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Chapter 11: Problem 37

The first-order rate constant for the decomposition of a certain hormone in water at \(25^{\circ} \mathrm{C}\) is \(3.42 \times 10^{-4}\) day \(^{-1}\). (a) If a \(0.0200 \mathrm{M}\) solution of the hormone is stored at \(25^{\circ} \mathrm{C}\) for two months, what will its concentration be at the end of that period? (b) How long will it take for the concentration of the solution to drop from \(0.0200 M\) to \(0.00350 \mathrm{M} ?\) (c) What is the half-life of the hormone?

Short Answer

Expert verified
Answer: The concentration of the hormone after two months is approximately 0.010 M. The time it takes for the concentration to drop from 0.0200 M to 0.00350 M is approximately 3427 days. The half-life of the hormone is approximately 2026 days.

Step by step solution

01

(Part a: Finding the concentration after two months)

You are given the initial concentration \([A]_0 = 0.0200M\) and the rate constant \(k=3.42\times10^{-4} \text{ day}^{-1}\). First, identify the time period (two months) and convert it to days: $$\text{2 months} = 2\times30 = 60 \text{ days}$$ Then, apply the first-order reaction formula, using the initial concentration, rate constant, and time: $$[A]_t = [A]_0 e^{-kt}$$ $$[A]_t = 0.0200 \times e^{-3.42 \times 10^{-4} \times 60}$$ Now, calculate the concentration after 60 days: $$[A]_t \approx 0.010 \; \text{M}$$ So, the concentration of the hormone after two months will be approximately \(0.010 \: \text{M}\).
02

(Part b: Finding the time for the concentration to drop)

We are given the initial concentration \([A]_0 = 0.0200 \: \text{M}\) and the final concentration \([A]_t = 0.00350 \: \text{M}\). Using the first-order reaction formula, we can find the time \(t\) required for this change: $$[A]_t = [A]_0 e^{-kt}$$ Next, rearrange the formula to solve for \(t\): $$t = \frac{\ln{\frac{[A]_t}{[A]_0}}}{-k}$$ Now, plug in the given values and calculate the time: $$t = \frac{\ln{\frac{0.00350}{0.0200}}}{-3.42 \times 10^{-4}} \approx 3427 \; \text{days}$$ Therefore, it will take approximately \(3427 \: \text{days}\) for the concentration of the hormone solution to drop from \(0.0200 \: \text{M}\) to \(0.00350 \: \text{M}\).
03

(Part c: Finding the half-life of the hormone)

To determine the half-life \(t_{1/2}\) for the hormone, use the first-order reaction half-life formula: $$t_{1/2} = \frac{\ln{2}}{k}$$ Now, plug in the given rate constant and calculate the half-life: $$t_{1/2} = \frac{\ln{2}}{3.42 \times 10^{-4}} \approx 2026 \; \text{days}$$ Thus, the half-life of the hormone is approximately \(2026 \: \text{days}\).

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Most popular questions from this chapter

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?

At high temperatures, the decomposition of cyclobutane is a first-order reaction. Its activation energy is \(262 \mathrm{~kJ} / \mathrm{mol}\). At \(477^{\circ} \mathrm{C}\), its half-life is \(5.00 \mathrm{~min}\). What is its half-life (in seconds) at \(527^{\circ} \mathrm{C}\) ?

The reaction $$\mathrm{ICl}(g)+\frac{1}{2} \mathrm{H}_{2}(g) \longrightarrow \frac{1}{2} \mathrm{I}_{2}(g)+\mathrm{HCl}(g)$$ is first-order in both reactants. The rate of the reaction is \(4.89 \times 10^{-5} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) when the ICl concentration is \(0.100 M\) and that of the hydrogen gas is \(0.030 \mathrm{M}\) (a) What is the value of \(k\) ? (b) At what concentration of hydrogen is the rate \(5.00 \times 10^{-4} \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) and \([\mathrm{ICl}]=0.233 \mathrm{M?}\) (c) At what concentration of iodine chloride is the rate \(0.0934 \mathrm{~mol} / \mathrm{L} \cdot \mathrm{s}\) if the hydrogen concentration is three times that of ICl?

For the reaction $$\mathrm{X}+\mathrm{Y} \longrightarrow \mathrm{R}+\mathrm{Z} \quad \Delta H=+295 \mathrm{~kJ},$$ draw a reaction-energy diagram for the reaction if its activation energy is \(378 \mathrm{~kJ} .\)

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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