Q105CP - Chemistry: Molecular Nature Of Matter And Change

Question

Question:Many drugs decompose in blood by a first-order process. 

(a) Two tablets of aspirin supply 0.60 g of the active compound. After 30 min, this compound reaches a maximum concentration of 2 mg/100 mL of blood. If the half-life for its breakdown is 90 min, what is its concentration (in mg/100 mL) 2.5 h after it reaches its maximum concentration? 

(b) For the decomposition of an antibiotic in a person with a normal temperature $\mathbf{(}\mathbf{98}\mathbf{.}\mathbf{6}\mathbf{°}\mathbf{ }\mathbf{F}\mathbf{)}\mathbf{,}\mathbf{ }\mathbf{k}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{1}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$ ; for a person with a fever at $\mathbf{101}\mathbf{.}\mathbf{9}\mathbf{°}\mathbf{ }\mathbf{F}$, $\mathbf{k}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{9}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}{\mathbf{s}}^{\mathbf{-}\mathbf{1}}$ . If the person with the fever must take another pill when of the first pill has decomposed, how many hours should she wait to take a second pill? A third pill? (Assume the pill is effective immediately.)

(c) Calculate Ea for decomposition of the antibiotic in part (b).

Step-by-Step Solution

Answer

(a)   The concentration of aspirin is 0.63 mg/100 mL.

(b)   The number of hours should be taken for a third pill is 9.9 h.

(c)   The activation energy is 101kJ/mol.

Step 1:What is the concentration of aspirin

Using the formula below, get the rate constant for the first order process:

${\mathrm{t}}_{1/2}=\frac{\mathrm{In}2}{\mathrm{k}}$

The rate constant is k, and the half-life time for the first order process is ${\mathrm{t}}_{1/2}$.

In the given formula, substitute  =90 min to get

$$\begin{gathered} \mathrm{k}=\frac{\mathrm{In}2}{{\mathrm{t}}_{1/2}} \\ =\frac{0.693}{90 \min } \\ =7.7\times {10}^{-3}{\min }^{-1} \end{gathered}$$

Thus, the rate constant is $7.7\times {10}^{-3}{\min }^{-1}$.

Consider ${\left[A\right]}_0$ for the initial concentration and ${\left[A\right]}_t$ for the concentration at time t.

For a first-order reaction, use the integrated rate law to get the reactant concentration.

                        $\mathrm{IN}\frac{{\left[\mathrm{A}\right]}_{\mathrm{t}}}{{\left[\mathrm{A}\right]}_0}=-\mathrm{kt}$

$$\begin{gathered} \mathrm{In}\frac{{\left[\mathrm{A}\right]}_{\mathrm{t}}}{{\left[2 \mathrm{mg}/ 100 \mathrm{mL}\right]}_0}=-\left(7.7\times {10}^{-3}{\min }^{-1}\right)\left(2.5 \mathrm{h}\times \frac{60 \min }{1 \mathrm{h}}\right) \\ \mathrm{In}\frac{{\left[\mathrm{A}\right]}_{\mathrm{t}}}{{\left[2 \mathrm{mg}/ 100 \mathrm{mL}\right]}_0}=-1.155 \\ \mathrm{In}\frac{{\left[\mathrm{A}\right]}_{\mathrm{t}}}{{\left[2 \mathrm{mg}/ 100 \mathrm{mL}\right]}_0}={\mathrm{e}}^{-1.155} \\ \mathrm{In}\frac{{\left[\mathrm{A}\right]}_{\mathrm{t}}}{{\left[2 \mathrm{mg}/ 100 \mathrm{mL}\right]}_0}=0.315 \\ {\left[\mathrm{A}\right]}_{\mathrm{t}}=0.315\times \left(2 \mathrm{mg}/100 \mathrm{mL}\right) \\ =0.63 \mathrm{mg}/100 \mathrm{mL} \end{gathered}$$

Therefore, the concentration of aspirin is 0.63 mg/100 mL.

Step2:The number of hours should be taken for a third pill

Determine the time interval using integrated rate law for first order reaction.

Use the rate constant for the fever. That is $3.9\times {10}^{-5}{s}^{-1}$.

For second pill: 2/3 of the first fill has decomposed.

Therefore, ${\left[\mathrm{A}\right]}_{\mathrm{t}}=\left(1-\frac{2}{3}\right){\left[\mathrm{A}\right]}_0 \mathrm{and}{\left[\mathrm{A}\right]}_0={\left[\mathrm{A}\right]}_0$

Substitute the all known values in first-order reaction, get

$$\begin{gathered} \mathrm{In}\frac{{\left[\mathrm{A}\right]}_{\mathrm{t}}}{{\left[\mathrm{A}\right]}_0}=-\mathrm{kt} \\ \mathrm{In}\frac{\left(1/3\right){\left[\mathrm{A}\right]}_0}{{\left[\mathrm{A}\right]}_0}=-\left(3.9\times {10}^{-5}{\mathrm{s}}^{-1}\right)\times \mathrm{t}\times \left(\frac{3600 \mathrm{s}}{1 \mathrm{h}}\right) \\ \mathrm{In}\left(\frac{1}{3}\right)=-0.1404\times \mathrm{t}\times \frac{1}{\mathrm{h}} \\ \mathrm{t}=7.8 \mathrm{h} \end{gathered}$$

Therefore, the number of hours should be taken for a second pill is 7.8h.

For third pill:   ${\left[\mathrm{A}\right]}_t=\frac{1}{3}{\left[\mathrm{A}\right]}_0 \mathrm{and} {\left[\mathrm{A}\right]}_0=\left(1+\frac{1}{3}\right){\left[\mathrm{A}\right]}_0=\frac{4}{3}{\left[\mathrm{A}\right]}_0$

Substitute the all known values in first-order reaction, get

$$\begin{gathered} \mathrm{In}\frac{{\left[\mathrm{A}\right]}_{\mathrm{t}}}{{\left[\mathrm{A}\right]}_0}=-\mathrm{kt} \\ \mathrm{In}\frac{\left(1/3\right){\left[\mathrm{A}\right]}_0}{\left(4/3\right){\left[\mathrm{A}\right]}_0}=-\left(3.9\times {10}^{-5}{\mathrm{s}}^{-1}\right)\times \mathrm{t}\times \left(\frac{3600 \mathrm{s}}{1 \mathrm{h}}\right) \\ \mathrm{In}\left(\frac{1}{4}\right)=-0.1404\times \mathrm{t}\times \frac{1}{\mathrm{h}} \\ \mathrm{t}=9.9 \mathrm{h} \end{gathered}$$

Therefore, the number of hours should be taken for a third pill is 9.9 h.

Step3:Calculate Ea

The Arrhenius equation can be expressed as follows:

$$\begin{gathered} \mathrm{k}={\mathrm{Ae}}^{-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT}} \\ \mathrm{In} \mathrm{k}=\mathrm{In} \mathrm{A}-{\mathrm{E}}_{\mathrm{a}}/\mathrm{RT} \end{gathered}$$

Here, k is the rate constant, A is the frequency factor, ${\mathrm{E}}_{\mathrm{a}}$ is the activation energy of the reaction at a specified temperature T and R is the gas constant.

Write the equation for the two rate constants, ${\mathrm{k}}_1 \mathrm{and} {\mathrm{k}}_2$ and at temperatures ${\mathrm{T}}_1 \mathrm{and} {\mathrm{T}}_2$and respectively and obtain new expression:

$$\begin{gathered} \mathrm{In} {\mathrm{k}}_1=\mathrm{In} \mathrm{A}‐{\mathrm{E}}_{\mathrm{a}}/{\mathrm{RT}}_1 \\ \mathrm{In} {\mathrm{k}}_2=\mathrm{In} \mathrm{A}‐{\mathrm{E}}_{\mathrm{a}}/{\mathrm{RT}}_1 \\ \mathrm{In} {\mathrm{k}}_2-\mathrm{In} {\mathrm{k}}_1=-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right) \\ \mathrm{In}\left(\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}\right)=-\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right) \end{gathered}$$

Covert the units of temperature from degrees of foreign heat (F) to Kelvin (K) scale:

$$\begin{gathered} {\mathrm{T}}_1=\left(273.15+\left(\frac{5}{9}\times \left[98.6°\mathrm{F}‐32\right]\right)\right)\mathrm{K} \\ =310.15 \mathrm{K} \\ {\mathrm{T}}_2=\left(273.15+\left(\frac{5}{9}\times \left[101.9°\mathrm{F}‐32\right]\right)\right)\mathrm{K} \\ =311.98 \mathrm{K} \end{gathered}$$

Also have, ${\mathrm{k}}_1=3.1\times {10}^{-5}{\mathrm{s}}^{-1} \mathrm{and} {\mathrm{k}}_2=3.9\times {10}^{-5}{\mathrm{s}}^{-1}$

Substitute all given values in Arrhenius equation, to getEa

$$\begin{gathered} \mathrm{In}\left(\frac{3.9\times {10}^{-5}{\mathrm{s}}^{-1}}{3.1\times {10}^{-5}{\mathrm{s}}^{-1}}\right)=-\frac{{\mathrm{E}}_{\mathrm{a}}}{8.314 \mathrm{J}/\mathrm{mol}.\mathrm{K}}\left(\frac{1}{311.98 \mathrm{K}}-\frac{1}{310.15 \mathrm{K}}\right) \\ {\mathrm{E}}_{\mathrm{a}}=\frac{\left(8.314 \mathrm{J}/\mathrm{mol}.\mathrm{K}\right)\times \mathrm{In}\left(\frac{3.9\times {10}^{-5}{\mathrm{s}}^{-1}}{3.1\times {10}^{-5}{\mathrm{s}}^{-1}}\right)}{\left(\frac{1}{311.98 \mathrm{K}}-\frac{1}{310.15 \mathrm{K}}\right)} \\ =1.01\times 105 \mathrm{J}/\mathrm{mol} \\ =101 \mathrm{kJ}/\mathrm{mol} \end{gathered}$$

Therefore, the activation energy is 101kJ/mol.