Chapter 4: Stoichiometry

Rationale For Chapter 4

In chapter 2 we saw that if we had –rA as a function of X, [–rA= f(X)] we could size many reactors and reactor sequences and systems.

Three diagrams comparing the volumes of reactors (PFR, CSTR, and a combination of CSTR and PFR) for a given reaction. The x-axis in each diagram is labeled X (conversion), and the y-axis is labeled FA0/-rA. The first diagram shows a shaded area representing the volume of a plug flow reactor (VPFR) with a curved line increasing towards the right. The second diagram shows a shaded rectangle representing the volume of a continuous stirred-tank reactor (VCSTR). The third diagram combines a shaded rectangle (CSTR) and a shaded curved area (PFR), indicating a combined CSTR + PFR system.

How do we obtain –rA = f(X)?

We do this in two steps

1. Part 1 - Chapter 3 Rate Law – Find the rate as a function of concentration,

          –rA = k fn (CA, CB …)

2. Part 2 - Chapter 4 Stoichiometry – Find the concentration as a function of conversion

          CA = g(X)

Combine Part 1 and Part 2 to get -rA=f(X)


Stoichiometry


We shall set up Stoichiometric Tables using A as our basis of calculation in the following reaction. We will use the stoichiometric tables to express the concentration as a function of conversion. We will combine Ci = f(X) with the appropriate rate law to obtain -rA = f(X).

\( A + \frac{b}{a} B \rightarrow \frac{c}{a} C + \frac{d}{a} D \)


Topics

  1. Batch System Stoichiometric Table
  2. Flow System Stoichiometric Table

Batch System Stoichiometric Table

Top

Species Symbol Initial Change Remaining
A A \( N_{A0} \) \( -N_{A0}X \) \( N_A = N_{A0}(1 - X) \)
B B \( N_{B0} = N_{A0} \Theta_B \) \( -\frac{b}{a} N_{A0}X \) \( N_B = N_{A0} \left( \Theta_B - \frac{b}{a} X \right) \)
C C \( N_{C0} = N_{A0} \Theta_C \) \( +\frac{c}{a} N_{A0}X \) \( N_C = N_{A0} \left( \Theta_C + \frac{c}{a} X \right) \)
D D \( N_{D0} = N_{A0} \Theta_D \) \( +\frac{d}{a} N_{A0}X \) \( N_D = N_{A0} \left( \Theta_D + \frac{d}{a} X \right) \)
Inert I \( N_I = N_{A0} \Theta_I \) – \( N_I = N_{A0} \Theta_I \)
         \( N_{T0} \) \( N_T = N_{T0} + \delta N_{A0}X \)

Where:
\( \Theta_i = \frac{N_{i0}}{N_{A0}} = \frac{C_{i0}}{C_{A0}} = \frac{Y_{i0}}{Y_{A0}} \quad \text{and} \quad \delta = \frac{d}{a} + \frac{c}{a} - \frac{b}{a} - 1 \)

Concentration -- Batch System: \( C_A = \frac{N_A}{V} \)

Constant Volume Batch: Note: if the reaction occurs in the liquid phase or if a gas phase reaction occurs in a rigid (e.g., steel) batch reactor, then:

\(V = V_0\)

\(C_A = \frac{N_A}{V} = \frac{N_{A0}(1 - X)}{V_0} = C_{A0}(1 - X)\)

\( C_B = \frac{N_B}{V} = \frac{N_{A0}}{V_0} \left( \Theta_B - \frac{b}{a} X \right) = C_{A0} \left( \Theta_B - \frac{b}{a} X \right) \)

etc.

if \(-r_A = k_A C_A^2 C_B\) then

\(-r_A = C_{A0}^3 (1 - X)^2 \left( \Theta_B - \frac{b}{a} X \right)\)

Constant Volume Batch

and we have -rA=f(x)

Self Test Writing -rA solely as a function of X


Flow System Stoichiometric Table

Top


A flow diagram of a reactor with multiple feed streams entering from the left and multiple product streams exiting on the right. The inputs on the left are labeled FAO, FBO, FCO, FDO, and FIO, representing the initial molar flow rates of different components. The outputs on the right are labeled FA, FB, FC, FD, and FI, representing the final molar flow rates. An additional variable X, representing conversion, is listed on the right.
Species Symbol Reactor Feed Change Reactor Effluent
A A \( F_{A0} \) \( -F_{A0}X \) \( F_A = F_{A0}(1 - X) \)
B B \( F_{B0} = F_{A0}\Theta_B \) \( -\frac{b}{a} F_{A0}X \) \( F_B = F_{A0}\left(\Theta_B - \frac{b}{a} X\right) \)
C C \( F_{C0} = F_{A0}\Theta_C \) \( +\frac{c}{a} F_{A0}X \) \( F_C = F_{A0}\left(\Theta_C + \frac{c}{a} X\right) \)
D D \( F_{D0} = F_{A0}\Theta_D \) \( +\frac{d}{a} F_{A0}X \) \( F_D = F_{A0}\left(\Theta_D + \frac{d}{a} X\right) \)
Inert I \( F_{I0} = F_{A0}\Theta_I \) — \( F_I = F_{A0}\Theta_I \)
         \( F_{T0} \) \( F_T = F_{T0} + \delta F_{A0}X \)

Where:

\( \Theta_i = \frac{F_{i0}}{F_{A0}} = \frac{C_{i0} u_0}{C_{A0} u_0} = \frac{C_{i0}}{C_{A0}} = \frac{y_{i0}}{y_{A0}} \quad \text{and} \quad \delta = \frac{d}{a} + \frac{c}{a} - \frac{b}{a} - 1 \)



Concentration -- Flow System: \( C_A = \frac{F_A}{u} \)


Liquid Phase Flow System: \( u = u_0 \)


\( C_A = \frac{F_A}{u} = \frac{F_{A0}(1 - X)}{u_0} = C_{A0}(1 - X) \)      Flow Liquid Phase

\( C_B = \frac{F_B}{u} = \frac{F_{A0}}{u_0} \left( \Theta_B - \frac{b}{a} X \right) = C_{A0} \left( \Theta_B - \frac{b}{a} X \right) \)

etc.

 

If the rate of reaction were -rA = kCACB then we would have
\( -r_A = k C_{A0}^2 (1 - X) \left( \Theta_B - \frac{b}{a} X \right) \)
This gives us -rA = f(X). Consequently, we can use the methods discussed in Chapter 2 to size a large number of reactors, either alone or in series.

Gas Phase Flow System: \( \upsilon = \upsilon_0 (1 + \epsilon X) \left( \frac{T}{T_0} \right) \left( \frac{P_0}{P} \right) \)

Derive Derive this equation

\(C_A = \frac{F_A}{\upsilon} = \frac{F_{A0}(1 - X)}{\upsilon_0 (1 + \epsilon X)} \left(\frac{T_0}{T}\right) \left(\frac{P}{P_0}\right) = C_{A0} \left(\frac{(1 - X)}{(1 + \epsilon X)}\right) \left(\frac{T_0}{T}\right) \left(\frac{P}{P_0}\right)\)

\( C_B = \frac{F_B}{\upsilon} = \frac{F_{A0} \left( \Theta_B - \frac{b}{a} X \right)}{\upsilon_0 (1 + \epsilon X)} \left( \frac{T_0}{T} \right) \left( \frac{P}{P_0} \right) = C_{A0} \left( \frac{\Theta_B - \frac{b}{a} X}{1 + \epsilon X} \right) \left( \frac{T_0}{T} \right) \left( \frac{P}{P_0} \right) \)

etc.

Again, these equations give us information about -rA = f(X), which we can use to size reactors.


For example if the gas phase reaction has the rate law 

\( -r_A = k_A C_A^2 C_B \)

then 

Flow Gas Phase

\(-r_A = \frac{C_{A0}^3 (1 - X)^2 \left( \Theta_B - \frac{b}{a} X \right)}{(1 + \epsilon X)^3}\)


with
\( C_{A0} = \frac{P_{A0}}{RT} \) \( \Theta_B = \frac{y_{B0}}{y_{A0}} = \frac{C_{B0}}{C_{A0}} = \frac{F_{B0}}{F_{A0}} \)

Example Rate Law in terms of Partial Pressures
Example Stoichiometry Table for Conversion
Example Oxidation of Naphthalene to Phtahalic Anhydride
Critical Critical Thinking Questions for the Oxidation of Naphthalene
Example Production of Nitric Acid
Example Expressing a Catalytic Rate Law in Terms of Conversion


Diagram showing expressions for concentration as a function of conversion for a reaction A + (b/a)B → (c/a)C + (d/a)D. The diagram splits into two branches: 'Liquid Phase' and 'Gas Phase.' In the Liquid Phase, the paths further split into 'Flow' and 'Batch' processes with no phase change, and the concentration of B (CB) is expressed in terms of the initial concentration CA0, the stoichiometric coefficient (b/a), and conversion (X). The expressions are CB = CA0 (θB - (b/a)X). In the Gas Phase, the paths split into 'Flow' and 'Batch' processes, and equations include pressure (P), temperature (T), and molar flows. The gas phase branch also accounts for volume changes due to pressure and temperature variations, with expressions for CB incorporating terms like NT, P0, T0, and ε. The gas phase paths account for cases with and without phase changes or pressure drops, with conditions labeled as 'Isothermal' and 'Neglect Pressure Drop.' The bottom left box defines parameters like ε, δ, CT0, and CA0. The figure caption states: 'Expressing concentration as a function of conversion.'

 

Calculating the equilibrium conversion for gas phase reaction

Consider the following elementary reaction with KC and = 20 dm3/mol and CA0 = 0.2 mol/dm3. Pure A fed. Calculate the equilibrium conversion, Xe, for both a batch reactor and a flow reactor.

\( \text{2A} \rightleftharpoons \text{B} \)

\( -r_A = k_A \left[ C_A^2 - \frac{C_B}{K_C} \right] \)

Solution

At equilibrium

\( -r_A \equiv 0 \equiv k_A \left[ C_{Ae}^2 - \frac{C_{Be}}{K_C} \right] \)

\( K_C = \frac{C_{Be}}{C_{Ae}^2} \)

Stoichiometry

\( \text{A} \rightarrow \frac{\text{B}}{2} \)

Batch

Species Initial Change Remaining
A \( N_{A0} \) \( -N_{A0}X \) \( N_A = N_{A0}(1 - X) \)
B \( 0 \) \( +\frac{N_{A0}X}{2} \) \( N_B = \frac{N_{A0}X}{2} \)
\( N_{T0} = N_{A0} \) \( N_T = N_{A0} - \frac{N_{A0}X}{2} \)

Constant Volume V = V0

\( C_A = \frac{N_A}{V} = \frac{N_A}{V_0} = \frac{N_{A0}(1 - X)}{V_0} = C_{A0}(1 - X) \) \( C_B = \frac{N_B}{V} = \frac{N_{A0}X/2}{V_0} = \frac{C_{A0}X}{2} \) \( K_C = \frac{C_{Be}}{C_{Ae}^2} = \frac{C_{A0} \frac{X_e}{2}}{C_{A0}^2 (1 - X_e)^2} \) \( 2K_C C_{A0} = \frac{X_e}{(1 - X_e)^2} = \left(2\right)\left(20 \, \frac{\text{dm}^3}{\text{mol}}\right)\left(0.2 \, \frac{\text{mol}}{\text{dm}^3}\right) = 8 \) \( 8X_e^2 - 17X_e + 8 = 0 \)

Solving

Batch: X = 0.7

Flow

Species Fed Change Remaining
A \( F_{A0} \) \( -F_{A0}X \) \( F_A = F_{A0}(1 - X) \)
B \( 0 \) \( +\frac{F_{A0}X}{2} \) \( F_B = \frac{F_{A0}X}{2} \)
\( F_{T0} = F_{A0} \) \( F_T = F_{A0} - \frac{F_{A0}X}{2} \)

\( C_{Ae} = \frac{C_{A0}(1 - X_e)}{1 + \varepsilon X_e} \) \( C_{Be} = \frac{C_{A0} X_e}{2 (1 + \varepsilon X_e)} \) \( K_C = \frac{\frac{C_{A0} X_e}{2 (1 + \varepsilon X_e)}}{\left(\frac{C_{A0}(1 - X_e)}{1 + \varepsilon X_e}\right)^2} = \frac{X_e (1 + \varepsilon X_e)}{2 C_{A0} (1 - X_e)^2} \) \( 2K_C C_{A0} = \frac{X_e + \varepsilon X_e^2}{1 - 2X_e + X_e^2} = 2 \left( 20 \, \frac{\text{dm}^3}{\text{mol}} \right) \left( 0.2 \, \frac{\text{mol}}{\text{dm}^3} \right) = 8 \) \( \varepsilon = y_{A0} \delta = 1 \left( \frac{1}{2} - 1 \right) = -\frac{1}{2} \) \( 8 = \frac{X_e - 0.5 X_e^2}{1 - 2 X_e + X_e^2} \) \( 8.5 X_e^2 - 17 X_e + 8 = 0 \)
Flow : \(X_e = 0.757\)
Recall
Batch : \(X_e = 0.70\)
Link The following video on limiting reactants was made by Professor Lane's 2008 Reaction Engineering class at the University of Alabama, Tuscaloosa.
Self Test Old Exam Questions
Self Test Matching CA versus X curves
Self Test What is wrong with this solution?
Self assess Assessment of Objective for Chapter 4
 

* All chapter references are for the 1st Edition of the text Essentials of Chemical Reaction Engineering .