Chapter 20, Problem 20.57E

An interesting pair of consecutive reactions involve the absorption of ethyl alcohol by the body, which is a first-order process, and the consequent oxidation of alcohol to acetaldehyde by liver alcohol dehydrogenase (LADH), which is a zeroth order process. The differential changes in the three states of ethanol can therefore be described as

$\begin{array}{rll}-\frac{d\left[\text{A}\right]}{dt}& =& k_1{\left[\text{A}\right]}_t\\ \frac{d\left[\text{B}\right]}{dt}& =& k_1{\left[\text{A}\right]}_t-k_2\\ \frac{d\left[\text{C}\right]}{dt}& =& k_2\end{array}$

which are slightly modified from equations $20.46$. The integrated form of the first equation is the same as for two consecutive first-order reactions, but for the second and third reactions, they will not be. (a) What do $\text{A}$, $\text{B}$ and $\text{C}$ stand for in this example? (b) Determine an integrated form for $\left[\text{B}\right]$ over time. Do this by finding an integrated expression for ${\left[\text{A}\right]}_t$ (Hint: Refer to the chapter!), substitute for ${\left[\text{A}\right]}_t$ in the second expression, rearrange the infinitesimals, and integrate one side over $\left[\text{B}\right]$ as the variable and the other side over time, $t$, as the variable. The integrations are actually simpler than for the consecutive first-order reaction example in the chapter itself. (c) Determine an integrated form for $\left[\text{C}\right]$ over time. (d) Rough values of $k_1$ and $k_2$ for people are $3.00\times {10}^{-3}{\text{s}}^{-1}$ and $4.44\times {10}^{-5}\text{mol}/\text{s}$, respectively. Use your expressions for the amounts over time to plot a graph of ${\left[\text{A}\right]}_t$, ${\left[\text{B}\right]}_t$, and ${\left[\text{C}\right]}_t$ versus time. (A graphing calculator or a computer with a graphing program would be useful.) Use $1.00\text{mol}$ of ${\text{C}}_{\text{2}}{\text{H}}_{\text{5}}\text{OH}$ as ${\left[\text{A}\right]}_0$. Vary this value and see how it affects the graphs of ${\left[\text{A}\right]}_t$, ${\left[\text{B}\right]}_t$, and ${\left[\text{C}\right]}_t$.

Interpretation Introduction

(a)

Interpretation:

The species $\text{A}$, $\text{B}$ and $\text{C}$ are to be predicted from the consecutive reaction involving alcohol and acetaldehyde.

Concept introduction:

In the consecutive reactions the product of first reaction acts as the reactant for the second reaction and so on. The radioactive decays are one of the examples of the consecutive reactions. The simple two step consecutive reaction is shown below.

$\text{A}\stackrel{k_1}{\to }\text{B}\stackrel{k_2}{\to }\text{C}$

Answer to Problem 20.57E

The species $\text{A}$, $\text{B}$ and $\text{C}$ are, ethanol, intermediate between ethanol and acetaldehyde and acetaldehyde respectively.

Explanation of Solution

The given consecutive reaction involves the absorption of ethyl alcohol by the body which is a first order reaction followed by the oxidation of ethyl alcohol to acetaldehyde in the presence of liver alcohol dehydrogenase (LADH).

The rate of change of concentration of the three species is given below.

$\begin{array}{rll}-\frac{d\left[\text{A}\right]}{dt}& =& k_1{\left[\text{A}\right]}_t\\ \frac{d\left[\text{B}\right]}{dt}& =& k_1{\left[\text{A}\right]}_t-k_2\\ \frac{d\left[\text{C}\right]}{dt}& =& k_2\end{array}$

From the above consecutive reaction it is concluded that $\text{A}$ is ethanol and $\text{C}$ is acetaldehyde. The concentration of $\text{B}$ depends upon the concentration of $\text{A}$ at time $t$ and rate constant $k_2$. Therefore, $\text{B}$ is the intermediate between ethanol and acetaldehyde.

Conclusion

The species $\text{A}$, $\text{B}$ and $\text{C}$ are, ethanol, intermediate between ethanol and acetaldehyde and acetaldehyde respectively.

Interpretation Introduction

(b)

Interpretation:

The integrated form for $\left[\text{B}\right]$ over time is to be stated.

Concept introduction:

In the consecutive reactions, the product of first reaction acts as the reactant for the second reaction and so on. The radioactive decays are one of the examples of the consecutive reactions. The simple two step consecutive reaction is shown below.

$\text{A}\stackrel{k_1}{\to }\text{B}\stackrel{k_2}{\to }\text{C}$

Answer to Problem 20.57E

The integrated form for $\left[\text{B}\right]$ over time is shown below.

$\left[\text{B}\right]=-{\left[\text{A}\right]}_0\cdot e^{-k_{1}t}-k_{2}t$

Explanation of Solution

The differential change in the concentration of $\text{B}$ is shown below.

$\frac{d\left[\text{B}\right]}{dt}=k_1{\left[\text{A}\right]}_t-k_2$…(1)

The integrated rate law for the first order reaction of ${\left[\text{A}\right]}_t$ is shown below.

$\begin{array}{rll}\ln {\left[\text{A}\right]}_t& =& \ln {\left[\text{A}\right]}_0-k_{1}t\\ {\left[A\right]}_t& =& {\left[\text{A}\right]}_0\cdot e^{-k_{1}t}\end{array}$…(2)

Substitute equation (2) in equation (1).

$\frac{d\left[\text{B}\right]}{dt}=k_1{\left[\text{A}\right]}_0\cdot e^{-k_{1}t}-k_2$

Integrate the above equation.

$\begin{array}{rll}{\displaystyle \int d\left[\text{B}\right]}& =& {\displaystyle \int \left(k_1{\left[\text{A}\right]}_0\cdot e^{-k_{1}t}-k_2\right)dt}\\ {\displaystyle \int d\left[\text{B}\right]}& =& k_1{\left[\text{A}\right]}_0{\displaystyle \int e^{-k_{1}t}dt}-k_2{\displaystyle \int dt}\\ \left[\text{B}\right]& =& -k_1{\left[\text{A}\right]}_0.\frac{e^{-k_{1}t}}{k_1}-k_{2}t\\ & =& -{\left[\text{A}\right]}_0\cdot e^{-k_{1}t}-k_{2}t\end{array}$

Conclusion

The integrated form for $\left[\text{B}\right]$ over time is shown below.

$\left[\text{B}\right]=-{\left[\text{A}\right]}_0\cdot e^{-k_{1}t}-k_{2}t$

Interpretation Introduction

(c)

Interpretation:

The integrated form for $\left[\text{C}\right]$ over time is to be determined.

Concept introduction:

In the consecutive reactions the product of first reaction acts as the reactant for the second reaction and so on. The radioactive decays are one of the examples of the consecutive reactions. The simple two step consecutive reaction is shown below.

$\text{A}\stackrel{k_1}{\to }\text{B}\stackrel{k_2}{\to }\text{C}$

Answer to Problem 20.57E

The integrated form for $\left[\text{C}\right]$ over time is shown below.

$\left[\text{C}\right]=-\left[\text{B}\right]\ln t-{\left[\text{A}\right]}_0\cdot \left(\ln \left|t\right|+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-k_{1}t\right)}^n}{n\cdot n!}}\right)$

Explanation of Solution

The differential change in the concentration of $\text{C}$ is shown below.

$\frac{d\left[\text{C}\right]}{dt}=k_2$…(3)

The integrated form for $\left[\text{B}\right]$ over time is shown below.

$\begin{array}{rll}\left[\text{B}\right]& =& -{\left[\text{A}\right]}_0\cdot e^{-k_{1}t}-k_{2}t\\ k_2& =& -\left(\frac{\left[\text{B}\right]+{\left[\text{A}\right]}_0\cdot e^{-k_{1}t}}{t}\right)\end{array}$…(4)

Substitute equation (4) in equation (3).

$\begin{array}{rll}\frac{d\left[\text{C}\right]}{dt}& =& -\left(\frac{\left[\text{B}\right]+{\left[\text{A}\right]}_0\cdot e^{-k_{1}t}}{t}\right)\\ & =& -\frac{\left[\text{B}\right]}{t}-\frac{{\left[\text{A}\right]}_0\cdot e^{-k_{1}t}}{t}\\ d\left[\text{C}\right]& =& -\frac{\left[\text{B}\right]}{t}dt-\frac{{\left[\text{A}\right]}_0\cdot e^{-k_{1}t}}{t}dt\end{array}$

Integrate the above equation.

$\begin{array}{rll}d\left[\text{C}\right]& =& -{\displaystyle \int \frac{\left[\text{B}\right]}{t}dt}-{\displaystyle \int \frac{{\left[\text{A}\right]}_0\cdot e^{-k_{1}t}}{t}dt}\\ \left[\text{C}\right]& =& -\left[\text{B}\right]\ln t-{\left[\text{A}\right]}_0{\displaystyle \int \frac{1}{t}\cdot e^{-k_{1}t}dt}\end{array}$

The above exponential function is integrated using the identity shown below.

${\displaystyle \int \frac{e^{cx}}{x}dx}=\ln \left|x\right|+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(cx\right)}^n}{n\cdot n!}}$

On applying this identity on the exponential function the expression for the integrated form for $\left[\text{C}\right]$ over time is shown below.

$\left[\text{C}\right]=-\left[\text{B}\right]\ln t-{\left[\text{A}\right]}_0\cdot \left(\ln \left|t\right|+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-k_{1}t\right)}^n}{n\cdot n!}}\right)$

Conclusion

The integrated form for $\left[\text{C}\right]$ over time is shown below.

$\left[\text{C}\right]=-\left[\text{B}\right]\ln t-{\left[\text{A}\right]}_0\cdot \left(\ln \left|t\right|+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-k_{1}t\right)}^n}{n\cdot n!}}\right)$

Interpretation Introduction

(d)

Interpretation:

The graph for ${\left[\text{A}\right]}_t$, ${\left[\text{B}\right]}_t$ and ${\left[\text{C}\right]}_t$ versus time is to be plotted.

Concept introduction:

In the consecutive reactions the product of first reaction acts as the reactant for the second reaction and so on. The radioactive decays are one of the examples of the consecutive reactions. The simple two step consecutive reaction is shown below.

$\text{A}\stackrel{k_1}{\to }\text{B}\stackrel{k_2}{\to }\text{C}$

Answer to Problem 20.57E

The graph for ${\left[\text{A}\right]}_t$, ${\left[\text{B}\right]}_t$ and ${\left[\text{C}\right]}_t$ versus time is shown below.

Explanation of Solution

The expressions for ${\left[\text{A}\right]}_t$, ${\left[\text{B}\right]}_t$, and ${\left[\text{C}\right]}_t$ are shown below.

$\begin{array}{l}{\left[A\right]}_t={\left[\text{A}\right]}_0\cdot e^{-k_{1}t}\\ \left[\text{B}\right]=-{\left[\text{A}\right]}_0\cdot e^{-k_{1}t}-k_{2}t\\ \left[\text{C}\right]=-\left[\text{B}\right]\ln t-{\left[\text{A}\right]}_0\cdot \left(\ln \left|t\right|+{\displaystyle \sum _{n=1}^{\infty }\frac{{\left(-k_{1}t\right)}^n}{n\cdot n!}}\right)\end{array}$

The value of ${\left[\text{A}\right]}_0$ is given as $1.00\text{mol}$ and the rate constant is $k_1$ and $k_2$ for people are $3.00\times {10}^{-3}{\text{s}}^{-1}$ and $4.44\times {10}^{-5}\text{mol}/\text{s}$, respectively

The plot for ${\left[\text{A}\right]}_t$, ${\left[\text{B}\right]}_t$ and ${\left[\text{C}\right]}_t$ versus time is shown below.

Figure 1

The concentration of ${\left[\text{A}\right]}_t$ decreases over time exponentially whereas for ${\left[\text{B}\right]}_t$, the concentration increases and then decreases and for ${\left[\text{C}\right]}_t$ the concentration increases over time.

Conclusion

The graph for ${\left[\text{A}\right]}_t$, ${\left[\text{B}\right]}_t$ and ${\left[\text{C}\right]}_t$ versus time is shown in Figure 1.