91影视

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

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

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

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.

role="math" localid="1654928290095" $\mathrm{IN}\frac{{\left[\mathrm{A}\right]}_{\mathrm{t}}}{{\left[\mathrm{A}\right]}_0}=-\mathrm{kt}$

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

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

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

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

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

role="math" localid="1654928916433" $$\begin{gathered} \mathrm{In}\frac{{\left[\mathrm{A}\right]}_{\mathrm{t}}}{{\left[\mathrm{A}\right]}_0}=-\mathrm{kt} \\ \mathrm{In}\frac{\left(1/3\right){\left[\mathrm{A}\right]}_0}{{\left[\mathrm{A}\right]}_0}=-\left(3.9脳{10}^{-5}{\mathrm{s}}^{-1}\right)脳\mathrm{t}脳\left(\frac{3600\mathrm{s}}{1\mathrm{h}}\right) \\ \mathrm{In}\left(\frac{1}{3}\right)=-0.1404脳\mathrm{t}脳\frac{1}{\mathrm{h}} \\ \mathrm{t}=7.8\mathrm{h} \end{gathered}$$

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

For third pill: role="math" localid="1654929379501" ${\left[\mathrm{A}\right]}_t=\frac{1}{3}{\left[\mathrm{A}\right]}_0\mathrm{and}{\left[\mathrm{A}\right]}_0=\left(1+\frac{1}{3}\right){\left[\mathrm{A}\right]}_0=\frac{4}{3}{\left[\mathrm{A}\right]}_0$

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

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

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

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 ${\mathrm{T}}_1\mathrm{and}{\mathrm{T}}_2$andrespectively and obtain new expression:

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

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

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

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

Substitute all given values in Arrhenius equation, to getEa

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

Therefore, the activation energy is 101kJ/mol.