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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Saturation Value
In the context of a first-order reaction, the saturation value is the point at which the accumulation of a reactant reaches a steady state and no longer increases significantly. This happens when the rate of input of the reactant balances out with the rate of its elimination. In pharmaceutical terms, this concept is applied to drugs to predict the highest concentration that the drug will reach in the body under regular dosage regimes.
The equation to find the saturation value in such a scenario is:\[ \text{saturation value} = \frac{X}{1-10^{-a}} \]Where:
  • \(X\) is the dose of the drug administered regularly.
  • \(a\) is a factor derived from the time between doses and the drug's half-life, calculated as \(a = 0.30 \frac{\Delta t}{t_{1/2}}\).
A greater understanding of the saturation value helps in ensuring that medication dosage is safe and does not inadvertently lead to toxicity by accumulating excessively in the body.
Half-life
Half-life (\(t_{1/2}\)) is a critical concept in pharmacokinetics—the study of how drugs move within the body. It refers to the time required for the concentration of a drug in the body to reduce by half. Understanding the half-life of a drug is essential for making recommendations on how frequently a medication should be administered to maintain its therapeutic effect without overdose.The half-life depends on both the rate of elimination of the drug from the body and its clearance. For the given exercise, the drug's half-life is 2 days (or 48 hours), meaning it takes that amount of time for half of the drug's concentration to be eliminated through metabolism and excretion.Knowing the half-life helps pharmacologists and medical professionals calculate "\(a\)" which is further used to find the saturation value. This understanding aids in optimizing dosage regimens for various patients, taking into account factors like kidney and liver function that might affect drug clearance.
Dosage Calculation
Dosage calculation is a fundamental part of ensuring the effectiveness and safety of drug therapy. In this exercise, the calculation involves determining the amount of drug that can be safely administered within a given time frame without exceeding the saturation point and risking side effects.To calculate the dosage, consider the regularity of drug administration and its half-life, to derive the value of "\(a\)" for use in the saturation formula. Then, figure out how many times the saturation value can be safely maintained to avoid exceeding a threshold after which side effects like toxicities may begin to manifest.
  • The goal is to manage the amount of drug in the body so it stays below the level that causes side effects—here specified as when \(0.500\,\mathrm{g}\) of drug accumulates.
  • Such calculations ensure that the therapy is both effective and safe, focusing on the patient's wellbeing without adverse effects.
Drug Accumulation
Drug accumulation refers to the build-up of a drug within the body over time, typically as a consequence of repeated dosing. When a drug is administered continuously before the body can entirely eliminate the previous dose, its levels can compound, reaching a potentially harmful saturation. The problem showcases how understanding drug accumulation is integral to tailoring medication dosages. As medicine is taken repeatedly throughout the day, it is essential to predict the maximum amount that might build up in the body. Using the saturation value in conjunction with half-life allows for the plotting of a safe dosing schedule to ensure that drug levels stabilize before reaching a toxic concentration. This exercise demonstrates the importance of managing accumulated doses carefully, adjusting whenever necessary to prevent symptoms like headaches, dizziness, or more severe side effects associated with high concentrations of pharmaceuticals in the body.

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Most popular questions from this chapter

The gas-phase reaction between hydrogen and iodine $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)$$ proceeds with a rate constant for the forward reaction at \(700^{\circ} \mathrm{C}\) of \(138 \mathrm{~L} / \mathrm{mol} \cdot \mathrm{s}\) and an activation energy of \(165 \mathrm{~kJ} / \mathrm{mol}\). (a) Calculate the activation energy of the reverse reaction given that \(\Delta H_{i}^{\circ}\) for HI is \(26.48 \mathrm{~kJ} / \mathrm{mol}\) and \(\Delta H_{i}^{\circ}\) for \(\mathrm{I}_{2}(\mathrm{~g})\) is \(62.44 \mathrm{~kJ} / \mathrm{mol}\). (b) Calculate the rate constant for the reverse reaction at \(700^{\circ} \mathrm{C}\). (Assume \(\mathrm{A}\) in the equation \(k=\mathrm{Ae}^{-\mathcal{P}_{2} / \mathrm{RT}^{\prime}}\) is the same for both forward and reverse reactions.) (c) Calculate the rate of the reverse reaction if the concentration of HI is \(0.200 M\). The reverse reaction is second-order in HI.

For the zero-order decomposition of ammonia on tungsten $$\mathrm{NH}_{3}(\mathrm{~g}) \stackrel{\mathrm{w}}{\longrightarrow} \frac{1}{2} \mathrm{~N}_{2}(g)+\frac{3}{2} \mathrm{H}_{2}(g)$$ the rate constant is \(2.08 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\). (a) What is the half-life of a \(0.250 \mathrm{M}\) solution of ammonia? (b) How long will it take for the concentration of ammonia to drop from \(1.25 M\) to \(0.388 M ?\)

Express the rate of the reaction $$2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ in terms of (a) \(\Delta\left[\mathrm{C}_{2} \mathrm{H}_{6}\right]\) (b) \(\Delta\left[\mathrm{CO}_{2}\right]\)

Hypofluorous acid, HOF, is extremely unstable at room temperature. The following data apply to the decomposition of HOF to \(\mathrm{HF}\) and \(\mathrm{O}_{2}\) gases at a certain temperature. $$\begin{array}{cl}\hline \text { Time (min) } & \text { [HOF] } \\\\\hline 1.00 & 0.607 \\\2.00 & 0.223 \\\3.00 & 0.0821 \\\4.00 & 0.0302 \\\5.00 & 0.0111 \\\\\hline \end{array}$$ (a) By plotting the data, show that the reaction is first-order. (b) From the graph, determine \(k\). (c) Using \(k\), find the time it takes to decrease the concentration to \(0.100 \mathrm{M}\) (d) Calculate the rate of the reaction when [HOF] \(=0.0500 M\).

Derive the integrated rate law, \([\mathrm{A}]=[\mathrm{A}]_{0}-k t\), for a zero-order reaction. (Hint: Start with the relation \(-\Delta[\mathrm{A}]=k \Delta \mathrm{t}\).)

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