Chapter 18, Problem 18.3P

The second-order liquid-phase reaction

$\text{A}\stackrel{}{\to }\text{B}+\text{C}$

is to be carried out isothermally. The entering concentration of A is 1.0 mol/dm³. The specific reaction rate is 1.0 dm³/mol·min. A number of used reactors (shown below) are available, each of which has been characterized by an RTD. There are two crimson and white reactors, and three maize and blue reactors available.

1. (a) You have $50,000 available to spend. What is the greatest conversion you can achieve with the available money and reactors?
2. (b) How would your answer to (a) change if you had an extra $75.000 available to spend?
3. (c) From which cities do you think the various used reactors came from?

Interpretation Introduction

Interpretation:

The greatest amount of the conversion that can be attained with the available money and reactors is to be stated.

Concept introduction:

The conversion, X can be defined as the moles of any species A that are reacted per mole of A fed in the reactor.

Answer to Problem 18.3P

The greatest amount of the conversion that can be attained with the available money and reactors is $0.86$ that is present in both orange and blue reactor and silver and black reactor.

Explanation of Solution

It is given that some bromine present in a glass battery jar is dissolved in water and placed in the sunlight directly for the photochemical decay.

The given second-order liquid-phase reaction is as follows.

    $\text{A}\to \text{B}+\text{C}$

The initial entering concentration of reactant A, $C_{\text{A0}}$, is $1.0\text{mol}/{\text{dm}}^{\text{3}}$.

The given specific rate constant of the reaction, $k$, is $1.0{\text{dm}}^{\text{3}}\text{/mol}\cdot \min$.

The given reactors with their space time and area are shown as follows.

Reactors	$\sigma \left(\min \right)$	$\tau \left(\min \right)$	Cost
Maize and blue	$2$	$2$	$\$25,000$
Green and white	$4$	$4$	$\$50,000$
Scarlet and gray	$3.05$	$4$	$\$50,000$
Orange and blue	$2.31$	$4$	$\$50,000$
Purple and white	$5.17$	$4$	$\$50,000$
Silver and black	$2.5$	$4$	$\$50,000$
Crimson and white	$2.5$	$2$	$\$25,000$

The expression to calculate the number of tanks is given as follows.

    $n=\frac{{\tau }^2}{{\sigma }^2}$

Where,

• $n$ is the number of tanks.
• $\tau$ is the specific time.
• $\sigma$ is the cross-sectional area.

Thus, the new table for the reactors with calculated number of tanks is given as follows.

Reactors	$\sigma \left(\min \right)$	$\tau \left(\min \right)$	Cost	Number of tanks, $n=\frac{{\tau }^2}{{\sigma }^2}$
Maize and blue	$2$	$2$	$\$25,000$	1
Green and white	$4$	$4$	$\$50,000$	1
Scarlet and gray	$3.05$	$4$	$\$50,000$	2
Orange and blue	$2.31$	$4$	$\$50,000$	3
Purple and white	$5.17$	$4$	$\$50,000$	1
Silver and black	$2.5$	$4$	$\$50,000$	3
Crimson and white	$2.5$	$2$	$\$25,000$	1

The conversion for the second order reaction is calculated by the expression given below.

    $X=\frac{2D_{\text{a}}\sqrt{4D_{\text{a}}+1}}{2D_{\text{a}}}$

Where,

• $D_{\text{a}}$ is the equals to $k\tau C_{\text{A0}}$.
• $C_{\text{A0}}$ is the initial concentration of A.

The expression to calculate the final concentration is as follows.

    $C_{\text{A}}=C_{\text{A0}}\left(1-X\right)$

The single tank reactors are maize and blue; green and white; purple and white; and crimson and white.

The values of conversion and the final concentration corresponding to the single tank reactor are given in the table as follows.

Reactors	$\sigma \left(\min \right)$	$\tau \left(\min \right)$	Number of tanks, $n=\frac{{\tau }^2}{{\sigma }^2}$	$D_a$=$k\tau C_{\text{A0}}$	$X_1=\frac{2D_a\sqrt{4D_a+1}}{2D_a}$	$C_{\text{A1}}=C_{\text{A0}}\left(1-X\right)$
Maize and blue	$2$	$2$	1	$2$	$0.5$	$0.5$
Green and white	$4$	$4$	1	$4$	$0.61$	$0.39$
Purple and white	$5.17$	$4$	1	$4$	$0.61$	$0.39$
Crimson and white	$2.5$	$2$	1	2	$0.5$	$0.5$

The double and triple tank reactors are scarlet and gray; orange and blue; and silver and black.

The values of conversion and the final concentration corresponding to the single tank reactor are given in the table as follows.

Reactors	$\sigma \left(\min \right)$	$\tau \left(\min \right)$	Number of tanks, $n=\frac{{\tau }^2}{{\sigma }^2}$	$D_a$=$k\tau C_{\text{A0}}$	$X_1=\frac{2D_a\sqrt{4D_a+1}}{2D_a}$	$C_{\text{A1}}=C_{\text{A0}}\left(1-X\right)$
Scarlet and gray	$3.05$	$4$	$2$	$4$	$0.61$	$0.39$
Orange and blue	$2.31$	$4$	$3$	$4$	$0.61$	$0.39$
Silver and black	$2.5$	$4$	$3$	$4$	$0.61$	$0.39$

The final concentration, $C_{\text{A1}}$, for the tank-1 of scarlet and gray reactor is $0.3\text{9}{\text{dm}}^{\text{3}}\text{/mol}$.

For the second tank of scarlet and gray reactor, the value of $D_{\text{a2}}$ is calculated as follows.

    $D_{\text{a1}}=k\tau C_{\text{A1}}$

Substitute the value of $k$ as $1.0{\text{dm}}^{\text{3}}\text{/mol}\cdot \min$, $\tau$ as $4$ and $C_{\text{A1}}$ as $0.3\text{9}{\text{dm}}^{\text{3}}\text{/mol}$ in the above equation.

    $\begin{array}{c}{D}_{\text{a1}}=1.0{\text{dm}}^{\text{3}}\text{/mol}\cdot \min \times 4\min \times 0.3\text{9}{\text{dm}}^{\text{3}}\text{/mol}\\ =\text{1}\text{.56}\end{array}$

For the second tank of scarlet and gray reactor, the value of $X_2$ that is conversion is calculated as follows.

    $X_2=\frac{2D_{\text{a1}}\sqrt{4D_{\text{a1}}+1}}{2D_{\text{a1}}}$

Substitute the value of $D_{\text{a1}}$ as $\text{1}\text{.56}$ in the above equation.

    $\begin{array}{c}{X}_2=\frac{2\times \text{1}\text{.56}\sqrt{4\times \text{1}\text{.56}+1}}{2\times \text{1}\text{.56}}\\ =0.46\end{array}$

Thus, the final concentration for the second tank of the reactor is calculated by the expression below.

    $C_{\text{A2}}=C_{\text{A1}}\left(1-X_2\right)$

Substitute the value of $C_{\text{A1}}$ as $0.3\text{9}{\text{dm}}^{\text{3}}\text{/mol}$ and $X_2$ as $0.46$ in the above equation.

    $\begin{array}{c}{C}_{\text{A2}}=0.3\text{9}{\text{dm}}^{\text{3}}\text{/mol}\left(1-0.46\right)\\ =0.21{\text{dm}}^{\text{3}}\text{/mol}\end{array}$

The overall conversion of the scarlet and gray reactor is calculated by expression below.

    $X_{overall}=1-\frac{C_{\text{A2}}}{C_{\text{A0}}}$

Substitute the value of $C_{\text{A2}}$ as $0.21{\text{dm}}^{\text{3}}\text{/mol}$ and $C_{\text{A0}}$ as $1.0\text{mol}/{\text{dm}}^{\text{3}}$ in the above equation.

    $\begin{array}{c}{X}_{overall}=1-\frac{0.21{\text{dm}}^{\text{3}}\text{/mol}}{1.0\text{mol}/{\text{dm}}^{\text{3}}}\\ =0.79\end{array}$

Thus, the total conversion for scarlet and gray reactor is $0.79$.

The final concentration, $C_{\text{A2}}$, for the tank-2 of orange and blue reactor is $0.21{\text{dm}}^{\text{3}}\text{/mol}$.

For the third tank of orange and blue reactor, the value of $D_{\text{a2}}$ is calculated as follows.

    $D_{\text{a2}}=k\tau C_{\text{A2}}$

Substitute the value of $k$ as $1.0{\text{dm}}^{\text{3}}\text{/mol}\cdot \min$, $\tau$ as $4$ and $C_{\text{A2}}$ as $0.21{\text{dm}}^{\text{3}}\text{/mol}$ in the above equation.

    $\begin{array}{c}{D}_{\text{a2}}=1.0{\text{dm}}^{\text{3}}\text{/mol}\cdot \min \times 4\min \times 0.21{\text{dm}}^{\text{3}}\text{/mol}\\ =\text{0}\text{.84}\end{array}$

For the third tank of orange and blue reactor, the value of $X_2$ that is conversion is calculated as follows.

    $X_2=\frac{2D_{\text{a2}}\sqrt{4D_{\text{a2}}+1}}{2D_{\text{a2}}}$

Substitute the value of $D_{\text{a2}}$ in the above equation.

    $\begin{array}{c}{X}_2=\frac{2\times 0.84\sqrt{4\times 0.84+1}}{2\times 0.84}\\ =0.35\end{array}$

Thus, the final concentration for the third tank of the reactor is calculated by the expression below.

    $C_{\text{A3}}=C_{\text{A2}}\left(1-X_2\right)$

Substitute the value of $C_{\text{A1}}$ as $0.3\text{9}{\text{dm}}^{\text{3}}\text{/mol}$ and $X_2$ as $0.35$ in the above equation.

    $\begin{array}{c}{C}_{\text{A3}}=0.21{\text{dm}}^{\text{3}}\text{/mol}\left(1-0.35\right)\\ =0.14{\text{dm}}^{\text{3}}\text{/mol}\end{array}$

The overall conversion of the orange and blue reactor is calculated by expression below.

    $X_{overall}=1-\frac{C_{\text{A3}}}{C_{\text{A0}}}$

Substitute the value of $C_{\text{A3}}$ as $0.14{\text{dm}}^{\text{3}}\text{/mol}$ and $C_{\text{A0}}$ as $1.0\text{mol}/{\text{dm}}^{\text{3}}$ in the above equation.

    $\begin{array}{c}{X}_{overall}=1-\frac{0.14{\text{dm}}^{\text{3}}\text{/mol}}{1.0\text{mol}/{\text{dm}}^{\text{3}}}\\ =0.86\end{array}$

Thus, the total conversion for orange and blue reactor is $0.86$.

For silver and black reactors, they contain three tanks as well. Thus, the overall conversion of silver and black is also $0.86$.

The table that is showing the overall conversion and cost of all the reactors is as follows.

Reactors	$n$	Cost	$X_{\text{overall}}$
Maize and blue	$1$	$\$25,000$	0.5
Green and white	$1$	$\$50,000$	0.61
Scarlet and gray	$2$	$\$50,000$	0.79
Orange and blue	$3$	$\$50,000$	0.86
Purple and white	$1$	$\$50,000$	0.61
Silver and black	$3$	$\$50,000$	0.86
Crimson and white	$1$	$\$25,000$	0.5

The available money to spend is $\$50,000$.

According to the above table, the greatest conversion corresponds to the orange and blue reactor and silver and black reactor which is $0.86$.

Interpretation Introduction

Interpretation:

The greatest amount of the conversion that can be attained if an individual had the availability of extra $\$75000$ to spend is to be stated.

Concept introduction:

The conversion, X can be defined as the moles of any species A that are reacted per mole of A fed in the reactor.

Answer to Problem 18.3P

The greatest conversion can be attained with availability of extra $\$75000$ to spend by the help of both maize and blue reactor and orange and blue reactor.

Explanation of Solution

The table that is showing the overall conversion and cost of all the reactors is as follows.

Reactors	$n$	Cost	$X_{\text{overall}}$
Maize and blue	$1$	$\$25,000$	0.5
Green and white	$1$	$\$50,000$	0.61
Scarlet and gray	$2$	$\$50,000$	0.79
Orange and blue	$3$	$\$50,000$	0.86
Purple and white	$1$	$\$50,000$	0.61
Silver and black	$3$	$\$50,000$	0.86
Crimson and white	$1$	$\$25,000$	0.5

The previous available money to spend is $\$50,000$.

The available money to spend is $\$75,000$.

As, $\$25,000$ extra money available, thus, with the help of orange and blue reactor having cost $\$50,000$ and maize and blue reactor having cost $\$25,000$, the greatest conversion can be achieved.

Interpretation Introduction

Interpretation:

The cities from which different types of used reactors came are to be stated.

Concept introduction:

The conversion, X can be defined as the moles of any species A that are reacted per mole of A fed in the reactor.

Answer to Problem 18.3P

The cities from which different types of used reactors came are stated below.

Explanation of Solution

The cities from which different types of used reactors came are follows.

• Evanston, Illinois
• Columbus, OR
• Ann Arbor, Michigan
• East Lansing, Michigan
• Urbana, Illinois
• West Lafayette, Indiana
• Madison, Wisconsin