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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

The equation for the iodination of acetone in acidic solution is $$\mathrm{CH}_{3} \mathrm{COCH}_{3}(a q)+\mathrm{I}_{2}(a q) \longrightarrow \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{I}(a q)+\mathrm{H}^{+}(a q)+\mathrm{I}^{-}(a q)$$ The rate of the reaction is found to be dependent not only on the concentration of the reactants but also on the hydrogen ion concentration. Hence the rate expression of this reaction is $$\text { rate }=k\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]^{m}\left[\mathrm{I}_{2}\right]^{n}\left[\mathrm{H}^{+}\right]^{p}$$ The rate is obtained by following the disappearance of iodine using starch as an indicator. The following data are obtained: $$ \begin{array}{cccc} \hline\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right] & \left.\mathrm{[H}^{+}\right] & {\left[\mathrm{I}_{2}\right]} & \text { Initial Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{s}) \\ \hline 0.80 & 0.20 & 0.001 & 4.2 \times 10^{-6} \\ 1.6 & 0.20 & 0.001 & 8.2 \times 10^{-6} \\ 0.80 & 0.40 & 0.001 & 8.7 \times 10^{-6} \\ 0.80 & 0.20 & 0.0005 & 4.3 \times 10^{-6} \\ \hline\end{array}$$ (a) What is the order of the reaction with respect to each reactant? (b) Write the rate expression for the reaction. (c) Calculate \(k\). (d) What is the rate of the reaction when \(\left[\mathrm{H}^{+}\right]=0.933 M\) and \(\left[\mathrm{CH}_{3} \mathrm{COCH}_{3}\right]=3\left[\mathrm{H}^{+}\right]=10\left[\mathrm{I}^{-}\right] ?\)

Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) hydrolyzes into glucose and fructose. The hydrolysis is a first-order reaction. The half-life for the hydrolysis of sucrose is \(64.2\) min at \(25^{\circ} \mathrm{C}\). How many grams of sucrose in \(1.25 \mathrm{~L}\) of a \(0.389 \mathrm{M}\) solution are hydrolyzed in \(1.73\) hours?

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A reaction has two reactants \(\mathrm{A}\) and \(\mathrm{B}\). What is the order with respect to each reactant and the overall order of the reaction described by each of the following rate expressions? (a) rate \(=k_{1}[\mathrm{~A}]^{3}\) (b) rate \(=k_{2}[\mathrm{~A}] \times[\mathrm{B}]\) (c) rate \(=k_{3}[\mathrm{~A}] \times[\mathrm{B}]^{2}\) (d) rate \(=k_{4}[\mathrm{~B}]\)
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