (1) \(0.10 M \mathrm{~A}_{\mathrm{i}} 0.10 \mathrm{M} \mathrm{B}\) (2) \(0.15 \mathrm{MA}_{;} 0.15 \mathrm{M} \mathrm{B}\) (3) \(0.06 M \mathrm{~A} ; 1.0 M \mathrm{~B}\) (b) If the temperature is increased, describe (using the words increases, decreases, or remains the same) what happens to the rate, the value of \(k\), and \(E_{x}\)

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WEB When boron trifluoride reacts with ammonia, the following \(T\) reaction occurs: for $$\mathrm{BF}_{3}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{BF}_{3} \mathrm{NH}_{3}(g)$$ The following data are obtained at a particular temperature: $$\begin{array}{cccc}\hline \text { Expt. } & {\left[\mathrm{BF}_{3}\right]} & {\left[\mathrm{NH}_{3}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\\\\hline 1 & 0.100 & 0.100 & 0.0341 \\ 2 & 0.200 & 0.233 & 0.159 \\ 3 & 0.200 & 0.0750 & 0.0512 \\ 4 & 0.300 & 0.100 & 0.102 \\\\\hline\end{array}$$

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