Chapter 5, Problem 5.4Q

Example 5-1. (1) What would be the error in $k$ if the batch reactor were only $80\%$ filled with the same concentrations of reactants, instead of being completely filled as in the example? (2) What generalizations can you draw from this example?

Interpretation Introduction

Interpretation:

The error in the value of $k$ is to be calculated if the given batch reactor in example 5.1 is filled 80% with same concentration of reactants rather than completely filled.

Concept Introduction:

For a Continuous Stirred Tank Reactor (CSTR), the design equation to be used is:

  $V=\frac{F_{A0}X}{{\left(-r_A\right)}_{exit}}$   ....... (1)

Here, $V$ is the volume of the reactor, $F_{A0}$ is the entering molar flowrate of species A , $X$ is the exit conversion of A , and ${\left(-r_A\right)}_{exit}$ is the rate of reaction of species A at exit.

Answer to Problem 5.4Q

The new value of $k$ is same as the original value calculated since its value is not dependent on the reactor volume and depends on the concentration which is not altered.

There is no error in the new value.

Explanation of Solution

Given information:

Example 5.1 data:

In a batch reactor, 500 mL of 2 M $\left(2\text{kmol}/{\text{m}}^3\right)$ solution of ethylene oxide (A) in water and 500 mL of water (B) having 0.9 wt% sulfuric acid as catalyst is mixed originally. Reactor temperature is maintained at ${55}^{\circ }\text{C}$ . Since water is present in excess, its concentration is considered to be $55.5\text{mol}/{\text{dm}}^3$ . The first order reaction that takes place in the reactor is:

  $A+B\stackrel{catalyst}{\to }C$

Here

  $\begin{array}{l}A=\text{Ethylene oxide}\\ B=\text{Water}\\ C=\text{Ethylene glycol}\end{array}$

Since the reaction is first order in A, rate law is:

  $-r_A=k{C}_A$

Concentration time data for ethylene glycol (C) is given as:

Time (min)	Concentration of ethylene glycol   $C_C\left(\text{kmol}/{\text{m}}^3\right)$
0.0	0.000
0.5	0.145
1.0	0.270
1.5	0.376
2.0	0.467
3.0	0.610
4.0	0.715
6.0	0.848
10.0	0.957

Example- 5.1 when the batch reactor is filled completely.

For constant volume batch reactor, the equation that relates concentration, time and rate constant for first order reaction is:

  $\ln \frac{C_{A0}-C_C}{C_{A0}}=-kt$   ....... (2)

Table prepared for processed data for $\ln \left(\frac{C_{A0}-C_C}{C_{A0}}\right)$ is:

Time (min)	Concentration of ethylene glycol   $C_C\left(\text{kmol}/{\text{m}}^3\right)$	$\ln \left(\frac{C_{A0}-C_C}{C_{A0}}\right)$
0.0	0.000	-0.0000
0.5	0.145	-0.1570
1.0	0.270	-0.3150
1.5	0.376	-0.4720
2.0	0.467	-0.6290
3.0	0.610	-0.9420
4.0	0.715	-0.1.2550
6.0	0.848	-1.08840
10.0	0.957	-3.1470

The graph of $\ln \left(\frac{C_{A0}-C_C}{C_{A0}}\right)$ versus t gives the value of slope as:

  $k=0.311{\text{ min}}^{-1}$

When the reactor is filled 80% with the same concentration of reactants, equation (2) remains unaffected. Thus, the value of k does not depend on the volume of the reactor and is dependent on the concentration of the reactants and products. Hence, the new value of $k$ will be:

  $k_{New}=0.311{\text{ min}}^{-1}$

Therefore, there is no error in the value of specific reaction rate.

Interpretation Introduction

Interpretation:

The generalization from the results of part (1) is to be inferred.

Concept Introduction:

For a Continuous Stirred Tank Reactor (CSTR), the design equation to be used is:

  $V=\frac{F_{A0}X}{{\left(-r_A\right)}_{exit}}$   ....... (1)

Here, $V$ is the volume of the reactor, $F_{A0}$ is the entering molar flowrate of species A , $X$ is the exit conversion of A , and ${\left(-r_A\right)}_{exit}$ is the rate of reaction of species A at exit.

Explanation of Solution

From the results of given example-5.1 and part (1), it can be generalized that for the first order reactions, specific reaction rate is independent of the reactor volume and is dependent on the concentration of ethylene oxide (A) and ethylene glycol (C). Any change in the volume of the reactor without altering the concentration will not affect the specific reaction rate.