Chapter 6, Problem 6.3P

(a)

Interpretation Introduction

Interpretation:

The conversion after $10$, $50$ and $100$ minutes 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.

A batch reactor is a closed system with no continuous flow of reactants entering the system or products leaving the system while the reaction takes place.

In most batch reactors, the longer a reactant stays in the reactor, the more the reactant is converted to product until either equilibrium is reached or the reactant is exhausted. Consequently, in batch systems the conversion X is a function of the time the reactants spend in the reactor.

Answer to Problem 6.3P

The conversion after $10$, $50$ and $100$ minutes is $47.1\%$, $81.5\%$ and $89.9\%$ respectively.

Explanation of Solution

The given second-order liquid phase reaction which takes place in a batch reactor is as follows.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COCH}}_{\text{2}}\text{Br}+{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{N}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COCH}}_{\text{2}}{\text{NC}}_{\text{5}}{\text{H}}_{\text{5}}\text{Br}$

The given specific rate constant for the reaction is $0.0445{\text{dm}}^{\text{3}}\text{/mol/min}$.

The given volume of reactor 1 is $\text{1000}{\text{dm}}^{\text{3}}$.

The concentration of each reactant after the mixing is $\text{2}\text{M}$.

The temperature of the batch reactor is $\text{35°C}$.

The given two reactors are shown below.

Figure 1

The concentration of the reactant is assumed to be $x{\text{mol/dm}}^{\text{3}}$ which gives $x{\text{mol/dm}}^{\text{3}}$ of the product. The second order phase reaction is given in the form as follows.

    $\frac{dx}{dt}=-k{x}^2$

Where,

$k$ is the rate constant with value $0.0445{\text{dm}}^{\text{3}}\text{/mol/min}$.

The expression for the integrated general concentration of the reactant before mixing the $x_f$ is given below.

    $\begin{array}{c}{\displaystyle \underset{x_i}{\overset{x_f}{\int }}\frac{dx}{x^2}}=-k{\displaystyle \underset{t_0}{\overset{t_f}{\int }}dt}\\ {\left[-\frac{1}{x}\right]}_{x_i}^{x_f}=-k{\left[t\right]}_0^t\\ \left[-\frac{1}{x_f}+\frac{1}{x_i}\right]=-kt\\ \frac{1}{x_f}=kt+\frac{1}{x_i}\end{array}$        (1)

Where,

$x_i$ is the initial concentration.

Substitute the value of $k$ as $0.0445$ and $x_i$ as $\text{2}$ in equation (1).

    $\begin{array}{l}\frac{1}{x_f}=0.0445\times t+\frac{1}{2}\\ \frac{1}{x_f}=0.0445\times t+0.5\\ x_f=\frac{1}{0.0445\times t+0.5}\end{array}$        (2)

Substitute $t=10\min$ in equation (2).

    $\begin{array}{c}{x}_f=\frac{1}{0.0445\times 10+0.5}\\ =\text{1}\text{.058}{\text{mol/dm}}^{\text{3}}\end{array}$

The expression to calculate the conversion after $10\min$ is given as follows.

    $X=\frac{x_i-x_f}{x_i}\times 100$        (3)

Substitute $x_i=2$, $x_f=1.058$ in equation (3).

    $\begin{array}{c}X=\frac{2-1.058}{2}\times 100\\ =47.1\%\end{array}$

For the conversion after $50\min$, Substitute $t=50\min$ in equation (2).

    $\begin{array}{c}{x}_f=\frac{1}{0.0445\times 50+0.5}\\ =\text{0}\text{.37}{\text{mol/dm}}^{\text{3}}\end{array}$

Substitute $x_i=2$, $x_f=\text{0}\text{.37}$ in equation (3).

    $\begin{array}{c}X=\frac{2-\text{0}\text{.37}}{2}\times 100\\ =81.5\%\end{array}$

For the conversion after $100\min$, Substitute $t=100\min$ in equation (2).

    $\begin{array}{c}{x}_f=\frac{1}{0.0445\times 100+0.5}\\ =\text{0}\text{.202}{\text{mol/dm}}^{\text{3}}\end{array}$

Substitute $x_i=2$, $x_f=\text{0}\text{.202}$ in equation (3).

    $\begin{array}{c}X=\frac{2-\text{0}\text{.202}}{2}\times 100\\ =89.9\%\end{array}$

Therefore, the conversion after $10$, $50$ and $100$ minutes is $47.1\%$, $81.5\%$ and $89.9\%$ respectively.

(b)

Interpretation Introduction

Interpretation:

The conversion and concentration of each species present in the reactor 1 after $10$, $50$ and $80$ minutes 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.

A batch reactor is a closed system with no continuous flow of reactants entering the system or products leaving the system while the reaction takes place.

In most batch reactors, the longer a reactant stays in the reactor, the more the reactant is converted to product until either equilibrium is reached or the reactant is exhausted. Consequently, in batch systems the conversion X is a function of the time the reactants spend in the reactor.

Answer to Problem 6.3P

The concentration after 10, $50$ and $80$ minutes is $0.942{\text{mol/dm}}^{\text{3}}$, $1.63{\text{mol/dm}}^{\text{3}}$ and $1.76{\text{mol/dm}}^{\text{3}}$ respectively.  The conversion after $10$, $50$ and $80$ minutes is $47.1\%$, $81.5\%$ and $87.7\%$ respectively.

Explanation of Solution

The given second-order liquid phase reaction which takes place in a batch reactor is as follows.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COCH}}_{\text{2}}\text{Br}+{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{N}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COCH}}_{\text{2}}{\text{NC}}_{\text{5}}{\text{H}}_{\text{5}}\text{Br}$

The given specific rate constant for the reaction is $0.0445{\text{dm}}^{\text{3}}\text{/mol/min}$.

The given volume of reactor 1 is $\text{1000}{\text{dm}}^{\text{3}}$.

The concentration of each reactant after the mixing is $\text{2}\text{M}$.

The temperature of the batch reactor is $\text{35°C}$.

The given two reactors are shown below.

Figure 1

It is given that after 10 minutes, the species in the reactor 1 are drained out at the rate of $10{\text{dm}}^{\text{3}}\text{/min}$ to the reactor 2.

The solution that is left out in 10 minutes is $10\times 10=\text{100}{\text{dm}}^{\text{3}}$.

The total solution that is remaining in the reactor 1 is $1000-100=900{\text{dm}}^{\text{3}}$.

If the solution is $\text{1000}{\text{dm}}^3$ then there are $x\text{mol}$ of each species is present.

So, in $\text{900}{\text{dm}}^3$, the number of total moles are $\frac{x\times \text{900}}{1000}=0.9x\text{mol}$.

The concentration of each species is calculated below given below.

    $\frac{x\times 0.\text{9}}{900}\times 100\left(\text{M}\right)=x\text{M}$

Thus, the concentration of each species remains the same in $900{\text{dm}}^{\text{3}}$ as well as calculated in part (a).

So, the concentration of each species after $10\text{minutes}$ is $\text{1}\text{.058}{\text{mol/dm}}^{\text{3}}$.

The concentration of the product after $10\text{minutes}$ is $\left(\text{2}-\text{1}\text{.058}\right){\text{mol/dm}}^{\text{3}}=0.942{\text{mol/dm}}^{\text{3}}$

So, the concentration of each species after $50\text{minutes}$ is $\text{0}\text{.37}{\text{mol/dm}}^{\text{3}}$.

The concentration of the product after $50\text{minutes}$ is $\left(\text{2}-\text{0}\text{.37}\right){\text{mol/dm}}^{\text{3}}=1.63{\text{mol/dm}}^{\text{3}}$.

The expression to calculate the conversion after $10\min$ is given as follows.

    $X=\frac{x_i-x_f}{x_i}\times 100$        (3)

Substitute $x_i=2$, $x_f=1.058$ in equation (3).

    $\begin{array}{c}X=\frac{2-1.058}{2}\times 100\\ =47.1\%\end{array}$

Substitute $x_i=2$, $x_f=\text{0}\text{.37}$ in equation (3).

    $\begin{array}{c}X=\frac{2-\text{0}\text{.37}}{2}\times 100\\ =81.5\%\end{array}$

For the conversion after $80\min$, Substitute $t=100\min$ in equation (2).

    $\begin{array}{c}{x}_f=\frac{1}{0.0445\times 80+0.5}\\ =\text{0}\text{.24}{\text{mol/dm}}^{\text{3}}\end{array}$

The concentration of the product after $80\text{minutes}$ is $\left(\text{2}-\text{0}\text{.24}\right){\text{mol/dm}}^{\text{3}}=1.76{\text{mol/dm}}^{\text{3}}$.

Therefore, the concentration after $10$, $50$ and $80$ minutes is $0.942{\text{mol/dm}}^{\text{3}}$, $1.63{\text{mol/dm}}^{\text{3}}$ and $1.76{\text{mol/dm}}^{\text{3}}$ respectively.

Substitute $x_i=2$, $x_f=\text{0}\text{.24}$ in equation (3).

    $\begin{array}{c}{X}_{10\min }=\frac{2-\text{0}\text{.24}}{2}\times 100\\ =87.7\%\end{array}$

Therefore, the conversion after $10$, $50$ and $80$ minutes is $47.1\%$, $81.5\%$ and $87.7\%$ respectively.

(c)

Interpretation Introduction

Interpretation:

The conversion and concentration of each species present in the reactor 2 that is filling up from reactor 1 after $10$ and $50$ minutes 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.

A batch reactor is a closed system with no continuous flow of reactants entering the system or products leaving the system while the reaction takes place.

In most batch reactors, the longer a reactant stays in the reactor, the more the reactant is converted to product until either equilibrium is reached or the reactant is exhausted. Consequently, in batch systems the conversion X is a function of the time the reactants spend in the reactor.

Answer to Problem 6.3P

The conversion of each species present in the reactor 2 that is filling up from reactor 1 after $10$ and $50$ minutes are $47.1\%$ and $81.5\%$. The concentration of each species present in the reactor 2 that is filling up from reactor 1 after $10$ and $50$ minutes are $0.942{\text{mol/dm}}^{\text{3}}$ and $1.63{\text{mol/dm}}^{\text{3}}$ respectively.

Explanation of Solution

The given second-order liquid phase reaction which takes place in a batch reactor is as follows.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COCH}}_{\text{2}}\text{Br}+{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{N}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COCH}}_{\text{2}}{\text{NC}}_{\text{5}}{\text{H}}_{\text{5}}\text{Br}$

The given specific rate constant for the reaction is $0.0445{\text{dm}}^{\text{3}}\text{/mol/min}$.

The given volume of reactor 1 is $\text{1000}{\text{dm}}^{\text{3}}$.

The concentration of each reactant after the mixing is $\text{2}\text{M}$.

The temperature of the batch reactor is $\text{35°C}$.

The given two reactors are shown below.

Figure 1

It is given that after 10 minutes, the species in the reactor 1 are drained out at the rate of $10{\text{dm}}^{\text{3}}\text{/min}$ to the reactor 2.

The solution that is left in 10 minutes is $10\times 10=\text{100}{\text{dm}}^{\text{3}}$.

The total solution that is remaining in the reactor 1 is $1000-100=900{\text{dm}}^{\text{3}}$.

If the solution is $\text{1000}{\text{dm}}^3$ then there are $x\text{mol}$ of each species is present.

So, in $\text{900}{\text{dm}}^3$, the number of total moles are $\frac{x\times \text{900}}{1000}=0.9x\text{mol}$.

The concentration of each species is calculated below given below.

    $\frac{x\times 0.\text{9}}{900}\times 100\left(\text{M}\right)=x\text{M}$

Thus, the concentration of each species remains the same in $900{\text{dm}}^{\text{3}}$ as well as calculated in part (a).

So, the concentration of each species after $10\text{minutes}$ is $\text{1}\text{.058}{\text{mol/dm}}^{\text{3}}$.

The concentration of the product after $10\text{minutes}$ is $\left(\text{2}-\text{1}\text{.058}\right){\text{mol/dm}}^{\text{3}}=0.942{\text{mol/dm}}^{\text{3}}$

At the end of 50 minutes, the volume of reactor 1 is $\text{500}{\text{dm}}^{\text{3}}$ and the concentration of each species after $50\text{minutes}$ is $\text{0}\text{.37}{\text{mol/dm}}^{\text{3}}$.

The concentration of the product after $50\text{minutes}$ is $\left(\text{2}-\text{0}\text{.37}\right){\text{mol/dm}}^{\text{3}}=1.63{\text{mol/dm}}^{\text{3}}$.

At the end of 50 minutes, the volume of reactor 2 has $\text{500}{\text{dm}}^{\text{3}}$ with same concentration of each species and the conversion rate is same as reactor 1.

Therefore, the concentration after $10$ and $50$ minutes is $0.942{\text{mol/dm}}^{\text{3}}$ and $1.63{\text{mol/dm}}^{\text{3}}$ respectively.

Therefore, the conversion after $10$ and $50$ minutes is $47.1\%$, and $81.5\%$ respectively.

(d)

Interpretation Introduction

Interpretation:

The overall conversion and concentration when the contents of the reactants are mixed together at the end of $50$ minutes 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.

A batch reactor is a closed system with no continuous flow of reactants entering the system or products leaving the system while the reaction takes place.

In most batch reactors, the longer a reactant stays in the reactor, the more the reactant is converted to product until either equilibrium is reached or the reactant is exhausted. Consequently, in batch systems the conversion X is a function of the time the reactants spend in the reactor.

Answer to Problem 6.3P

The overall conversion of the two reactor after mixing the contents is $81.5\%$.

Explanation of Solution

The given second-order liquid phase reaction which takes place in a batch reactor is as follows.

    ${\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COCH}}_{\text{2}}\text{Br}+{\text{C}}_{\text{6}}{\text{H}}_{\text{5}}\text{N}\to {\text{C}}_{\text{6}}{\text{H}}_{\text{5}}{\text{COCH}}_{\text{2}}{\text{NC}}_{\text{5}}{\text{H}}_{\text{5}}\text{Br}$

The given specific rate constant for the reaction is $0.0445{\text{dm}}^{\text{3}}\text{/mol/min}$.

The given volume of reactor 1 is $\text{1000}{\text{dm}}^{\text{3}}$.

The concentration of each reactant after the mixing is $\text{2}\text{M}$.

The temperature of the batch reactor is $\text{35°C}$.

The given two reactors are shown below.

Figure 1

At the end of 50 minutes, the volume of reactor 1 is $\text{500}{\text{dm}}^{\text{3}}$ and the concentration of each species after $50\text{minutes}$ is $\text{0}\text{.37}{\text{mol/dm}}^{\text{3}}$.

The concentration of the product after $50\text{minutes}$ is $\left(\text{2}-\text{0}\text{.37}\right){\text{mol/dm}}^{\text{3}}=1.63{\text{mol/dm}}^{\text{3}}$.

The number of moles of reactant are $0.37{\text{mol/dm}}^{\text{3}}\times 500{\text{dm}}^{\text{3}}=185\text{mol}$.

The number of moles of the product are $1.63{\text{mol/dm}}^{\text{3}}\times 500{\text{dm}}^{\text{3}}=815\text{mol}$.

At the end of 50 minutes, the volume of reactor 2 is $\text{500}{\text{dm}}^{\text{3}}$ with same concentration.

So, the total volume after the mixing of the species of reactor 1 and 2 is $500+500=1000{\text{dm}}^{\text{3}}$.

The total moles of reactants after the mixing of the species of reactor 1 and 2 is

$185+185=370\text{mol}$.

The total concentration of reactants after the mixing of the species of reactor 1 and 2 is

$\frac{\text{370}\text{mol}}{\text{1000}{\text{dm}}^{\text{3}}}=0.37\frac{\text{mol}}{{\text{dm}}^{\text{3}}}$.

The overall conversion after mixing in reactor,

    $\begin{array}{c}X=\frac{0.37}{2}\times 100\\ =\text{18}\text{.5}\text{\%}\end{array}$

The total moles of product after the mixing of the species of reactor 1 and 2 is $815+815=1630\text{mol}$.

The total concentration of product after the mixing of the species of reactor 1 and 2 is $\frac{1630\text{mol}}{\text{1000}{\text{dm}}^{\text{3}}}=1.63\frac{\text{mol}}{{\text{dm}}^{\text{3}}}$.

The conversion,

    $\begin{array}{c}X=\frac{1.63}{2}\times 100\\ =\text{81}\text{.5}\text{\%}\end{array}$

Therefore, the overall conversion of the two reactor after mixing the contents is $81.5\%$.

(e)

Interpretation Introduction

Interpretation:

The application of the six ideas that is given in table on this problem 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.

The reactors that consist of the particles of solid catalysts which are packed in the form of bed are known as fixed bed reactors.

Answer to Problem 6.3P

The application of the six ideas that is given in table on this problem is not appropriate.

Explanation of Solution

The given six ideas are as follows.

• Brainstorm ideas to ask another question or advice another calculation for the homework problem.
• Brainstorm ways to work this homework problem incorrectly.
• Brainstorm ways to make this problem highly easy or more difficult.
• Brainstorm a list of things that can be learned from working this homework problem.
• Brainstorm the reason why the calculations over predicted the conversion that measured if the reactor was kept on stream.
• The “What if…” questions are particularly effective when they are used with the living example problems in which variation in parameters to explore the problem is carried out.

These six ideas are not relevant according to the demand of the question. Thus, The application of the six ideas that is given in table on this problem cannot be possible according to the demand of the question.