12.3 Multiple Reactions, Catalysis, and Heterogeneous Kinetics

In industrial chemical processes, reactions rarely occur in isolation. Multiple reactions (occurring in series or parallel) are the norm, and catalysts are widely used to accelerate rates and steer selectivity. On the FE Chemical exam, you must understand how to select and operate reactors to maximize selectivity for multiple reactions, identify the steps in heterogeneous catalytic systems, apply packed bed design equations, and analyze mass transfer limitations using the Thiele modulus and effectiveness factor.

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Multiple Reactions and Reactor Selectivity

When multiple reactions occur, the primary objective is to maximize the formation of the desired product ($D$) while minimizing the undesired product ($U$).

1. Parallel Reactions

Consider the competing parallel reactions: $\text{Reaction 1 (Desired): } \text{A} \xrightarrow{k_1} \text{D} \quad (-r_{\text{A1}} = r_{\text{D}} = k_1 C_{\text{A}}^{a_1})$ $\text{Reaction 2 (Undesired): } \text{A} \xrightarrow{k_2} \text{U} \quad (-r_{\text{A2}} = r_{\text{U}} = k_2 C_{\text{A}}^{a_2})$ The instantaneous selectivity of desired product D with respect to undesired product U is: $S_{\text{D/U}} = \frac{r_{\text{D}}}{r_{\text{U}}} = \frac{k_1}{k_2} C_{\text{A}}^{a_1 - a_2}$ To maximize this selectivity, the concentration of reactant A ($C_{\text{A}}$) must be controlled:

• If $a_1 > a_2$: The exponent $(a_1 - a_2)$ is positive. To maximize selectivity, we must keep $C_{\text{A}}$ as high as possible. This is achieved by using a Plug Flow Reactor (PFR) or a Batch Reactor, where reactant concentration starts high and decreases gradually, rather than a CSTR. Feeding reactants without diluents also helps.
• If $a_1 < a_2$: The exponent $(a_1 - a_2)$ is negative. To maximize selectivity, we must keep $C_{\text{A}}$ as low as possible. This is achieved by using a Continuous Stirred Tank Reactor (CSTR), which operates entirely at the low exit concentration, or by diluting the feed.

2. Series Reactions

Consider series reactions where the desired product is the intermediate, R: $\text{A} \xrightarrow{k_1} \text{R} \xrightarrow{k_2} \text{S}$ Reactant A is converted to the desired product R, which then reacts to form the undesired waste product S.

• The concentration of R reaches a maximum at a specific time (in a batch reactor) or space time (in a flow reactor).
• Because a CSTR exhibits back-mixing (where exiting fluid is mixed with incoming reactant, and some fluid leaves immediately while some remains for a long time), the maximum concentration of the intermediate R is significantly lower in a CSTR than in a PFR. Therefore, a PFR or Batch Reactor is preferred to maximize the yield of the intermediate R.

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Homogeneous vs. Heterogeneous Kinetics and Catalysis

• Homogeneous reactions take place in a single phase, and their rates are based on reactor fluid volume.
• Heterogeneous reactions involve more than one phase, most commonly a fluid reactant reacting on the surface of a solid catalyst. In these systems, the reaction rate is defined relative to the mass of the catalyst ($W$) rather than fluid volume: $-r_{\text{A}}' = -\frac{1}{W} \frac{dN_{\text{A}}}{dt}$ where $-r_{\text{A}}'$ has units of $\text{mol}/(\text{g}{\text{cat}}\cdot\text{s})$ or $\text{mol}/(\text{kg}{\text{cat}}\cdot\text{s})$.

Packed Bed Reactor (PBR) Design Equation

For a PBR (also called a fixed-bed catalytic reactor), the catalyst is packed inside tubes. The design equation relates the mass of catalyst $W$ required to achieve a target conversion $X_{\text{A}}$: $W = F_{\text{A}0} \int_0^{X_{\text{A}}} \frac{dX_{\text{A}}}{-r_{\text{A}}'}$ This is the heterogeneous analog of the PFR volume design equation.

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Steps in a Heterogeneous Catalytic Reaction

For a catalytic reaction to occur on a solid catalyst, reactants must transport from the bulk fluid to the active sites inside the catalyst pores, react, and then the products must transport back. The process consists of seven sequential steps:

1. External diffusion of reactants from the bulk fluid phase to the external surface of the catalyst pellet.
2. Internal diffusion of reactants through the catalyst pores to the active sites.
3. Adsorption of reactants onto the catalyst active sites (often modeled by Langmuir adsorption isotherms).
4. Surface reaction of adsorbed species to form adsorbed products.
5. Desorption of products from the active sites.
6. Internal diffusion of products out of the pores to the external surface.
7. External diffusion of products from the external surface back into the bulk fluid phase.

If any of these steps is significantly slower than the others, it becomes the Rate-Limiting Step (RLS).

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Mass Transfer Limitations: Thiele Modulus and Effectiveness Factor

Pore diffusion and external mass transfer can severely limit the overall reaction rate, making the catalyst under-utilized.

1. The Thiele Modulus ($\phi$)

The Thiele modulus is a dimensionless number that represents the ratio of the intrinsic surface reaction rate to the rate of internal diffusion through the catalyst pores. For a first-order reaction in a flat plate or spherical catalyst pellet: $\phi = L \sqrt{\frac{k}{D_e}}$ where:

• $L$ is the characteristic length of the catalyst pellet (volume of pellet divided by external surface area; for a sphere, $L = R_p / 3$).
• $k$ is the intrinsic first-order reaction rate constant (based on pellet volume).
• $D_e$ is the effective diffusivity of the reactant inside the pores.

2. The Internal Effectiveness Factor ($\eta$)

The internal effectiveness factor $\eta$ is defined as the ratio of the actual overall reaction rate to the rate that would be achieved if the entire interior of the pellet were exposed to the reactant concentration and temperature at the external pellet surface: $\eta = \frac{\text{Actual Reaction Rate}}{\text{Reaction Rate at Surface Conditions}}$ For isothermal systems, $\eta$ is related to the Thiele modulus:

• Reaction-Controlled Regime ($\phi < 1$): When diffusion is very fast compared to reaction, $\phi$ is small and $\eta \approx 1$. The reactant concentration is uniform throughout the pellet, and the catalyst is fully utilized.
• Diffusion-Controlled Regime ($\phi > 3$ or large): When diffusion is slow compared to reaction, $\phi$ is large and $\eta \approx 1/\phi$. The reactant is consumed near the external surface of the pellet, leaving the interior of the catalyst unutilized.

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Worked Example: Packed Bed Reactor Design

Problem: A second-order heterogeneous catalytic reaction $\text{A} \rightarrow \text{B}$ is carried out in a Packed Bed Reactor (PBR). The rate law is $-r_{\text{A}}' = k' C_{\text{A}}^2$, where $k' = 0.15\text{ L}^2/(\text{mol}\cdot\text{g}{\text{cat}}\cdot\text{min})$. The feed is a liquid mixture entering at a volumetric flow rate $v_0 = 5\text{ L/min}$ with an inlet concentration $C{\text{A}0} = 1.0\text{ mol/L}$. Calculate the mass of catalyst $W$ (in grams) required to achieve $75%$ conversion ($X_{\text{A}} = 0.75$). Assume no pressure drop and constant density.

Solution:

1. State the parameters:

   • $v_0 = 5\text{ L/min}$
   • $C_{\text{A}0} = 1.0\text{ mol/L}$
   • $F_{\text{A}0} = v_0 C_{\text{A}0} = 5.0\text{ mol/min}$
   • $k' = 0.15\text{ L}^2/(\text{mol}\cdot\text{g}_{\text{cat}}\cdot\text{min})$
   • $X_{\text{A}} = 0.75$

2. Express concentration in terms of conversion (for constant density): $C_{\text{A}} = C_{\text{A}0}(1 - X_{\text{A}})$ $-r_{\text{A}}' = k' C_{\text{A}}^2 = k' C_{\text{A}0}^2 (1 - X_{\text{A}})^2$

3. Set up the PBR design equation: $W = F_{\text{A}0} \int_0^{X_{\text{A}}} \frac{dX_{\text{A}}}{k' C_{\text{A}0}^2 (1 - X_{\text{A}})^2} = \frac{F_{\text{A}0}}{k' C_{\text{A}0}^2} \int_0^{X_{\text{A}}} \frac{dX_{\text{A}}}{(1 - X_{\text{A}})^2}$

4. Integrate the expression: $\int_0^{X_{\text{A}}} \frac{dX_{\text{A}}}{(1 - X_{\text{A}})^2} = \left[ \frac{1}{1 - X_{\text{A}}} \right]_0^{X_{\text{A}}} = \frac{1}{1 - X_{\text{A}}} - 1 = \frac{X_{\text{A}}}{1 - X_{\text{A}}}$

5. Substitute the values: $W = \frac{5.0}{0.15 \cdot (1.0)^2} \cdot \left( \frac{0.75}{1 - 0.75} \right)$ $W = \frac{5.0}{0.15} \cdot \left( \frac{0.75}{0.25} \right) = 33.33 \cdot 3 = 100\text{ g}$

Thus, $100\text{ g}$ of catalyst is required to achieve the desired conversion of $75%$.