Chapter 4 Stoichiometry

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4. Stoichiometry

If you are thinking about someone while reading this book, you are most definitely in love.

—Tim Newberger
Undergraduate ChE Student W2013

Overview. In Chapter 2 we carried out the sizing and sequencing of flow reactors given the reaction rate, –r_A, as a function of conversion X. To find –r_A = f(X) is a two-step process. In Chapter 3, we described (Step 1) how the rate of reaction, –r_A, is related to concentration and temperature. In this chapter, we show how concentration can be related to conversion (Step 2), and once we do that, we will have –r_A = f(X) and can design a multitude of reaction systems. We will use stoichiometric tables, along with the definitions of concentration, to find the concentration as a function of conversion.

Two sets of equations are shown, at the left represent "Batch" and at the right represents "Flow."

• For batch systems the reactor is rigid, so V = V₀, and one then uses the stoichiometric table to express concentration as a function of conversion: C_A = N_A / V₀ = C_A0(1 – X).

• For liquid-phase flow systems, the volumetric flow rate is considered to be constant, υ=υ₀, and C_A = (F_A0/υ₀)( 1 – X) = C_A0(1 – X).

• For gas-phase flow systems, the process becomes more complicated, as the volumetric flow rate for gases can vary with conversion, temperature, and pressure, and we need to develop the relationship relating υ and X, i.e., υ = υ₀ (1 + εΧ)(P₀ / P)(T/T₀) and thus

$C_{A} = \frac{C_{A 0} (1 - X)}{(1 + ɛ X)} \frac{P}{P_{0}} \frac{T_{0}}{T} = C_{A 0} \frac{(1 - X)}{(1 + ɛ X)} p \frac{T_{0}}{T}$ $C_{A} = \frac{C_{A 0} (1 - X)}{(1 + ɛ X)} \frac{P}{P_{0}} \frac{T_{0}}{T} = C_{A 0} \frac{(1 - X)}{(1 + ɛ X)} p \frac{T_{0}}{T}$

• For most gas-solid catalyst reactions, the rate laws are written in terms of partial pressures, which can also be written in terms of conversion.

$P_{A} = C_{A} R T = P_{A 0} \frac{(1 - X)}{(1 + ɛ X)} \frac{P}{P_{0}} = P_{A 0} \frac{(1 - X)}{(1 + ɛ X)} p$ $P_{A} = C_{A} R T = P_{A 0} \frac{(1 - X)}{(1 + ɛ X)} \frac{P}{P_{0}} = P_{A 0} \frac{(1 - X)}{(1 + ɛ X)} p$

After completing this chapter, you will be able to write the rate of reaction as a function of conversion and to calculate the equilibrium conversion for both batch and flow reactors.

Now that we have shown how the rate law can be expressed as a function of concentrations, we need only express concentration as a function of conversion in order to carry out calculations similar to those presented in Chapter 2 to size reactors. If the rate law depends on more than one species, we must relate the concentrations of the different species to each other. This relationship is most easily established with the aid of a stoichiometric table. This table presents the stoichiometric relationships between reacting molecules for a single reaction. That is, it tells us how many molecules of one species will be formed during a chemical reaction when a given number of molecules of another species disappears. These relationships will be developed for the general reaction

$\begin{matrix} a A + b B ⇆ c C + d D & (2-1) \end{matrix}$ $\begin{matrix} a A + b B ⇆ c C + d D & (2-1) \end{matrix}$

Recall that we have already used stoichiometry to relate the relative rates of reaction for Equation (2-1):

\frac{r_{A}}{- a} = \frac{r_{B}}{- b} = \frac{r_{C}}{c} = \frac{r_{D}}{d} (3 - 1)

$\frac{r_{A}}{- a} = \frac{r_{B}}{- b} = \frac{r_{C}}{c} = \frac{r_{D}}{d} (3 - 1)$

This stoichiometric relationship relating reaction rates will be used in Chapters 6 and 8.

In formulating our stoichiometric table, we shall take species A as our basis of calculation (i.e., the limiting reactant) and then divide through by the stoichiometric coefficient of A

$\begin{matrix} A + \frac{b}{a} B \to \frac{c}{a} C + \frac{d}{a} D & (2-2) \end{matrix}$ $\begin{matrix} A + \frac{b}{a} B \to \frac{c}{a} C + \frac{d}{a} D & (2-2) \end{matrix}$

in order to put everything on a basis of “per mole of A."

Next, we develop the stoichiometric relationships for reacting species that give the change in the number of moles of each species (i.e., A, B, C, and D).

A spherical batch reactor is shown with number of moles placed initially and the moles remaining at time t.

Figure 4-1 Batch reactor. (Schematic with special permission by Renwahr.) (http://encyclopedia.che.engin.umich.edu/Pages/Reactors/Batch/Batch.html)

4.1 Batch Systems

Batch reactors are primarily used for the production of specialty chemicals and to obtain reaction rate data in order to determine reaction rate laws and rate-law parameters such as k, the specific reaction rate.

Figure 4-1 shows a starving artist’s rendition of a batch system in which we will carry out the reaction given by Equation (2-2). At time t = 0, we will open the reactor and place a number of moles of species A, B, C, and D, and inerts I (N_A0, N_B0, N_C0, N_D0, and N_I0, respectively) into the reactor.

Species A is our basis of calculation, and N_A0 is the number of moles of A initially present in the reactor. After a time t, N_A0X moles of A are consumed in the system as a result of the chemical reaction, leaving (N_A0 – N_A0X) moles of A in the system. That is, the number of moles of A remaining in the reactor after a conversion X has been achieved is

N_A = N_A0 – N_A0X = N_A0(1 – X)

We now will use conversion in a similar fashion to express the number of moles of B, C, and D in terms of conversion.

To determine the number of moles of each species remaining after N_A0X moles of A have reacted, we form the stoichiometric table (Table 4-1). This stoichiometric table presents the following information:

TABLE 4-1 STOICHIOMETRIC TABLE FOR A BATCH SYSTEM

A cylindrical batch reactor is shown with number of moles placed initially and the moles remaining at time t.

Components of the stoichiometric table

Column 1: the species in the reaction

Column 2: the number of moles of each species initially present

Column 3: the change in the number of moles brought about by reaction

Column 4: the number of moles remaining in the system at time t

To calculate the number of moles of species B remaining at time t, we recall that at time t the number of moles of A that have reacted is N_A0X. For every mole of A that reacts, b / a moles of B must react; therefore, the total number of moles of B that have reacted is

\begin{matrix} Moles B reacted & = & \frac{Moles B reacted}{Moles A reacted} \cdot Moles A reacted \\ = & \frac{b}{a} (N_{A 0} X) \end{matrix}

$\begin{matrix} Moles B reacted & = & \frac{Moles B reacted}{Moles A reacted} \cdot Moles A reacted \\ = & \frac{b}{a} (N_{A 0} X) \end{matrix}$

Because B is disappearing from the system, the sign of the “change” is negative. N_B0 is the number of moles of B initially in the system. Therefore, the number of moles of B remaining in the system, N_B, at a time t, is given in the last column of Table 4-1 as

N_{B} = N_{B0} - \frac{b}{a} N_{A0} X

$N_{B} = N_{B0} - \frac{b}{a} N_{A0} X$

The complete stoichiometric table delineated in Table 4-1 is for all species in the general reaction

\begin{matrix} A + \frac{b}{a} B \to \frac{c}{a} C + \frac{d}{a} D \end{matrix} (2 - 2)

$\begin{matrix} A + \frac{b}{a} B \to \frac{c}{a} C + \frac{d}{a} D \end{matrix} (2 - 2)$

Let’s take a look at the totals in the last column of Table 4-1. The stoichiometric coefficients in parentheses (d/a + c/a – b/a – 1) represent the change in the total number of moles per mole of A reacted. Because this term occurs so often in our calculations, it is given the symbol δ:

\begin{matrix} δ = \frac{d}{a} + \frac{c}{a} - \frac{b}{a} - 1 \end{matrix} (4 - 1)

$\begin{matrix} δ = \frac{d}{a} + \frac{c}{a} - \frac{b}{a} - 1 \end{matrix} (4 - 1)$

Definition of δ

The parameter δ

\begin{matrix} δ = \frac{Change in the total number of moles}{Mole of A reacted} \end{matrix}

$\begin{matrix} δ = \frac{Change in the total number of moles}{Mole of A reacted} \end{matrix}$

The total number of moles can now be calculated from the equation

N_T = N_T0 + δ(N_A0X)

We want C_j = h_j (X)

We recall from Chapter 1 and Chapter 3 that the kinetic rate law (e.g., $- r_{A =} k C_{A}^{2}$ $- r_{A =} k C_{A}^{2}$ ) is a function solely of the intensive properties of the reacting system (e.g., temperature, pressure, concentration, and catalysts, if any). The reaction rate, –r_A, usually depends on the concentration of the reacting species raised to some power. Consequently, to determine the reaction rate as a function of conversion, Χ, we need to know the concentrations of the reacting species as a function of conversion, Χ.

4.1.1 Batch Concentrations for the Generic Reaction, Equation (2-2)

Batch concentration

The concentration of A is the number of moles of A per unit volume

C_{A} = \frac{N_{A}}{V}

$C_{A} = \frac{N_{A}}{V}$

After writing similar equations for B, C, and D, we use the stoichiometric table to express the concentration of each component in terms of the conversion Χ:

\begin{matrix} C_{A} = \frac{N_{A}}{V} = \frac{N_{A0} (1 - X)}{V} & (4-2) \end{matrix}

$\begin{matrix} C_{A} = \frac{N_{A}}{V} = \frac{N_{A0} (1 - X)}{V} & (4-2) \end{matrix}$

\begin{matrix} C_{B} = \frac{N_{B}}{V} = \frac{N_{B 0} - (b / a) N_{A0} X}{V} & (4-3) \end{matrix}

$\begin{matrix} C_{B} = \frac{N_{B}}{V} = \frac{N_{B 0} - (b / a) N_{A0} X}{V} & (4-3) \end{matrix}$

\begin{matrix} C_{C} = \frac{N_{C}}{V} = \frac{N_{C 0} + (c / a) N_{A0} X}{V} & (4-4) \end{matrix}

$\begin{matrix} C_{C} = \frac{N_{C}}{V} = \frac{N_{C 0} + (c / a) N_{A0} X}{V} & (4-4) \end{matrix}$

\begin{matrix} C_{D} = \frac{N_{D}}{V} = \frac{N_{D 0} + (d / a) N_{A0} X}{V} & (4-5) \end{matrix}

$\begin{matrix} C_{D} = \frac{N_{D}}{V} = \frac{N_{D 0} + (d / a) N_{A0} X}{V} & (4-5) \end{matrix}$

Because almost all batch reactors are solid vessels, the reactor volume is constant, so we can take V = V₀, then

\begin{matrix} C_{A} = \frac{N_{A}}{V_{0}} = \frac{N_{A0} (1 - X)}{V_{0}} \end{matrix}

$\begin{matrix} C_{A} = \frac{N_{A}}{V_{0}} = \frac{N_{A0} (1 - X)}{V_{0}} \end{matrix}$

C_{A} \begin{matrix} = C_{A0} (1 - X) & (4-6 \end{matrix})

$C_{A} \begin{matrix} = C_{A0} (1 - X) & (4-6 \end{matrix})$

We will soon see that Equation (4-6) for constant volume batch reactors also applies to continuous-flow liquid-phase systems.

We further simplify these equations by defining the parameter Θ_i, which allows us to factor out N_A0 in each of the expressions for concentration

\begin{matrix} Θ_{i} = \frac{N_{i 0}}{N_{A 0}} = \frac{C_{i 0}}{C_{A 0}} = \frac{y_{i 0}}{y_{A 0}}, \\ Θ_{i} = \frac{Moles od species "i" initially}{Moles od species A initially} \\ C_{B} = \frac{N_{{A0[N}_{B0} / N_{A0} - (b / a) X}}{V_{0}} = \frac{N_{A 0 [Θ_{B} - (b / a) X]}}{V_{0}} \end{matrix}

$\begin{matrix} Θ_{i} = \frac{N_{i 0}}{N_{A 0}} = \frac{C_{i 0}}{C_{A 0}} = \frac{y_{i 0}}{y_{A 0}}, \\ Θ_{i} = \frac{Moles od species "i" initially}{Moles od species A initially} \\ C_{B} = \frac{N_{{A0[N}_{B0} / N_{A0} - (b / a) X}}{V_{0}} = \frac{N_{A 0 [Θ_{B} - (b / a) X]}}{V_{0}} \end{matrix}$

\begin{matrix} C_{B} = C_{A0} (Θ_{B} - \frac{b}{a} X) & (4-7) \end{matrix}

$\begin{matrix} C_{B} = C_{A0} (Θ_{B} - \frac{b}{a} X) & (4-7) \end{matrix}$

with Θ_{B} = \frac{N_{B 0}}{N_{A0}}

$with Θ_{B} = \frac{N_{B 0}}{N_{A0}}$

Feed

Equalmolar:

Θ_B = 1

Stoichiometric:

Θ_{B} = \frac{b}{a}

$Θ_{B} = \frac{b}{a}$

for an equalmolar feed Θ_B = 1 and for a stoichiometric feed Θ_B = ba. Continuing for species C and D

\begin{matrix} C_{C} = \frac{N_{A0} [Θ_{C} + (c / a) X]}{V_{0}} \\ C_{C} = C_{A0} (Θ_{C} + \frac{c}{a} X) & (4-8) \end{matrix}

$\begin{matrix} C_{C} = \frac{N_{A0} [Θ_{C} + (c / a) X]}{V_{0}} \\ C_{C} = C_{A0} (Θ_{C} + \frac{c}{a} X) & (4-8) \end{matrix}$

with Θ_{C} = \frac{N_{C0}}{N_{A0}}

$with Θ_{C} = \frac{N_{C0}}{N_{A0}}$

Concentration for constant-volume batch reactors and for liquid phase continuous-flow reactors

\begin{matrix} C_{D} = \frac{N_{A0} [Θ_{D} + (d / a) X]}{V_{0}} \\ C_{D} = C_{A0} (Θ_{D} + \frac{d}{a} X) & (4-9) \end{matrix}

$\begin{matrix} C_{D} = \frac{N_{A0} [Θ_{D} + (d / a) X]}{V_{0}} \\ C_{D} = C_{A0} (Θ_{D} + \frac{d}{a} X) & (4-9) \end{matrix}$

with Θ_{D} = \frac{N_{D 0}}{N_{A0}}

$with Θ_{D} = \frac{N_{D 0}}{N_{A0}}$

For constant-volume batch reactors, e.g., steel containers V = V₀, we now have concentration as a function of conversion. if we know the rate law, we can now obtain –r_A = f(X) to couple with the differential mole balance in terms of conversion in order to solve for the reaction time, t.

For liquid-phase reactions taking place in solution, the solvent usually dominates the situation. For example, most liquid-phase organic reactions do not change density during the course of the reaction and represent still another case for which the constant-volume simplifications apply. As a result, changes in the density of the solute do not affect the overall density of the solution significantly and therefore it is essentially a constant-volume reaction process, V = V₀ and υ = υ₀. Consequently, Equations (4-6) through (4-9) can be used for liquid-phase reactions as well. An important exception to this general rule exists for polymerization processes.

For liquids V = V₀ and υ = υ₀

To summarize for constant-volume batch systems and for liquid-phase reactions, we can use a rate law for reaction (2-2) such as –r_A = k_AC_AC_B to obtain –r_A = f(X); that is,

A graph is shown with the horizontal axis representing "X" and the vertical axis representing "F subscript A0 over negative r subscript A." A curve representing V subscript PFR is drawn from the bottom of the vertical axis and increases gradually up to top right of the graph.

- r_{A} = k C_{A} C_{B} = k C_{A0}^{2} (1-X) (Θ_{B} - \frac{b}{a} X) = f (X)

$- r_{A} = k C_{A} C_{B} = k C_{A0}^{2} (1-X) (Θ_{B} - \frac{b}{a} X) = f (X)$

Substituting for the given parameters k, C_A0, and Θ_B, we can now use the techniques in Chapter 2 to size the CSTRs and PFRs for liquid-phase reactions.

Choosing a basis of calculation

Example 4-1 Expressing C_j = h_j (X) for a Liquid-Phase Batch Reaction

Soap is necessary to clean such things as dirty softball and soccer players’ uniforms, dirty hands, and foul mouths. Soap consists of the sodium and potassium salts of various fatty acids, such as oleic, stearic, palmitic, lauric, and myristic acids. The saponification reaction for the formation of soap from aqueous caustic soda and glyceryl stearate is

3NaOH(aq) + (C₁₇H₃₅COO)₃C₃H₅ → 3C₁₇ H₃₅COONa + C₃H₅(OH)₃

Letting Χ represent the conversion of sodium hydroxide (the moles of sodium hydroxide reacted per mole of sodium hydroxide initially present), set up a stoichio-metric table expressing the concentration of each species in terms of its initial concentration and the conversion, Χ.

Solution

Because we have taken sodium hydroxide as our basis of calculation, we divide through by the stoichiometric coefficient of sodium hydroxide to put the reaction expression in the form

\begin{matrix} NaOH + \frac{1}{3} (C_{17} H_{35} {COO)}_{3} C_{3} H_{5} \to C_{17} H_{35} COONa+ \frac{1}{3} C_{3} H_{5} (OH)_{3} \\ A + \frac{1}{3} B \to C + \frac{1}{3} D \end{matrix}

$\begin{matrix} NaOH + \frac{1}{3} (C_{17} H_{35} {COO)}_{3} C_{3} H_{5} \to C_{17} H_{35} COONa+ \frac{1}{3} C_{3} H_{5} (OH)_{3} \\ A + \frac{1}{3} B \to C + \frac{1}{3} D \end{matrix}$

We may then perform the calculations shown in Table E4-1.1. Because this is a liquid-phase reaction, the density ρ is considered to be constant; therefore, V = V₀.

\begin{matrix} C_{A} = \frac{N_{A}}{V} = \frac{N_{A}}{V_{0}} = \frac{N_{A0} (1 - X)}{V_{0}} = C_{A0} (1 - X) \\ Θ_{B} = \frac{C_{B0}}{C_{A0}} Θ_{C} = \frac{C_{C 0}}{C_{A0}} Θ_{D} = \frac{C_{D0}}{C_{A0}} \end{matrix}

$\begin{matrix} C_{A} = \frac{N_{A}}{V} = \frac{N_{A}}{V_{0}} = \frac{N_{A0} (1 - X)}{V_{0}} = C_{A0} (1 - X) \\ Θ_{B} = \frac{C_{B0}}{C_{A0}} Θ_{C} = \frac{C_{C 0}}{C_{A0}} Θ_{D} = \frac{C_{D0}}{C_{A0}} \end{matrix}$

TABLE E4-1.1 STOICHIOMETRIC TABLE FOR LIQUID-PHASE SOAP REACTION

Analysis: The purpose of this example was to show how the generic reaction in Table 4-1 is applied to a real reaction.

Stoichiometric table (batch)

Example 4-2 What Is the Limiting Reactant?

Having set up the stoichiometric table in Example 4-1, one can now readily use it to calculate the concentrations at a given conversion. if the initial mixture consists of sodium hydroxide at a concentration of 10 mol/dm³ (i.e., 10 mol/L or 10 kmol/m³) and glyceryl stearate at a concentration of 2 mol/dm³, what are the concentrations of glycerol stearate, B, and of glycerine, D, when the conversion of sodium hydroxide is (a) 20% and (b) 90%?

Solution

Only the reactants NaOH and (C₁₇H₃₅COO)₃C₃H₅ are initially present; therefore, Θ_C = Θ_D = 0 .

(a) For 20% conversion of NaOH

\begin{matrix} C_{D} = C_{A0} (\frac{X}{3}) = (10) (\frac{0.2}{3}) = 0.67 mol/L = 0.67 mol/d m^{3} \\ C_{B} = C_{A0} (Θ_{B} - \frac{X}{3}) = 10 (\frac{2}{10} - \frac{0.2}{3}) = 10 (0.133) = 1.33 mol/d m^{3} \end{matrix}

$\begin{matrix} C_{D} = C_{A0} (\frac{X}{3}) = (10) (\frac{0.2}{3}) = 0.67 mol/L = 0.67 mol/d m^{3} \\ C_{B} = C_{A0} (Θ_{B} - \frac{X}{3}) = 10 (\frac{2}{10} - \frac{0.2}{3}) = 10 (0.133) = 1.33 mol/d m^{3} \end{matrix}$

(b) For 90% conversion of NaOH

C_{D} = C_{A0} (\frac{X}{3}) = 10 (\frac{0.9}{3}) = 3 mol/d m^{3}

$C_{D} = C_{A0} (\frac{X}{3}) = 10 (\frac{0.9}{3}) = 3 mol/d m^{3}$

Let us find C_B

C_{B} = 10 (\frac{2}{10} - \frac{0.9}{3}) = 10 (0.2 - 0.3) = - 1 mol/d m^{3}

$C_{B} = 10 (\frac{2}{10} - \frac{0.9}{3}) = 10 (0.2 - 0.3) = - 1 mol/d m^{3}$

Oops!! Negative concentration—impossible! What went wrong?

Analysis: We chose the wrong basis of calculation! We must choose the limiting reactant as our basis of calculation. Ninety-percent conversion of NaOH is not possible because glyceryl stearate is the limiting reactant and is used up before 90% of the NaOH can be reacted. Glyceryl stearate should have been our basis of calculation and therefore we should not have divided the reaction as written by the stoichi-ometric coefficient of 3. It’s always important to make sure you learn something from a mistake, such as choosing the wrong basis of calculation. There is no such thing as failure, unless one fails to learn from it.

The basis of calculation must be the limiting reactant.

4.2 Flow Systems

The form of the stoichiometric table for a continuous-flow system (see Figure 4-2) is virtually identical to that for a batch system (Table 4-1), except that we replace N_j0 by F_j0 and N_j by F_j (Table 4-2). Again taking A as the basis, we divide Equation (2-1) through by the stoichiometric coefficient of A to obtain

\begin{matrix} A + \frac{b}{a} B \to \frac{c}{a} C + \frac{d}{a} D & (2-2) \end{matrix}

$\begin{matrix} A + \frac{b}{a} B \to \frac{c}{a} C + \frac{d}{a} D & (2-2) \end{matrix}$

In the Flow reactor, Feed rates are shown at left and effluent rates are shown at right with an equation placed between them.

Figure 4-2 Flow reactor.

TABLE 4-2 STOICHIOMETRIC TABLE FOR A FLOW SYSTEM

Species	Feed Rate to Reactor (mol/time)	Change within Reactor (mol/time)	Effluent Rate from Reactor (mol/time)
A	F_A0	-F_A0E	F_A = F_A0 (1 – X)
B	F_B0 = Θ_BF_A0	$- \frac{b}{a} F_{A0} X$ $- \frac{b}{a} F_{A0} X$	$F_{B} = F_{A0} (Θ_{B} - \frac{b}{a} X)$ $F_{B} = F_{A0} (Θ_{B} - \frac{b}{a} X)$
C	F_C0 = Θ_CF_A0	$\frac{c}{a} F_{A0} X$ $\frac{c}{a} F_{A0} X$	$F_{C} = F_{A0} (Θ_{C} + \frac{c}{a} X)$ $F_{C} = F_{A0} (Θ_{C} + \frac{c}{a} X)$
D	F_D0 = Θ_DF_A0	$\frac{d}{a} F_{A0} X$ $\frac{d}{a} F_{A0} X$	$F_{D} = F_{A0} (Θ_{D} + \frac{d}{a} X)$ $F_{D} = F_{A0} (Θ_{D} + \frac{d}{a} X)$
I	F_I0 = Θ_IF_A0	–	F_I = F_A0Θ_I
Totals	F_T0		$\begin{matrix} F_{T} = F_{T0} + (\frac{d}{a} + \frac{c}{a} - \frac{b}{a} - 1) F_{A0} X \\ F_{T} = F_{T0} + δ F_{A0} X \end{matrix}$ $\begin{matrix} F_{T} = F_{T0} + (\frac{d}{a} + \frac{c}{a} - \frac{b}{a} - 1) F_{A0} X \\ F_{T} = F_{T0} + δ F_{A0} X \end{matrix}$

where

\begin{matrix} Θ_{i} = \frac{F_{B0}}{F_{A 0}} = \frac{C_{B 0} ν_{0}}{C_{A 0 ν_{0}}} = \frac{C_{B0}}{C_{A 0}} = \frac{y_{B0}}{y_{A0}} \end{matrix}

$\begin{matrix} Θ_{i} = \frac{F_{B0}}{F_{A 0}} = \frac{C_{B 0} ν_{0}}{C_{A 0 ν_{0}}} = \frac{C_{B0}}{C_{A 0}} = \frac{y_{B0}}{y_{A0}} \end{matrix}$

Stoichiometric table for a flow system

and Θ_C, Θ_D, and Θ_I, are defined similarly

and, as before,

\begin{matrix} δ = \frac{d}{a} + \frac{c}{a} - \frac{b}{a} - 1 \end{matrix} (4 - 1)

$\begin{matrix} δ = \frac{d}{a} + \frac{c}{a} - \frac{b}{a} - 1 \end{matrix} (4 - 1)$

4.2.1 Equations for Concentrations in Flow Systems

For a flow system, the concentration C_A at a given point can be determined from the molar flow rate F_A and the volumetric flow rate υ at that point:

Definition of concentration for a flow system

\begin{matrix} C_{A} = \frac{F_{A}}{ν} = \frac{moles/time}{liters/time} = \frac{moles}{liter} \end{matrix} (4 - 10)

$\begin{matrix} C_{A} = \frac{F_{A}}{ν} = \frac{moles/time}{liters/time} = \frac{moles}{liter} \end{matrix} (4 - 10)$

Units of υ are typically given in terms of liters per second, cubic decimeters per second, or cubic feet per minute. We now can write the concentrations of A, B, C, and D for the general reaction given by Equation (2-2) in terms of their respective entering molar flow rates ( F_A0, F_B0, F_C0, F_D0 ), the conversion, X, and the volumetric flow rate, υ.

\begin{matrix} C_{A} = \frac{F_{A}}{υ} = \frac{F_{A0}}{υ} (1 - X) & C_{B} = \frac{F_{B}}{υ} = \frac{F_{B0} - (b / a) F_{A0} X}{υ} \\ C_{C} = \frac{F_{C}}{υ} = \frac{F_{C 0} + (c / a) F_{A0} X}{υ} & C_{D} = \frac{F_{D}}{υ} = \frac{F_{D 0} - (d / a) F_{A0} X}{υ} & (4-11) \end{matrix}

$\begin{matrix} C_{A} = \frac{F_{A}}{υ} = \frac{F_{A0}}{υ} (1 - X) & C_{B} = \frac{F_{B}}{υ} = \frac{F_{B0} - (b / a) F_{A0} X}{υ} \\ C_{C} = \frac{F_{C}}{υ} = \frac{F_{C 0} + (c / a) F_{A0} X}{υ} & C_{D} = \frac{F_{D}}{υ} = \frac{F_{D 0} - (d / a) F_{A0} X}{υ} & (4-11) \end{matrix}$

We now focus on determining the volumetric flow rate, υ.

4.2.2 Liquid-Phase Concentrations

For liquids, the fluid volume change with reaction is negligible when no phase changes are taking place. Consequently, we can take

υ = υ₀

For liquids

C_A = C_A0 (1 – X)

C_{B} = C_{A0} (Θ_{B} - \frac{b}{a} X)

$C_{B} = C_{A0} (Θ_{B} - \frac{b}{a} X)$

Therefore, for a given rate law we have –r_A = g(X)

Then

\begin{matrix} C_{A} = \frac{F_{A0}}{υ_{0}} (1 - X) = C_{A0} (1 - X) & (4-12) \end{matrix}

$\begin{matrix} C_{A} = \frac{F_{A0}}{υ_{0}} (1 - X) = C_{A0} (1 - X) & (4-12) \end{matrix}$

\begin{matrix} C_{B} = C_{A0} (Θ_{B} - \frac{b}{a} X) & (4-13) \end{matrix}

$\begin{matrix} C_{B} = C_{A0} (Θ_{B} - \frac{b}{a} X) & (4-13) \end{matrix}$

and so forth for C_C and C_D.

Consequently, using any one of the rate laws in Chapter 3, we can now find –r_A = f(X) for liquid-phase reactions. However, for gas-phase reactions the volumetric flow rate most often changes during the course of the reaction because of a change in the total number of moles or a change in temperature or pressure. Hence, one cannot always use Equation (4-13) to express concentration as a function of conversion for gas-phase reactions.

4.2.3 Gas-Phase Concentrations

In our previous discussions, we considered primarily systems in which the reaction volume or volumetric flow rate did not vary as the reaction progressed. Most batch and liquid-phase and some gas-phase systems fall into this category. There are other systems, though, in which either V or υ does vary, and these will now be considered.

A situation where one encounters a varying flow rate occurs quite frequently in gas-phase reactions that do not have an equal number of product and reactant moles. For example, in the synthesis of ammonia

N_{2} + 3 H_{2} ⇄ 2 {NH}_{3}

$N_{2} + 3 H_{2} ⇄ 2 {NH}_{3}$

4 mol of reactants gives 2 mol of product. In flow systems where this type of reaction occurs, the molar flow rate will be changing as the reaction progresses. Because equal numbers of moles occupy equal volumes in the gas phase at the same temperature and pressure, the volumetric flow rate will also change.

In the stoichiometric tables presented on the preceding pages, it was not necessary to make assumptions concerning a volume change in the first four columns of the table (i.e., the species, initial number of moles or molar feed rate, change within the reactor, and the remaining number of moles or the molar effluent rate). All of these columns of the stoichiometric table are independent of the volume or density, and they are identical for constant-volume (constant-density) and varying-volume (varying-density) situations. Only when concentration is expressed as a function of conversion does variable density enter the picture.

Flow Reactors with Variable Volumetric Flow Rate. To derive the concentrations of each species in terms of conversion for a gas-phase flow system, we shall start by using the relationships for the total concentration. The total concentration, C_T, at any point in the reactor is the total molar flow rate, F_T, divided by the volumetric flow rate, υ [cf. Equation (4-10)]. In the gas phase, the total concentration is also found from the gas law, C_T = P/ZRT. Equating these two relationships gives

\begin{matrix} C_{T} = \frac{F_{T}}{υ} = \frac{P}{Z R T} & (4-14) \end{matrix}

$\begin{matrix} C_{T} = \frac{F_{T}}{υ} = \frac{P}{Z R T} & (4-14) \end{matrix}$

At the entrance to the reactor

\begin{matrix} C_{T0} = \frac{F_{T0}}{υ_{0}} = \frac{P_{0}}{Z_{0} R T_{0}} & (4-15) \end{matrix}

$\begin{matrix} C_{T0} = \frac{F_{T0}}{υ_{0}} = \frac{P_{0}}{Z_{0} R T_{0}} & (4-15) \end{matrix}$

Taking the ratio of Equation (4-14) to Equation (4-15) and assuming negligible changes in the compressibility factor occur, i.e., Z ≌ Z₀, during the course of the reaction, we have upon rearrangement

Gas-phase reactions

\begin{matrix} ν = ν_{0} (\frac{F_{T}}{F_{T 0}}) \frac{P_{0}}{P} (\frac{T}{T_{0}}) \end{matrix} (4 - 16)

$\begin{matrix} ν = ν_{0} (\frac{F_{T}}{F_{T 0}}) \frac{P_{0}}{P} (\frac{T}{T_{0}}) \end{matrix} (4 - 16)$

We can now express the concentration of species j for a flow system in terms of its flow rate, F_j, the temperature, T, and total pressure, P.

C_{j} = \frac{F_{j}}{υ} = \frac{F_{j}}{υ_{0} (\frac{F_{T}}{F_{T0}} \frac{P_{0}}{P} \frac{T}{T_{0}})} = (\frac{F_{T0}}{υ_{0}}) (\frac{F_{j}}{F_{T}}) (\frac{P}{P_{0}}) (\frac{T_{0}}{T})

$C_{j} = \frac{F_{j}}{υ} = \frac{F_{j}}{υ_{0} (\frac{F_{T}}{F_{T0}} \frac{P_{0}}{P} \frac{T}{T_{0}})} = (\frac{F_{T0}}{υ_{0}}) (\frac{F_{j}}{F_{T}}) (\frac{P}{P_{0}}) (\frac{T_{0}}{T})$

\begin{matrix} C_{j} {= C}_{T 0} (\frac{F_{j}}{F_{T}}) (\frac{P_{0}}{P}) (\frac{T}{T_{0}}) \end{matrix} (4 - 17)

$\begin{matrix} C_{j} {= C}_{T 0} (\frac{F_{j}}{F_{T}}) (\frac{P_{0}}{P}) (\frac{T}{T_{0}}) \end{matrix} (4 - 17)$

Use this concentration equation for membrane reactors (Chapter 6) and for multiple reactions (Chapter 8).

The total molar flow rate is just the sum of the molar flow rates of each of the species in the system and is

\begin{matrix} \begin{matrix} F_{T} = F_{A} + F_{B} + F_{C} + F_{D} + F_{I} + ... = Σ_{j = 1}^{n} & F_{j} \end{matrix} & (4-18) \end{matrix}

$\begin{matrix} \begin{matrix} F_{T} = F_{A} + F_{B} + F_{C} + F_{D} + F_{I} + ... = Σ_{j = 1}^{n} & F_{j} \end{matrix} & (4-18) \end{matrix}$

We can also write Equation (4-17) in terms of the mole fraction of species j, y_j, and the pressure ratio, p, with respect to the initial or entering conditions, i.e., sub “0”

\begin{matrix} y_{i} = \frac{F_{j}}{F_{T}} \\ p = \frac{P}{P_{0}} \end{matrix}

$\begin{matrix} y_{i} = \frac{F_{j}}{F_{T}} \\ p = \frac{P}{P_{0}} \end{matrix}$

\begin{matrix} C_{j} = C_{T0} y_{j} p \frac{T_{0}}{T} & (4-19) \end{matrix}

$\begin{matrix} C_{j} = C_{T0} y_{j} p \frac{T_{0}}{T} & (4-19) \end{matrix}$

The molar flow rates, F_j, are found by solving the mole balance equations. The concentration given by Equation (4-17) will be used for measures other than conversion when we discuss membrane reactors (Chapter 6) and multiple gas-phase reactions (Chapter 8).

Now, let’s express the concentration in terms of conversion for gas flow systems. From Table 4-2, the total molar flow rate can be written in terms of conversion and is

F_T = F_T0 + F_A0 δX

We divide this equation through by F_T0

\frac{F_{T}}{F_{T 0}} = 1 + \frac{F_{A0}}{F_{T 0}} δ X = 1 + \overset{ɛ}{\overset{︷}{y_{A 0} δ}} X

$\frac{F_{T}}{F_{T 0}} = 1 + \frac{F_{A0}}{F_{T 0}} δ X = 1 + \overset{ɛ}{\overset{︷}{y_{A 0} δ}} X$

Then

$\begin{matrix} \frac{F_{T}}{F_{T0}} = 1 + ɛ X & (4-20) \end{matrix}$ $\begin{matrix} \frac{F_{T}}{F_{T0}} = 1 + ɛ X & (4-20) \end{matrix}$

where y_A0 is the mole fraction of A at the inlet (i.e., (F_A0/F_T0)), and where δ is given by Equation (4-1) and ε is given by

Relationship between δ and ε

$ε = (\frac{d}{a} + \frac{c}{a} - \frac{b}{a} - 1) \frac{F_{A0}}{F_{T0}} = y_{A0} δ$ $ε = (\frac{d}{a} + \frac{c}{a} - \frac{b}{a} - 1) \frac{F_{A0}}{F_{T0}} = y_{A0} δ$

\begin{matrix} ε = y_{A0} δ \end{matrix} (4 - 21)

$\begin{matrix} ε = y_{A0} δ \end{matrix} (4 - 21)$

Equation (4-21) holds for both batch and flow systems. To interpret ε, let’s rearrange Equation (4-20) at complete conversion (i.e., X = 1 and F_T = F_Tf)

$ε = \frac{F_{T f} - F_{T0}}{F_{T0}}$ $ε = \frac{F_{T f} - F_{T0}}{F_{T0}}$

Interpretation of ε

\begin{matrix} ε = \frac{change in total number of moles for complete conversion}{Total moles fed} \end{matrix} (4 - 22)

$\begin{matrix} ε = \frac{change in total number of moles for complete conversion}{Total moles fed} \end{matrix} (4 - 22)$

Gas-phase volumetric flow rate

Substituting for (F_T/F_T0 ) in Equation (4-16) for the volumetric flow rate, υ, we have

\begin{matrix} ν = ν_{0} (1 + ε X) \frac{P_{0}}{P} (\frac{T}{T_{0}}) \end{matrix} (4 - 23)

$\begin{matrix} ν = ν_{0} (1 + ε X) \frac{P_{0}}{P} (\frac{T}{T_{0}}) \end{matrix} (4 - 23)$

The concentration of species j in a flow system is

$\begin{matrix} C_{j} = \frac{F_{j}}{v} & (4-24) \end{matrix}$ $\begin{matrix} C_{j} = \frac{F_{j}}{v} & (4-24) \end{matrix}$

The molar flow rate of species j is

$F_{j} = F_{j 0} + v_{j} (F_{A 0} X) = F_{A0} (Θ_{j} + v_{j} X)$ $F_{j} = F_{j 0} + v_{j} (F_{A 0} X) = F_{A0} (Θ_{j} + v_{j} X)$

where V_j is the stoichiometric coefficient for species j, which is negative for reactants and positive for products. For example, for the reaction

$\begin{matrix} A + \frac{b}{a} B \to \frac{c}{a} C + \frac{d}{a} D & (2-2) \end{matrix}$ $\begin{matrix} A + \frac{b}{a} B \to \frac{c}{a} C + \frac{d}{a} D & (2-2) \end{matrix}$

\begin{matrix} V_{A} = - 1, & V_{B} = - b / a, & V_{C} = c / a, & V_{D} = d / a, & and Θ_{j} = F_{j 0} / F_{A 0} \end{matrix} .

$\begin{matrix} V_{A} = - 1, & V_{B} = - b / a, & V_{C} = c / a, & V_{D} = d / a, & and Θ_{j} = F_{j 0} / F_{A 0} \end{matrix} .$

Substituting for the gas volumetric flow rate, υ, using Equation (4-23) and for F_j, we have

$C_{j} = \frac{F_{A0} (Θ_{j} + v_{j} X)}{v_{0} ((1 + ε X) \frac{P_{0}}{P} + \frac{T}{T_{0}})}$ $C_{j} = \frac{F_{A0} (Θ_{j} + v_{j} X)}{v_{0} ((1 + ε X) \frac{P_{0}}{P} + \frac{T}{T_{0}})}$

Gas-phase concentration as a function of conversion

Rearranging

\begin{matrix} C_{j} = \frac{C_{A 0} (Θ_{j} + v_{j} X)}{1 + ε X} (\frac{P}{P_{0}}) \frac{T_{0}}{T} & (4 - 25) \end{matrix}

$\begin{matrix} C_{j} = \frac{C_{A 0} (Θ_{j} + v_{j} X)}{1 + ε X} (\frac{P}{P_{0}}) \frac{T_{0}}{T} & (4 - 25) \end{matrix}$

Recall that y_A0 = F_A0/F_T0, C_A0 = y_A0C_T0, and ε is given by Equation (4-21) (i.e., ε = y_A0δ).

The stoichiometric table for the gas-phase reaction (2-2) is given in Table 4-3.

TABLE 4-3 CONCENTRATIONS IN A VARIABLE-VOLUME GAS FLOW SYSTEM

\begin{array}{l} C_{A} = \frac{F_{A}}{υ} = \frac{F_{A 0} (1 - X)}{υ} & = \frac{F_{A 0} (1 - X)}{υ_{0} (1 + ε X)} (\frac{T_{0}}{T}) \frac{P}{P_{0}} & = C_{A 0} (\frac{1 - X}{1 + ε X}) \frac{T_{0}}{T} (\frac{P}{P_{0}}) \\ C_{B} = \frac{F_{B}}{υ} = \frac{F_{A 0} [Θ_{B} - (b / a) X]}{υ} & = \frac{F_{A 0} [Θ_{B} - (b / a) X]}{υ_{0} (1 + ε X)} (\frac{T_{0}}{T}) \frac{P}{P_{0}} & = C_{A 0} (\frac{Θ_{B} - (b / a) X}{(1 + ε X)}) \frac{T_{0}}{T} (\frac{P}{P_{0}}) \\ C_{C} = \frac{F_{C}}{υ} = \frac{F_{A 0} [Θ_{c} + (c / a) X]}{υ} & = \frac{F_{A 0} [Θ_{C} + (c / a) X]}{υ_{0} (1 + ε X)} (\frac{T_{0}}{T}) \frac{P}{P_{0}} & = C_{A 0} (\frac{Θ_{C} + (c / a) X}{(1 + ε X)}) \frac{T_{0}}{T} (\frac{P}{P_{0}}) \\ C_{D} = \frac{F_{D}}{υ} = \frac{F_{A 0} [Θ_{D} + (d / a) X]}{υ} & = \frac{F_{A 0} [Θ_{D} + (d / a) X]}{υ_{0} (1 + ε X)} (\frac{T_{0}}{T}) \frac{P}{P_{0}} & = C_{A 0} (\frac{Θ_{D} + (d / a) X}{(1 + ε X)}) \frac{T_{0}}{T} (\frac{P}{P_{0}}) \\ C_{1} = \frac{F_{I}}{υ} = \frac{F_{A 0} Θ_{1}}{υ} & = \frac{F_{A 0} Θ_{I}}{υ_{0} (1 + ε X)} (\frac{T_{0}}{T}) \frac{P}{P_{0}} & = \frac{C_{A 0} Θ_{I}}{1 + ε X} (\frac{T_{0}}{T}) \frac{P}{P_{0}} \end{array}

$\begin{array}{l} C_{A} = \frac{F_{A}}{υ} = \frac{F_{A 0} (1 - X)}{υ} & = \frac{F_{A 0} (1 - X)}{υ_{0} (1 + ε X)} (\frac{T_{0}}{T}) \frac{P}{P_{0}} & = C_{A 0} (\frac{1 - X}{1 + ε X}) \frac{T_{0}}{T} (\frac{P}{P_{0}}) \\ C_{B} = \frac{F_{B}}{υ} = \frac{F_{A 0} [Θ_{B} - (b / a) X]}{υ} & = \frac{F_{A 0} [Θ_{B} - (b / a) X]}{υ_{0} (1 + ε X)} (\frac{T_{0}}{T}) \frac{P}{P_{0}} & = C_{A 0} (\frac{Θ_{B} - (b / a) X}{(1 + ε X)}) \frac{T_{0}}{T} (\frac{P}{P_{0}}) \\ C_{C} = \frac{F_{C}}{υ} = \frac{F_{A 0} [Θ_{c} + (c / a) X]}{υ} & = \frac{F_{A 0} [Θ_{C} + (c / a) X]}{υ_{0} (1 + ε X)} (\frac{T_{0}}{T}) \frac{P}{P_{0}} & = C_{A 0} (\frac{Θ_{C} + (c / a) X}{(1 + ε X)}) \frac{T_{0}}{T} (\frac{P}{P_{0}}) \\ C_{D} = \frac{F_{D}}{υ} = \frac{F_{A 0} [Θ_{D} + (d / a) X]}{υ} & = \frac{F_{A 0} [Θ_{D} + (d / a) X]}{υ_{0} (1 + ε X)} (\frac{T_{0}}{T}) \frac{P}{P_{0}} & = C_{A 0} (\frac{Θ_{D} + (d / a) X}{(1 + ε X)}) \frac{T_{0}}{T} (\frac{P}{P_{0}}) \\ C_{1} = \frac{F_{I}}{υ} = \frac{F_{A 0} Θ_{1}}{υ} & = \frac{F_{A 0} Θ_{I}}{υ_{0} (1 + ε X)} (\frac{T_{0}}{T}) \frac{P}{P_{0}} & = \frac{C_{A 0} Θ_{I}}{1 + ε X} (\frac{T_{0}}{T}) \frac{P}{P_{0}} \end{array}$

At last!

We now have C_j = h_j (X ) and

— r_A = g (X )

for variable-volume gas-phase reactions.

One of the major objectives of this chapter is to learn how to express any given rate law – r_A as a function of conversion. The schematic diagram in Figure 4-3 helps to summarize our discussion on this point. The concentration of the reactant species B in the generic reaction is expressed as a function of conversion in both flow and batch systems for various conditions of temperature, pressure, and volume.

Flow diagram expresses concentration as a function of conversion.

"The flow diagram starts from ""A plus (b over a) B forming (c over a) C plus (d over a) D is divided into as Liquid Phase and Gas Phase. The liquid phase is further divided into Flow and Batch and the gas phase is further divided into Batch and Flow. Under Flow of the Liquid Phase, the equation reads, “C subscript B equals F subscript B over upsilon.” Under Batch of the Liquid Phase, the equation reads, “C subscript B equals N subscript B over V.” When no phase change in the Flow, “upsilon equals upsilon subscript 0” and when no phase change in the batch “V equals V subscript 0.” The no phase change in flow and batch, the equation collectively forms ""C subscript B which equals C subscript A0 (theta subscript B minus (b over a) X).” In the box below the equation reads, “epsilon equals y subscript A0 delta, where delta equals ((d over a) plus (c over a) minus (b over a) minus 1),“C subscript T0 equals P subscript 0 over RT subscript 0,” and “C subscript A0 equals y subscript A0 C subscript T0” are given.

Under Batch of the Gas Phase, the equation reads, “C subscript B equals N subscript B over V.” The equation “V equals V subscript 0 times N subscript T over N subscript T0 times P subscript 0 over P times T over T subscript 0;” and ""C subscript B equals N subscript T0 over V subscript 0 times N subscript B over N subscript T times P over P subscript 0 times T subscript 0 over T.” Since N subscript T0 over V subscript 0 is simplified as C subscript T0. Then the equation becomes, C subscript B equals C subscript T0 times N subscript B over N subscript T times P over P subscript 0 times T subscript 0 over T.” Under Flow of the Gas Phase, the equation reads, “C subscript B equals F subscript B over upsilon.” The equation “Upsilon equals upsilon subscript 0 times F subscript T over F subscript T0 times P subscript 0 over P times T over T subscript 0;” and ""C subscript B equals F subscript T0 over upsilon subscript 0 times F subscript B over F subscript T times P over P subscript 0 times T subscript 0 over T.” since F subscript T0 over upsilon subscript 0 is simplified as C subscript T0. Then the equation becomes, C subscript B equals C subscript T0 times F subscript B over F subscript T times P over P subscript 0 times T subscript 0 over T.” When no phase change or no semipermeable membranes in the Batch and Flow of Gas Phase, the equation becomes, “upsilon equals upsilon subscript 0 (1 plus epsilon X) (P subscript 0 over P) (T over T subscript 0)” and “C subscript B equals C subscript A0 (theta subscript B minus (b over a) times X) over 1 plus epsilon X times (P over P subscript 0) times T subscript 0 over T.” At isothermal, the equation becomes “C subscript B equals C subscript A0 (theta subscript B minus (b over a) X) over 1 plus epsilon X times P over P subscript 0.” By neglecting pressure drop, the equation becomes “C subscript B equals C subscript A0 (theta subscript B minus (b over a) times X) over 1 plus epsilon X.”"

Figure 4-3 Expressing concentration as a function of conversion.

Example 4-3 Determining C_j = h_j (X) for a Gas-Phase Reaction

A mixture of 28% SO₂ and 72% air is charged to a flow reactor in which SO₂ is oxidized.

$2 SO_{2} + O_{2} \to 2 {SO}_{3}$ $2 SO_{2} + O_{2} \to 2 {SO}_{3}$

(a) First, set up a stoichiometric table using only the symbols (i.e., Θ_i, F_i).

(b) Next, prepare a second table evaluating the species concentrations as a function of conversion for the case when the total pressure is 1485 kPa (14.7 atm) and the temperature is constant at 227°C.

Solution

(a) Stoichiometric table. Taking SO₂ as the basis of calculation, we divide the reaction through by the stoichiometric coefficient of our chosen basis of calculation

$SO_{2} + \frac{1}{2} O_{2} \to {SO}_{3}$ $SO_{2} + \frac{1}{2} O_{2} \to {SO}_{3}$

The stoichiometric table is given in Table E4-3.1.

TABLE E4-3.1 STOICHIOMETRIC TABLE FOR $SO_{2} + \frac{1}{2} O_{2} \to {SO}_{3}$ $SO_{2} + \frac{1}{2} O_{2} \to {SO}_{3}$

Species	Symbol	Initially	Change	Remaining
SO₂	A	F_A0	−F_A0X	F_A = F_A0(1−X)
O₂	B	$F_{B 0} = Θ_{B} F_{A 0}$ $F_{B 0} = Θ_{B} F_{A 0}$	$- \frac{F_{A 0} X}{2}$ $- \frac{F_{A 0} X}{2}$	$F_{B} = F_{A 0} (Θ_{B} - \frac{1}{2} X)$ $F_{B} = F_{A 0} (Θ_{B} - \frac{1}{2} X)$
SO₃	C	0	+F_A0X	F_C = F_A0X
N₂	I	$F_{I 0} = Θ_{I} F_{A 0}$ $F_{I 0} = Θ_{I} F_{A 0}$	−	$F_{I} = F_{I 0} = Θ_{I} F_{A 0}$ $F_{I} = F_{I 0} = Θ_{I} F_{A 0}$

Totals		F_T0		$F_{T} = F_{T 0} - \frac{F_{A 0} X}{2}$ $F_{T} = F_{T 0} - \frac{F_{A 0} X}{2}$

(b) Expressing the concentration as a function of conversion.

From the definition of conversion, we substitute not only for the molar flow rate of SO₂ (A) in terms of conversion but also for the volumetric flow rate as a function of conversion

$C_{A} = \frac{F_{A}}{υ} = \frac{F_{A0} (1 - X)}{υ} (E4-3.1)$ $C_{A} = \frac{F_{A}}{υ} = \frac{F_{A0} (1 - X)}{υ} (E4-3.1)$

Neglecting pressure

drop, P = P₀,

$i.e., p = \frac{p}{p_{0}} \approx 1$ $i.e., p = \frac{p}{p_{0}} \approx 1$

Recalling Equation (4-23), we have

$v = v_{0} (1 + εX) \frac{P_{0}}{P} (\frac{T}{T_{0}}) (4 - 23)$ $v = v_{0} (1 + εX) \frac{P_{0}}{P} (\frac{T}{T_{0}}) (4 - 23)$

Neglecting pressure drop in the reaction, P = P₀ , yields

$\begin{matrix} ν = ν_{0} (1 + ɛX) (\frac{T}{T_{0}}) & (E 4 - 3.2) \end{matrix}$ $\begin{matrix} ν = ν_{0} (1 + ɛX) (\frac{T}{T_{0}}) & (E 4 - 3.2) \end{matrix}$

Isothermal operation, T = T₀

If the reaction is also carried out isothermally, T = T₀, we obtain

$υ = υ_{0} (1 + ε X)$ $υ = υ_{0} (1 + ε X)$

$C_{A} = \frac{F_{A 0} (1 - X)}{v_{0} (1 + ε X)} = C_{A 0} (\frac{1 - X}{1 + ε X})$ $C_{A} = \frac{F_{A 0} (1 - X)}{v_{0} (1 + ε X)} = C_{A 0} (\frac{1 - X}{1 + ε X})$

Similarly for B also with T = T₀ and P = P₀, (i.e., p = 1)

$C_{B} = C_{A0} \frac{(Θ_{B} - \frac{b}{a} X)}{1 + ε X} p \frac{T_{0}}{T} = \frac{C_{A0} (Θ_{B} - \frac{1}{2} X)}{1 + ε X}$ $C_{B} = C_{A0} \frac{(Θ_{B} - \frac{b}{a} X)}{1 + ε X} p \frac{T_{0}}{T} = \frac{C_{A0} (Θ_{B} - \frac{1}{2} X)}{1 + ε X}$

(c) Parameter evaluation and concentrations of a function of conversion plot. The inlet concentration of A is equal to the inlet mole fraction of A multiplied by the total inlet molar concentration. The total concentration can be calculated from an equation of state such as the ideal gas law. Recall that y_A0 = 0.28, T₀ = 500 K, and P₀ = 1485 kPa.

$\begin{matrix} C_{A0} & = y_{A0} C_{T0} = y_{A0} (\frac{P_{0}}{R T_{0}}) \\ = 0.28 [\frac{1485 kPa}{8.314 kPa \cdot {dm}^{3} / (mol \cdot K) \times 500 K}] \\ = 0.1 mol / {dm}^{3} \end{matrix}$ $\begin{matrix} C_{A0} & = y_{A0} C_{T0} = y_{A0} (\frac{P_{0}}{R T_{0}}) \\ = 0.28 [\frac{1485 kPa}{8.314 kPa \cdot {dm}^{3} / (mol \cdot K) \times 500 K}] \\ = 0.1 mol / {dm}^{3} \end{matrix}$

The total concentration at constant temperature and pressure is

The equation E4-3.3 under the parameter evaluation, the total concentration at constant temperature and pressure for C subscript T is given.

$C_{T0} = \frac{P_{o}}{R T_{o}} = \frac{1485 k P a}{[8.314 kPa \cdot {dm}^{3} / (mol \cdot K)] (500 K)} = 0.357 \frac{mol}{{dm}^{3}} (E4-3.4)$ $C_{T0} = \frac{P_{o}}{R T_{o}} = \frac{1485 k P a}{[8.314 kPa \cdot {dm}^{3} / (mol \cdot K)] (500 K)} = 0.357 \frac{mol}{{dm}^{3}} (E4-3.4)$

We now evaluate ε.

$ϵ = y_{A 0} δ = (0.28) (1-1 - \frac{1}{2}) = - 0.14 (E4-3.5)$ $ϵ = y_{A 0} δ = (0.28) (1-1 - \frac{1}{2}) = - 0.14 (E4-3.5)$

Initially, 72% of the total number of moles is air containing 21% O₂ and 79% N₂, along with 28% SO₂.

$\begin{array}{l} F_{A0} = (0.28) (F_{T0}) \\ F_{B 0} = (0.72) (0.21) (F_{T0}) \\ Θ_{B} = \frac{F_{B 0}}{F_{A0}} = \frac{(0.72) (0.21)}{0.28} = 0.54 \\ Θ_{I} = \frac{F_{I 0}}{F_{A0}} = \frac{(0.72) (0.79)}{0.28} = 2.03 \end{array}$ $\begin{array}{l} F_{A0} = (0.28) (F_{T0}) \\ F_{B 0} = (0.72) (0.21) (F_{T0}) \\ Θ_{B} = \frac{F_{B 0}}{F_{A0}} = \frac{(0.72) (0.21)}{0.28} = 0.54 \\ Θ_{I} = \frac{F_{I 0}}{F_{A0}} = \frac{(0.72) (0.79)}{0.28} = 2.03 \end{array}$

Substituting for C_A0 and ε in the species concentrations:

${SO}_{2 :} C_{A} = C_{A0} (\frac{1 - X}{1 + ε X}) = 0.1 (\frac{1 - X}{1–0.14 X}) {mol/dm}^{3} (E4-3.6)$ ${SO}_{2 :} C_{A} = C_{A0} (\frac{1 - X}{1 + ε X}) = 0.1 (\frac{1 - X}{1–0.14 X}) {mol/dm}^{3} (E4-3.6)$

Note: Because the volumetic flow rate varies with conversion, υ = υ₀(1 - 0.14X), the concentration of inerts (N₂) is not constant.

Now we could use techniques presented in Chapter 2 to size reactors.

The concentrations of different species at various conversions are calculated in Table E4-3.2 and plotted in Figure E4-3.1. Note that the concentration of N₂ is changing even though it is an inert species in this reaction!!

$O_{2 :} C_{B} = C_{A0} (\frac{Θ_{B} - \frac{1}{2} X}{1 + ε X}) = \frac{0.1 (0.54–0.5 X)}{1–0.14 X} {mol/dm}^{3} (E4-3.7)$ $O_{2 :} C_{B} = C_{A0} (\frac{Θ_{B} - \frac{1}{2} X}{1 + ε X}) = \frac{0.1 (0.54–0.5 X)}{1–0.14 X} {mol/dm}^{3} (E4-3.7)$

${S O}_{3 :} C_{c} = (\frac{C_{A0} X}{1 + ε X}) = \frac{0.1 X}{1–0.14 X} {mol/dm}^{3} (E4-3.8)$ ${S O}_{3 :} C_{c} = (\frac{C_{A0} X}{1 + ε X}) = \frac{0.1 X}{1–0.14 X} {mol/dm}^{3} (E4-3.8)$

$S O_{3 :} C_{I} = (\frac{C_{A0} X}{1 + ε X}) = \frac{0.1 X}{1–0.14 X} {mol/dm}^{3} (E4-3.8)$ $S O_{3 :} C_{I} = (\frac{C_{A0} X}{1 + ε X}) = \frac{0.1 X}{1–0.14 X} {mol/dm}^{3} (E4-3.8)$

TABLE E4-3.2 CONCENTRATION AS A FUNCTION OF CONVERSION

				C_i (mol/dm³)
Species		X = 0.0	X = 0.25	X = 0.5	X = 0.75	X = 1.0
SO₂	C_A =	0.100	0.078	0.054	0.028	0.000
O₂	C_B =	0.054	0.043	0.031	0.018	0.005
SO₃	C_C =	0.000	0.026	0.054	0.084	0.116
N₂	C_I =	0.203	0.210	0.218	0.227	0.236
Total	C_T =	0.357	0.357	0.357	0.357	0.357

A graph depicts concentration as a function of conversion.

Figure E4-3.1 Concentration as a function of conversion.

We are now in a position to express –r_A as a function of X and use the techniques in Chapter 2. However, we will use a better method to solve CRE problems, namely a Table of Integrals (Appendix A) or the Polymath or MATLAB software, discussed in the next chapter.

Analysis: In this example, we formed a stoichiometric table in terms of molar flow rates. We then showed how to express the concentrations of each species in a gas-phase reaction in which there is a change in the total number of moles. Next, we plotted each species concentration as a function of conversion and noted that the concentration of the inert, N₂, was not constant but increased with increasing conversion because of the decrease in the total molar flow rate, F_T, with conversion.

As previously mentioned many, if not most, rate laws for catalyst reactions are given in terms of partial pressures. Luckily, partial pressures are easily related to conversion with the aid of the ideal gas law and Equation (4-17).

The equation representing the ideal gas law P subscript i is shown.

P_{i} = C_{i} R T = C_{T 0} \frac{F_{i}}{F_{T 0}} {\overset{︷}{(\frac{P}{P_{0}})}}_{}^{P} \frac{T_{0}}{T} R T = \overset{P_{T 0}}{\overset{︷}{C_{T 0} {R T}_{0}}} \frac{\overset{y_{i}}{\overset{︷}{F_{i}}}}{F_{T 0}} P

$P_{i} = C_{i} R T = C_{T 0} \frac{F_{i}}{F_{T 0}} {\overset{︷}{(\frac{P}{P_{0}})}}_{}^{P} \frac{T_{0}}{T} R T = \overset{P_{T 0}}{\overset{︷}{C_{T 0} {R T}_{0}}} \frac{\overset{y_{i}}{\overset{︷}{F_{i}}}}{F_{T 0}} P$

The following equation is used when the mole balance is written in terms of molar flow rates

\begin{matrix} P_{i} = P_{T 0} (\frac{F_{i}}{F_{T 0}}) p & (4 - 26) \end{matrix}

$\begin{matrix} P_{i} = P_{T 0} (\frac{F_{i}}{F_{T 0}}) p & (4 - 26) \end{matrix}$

However, when the mole balance is written in terms of conversion, we use Equation (4-25)

$\begin{matrix} P_{i} & = C_{i} R T = C_{A 0} \frac{(Θ_{i} + υ_{i} X)}{(1 + ε X)} \frac{P}{P_{0}} \frac{T_{0}}{T} R T \\ = C_{A 0} R T_{0} \frac{(Θ_{i} + υ_{i} X)}{(1 + ε X)} p \end{matrix}$ $\begin{matrix} P_{i} & = C_{i} R T = C_{A 0} \frac{(Θ_{i} + υ_{i} X)}{(1 + ε X)} \frac{P}{P_{0}} \frac{T_{0}}{T} R T \\ = C_{A 0} R T_{0} \frac{(Θ_{i} + υ_{i} X)}{(1 + ε X)} p \end{matrix}$

\begin{matrix} P_{i} = P_{A 0} (\frac{Θ_{i} + v_{i} X}{1 + ε X}) p & (4 - 27) \end{matrix}

$\begin{matrix} P_{i} = P_{A 0} (\frac{Θ_{i} + v_{i} X}{1 + ε X}) p & (4 - 27) \end{matrix}$

For example, the rate law for the hydrodemethylation of toluene (T) to form methane (M) and benzene (B) given by Equation (10-80) on page 461 can now be written in terms of conversion.

T + H₂ → M + B

$- {r_{T}}^{'} = \frac{k P_{A 0}^{2} (1 - X) (Θ_{H_{2}} - X)}{1 + K_{B} P_{A 0} X + K_{B} P_{A 0} (1 - X)} (p)^{2}$ $- {r_{T}}^{'} = \frac{k P_{A 0}^{2} (1 - X) (Θ_{H_{2}} - X)}{1 + K_{B} P_{A 0} X + K_{B} P_{A 0} (1 - X)} (p)^{2}$

If you haven’t decided which computer to buy or borrow, or don’t have your integral tables, I suppose you could resort to the graphical techniques in Chapter 2 and use a Levenspiel plot, (F_A0/— r’_A) versus X, to achieve a specified conversion of toluene.

Example 4-4 Expressing the Rate Law for SO₂ Oxidation in Terms of Partial Pressures and Conversions

The SO₂ oxidation discussed in Example 4-3 is to be carried out over a solid platinum catalyst. As with almost all gas-solid catalytic reactions, the rate law is expressed in terms of partial pressures instead of concentrations. The rate law for this SO₂ oxidation was found experimentally to be¹

¹ Uychara, O.A., and K. M. Watson, Ind. Engrg. Chem. 35 p.541.

$- r_{S 0_{2}}^{'} = \frac{k [P_{{SO}_{2}} \sqrt{P_{O_{2}}} - \frac{P_{{SO}_{3}}}{K_{P}}]}{{(1 + \sqrt{P_{O_{2}} K_{O_{2}}} + P_{{SO}_{3}}, K_{{SO}_{3}})}^{2}}, {mol SO}_{2} oxidized/ (h) / (g-cat) (E4-4.1)$ $- r_{S 0_{2}}^{'} = \frac{k [P_{{SO}_{2}} \sqrt{P_{O_{2}}} - \frac{P_{{SO}_{3}}}{K_{P}}]}{{(1 + \sqrt{P_{O_{2}} K_{O_{2}}} + P_{{SO}_{3}}, K_{{SO}_{3}})}^{2}}, {mol SO}_{2} oxidized/ (h) / (g-cat) (E4-4.1)$

where P_i (atm) is the partial pressure of species.

The reaction is to be carried out isothermally at 400°C. At this temperature the rate constant k per gram of catalyst (g-cat), the adsorption constants for O₂ (K_O ) and SO₂ ( K_SOi₂ ), and the pressure equilibrium constant, K_P were experimentally found to be:

k = 9.7 mol SO₂/atm^3/2/h/g-cat

K_O₂ = 38.5 atm^-1, K_SO₃ = 42.5 atm^-1, and K_P = 930 atm^-1/2

The total pressure and the feed composition (e.g., 28% SO₂) are the same as in Example 4-3. Consequently, the entering partial pressure of SO₂ is 4.1 atm. There is no pressure drop.

(a) Write the rate law as a function of conversion evaluating all other parameters.

(b) Prepare a Levenspiel plot by sketching the reciprocal rate [1 /(— r_A)] as a function of X up to the equilibrium conversion X_e, and note where the rate goes to zero and [1 /( — r_A)] goes to infinity.

Solution

Part (a)

No Pressure Drop and Isothermal Operation

For SO₂

First we need to recall the relationship between partial pressure and concentration, followed by the relationship between concentration and conversion. Because we know how to express concentration as a function of conversion, we know how to express partial pressure as a function of conversion.

$P_{{SO}_{2}} = C_{{SO}_{2}} R T = \frac{F_{{SO}_{2}}}{υ} R T = \frac{F_{{SO}_{2, 0}} (1 - X) R T}{υ_{0} (1 + ε X) \frac{T}{T_{0}} \frac{P_{0}}{P}} = \frac{F_{{SO}_{2, 0}}}{υ_{0}} \frac{R T_{0 (1 - X)} \frac{P}{P_{0}}}{1 + ε X}$ $P_{{SO}_{2}} = C_{{SO}_{2}} R T = \frac{F_{{SO}_{2}}}{υ} R T = \frac{F_{{SO}_{2, 0}} (1 - X) R T}{υ_{0} (1 + ε X) \frac{T}{T_{0}} \frac{P_{0}}{P}} = \frac{F_{{SO}_{2, 0}}}{υ_{0}} \frac{R T_{0 (1 - X)} \frac{P}{P_{0}}}{1 + ε X}$

$P_{{SO}_{2}} = \frac{P_{{SO}_{2, 0}} (1 - X) \frac{P}{P_{0}}}{(1 + ε X)} = \frac{P_{{SO}_{2, 0}} (1 - X) P}{(1 + ε X)} (E4-4.2)$ $P_{{SO}_{2}} = \frac{P_{{SO}_{2, 0}} (1 - X) \frac{P}{P_{0}}}{(1 + ε X)} = \frac{P_{{SO}_{2, 0}} (1 - X) P}{(1 + ε X)} (E4-4.2)$

$P_{i} = P_{A 0} \frac{(Θ_{i} + v_{i} X)}{(1 + ϵ X)} \frac{P}{P_{0}} = P_{A 0} \frac{(Θ_{i} + v_{i} X)}{(1 + ϵ X)} p$ $P_{i} = P_{A 0} \frac{(Θ_{i} + v_{i} X)}{(1 + ϵ X)} \frac{P}{P_{0}} = P_{A 0} \frac{(Θ_{i} + v_{i} X)}{(1 + ϵ X)} p$

For no pressure drop P = P₀, i.e., p = 1

P_{{SO}_{2}} = \frac{P_{{SO}_{2}}, 0 (1 - X)}{(1 + ε X)}

$P_{{SO}_{2}} = \frac{P_{{SO}_{2}}, 0 (1 - X)}{(1 + ε X)}$

P_SO₂ ,0 = y_SO_2,0 P₀ = 4.1 atm. (E4-4.3)

For SO₃

\begin{matrix} P_{{SO}_{3}} = C_{{SO}_{3}} R T = \frac{C_{{SO}_{2}}, 0 R T_{0} X}{(1 + ε X)} = \frac{P_{{SO}_{2}}, 0 X}{1 + ε X} & (E 4 - 4.4) \end{matrix}

$\begin{matrix} P_{{SO}_{3}} = C_{{SO}_{3}} R T = \frac{C_{{SO}_{2}}, 0 R T_{0} X}{(1 + ε X)} = \frac{P_{{SO}_{2}}, 0 X}{1 + ε X} & (E 4 - 4.4) \end{matrix}$

For O₂

$P_{O_{2}} = C_{O_{2}} R T = C_{{SO}_{2, 0}} \frac{(Θ_{B} - \frac{1}{2} X) {RT}_{0}}{1 + ε X} = P_{{SO}_{2, 0}} \frac{(Θ_{B} - \frac{1}{2} X)}{1 + ε X} (E4-4.5)$ $P_{O_{2}} = C_{O_{2}} R T = C_{{SO}_{2, 0}} \frac{(Θ_{B} - \frac{1}{2} X) {RT}_{0}}{1 + ε X} = P_{{SO}_{2, 0}} \frac{(Θ_{B} - \frac{1}{2} X)}{1 + ε X} (E4-4.5)$

From Example 4-3

Θ_B = 0.54

Factoring out $\frac{1}{2}$ $\frac{1}{2}$ in Equation (E4-4.5) gives

\begin{matrix} P_{O_{2}} = P_{{SO}_{2},} 0 \frac{Θ_{B} - \frac{1}{2} X}{(1 + ε X)} = \frac{P_{{SO}_{2}}, 0 (1.08 - X)}{2 (1 + ε X)} & (E 4 - 4.6) \end{matrix}

$\begin{matrix} P_{O_{2}} = P_{{SO}_{2},} 0 \frac{Θ_{B} - \frac{1}{2} X}{(1 + ε X)} = \frac{P_{{SO}_{2}}, 0 (1.08 - X)}{2 (1 + ε X)} & (E 4 - 4.6) \end{matrix}$

From Equation (E4-3.5)

ε = -0.14

Substitute for the partial pressure in the rate-law equation (E4-4.1)

$- {r^{'}}_{{SO}_{2}}^{} = k [\frac{P_{{SO}_{2, 0}}^{3 / 2} (\frac{1 - x}{1–0.14 X}) \sqrt{\frac{(1.08 - X)}{2 (1–0.14 X)}} - \frac{P_{{SO}_{2, 0}} X}{(1–0.14 X)} (\frac{1}{930 {atm}^{- 1 / 2}})}{(1 + \sqrt{\frac{38.5 P_{{SO}_{2, 0}} (1.08 - X)}{2 (1–0.14 X)}} + \frac{42.5 P_{{SO}_{2, 0}} X}{1–0.14 X})^{2}}] (E4-4.7)$ $- {r^{'}}_{{SO}_{2}}^{} = k [\frac{P_{{SO}_{2, 0}}^{3 / 2} (\frac{1 - x}{1–0.14 X}) \sqrt{\frac{(1.08 - X)}{2 (1–0.14 X)}} - \frac{P_{{SO}_{2, 0}} X}{(1–0.14 X)} (\frac{1}{930 {atm}^{- 1 / 2}})}{(1 + \sqrt{\frac{38.5 P_{{SO}_{2, 0}} (1.08 - X)}{2 (1–0.14 X)}} + \frac{42.5 P_{{SO}_{2, 0}} X}{1–0.14 X})^{2}}] (E4-4.7)$

with k = 9.7 mol SO₂/atm^3/2/h/g-cat P_SO₂,0 = 4.1 atm, $P_{{SO}_{2, 0}}^{3 / 2}$ $P_{{SO}_{2, 0}}^{3 / 2}$ = 8.3 atm^3/2

\begin{matrix} - {r^{'}}_{{SO}_{2}} = 9.7 \frac{mol}{{h g-cat atm}^{3 / 2}} [\frac{\frac{8.3 {atm}^{3 / 2} (1 - X)}{(1 - 0.14 X)} \sqrt{\frac{1.08 - X}{2 (1 - 0.14 X)}} - \frac{0.0044 {atm}^{3 / 2} X}{(1 - 0.14 X)}}{{(1 + \frac{\sqrt{79 (1.08 - X)}}{(1 - 0.14 X)} + \frac{174 X}{1 - 0.14 X})}^{2}}] & (E4-4.8) \end{matrix}

$\begin{matrix} - {r^{'}}_{{SO}_{2}} = 9.7 \frac{mol}{{h g-cat atm}^{3 / 2}} [\frac{\frac{8.3 {atm}^{3 / 2} (1 - X)}{(1 - 0.14 X)} \sqrt{\frac{1.08 - X}{2 (1 - 0.14 X)}} - \frac{0.0044 {atm}^{3 / 2} X}{(1 - 0.14 X)}}{{(1 + \frac{\sqrt{79 (1.08 - X)}}{(1 - 0.14 X)} + \frac{174 X}{1 - 0.14 X})}^{2}}] & (E4-4.8) \end{matrix}$

Part (b)

We can now use Equation (E4-4.8) to plot the reciprocal rate as a function of X as shown in Figure E4-4). We note at the equilibrium conversion, X_e, the reciprocal rate goes to infinity. We could now use a Levenspiel plot to find the catalyst weight W in a packed-bed reactor (PBR) to achieve a specified conversion.

$F_{A0} \frac{d X}{d W} = - r_{A}' (2-17)$ $F_{A0} \frac{d X}{d W} = - r_{A}' (2-17)$

A graph depicts reciprocal rate of sulphur dioxide (SO subscript 2) oxidation as a function of conversion.

Figure E4-4.1 Reciprocal rate of SO₂ oxidation as a function of conversion.

However, we will see in the next chapter there is a much, much better way to solve for the catalysis weight, W, by using numerical software packages. For example, we would couple Equation (E4-4.8) with Equation (2-17) and use an ordinary differential equation (ODE) solver, such as Polymath, to find the conversion X as a function of catalyst weight W. So, be sure to buy or borrow a laptop computer before attempting to solve the problems in Chapter 5 and beyond.

Analysis: In most heterogeneous catalytic reactions, rate laws are expressed in terms of partial pressures instead of concentration. However, we see that through the use of the ideal gas law we could eas.ly express the partial pressure as a function of concentration then conversion in order to express the rate law as a function of conversion. In addition, for most all heterogeneous reactions you will usually find a term like (1 + K_AP_A + K_BP_B + . . .) in the denominator of the rate law, as will be explained in Chapter 10.

Need to first calculate X_e

4.3 Reversible Reactions and Equilibrium Conversion

Thus far in this chapter, we have focused mostly on irreversible reactions. The procedure one uses for the isothermal reactor design of reversible reactions is virtually the same as that for irreversible reactions, with one notable exception: the maximum conversion that can be achieved at the reaction temperature is the equilibrium conversion, X_e. In the following example, it will be shown how our algorithm for reactor design is easily extended to reversible reactions.

Example 4-4 Calculating the Equilibrium Conversion

The reversible gas-phase decomposition of nitrogen tetroxide, N₂O₄, to nitrogen dioxide, NO₂,

N_{2} O_{4} ⇄ 2 {NO}_{2}

$N_{2} O_{4} ⇄ 2 {NO}_{2}$

is to be carried out at constant temperature. The feed consists of pure N₂O₄ at 340 K and 202.6 kPa (2 atm). The concentration equilibrium constant, K_C, at 340 K is 0.1 mol/dm³ and the rate constant k_N₂o₄ is 0.5min^-1.

(a) Set up a stoichiometric table and then calculate the equilibrium conversion of N₂O₄ in a constant-volume batch reactor.

(b) Set up a stoichiometric table and then calculate the equilibrium conversion of N₂O₄ in a flow reactor.

(c) Assuming the reaction is elementary, express the rate of reaction solely as a function of conversion for a flow system and for a batch system.

(d) Determine the CSTR volume necessary to achieve 80% of the equilibrium conversion.

Solution

\begin{matrix} N_{2} O_{4} ⇄ 2 {NO}_{2} \\ A ⇄ 2 B \end{matrix}

$\begin{matrix} N_{2} O_{4} ⇄ 2 {NO}_{2} \\ A ⇄ 2 B \end{matrix}$

At equilibrium, the concentrations of the reacting species are related by the relationship dictated by thermodynamics (see Equation (3-10) and Appendix C).

$K_{C} = \frac{C_{Be}^{2}}{C_{Ae}} (E4-5.1)$ $K_{C} = \frac{C_{Be}^{2}}{C_{Ae}} (E4-5.1)$

(a) Batch system—constant volume, V = V₀.

TABLE E4-5.1 STOICHIOMETRIC TABLE FOR BATCH REACTOR

Species	Symbol	Initial	Change	Remaining
N₂O₄	A	N_A0	−N_A0X	N_A = N_A0(1−X)
NO₂	B	0	+2N_A0X	N_B = 2N_A0X

		N_T0 = N_A0		N_T = N_T0 + N_A0X

Living Example Problem

There is a Polymath tutorial in the Summary Notes for Chapter 1 on the CRE Web site.

For batch systems C_i = N_i/V

$C_{A} = \frac{N_{A}}{V} = \frac{N_{A}}{V_{0}} = \frac{N_{A0} (1 - X)}{V_{0}} = C_{A0} (1 - X) (E4-5.2)$ $C_{A} = \frac{N_{A}}{V} = \frac{N_{A}}{V_{0}} = \frac{N_{A0} (1 - X)}{V_{0}} = C_{A0} (1 - X) (E4-5.2)$

$C_{B} = \frac{N_{B}}{V} = \frac{N_{B}}{V_{0}} = \frac{2 N_{A0} X}{V_{0}} = 2 C_{A0} X (E4-5.3)$ $C_{B} = \frac{N_{B}}{V} = \frac{N_{B}}{V_{0}} = \frac{2 N_{A0} X}{V_{0}} = 2 C_{A0} X (E4-5.3)$

$\begin{matrix} C_{A 0} & = \frac{y_{A 0} P_{0}}{R T_{0}} = \frac{(1) (2 atm)}{(0.082 atm \cdot {dm}^{3} / mol \cdot K) (340 K)} \\ = 0.07174 mol / {dm}^{3} \end{matrix}$ $\begin{matrix} C_{A 0} & = \frac{y_{A 0} P_{0}}{R T_{0}} = \frac{(1) (2 atm)}{(0.082 atm \cdot {dm}^{3} / mol \cdot K) (340 K)} \\ = 0.07174 mol / {dm}^{3} \end{matrix}$

At equilibrium in a batch reactor, X = X_eb, we substitute Equations (E4-5.2) and (E4-5.3) into Equation (E4-5.1)

$K_{c} = \frac{C_{Be}^{2}}{C_{Ae}} = \frac{4 C_{A0}^{2} X_{e b}^{2}}{C_{A0} (1 - X_{e b})} = \frac{4 C_{A0} X_{e b}^{2}}{1 - X_{e b}}$ $K_{c} = \frac{C_{Be}^{2}}{C_{Ae}} = \frac{4 C_{A0}^{2} X_{e b}^{2}}{C_{A0} (1 - X_{e b})} = \frac{4 C_{A0} X_{e b}^{2}}{1 - X_{e b}}$

(math-math-math-math) to get

$X_{e b} = \sqrt{\frac{K_{c} (1 - X_{eb})}{4 C_{A0}}} (E4-5.4)$ $X_{e b} = \sqrt{\frac{K_{c} (1 - X_{eb})}{4 C_{A0}}} (E4-5.4)$

We will use the software package Polymath to solve for the equilibrium conversion and let Xeb represent the equilibrium conversion in a constant-volume batch reactor. Equation (E4-5.4) written in Polymath format becomes

f(Xeb) = Xeb — [Kc^*(1 — Xeb)/(4^*Cao)]^0.5

The Polymath program and solution are given in Table E4-5.2.

When looking at Equation (E4-5.4), you probably asked yourself, "Why not use the quadratic formula to solve for the equilibrium conversion in both batch and flow systems?"

That is from Equation (E4-5.4) we would have

$B a t c h : X_{e b} = \frac{1}{8} [(- 1 + \sqrt{1 + 16 C_{A0} / K_{C})} / (C_{A0} / K_{C})]$ $B a t c h : X_{e b} = \frac{1}{8} [(- 1 + \sqrt{1 + 16 C_{A0} / K_{C})} / (C_{A0} / K_{C})]$

and from Equation (E4-5.8) we would have

$F l o w : X_{e f} = \frac{[(ε - 1) + \sqrt{(ε - 1)^{2} + 4 (ε + 4 C_{A0} / K_{C})}]}{2 (ε + 4 C_{A 0} / K_{C})}$ $F l o w : X_{e f} = \frac{[(ε - 1) + \sqrt{(ε - 1)^{2} + 4 (ε + 4 C_{A0} / K_{C})}]}{2 (ε + 4 C_{A 0} / K_{C})}$

TABLE E4-5.2 POLYMATH PROGRAM AND SOLUTION FOR BOTH BATCH AND FLOW SYSTEMS

The answer is that future problems will be nonlinear and require Polymath solutions; therefore, this simple exercise increases the reader’s ease in using Polymath.

The equilibrium conversion in a constant-volume batch reactor is

Polymath Tutorial Chapter 1

Images Summary Motes

Note: A tutorial on Polymath can be found in the summary notes for Chapter 1 on the CRE Web site (www.umich.edu/~elements/5e/index.html).

(b) Flow system. The stoichiometric table is the same as that for a batch system except that the number of moles of each species, N_i, is replaced by the molar flow rate of that species, F_i.

For constant temperature and pressure, the volumetric flow rate is υ = υ₀ ( 1 + εX), and the resulting concentrations of species A and B are

TABLE E4-5.2 STOICHIOMETRIC TABLE FOR CONTINUOUS FLOW REACTOR

Species	Symbol	Entering	Change	Leaving
N₂O₂	A	F_A0	−F_A0X	F_A = F_A0(1−X)
NO₂	B	0	+2F_A0X	F_B = 2F_A0X

		F_T0 = F_A0		F_T = F_T0 + F_A0X

$C_{A} = \frac{F A}{υ} = \frac{F_{A0} (1 - X)}{υ} = \frac{F_{A0} (1 - X)}{υ_{0} (1 + ε X)} = \frac{C_{A0} (1 - X)}{1 + ε X} (E4-5.5)$ $C_{A} = \frac{F A}{υ} = \frac{F_{A0} (1 - X)}{υ} = \frac{F_{A0} (1 - X)}{υ_{0} (1 + ε X)} = \frac{C_{A0} (1 - X)}{1 + ε X} (E4-5.5)$

$C_{B} = \frac{F_{B}}{υ} = \frac{2 F_{A0} X}{υ_{0} (1 + ε X)} = \frac{2 C_{A0} X}{(1 + ε X)} (E4-5.6)$ $C_{B} = \frac{F_{B}}{υ} = \frac{2 F_{A0} X}{υ_{0} (1 + ε X)} = \frac{2 C_{A0} X}{(1 + ε X)} (E4-5.6)$

At equilibrium, X = X_ef, we can substitute Equations (E4-5.5) and (E4-5.6) into Equation (E4-5.1) to obtain the expression

$K_{C} = \frac{{C_{B e}}^{2}}{C_{A e}} = \frac{[2 C_{A0} X_{e f} / (1 + ε X_{e f})]^{2}}{C_{A0} (1 - X_{e f}) / (1 + ε X_{e f})}$ $K_{C} = \frac{{C_{B e}}^{2}}{C_{A e}} = \frac{[2 C_{A0} X_{e f} / (1 + ε X_{e f})]^{2}}{C_{A0} (1 - X_{e f}) / (1 + ε X_{e f})}$

Simplifying gives

$K_{C} = \frac{4 C_{A0} {X_{e f}}^{2}}{(1 - X_{e f}) (1 + ε X_{e f})} (E4-5.7)$ $K_{C} = \frac{4 C_{A0} {X_{e f}}^{2}}{(1 - X_{e f}) (1 + ε X_{e f})} (E4-5.7)$

Rearranging to use Polymath yields

$X_{e f} = \sqrt{\frac{K_{C} (1 - X_{e f}) (1 + ε X_{e f})}{4 C_{A0}}} (E4-5.8)$ $X_{e f} = \sqrt{\frac{K_{C} (1 - X_{e f}) (1 + ε X_{e f})}{4 C_{A0}}} (E4-5.8)$

For a flow system with pure N₂0₄ feed, ε = y_A0 δ = 1 (2 — 1) = 1.

We shall let Xef represent the equilibrium conversion in a flow system. Equation (E4-5.8) written in the Polymath format becomes

f(Xef) = Xef -[kc*(1 — Xef)*( 1 + eps*Xef)/4/cao]^0.5

This solution is also shown in Table E4-5.2 $X_{ef} = 0.51$ $X_{ef} = 0.51$ .

Note that the equilibrium conversion in a flow reactor (i.e., X_ef = 0.51), with no pressure drop, is greater than the equilibrium conversion in a constant-volume batch reactor (X_eh = 0.44). Recalling Le Châtelier’s principle, can you suggest an explanation for this difference in X_e? This example is continued in problem P4-1_A (a).

(c) Rate laws. Assuming that the reaction follows an elementary rate law, then

$- r_{A} = k_{A} [C_{A} - \frac{{C_{B}}^{2}}{K_{C}}] (E4-5.9)$ $- r_{A} = k_{A} [C_{A} - \frac{{C_{B}}^{2}}{K_{C}}] (E4-5.9)$

1. For a constant volume (V = V₀) batch system

Here, C_A = N_A/V₀ and C_B = N_B/V₀. Substituting Equations (E4-5.2) and (E4-5.3) into the rate law, we obtain the rate of disappearance of A as a function of conversion

\begin{matrix} - r_{A} = k_{A} [C_{A} - \frac{C_{B}^{2}}{K_{C}}] = k_{A} [C_{A0} (1 - X) - \frac{4 C_{A0}^{2} X^{2}}{K_{C}}] & (E4-5.10) \end{matrix}

$\begin{matrix} - r_{A} = k_{A} [C_{A} - \frac{C_{B}^{2}}{K_{C}}] = k_{A} [C_{A0} (1 - X) - \frac{4 C_{A0}^{2} X^{2}}{K_{C}}] & (E4-5.10) \end{matrix}$

2. For a flow system

Here, C_A = F_A/v and C_B = F_B/v with υ = υ₀ (1 + eX). Consequently, we can substitute Equations (E4-5.5) and (E4-5.6) into Equation (E4-5.9) to obtain

\begin{matrix} - r_{A} = k_{A} [\frac{C_{A0} (1 - X)}{1 + ε X}] - [\frac{4 C_{A0}^{2} X^{2}}{K_{C} {(1 + ε X)}^{2}}] & (E4-5.11) \end{matrix}

$\begin{matrix} - r_{A} = k_{A} [\frac{C_{A0} (1 - X)}{1 + ε X}] - [\frac{4 C_{A0}^{2} X^{2}}{K_{C} {(1 + ε X)}^{2}}] & (E4-5.11) \end{matrix}$

-r_A = f(X) for a batch reactor with V = V₀

-r_A = f(X) for a flow reactor

As expected, the dependence of reaction rate on conversion, i.e., -r_A = f(X), for a constant-volume batch system [i.e., Equation (E4-5.10)] is different than that for a flow system [Equation (E4-5.11)] for gas-phase reactions.

If we substitute the values for C_A0, K_c, ε, and k_A = 0.5 min^-1 in Equation (E4-5.11), we obtain -r_A solely as a function of X for the flow system.

$- r_{A} = \frac{0.5}{\min} [0.072 \frac{mol (1 - X)}{{dm}^{3} (1 + X)} - \frac{4 (0.072 mol/dm3)^{2} X^{2}}{0.1 {mol/dm}^{3} (1 + X)^{2}}] (E4-5.9)$ $- r_{A} = \frac{0.5}{\min} [0.072 \frac{mol (1 - X)}{{dm}^{3} (1 + X)} - \frac{4 (0.072 mol/dm3)^{2} X^{2}}{0.1 {mol/dm}^{3} (1 + X)^{2}}] (E4-5.9)$

$- r_{A} = 0.036 [\frac{(1 - X)}{(1 + X)} - \frac{2.88 X^{2}}{(1 + X)^{2}}] (\frac{mol}{{dm}^{3} \cdot min}) (E4-5.12)$ $- r_{A} = 0.036 [\frac{(1 - X)}{(1 + X)} - \frac{2.88 X^{2}}{(1 + X)^{2}}] (\frac{mol}{{dm}^{3} \cdot min}) (E4-5.12)$

We can now prepare our Levenspiel plot to find either the cSTR or PFR volume.

A graph depicts Levenspiel plot for a flow system.

Figure E4-5.1 Levenspiel plot for a flow system.

We see as we approach equilibrium, X_e = 0.15, -r_A goes to zero and (1/-r_A) goes to infinity.

(d) CSTR volume. Just for fun (and this really is fun), let’s calculate the CSTR reactor volume necessary to achieve 80% of the equilibrium conversion of 51% (i.e., X = 0.8X_e = (0.8)(0.51) = 0.4) for a molar feed rate of A of 3 mol/min.

Plotting the reciprocal of Equation (E4-5.12) to obtain Levenspiel plot as we did in Chapter 2, it can be seen in Figure E4-5.1 when X = 0.4 then as

$(\frac{1}{- r_{A}})_{x = 0.4} = 143 \frac{{dm}^{3} \min}{mol}$ $(\frac{1}{- r_{A}})_{x = 0.4} = 143 \frac{{dm}^{3} \min}{mol}$

then for F_A0 = 3 mol/min

$\begin{matrix} V & = F_{A0} X (\frac{1}{- r_{A}})_{X = 0.4} \\ = (\frac{3 mol}{\min}) (0.4) (143) \frac{{dm}^{3} \min}{mol} \\ V & {=171 dm}^{3} \end{matrix}$ $\begin{matrix} V & = F_{A0} X (\frac{1}{- r_{A}})_{X = 0.4} \\ = (\frac{3 mol}{\min}) (0.4) (143) \frac{{dm}^{3} \min}{mol} \\ V & {=171 dm}^{3} \end{matrix}$

A quicker way to find the CSTR volume, rather than drawing a Levenspiel plot, is to simply evaluate Equation (E4-5.12) at X = 0.4

$- r_{A} = 0.036 [\frac{(1–0.4)}{(1 + 0.4)} - \frac{2.88 (0.4)^{2}}{(1 + (0.4))^{2}}] = 0.0070 {mol/dm}^{3} / \min$ $- r_{A} = 0.036 [\frac{(1–0.4)}{(1 + 0.4)} - \frac{2.88 (0.4)^{2}}{(1 + (0.4))^{2}}] = 0.0070 {mol/dm}^{3} / \min$

Using Equation (2-13), we can now find the reactor volume

\begin{matrix} V = & \frac{F_{A 0} X}{- r_{A} |_{X}} = \frac{F_{A0} (0.4)}{- r_{A} |_{0.4}} = \frac{(3 mol/min)(0.4)}{0.0070 \frac{mol}{{dm}^{3} \cdot min}} \\ V= & {\begin{matrix} 171 dm \end{matrix}}^{3} = 0.171 m^{3} \end{matrix}

$\begin{matrix} V = & \frac{F_{A 0} X}{- r_{A} |_{X}} = \frac{F_{A0} (0.4)}{- r_{A} |_{0.4}} = \frac{(3 mol/min)(0.4)}{0.0070 \frac{mol}{{dm}^{3} \cdot min}} \\ V= & {\begin{matrix} 171 dm \end{matrix}}^{3} = 0.171 m^{3} \end{matrix}$

The CSTR volume required to achieve 40% conversion is 0.171 m³. See Problem P4-1A (b) to calculate the PFR volume.

Analysis: The purpose of this example was to calculate the equilibrium conversion first for a constant volume batch reactor in part (a), and then for a constant pressure flow reactor in part (b). One notes that there is a change in the total number of moles in this reaction and, as a result, these two equilibrium conversions are not the same!! We next showed in part (c) how to express –r_A = ƒ(X) for a reversible gas-phase reaction. Here we noted that the equations for the reaction rate laws as a function of conversion, i.e., –r_A = ƒ(X), are different for batch and flow reactors. Finally, in Part (d) having –r_A = ƒ(X), we specified a molar flow rate of A (i.e., 3.0 mol A/min) and calculated the CSTR volume necessary to achieve 40% conversion. We did this calculation to give insight to the types of analyses we, as chemical reaction engineers, will carry out as we move into similar but more complex calculations in Chapters 5 and 6.

Closure. Having completed this chapter, you should be able to write the rate law solely in terms of conversion and the reaction-rate parameters, (e.g., k, K_C) for both liquid-phase and gas-phase reactions. Once expressing –r_A = ƒ(X) is accomplished, you can proceed to use the techniques in Chapter 2 to calculate reactor sizes and conversion for single CSTRs, PFRs, and PBRs, as well as those connected in series. However, in the next chapter we will show you how to carry out these calculations much more easily by instead using a Table of Integrals or software such as Polymath, MATLAB, or Wolfram without having to resort to Levenspiel plots. After studying this chapter you should also be able to calculate the equilibrium conversion for both constant-volume batch reactors and for constant-pressure flow reactors. We have now completed the following building blocks of our CRE tower. In Chapter 5 we will focus on the fourth and fifth blocks: Combine and Evaluate. In Chapter 5, we will

Illustration shows building blocks labeled, Stoichiometry, Rate Law, and Mode Balance from top to bottom.

focus on the combine and evaluation building blocks, which will then complete our algorithm for isothermal chemical reactor design.

The CRE Algorithm

• Mole Balance, Ch 1

• Rate Law, Ch 3

• Stoichiometry, Ch 4

• Combine, Ch 5

• Evaluate, Ch 5

• Energy Balance, Ch 11

Summary

1. The stoichiometric table for the reaction given by Equation (S4-1) being carried out in a flow system is

$\begin{matrix} A + \frac{b}{a} B \to \frac{c}{a} C + \frac{d}{a} D & (S4-1) \end{matrix}$ $\begin{matrix} A + \frac{b}{a} B \to \frac{c}{a} C + \frac{d}{a} D & (S4-1) \end{matrix}$

2. In the case of ideal gases, Equation (S4-3) relates volumetric flow rate to conversion. Batch constant volume:

$\begin{matrix} V = V_{0} & \begin{matrix} (S4-2) \end{matrix} \end{matrix}$ $\begin{matrix} V = V_{0} & \begin{matrix} (S4-2) \end{matrix} \end{matrix}$

Flow systems: Gas:

$\begin{matrix} v = v_{0} (\frac{P_{0}}{P}) (1 + ε X) \frac{T}{T_{0}} & (S4-3) \end{matrix}$ $\begin{matrix} v = v_{0} (\frac{P_{0}}{P}) (1 + ε X) \frac{T}{T_{0}} & (S4-3) \end{matrix}$

Liquid:

$\begin{matrix} v = v_{0} & (S4-4) \end{matrix}$ $\begin{matrix} v = v_{0} & (S4-4) \end{matrix}$

Species	Entering	Change	Leaving
A	F_A0	-F_A0X	F_A0(1 – X)
B	F_B0	$- (\frac{b}{a}) F_{A0} X$ $- (\frac{b}{a}) F_{A0} X$	$F_{A0} (Θ_{B} - \frac{b}{a} X)$ $F_{A0} (Θ_{B} - \frac{b}{a} X)$
C	F_C0	$(\frac{c}{a}) F_{A0} X$ $(\frac{c}{a}) F_{A0} X$	$F_{A0} (Θ_{C} + \frac{c}{a} X)$ $F_{A0} (Θ_{C} + \frac{c}{a} X)$
D	F_D0	$(\frac{d}{a}) F_{A0} X$ $(\frac{d}{a}) F_{A0} X$	$F_{A0} (Θ_{D} + \frac{d}{a} X)$ $F_{A0} (Θ_{D} + \frac{d}{a} X)$
I	F₁₀	---	F₁₀
________	________	________	________
Totals	^FT0	δF_A0X	F_T = F_T0 +δF_A0X

Definitions of δ and ɛ For the general reaction given by (S4-1), we have

\begin{matrix} δ = \frac{d}{a} + \frac{c}{a} - \frac{b}{a} - 1 & (S4-5) \end{matrix}

$\begin{matrix} δ = \frac{d}{a} + \frac{c}{a} - \frac{b}{a} - 1 & (S4-5) \end{matrix}$

δ = \frac{Change in total number of moles}{Mole of A reacted}

$δ = \frac{Change in total number of moles}{Mole of A reacted}$

and

\begin{matrix} ε = y_{A0} δ & (S 4 - 6) \end{matrix}

$\begin{matrix} ε = y_{A0} δ & (S 4 - 6) \end{matrix}$

ε = \frac{Change in total number of moles for complete conversion}{Total number of moles fed to the reactor}

$ε = \frac{Change in total number of moles for complete conversion}{Total number of moles fed to the reactor}$

3. For incompressible liquids or for batch gas phase reactions taking place in a constant volume, V = V₀, the concentrations of species A and C in the reaction given by Equation (S4-1) can be written as

$\begin{matrix} C_{A} = \frac{F_{A}}{v} = \frac{F_{A0}}{v_{0}} (1 - X) = C_{A0} (1 - X) & (S4-7) \end{matrix}$ $\begin{matrix} C_{A} = \frac{F_{A}}{v} = \frac{F_{A0}}{v_{0}} (1 - X) = C_{A0} (1 - X) & (S4-7) \end{matrix}$

$\begin{matrix} C_{c} = C_{A0} (Θ_{C} + \frac{c}{a} X) & (S4-8) \end{matrix}$ $\begin{matrix} C_{c} = C_{A0} (Θ_{C} + \frac{c}{a} X) & (S4-8) \end{matrix}$

Equations (S4-7) and (S4-8) also hold for gas-phase reactions carried out in constant-volume batch reactors.

4. For gas-phase reactions in continuous-flow reactors, we use the definition of concentration (C_A = F_A/υ) along with the stoichiometric table and Equation (S4-3) to write the concentration of A and C in terms of conversion.

$\begin{matrix} C_{A} = \frac{F_{A}}{v} = \frac{F_{A0} (1 - X)}{v} = C_{A0} [\frac{1 - X}{1 + \in X}] P (\frac{T_{0}}{T}) & (S4-9) \end{matrix}$ $\begin{matrix} C_{A} = \frac{F_{A}}{v} = \frac{F_{A0} (1 - X)}{v} = C_{A0} [\frac{1 - X}{1 + \in X}] P (\frac{T_{0}}{T}) & (S4-9) \end{matrix}$

$\begin{matrix} C_{c} = \frac{F_{c}}{v} = C_{A0} [\frac{Θ_{c} + (c / a) X}{1 + \in X}] P (\frac{T_{0}}{T}) & (S4-10) \end{matrix}$ $\begin{matrix} C_{c} = \frac{F_{c}}{v} = C_{A0} [\frac{Θ_{c} + (c / a) X}{1 + \in X}] P (\frac{T_{0}}{T}) & (S4-10) \end{matrix}$

with

$Θ_{c} = \frac{F_{c0}}{F_{A0}} = \frac{C_{c0}}{C_{A0}} = \frac{y_{c 0}}{y_{A 0}} and P= \frac{p}{p_{0}}$ $Θ_{c} = \frac{F_{c0}}{F_{A0}} = \frac{C_{c0}}{C_{A0}} = \frac{y_{c 0}}{y_{A 0}} and P= \frac{p}{p_{0}}$

5. In terms of gas-phase molar flow rates, the concentration of species i is

$\begin{matrix} C_{i} = C_{T 0} \frac{F_{i}}{F_{T}} \frac{P}{P_{0}} \frac{T_{0}}{T} & (S4-11) \end{matrix}$ $\begin{matrix} C_{i} = C_{T 0} \frac{F_{i}}{F_{T}} \frac{P}{P_{0}} \frac{T_{0}}{T} & (S4-11) \end{matrix}$

Equation (S4-11) must be used for membrane reactors (Chapter 6) and for multiple reactions (Chapter 8).

6. Many catalytic rate laws are given in terms of partial pressure, e.g.,

$\begin{matrix} - r_{A}^{'} = \frac{K_{A} P_{A}}{1 + K_{A} P_{A}} & (S4-12) \end{matrix}$ $\begin{matrix} - r_{A}^{'} = \frac{K_{A} P_{A}}{1 + K_{A} P_{A}} & (S4-12) \end{matrix}$

The partial pressure is related to conversion through the stoichiometric table. For any species “i” in the reaction

$\begin{matrix} P_{i} = P_{A 0} \frac{(Θ_{i} + v_{i} X)}{(1 + ε X)} & (S4-13) \end{matrix}$ $\begin{matrix} P_{i} = P_{A 0} \frac{(Θ_{i} + v_{i} X)}{(1 + ε X)} & (S4-13) \end{matrix}$

For species A

$P_{A} = P_{A 0} \frac{(1 - X)}{(1 + ε X)} P$ $P_{A} = P_{A 0} \frac{(1 - X)}{(1 + ε X)} P$

Substituting in the rate law

$\begin{matrix} - r_{A}^{'} = \frac{K_{A} P_{A 0} \frac{(1 - X)}{(1 + ε X)} P}{1 + K_{A} P_{A 0 \frac{(1 - X)}{(1 + ε X)} P}} = \frac{{k P}_{A 0} (1 - X) p}{(1 + ε X) + K_{A} p_{A 0} (1 - X) P} & (S4-14) \end{matrix}$ $\begin{matrix} - r_{A}^{'} = \frac{K_{A} P_{A 0} \frac{(1 - X)}{(1 + ε X)} P}{1 + K_{A} P_{A 0 \frac{(1 - X)}{(1 + ε X)} P}} = \frac{{k P}_{A 0} (1 - X) p}{(1 + ε X) + K_{A} p_{A 0} (1 - X) P} & (S4-14) \end{matrix}$

CRE Web Site Materials

• Expanded Materials (http://umich.edu/~elements/5e/04chap/expanded.html)

1. Web P4-1_B Puzzle Problem “What’s Four Things Are Wrong with This Solution?” (http://umich.edu/~elements/5e/04chap/expanded_ch04_puzzle.pddf)

• Learning Resources (http://umich.edu/~elements/5e/04chap/learn.html)

1. Summary Notes for Chapter 4 (http://umich.edu/~elements/5e/04chap/summary.html)

2. Self-Tests

A. Exercises
(http://www.umich.edu/~elements/5e/04chap/add.html)

B. i>clicker Questions (http://www.umich.edu/~elements/5e/04chap/iclicker_ch4_q1.html)

3. Interactive Computer Games (http://umich.edu/~elements/5e/icm/index.html) Quiz Show II (http://umich.edu/~elements/5e/icm/kinchal2.html)

Screenshot shows "Kinetics Challenge II" window.

"The content pane of the window reads as

For the reaction: A 2D

which is non-isothermal,

irreversible, gas phase with pure A fed to the reactor, what is the concentration of D?

Checkbox 1: C subscript D equals C subscript A0 times 2X over (1 plus X).

Checkbox 2: C subscript D equals C subscript A0 times 0.5X over (1 minus 0.5X) times [P over P subscript 0] [T over T subscript 0].

Checkbox 3: C subscript D equals C subscript A0 times 2X. Checkbox 4: C subscript D equals C subscript A0 times 2X over (1 plus X) times [P over P subscript 0] [T over T subscript 0] is selected.

Send Checked Answer button is given below the options with the text “Please check ONE box to the left of the desired answer and click on <Send Checked Answer>.” Host: Correct is displayed below it. Main Menu button is shown at the bottom left and Continue button is shown at bottom right. 00 is displayed in the navigation bar at bottom of the window."

4. Solved Problems

CDP4-B_B Microelectronics Industry and the Stoichiometric Table (http://umich.edu/~elements/5e/04chap/add-ahp03-b.html)

• Living Example Problem

1. Example 4-4 Rate Law for SO₂ Oxidation

2. Example 4-5 Calculating the Equilibrium Conversion (http://umich.edu/~elements/5e/04chap/live.html)

• FAQs (Frequently Asked Questions) Typos and Updates (http://umich.edu/~elements/5e/04chap/faq.html)

• Professional Reference Shelf (http://umich.edu/~elements/5e/04chap/prof.html)

Questions and Problems

The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.

A = • B = ▪ C = ♦ D = ♦♦

Questions

Q4-1_A i>clicker. Go to the Web site (http://www.umich.edu/~elements/5e/04chap/iclicker_ch4_q1.html) and view at leave five i>clicker questions. Choose one that could be used as is, or a variation thereof, to be included on the next exam. You also could consider the opposite case, explaining why the question should not be on the next exam. In either case, explain your reasoning.

Q4-2_A (a) List the important concepts that you learned from this chapter. Make a list of concepts that you are not clear about and ask your instructor or colleague about them.

(b) Explain the strategy to evaluate reactor design equations and how this chapter expands on Chapters 2 and 3.

Q4-3_A Example 4-1. Would the example be correct if water were considered an inert? Explain.

Q4-4_A Example 4-2. How would the answer change if the initial concentration of glyceryl stearate were 3 mol/dm³? Rework Example 4-2 correctly using the information given in the problem statement.

Q4-5_A Example 4-3. Under what conditions will the concentration of the inert nitrogen be constant? Plot Equation (E4-5.2) in terms of (1/–r_A) as a function of X up to value of X = 0.99. What did you find?

Computer Simulations and Experiments

P4-1_A (a) Example 4-4: SO₂ Oxidation

Wolfram

(i) What is the critical value of entering pressure of SO₂, such that the reaction rate first increases and then decreases at a given conversion? Explain this behavior.

(ii) Vary the parameters and list the parameters that have the greatest and the least effect on the reaction rate.

(iii) Vary Θ_B and observe the change in reaction rate. At what value of Θ_B does the rate of reaction become zero when 50% conversion is achieved?

(iv) Write a conclusion about the experiments you carried out by varying the sliders in parts (i) through (iii).

Polymath

(v) Using the molar flow rate of SO₂ (A) of 3 mol/h and Figure E4-4.1, calculate the fluidized bed (i.e., CSTR catalyst weight) necessary for 4o% conversion.

(vi) Next, consider the entering flow rate of SO₂ is 1,000 mol/h. Plot (F_A0/–r_A) as a function of X to determine the fluidized bed catalyst weight necessary to achieve 3o% conversion, and 99% of the equilibrium conversion, i.e., X = 0.99 X_e.

(b) Example 4-5: Equilibrium Conversion

(i) Using Figure E4-5.1 and the methods in Chapter 2, estimate the PFR volume for 40% conversion.

Wolfram

(ii) Plot the equilibrium conversion as a function of the mole fraction of inerts for both a PFR and a constant-volume batch reactor. Describe what you find.

(iii) What values of K_c and C_A0 cause Xf to be the farthest away from X_eb?

(iv) What values of K_c and C_A0 will cause and X_ef to be the closest together?

(v) Plot the ratio of $(\frac{X_{e b}}{X_{e r}})$ $(\frac{X_{e b}}{X_{e r}})$ as a function of y_A0 for different values of K_c and C_T0. What conclusions can you draw?

Interactive Computer Games

P4-2_A Load the Interactive Computer Games (ICG) Kinetic Challenge from the CRE Web site. Play the game and then record your performance number for the module that indicates your mastering of the material. Your professor has the key to decode your performance number. ICG Kinetics Challenge Performance #_________.

Kinetics Challenge II

Rate	Law	Stoich
100	100	100
200	200	200
300	300	300

Problems

P4-3_A The elementary reversible reaction

2 A ⇄ B

$2 A ⇄ B$

is carried out in a flow reactor where pure A is fed at a concentration of 4.0 mol/dm³. If the equilibrium conversion is found to be 60%,

(a) What is the equilibrium constant, K_C if the reaction is a gas phase reaction? (Ans.: Gas: K_C = 0.328 dm³/mol)

(b) What is the K_C if the reaction is a liquid-phase reaction? (Ans.: Liquid: K_C = 0.469 dm³/mol)

(c) Write — r_A solely as a function of conversion (i.e., evaluating all symbols) when the reaction is an elementary, reversible, gas-phase, isothermal reaction with no pressure drop with k_A = 2 dm⁶/mol•s and K_C = 0.5 all in proper units.

(d) Repeat (c) for a constant-volume batch reactor.

P4-4_B Stoichiometry. The elementary gas reaction

2A + B → C

is carried out isothermally in a PFR with no pressure drop. The feed is equal molar in A and B, and the entering concentration of A is 0.1 mol/dm³. Set up a stoichiometric table and then determine the following.

(a) What is the entering concentration (mol/dm³) of B?

(b) What are the concentrations of A and C (mol/dm³) at 25% conversion of A?

(c) What is the concentration of B (mol/dm³) at 25% conversion of A? (Ans.: C_B = 0.1 mol dm³)

(d) What is the concentration of B (mol/dm³) at 100% conversion of A?

(e) If at a particular conversion the rate of formation of C is 2 mol/min/dm³, what is the rate of formation of A at the same conversion?

(f) Write —r_A solely as a function of conversion (i.e., evaluating all symbols) when the reaction is an elementary, irreversible, gas-phase, isothermal reaction with no pressure drop with an equal molar feed and with C_A0 = 2.0 mol/dm³ at, k_A = 2 dm⁶/mol•s.

(g) What is the rate of reaction at X = 0.5?

P4-5_A Set up a stoichiometric table for each of the following reactions and express the concentration of each species in the reaction as a function of conversion, evaluating all constants (e.g., ε, Θ). Next, assume the reaction follows an elementary rate law, and write the reaction rate solely as a function of conversion, i.e., -rA = ƒ(X).

(a) For the liquid-phase reaction

the entering concentrations of ethylene oxide and water, after mixing the inlet streams, are 16.13 mol/dm³ and 55.5 mol/dm³, respectively. The specific reaction rate is K = 0.1 dm³/mol.s at 300 K with E = 12,500 cal/mol.

(1) After finding –r_A = ƒ(X), calculate the CSTR space-time, τ, for 90% conversion at 300 K and also at 350 K.

(2) If the volumetric flow rate is 200 liters per second, what are the corresponding reactor volumes? (Ans.: At 300 K: V = 439 dm³ and at 350 K: V = 22 dm³)

(b) For the isothermal gas-phase pyrolysis

C₂H₆ → C₂H₄ + H₂

pure ethane enters a flow reactor at 6 atm and 1100 K, with ΔP = 0. Set up a stoichiometric table and then write –r_A = ƒ(X). How would your equation for the concentration and reaction rate, i.e., –r_A = f(X), change if the reaction were to be carried out in a constant-volume batch reactor?

(c) For the isothermal, isobaric, catalytic gas-phase oxidation

The equation reads: Ethylene (C subscript 2 H subscript 4) reacts with half oxygen atom (O subscript 2) to give two singly bonded methylene (CH subscript 2) and in turn, it is single-bonded to an oxygen atom in a cyclic manner.

the feed enters a PBR at 6 atm and 260°C, and is a stoichiometric mixture of only oxygen and ethylene. Set up a stoichiometric table and then write — r’ as a function of partial pressures. Express the partial pressures and - r’ as a function of conversion for (1) a fluidized batch reactor and (2) a PBR. Finally, write - rA solely as a function of the rate constant and conversion.

(d) Set up a stoichiometric table for the isothermal, catalytic gas-phase reaction carried out in a fluid-ized CSTR.

Figure shows CSTR with fluidized catalyst pellets found inside the reactor with two arrows shown, that are acting in opposite direction. The inlet flow of fluid is shown at the bottom and the outlet flow of flow of fluid is shown at the top of the reactor.

The feed is stoichiometric and enters at 6 atm and 170°C. What catalyst weight is required to reach 80% conversion in a fluidized CSTR at 170°C and at 270 C? The rate constant is defined with respect to benzene and υ₀ = 50 dm³/min.

$K_{B} = \frac{53 = mol}{Kgcat \cdot \min \cdot {atm}^{3}} at 300 k with E = 80 kJ/mol$ $K_{B} = \frac{53 = mol}{Kgcat \cdot \min \cdot {atm}^{3}} at 300 k with E = 80 kJ/mol$

First write the rate law in terms of partial pressures and then express the rate law as a function of conversion. Assume Δ = 0 .

P4-6_A Orthonitroanaline (an important intermediate in dyes—called fast orange) is formed from the reaction of orthonitrochlorobenzene (ONCB) and aqueous ammonia (see explosion in Figure E13-2.1 in Example 13-2).

Equation depicts the formation of Orthonitroanaline.

The liquid-phase reaction is first order in both ONCB and ammonia with k = 0.0017 m³/kmol · min at 188°C with E = 11,273 cal/mol. The initial entering concentrations of ONCB and ammonia are 1.8 kmol/m³ and 6.6 kmol/m³, respectively (more on this reaction in Chapter 13).

(a) Set up a stoichiometric table for this reaction for a flow system.

(b) Write the rate law for the rate of disappearance of ONCB in terms of concentration.

(c) Explain how parts (a) and (b) would be different for a batch system.

(d) Write –r_A solely as a function of conversion.

(e) What is the initial rate of reaction (X = 0)

(f) What is the rate of reaction when X = 0.90

–r_A =_____

at 188°C? –r_A =_____

at 25°C? –r_A =_____

at 288°C? –r_A =_____

at 188°C? –r_A =_____

at 25°C? –r_A =_____

at 288°C? –r_A =_____

(g) What would be the corresponding CSTR reactor volume at 25°C to achieve 90% conversion and at 288°C for a feed rate of 2 dm³/min

at 25°C? V =_____

at 288°C? V =_____

P4-7_B Consider the following elementary gas-phase reversible reaction to be carried out isothermally with no pressure drop and for an equal molar feed of A and B with C_A0 = 2.0 mol/dm³.

2A + B ⇄ C

(a) What is the concentration of B initially? C_B0 =_____(mol/dm³)

(b) What is the limiting reactant?_____

(c) What is the exit concentration of B when the conversion of A is 25%? C_B = _____ (mol/dm³)

(d) Write -r_A solely as a function of conversion (i.e., evaluating all symbols) when the reaction is an elementary, reversible, gas-phase, isothermal reaction with no pressure drop with an equal molar feed and with C_A0 = 2.0 mol/dm³, k_A = 2 dm⁶/mol^2,⋅s, and K_C = 0.5 all in proper units –r_A = _____.

(e) What is the equilibrium conversion?

(f) What is the rate when the conversion is

(1) 0%?

(2) 50%?

(3) 0.99 X_e?

P4-8_B The gas-phase reaction

$\frac{1}{2} N_{2} + \frac{3}{2} H_{2} \to {NH}_{3}$ $\frac{1}{2} N_{2} + \frac{3}{2} H_{2} \to {NH}_{3}$

is to be carried out isothermally first in a flow reactor. The molar feed is 50% H₂ and 50% N₂, at a pressure of 16.4 atm and at a temperature of 227°C.

(a) Construct a complete stoichiometric table.

(b) Express the concentrations in mol/dm³ of each for the reacting species as a function of conversion. Evaluate C_A0, δ, and ε, and then calculate the concentrations of ammonia and hydrogen when the conversion of H₂ is 60%. (Ans.: C_H₂ = 0.1 mol/dm³)

(c) Suppose by chance the reaction is elementary with k_N₂ = 40 dm³/mol/s. Write the rate of reaction solely as a function of conversion for (1) a flow reactor and for (2) a constant-volume batch reactor.

P4-9_B The elementary reversible reaction

2A ⇆ B

is carried out isothermally in a flow reactor with no pressure drop and where pure A is fed at a concentration of 4.0 mol/dm³. If the equilibrium conversion is found to be 60%

(a) What is the equilibrium constant, K_C, if the reaction occurs in the gas phase?

(b) What is the K_C if the reaction is a liquid-phase reaction?

P4-10_B Consider the elementary gas-phase reversible reaction is carried out isothermically

A ⇆ 3C

Pure A enters at a temperature of 400 K and a pressure of 10 atm. At this temperature, K_C =

0.25(mol/dm³)². Write the rate law and then calculate the equilibrium conversion for each of the following situations:

(a) The gas-phase reaction is carried out in a constant-volume batch reactor.

(b) The gas-phase reaction is carried out in a constant-pressure batch reactor.

(c) Can you explain the reason why there would be a difference in the two values of the equilibrium conversion?

P4-11_b Repeat parts (a) through (c) of Problem P4-10_B for the reaction

3A ⇆ C

Pure A enters at 400 K, 10 atm and the equilibrium constant is K_C = 2.5 (dm³/mol)².

(d) Compare the equilibrium conversions in P4-10_B and P4-11_B. Explain the trends, e.g., is X_e for constant pressure always greater than X_e for constant volume? Explain the reasons for the similarities and differences in he equilibrium conversions in the two reactions.

P4-12_c Consider a cylindrical batch reactor that has one end fitted with a frictionless piston attached to a spring (Figure P4-11_C). The reaction

A + B → 8C

with the rate law

–r_A = K₁ C²_A C_B

is taking place in this type of reactor.

A cylindrical batch reactor is shown with one end of the reactor fitted with a frictionless piston attached to a spring with the reaction occurring in a closed zone.

Figure P4-11_C

(a) Write the rate law solely as a function of conversion, numerically evaluating all possible symbols. (Ans.: –r_A = 5.03 × 10^-9 [(1 - X)³/(1 + 3X)^3/2]lb mol/ft³· s.)

(b) What is the conversion and rate of reaction when V = 0.2 ft³? (Ans.: X = 0.259, –r_A = 8.63 X 10^-10 lb mol/ft³ · s.)

Additional information:

Equal moles of A and B are present at t = 0

Initial volume: 0.15 ft³

Value of k₁: 1.0 (ft³/lb mol)² · s^-1

The spring constant is such that the relationship between the volume of the reactor and pressure within the reactor is

V = (0.1)(P) (V in ft³, P in atm)

Temperature of system (considered constant): 140°F Gas constant: 0.73 ft³ · atm/lb mol. °R

Supplementary Reading

Further elaboration of the development of the general balance equation may be found on the CRE Web site www.umich.edu/~elements/5e/index.html and also may or may not be found in

KEILLOR, GARRISON and TIM RUSSELL, Dusty and Lefty: The Lives of the Cowboys (Audio CD), St. Paul, MN: Highbridge Audio, 2006.

FELDER, R. M., and R. W. ROUSSEAU, Elementary Principles of Chemical Processes, 4th ed. New York: Wiley, 2015, Chapter 4.

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Table of Contents for Chapter 4 Stoichiometry

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4. Stoichiometry

4.1 Batch Systems

4.1.1 Batch Concentrations for the Generic Reaction, Equation (2-2)

4.2 Flow Systems

4.2.1 Equations for Concentrations in Flow Systems

4.2.2 Liquid-Phase Concentrations

4.2.3 Gas-Phase Concentrations

4.3 Reversible Reactions and Equilibrium Conversion

Summary

CRE Web Site Materials

Questions and Problems

Questions

Computer Simulations and Experiments

Interactive Computer Games

Problems

Supplementary Reading

Table of Contents for
Chapter 4 Stoichiometry