14.1 Yield, Selectivity, and Multiple Reactions (Series and Parallel)

14.1 Yield, Selectivity, and Multiple Reactions (Series and Parallel)

In industrial chemical synthesis, single isolated reactions are rare. Most commercial processes involve multiple reactions occurring simultaneously, producing undesired byproducts alongside the target molecule. Downstream separation equipment (such as distillation columns, extractors, and crystallizers) often accounts for $60%$ to $80%$ of total plant capital and operating expenditures. Consequently, reactor design must prioritize selectivity and yield over raw conversion.

On the NCEES PE Chemical Exam, candidates must evaluate multiple reaction networks, choose the optimal reactor type (PFR vs. CSTR vs. batch/semi-batch), configure reactant contacting schemes, and calculate optimal residence times and maximum product concentrations for both parallel (competing) and series (consecutive) pathways governed by the NCEES PE Chemical Reference Handbook.

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1. Mathematical Definitions: Selectivity and Yield

To optimize multiple reaction systems, chemical engineers differentiate between instantaneous metrics (defined at a specific point in space or time) and overall metrics (integrated across the entire reactor effluent).

               +-------------------------------------------+
               |         MULTIPLE REACTION METRICS         |
               +-------------------------------------------+
               |  Instantaneous (Point/Local Rate Ratios)  |
               |  S_D/U = r_D / r_U                        |
               |  Y_D   = r_D / (-r_A)                     |
               +-------------------------------------------+
               |     Overall (Effluent Flow/Mole Ratios)   |
               |  S_hat = F_D / F_U = N_D / N_U            |
               |  Y_hat = F_D / (F_A0 - F_A)               |
               +-------------------------------------------+

Instantaneous Selectivity ($S_{D/U}$)

Consider reactant $A$ reacting to form desired product $D$ and undesired byproduct $U$. The instantaneous selectivity ($S_{D/U}$) is the ratio of the net rate of formation of $D$ to the net rate of formation of $U$:

$S_{D/U} = \frac{r_D}{r_U}$

Overall Selectivity ($\hat{S}_{D/U}$)

The overall selectivity ($\hat{S}_{D/U}$) is the ratio of total moles (or molar flow rate) of desired product $D$ leaving the reactor system to total moles of undesired product $U$:

$\hat{S}_{D/U} = \frac{F_D}{F_U} = \frac{N_D}{N_U} = \frac{\int_0^V r_D \, dV}{\int_0^V r_U \, dV}$

For a single ideal continuous stirred-tank reactor (CSTR), conditions throughout the vessel match the exit stream, meaning overall selectivity equals instantaneous selectivity evaluated at exit conditions: $\hat{S}{D/U} = S{D/U}(C_{A,exit})$. For a plug flow reactor (PFR) or batch reactor, overall selectivity is the integral average along the trajectory from inlet to outlet.

Instantaneous Yield ($Y_D$)

The instantaneous yield of desired product $D$ relative to key reactant $A$ is the rate of formation of $D$ divided by the rate of disappearance of $A$:

$Y_D = \frac{r_D}{-r_A} = \frac{r_D}{r_D + r_U}$

Overall Yield ($\hat{Y}_D$)

The overall yield ($\hat{Y}_D$) is the fraction of consumed reactant $A$ that was converted into desired product $D$:

$\hat{Y}_D = \frac{F_D}{F_{A0} - F_A} = \frac{N_D}{N_{A0} - N_A} = \frac{F_D}{F_{A0} X_A}$

For a PFR:

$\hat{Y}_D = \frac{1}{F_{A0} X_A} \int_0^V r_D \, dV = \frac{1}{X_A} \int_0^{X_A} Y_D \, dX_A$

For a CSTR:

$\hat{Y}_D = Y_D(C_{A,exit}) = \left. \frac{r_D}{-r_A} \right|_{exit}$

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2. Competing Parallel Reactions: Kinetic Order Rules

In parallel (competing) reactions, reactant $A$ is consumed along two or more independent simultaneous pathways:

A &\xrightarrow{k_1} D \quad (\text{Desired}), \quad &r_D &= k_1 C_A^{a_1} \\ A &\xrightarrow{k_2} U \quad (\text{Undesired}), \quad &r_U &= k_2 C_A^{a_2} \end{aligned}$$ The instantaneous selectivity of $D$ relative to $U$ is: $$S_{D/U} = \frac{r_D}{r_U} = \frac{k_1 C_A^{a_1}}{k_2 C_A^{a_2}} = \frac{k_1}{k_2} C_A^{a_1 - a_2}$$ To maximize the production of $D$ and minimize $U$, chemical engineers manipulate reactant concentration $C_A$ according to the difference in reaction orders $(a_1 - a_2)$: ``` KINETIC ORDER DECISION RULES | +----------------------+----------------------+ | | a1 > a2 a1 < a2 S_D/U ~ C_A^(+) S_D/U ~ 1 / C_A^(+) Maximize C_A Minimize C_A ------------------ ------------------ - PFR or Batch - CSTR - High pressure (gas) - High recycle ratio - Avoid inert diluents - Inert dilution - Do not dilute feed - Low pressure (gas) ``` ### Case 1: $a_1 > a_2$ (Desired Reaction Has Higher Order) - Because $(a_1 - a_2) > 0$, instantaneous selectivity $S_{D/U}$ increases monotonically with reactant concentration $C_A$. - **Reactor Strategy:** Maintain $C_A$ as high as possible throughout the reaction. - **Selection:** Use a **Plug Flow Reactor (PFR)** or a **Batch Reactor**. In these reactors, $C_A$ begins at its maximum feed value ($C_{A0}$) and decreases gradually along the reactor length or batch time. - **Operating Conditions:** Avoid inert diluents; operate at high system pressure if reactants are gaseous. ### Case 2: $a_1 < a_2$ (Undesired Reaction Has Higher Order) - Because $(a_1 - a_2) < 0$, instantaneous selectivity $S_{D/U}$ decreases as $C_A$ rises. Selectivity is maximized when $C_A$ is kept low. - **Reactor Strategy:** Maintain $C_A$ at a low, uniform concentration throughout the reaction zone. - **Selection:** Use a **Continuous Stirred-Tank Reactor (CSTR)**. The instant feed enters a CSTR, it is backmixed and diluted to the low exit concentration ($C_{A,exit}$). - **Alternative Strategy:** A PFR with a large recycle stream ($R \to \infty$) behaves like a CSTR, suppressing $C_A$. - **Operating Conditions:** Introduce inert diluents or operate gas systems under lower total pressures. ### Case 3: Temperature Sensitivity (Arrhenius Activation Energies) The ratio of rate constants depends exponentially on temperature through the Arrhenius equation ($k_i = A_i e^{-E_i / RT}$): $$\frac{k_1}{k_2} = \frac{A_1}{A_2} \exp\left( -\frac{E_1 - E_2}{RT} \right)$$ - If **$E_1 > E_2$**: The desired reaction has a higher activation energy. Increasing temperature increases $k_1$ much faster than $k_2$. **Operate at high temperature**. - If **$E_1 < E_2$**: The undesired reaction has a higher activation energy. High temperatures accelerate byproduct formation. **Operate at low temperature** (accepting a lower overall reaction rate or utilizing a more active catalyst). --- ## 3. Multi-Reactant Parallel Networks: Contacting Schemes When multiple reactants participate in competing reactions, such as: $$\begin{aligned} A + B &\xrightarrow{k_1} D \quad (\text{Desired}), \quad &r_D &= k_1 C_A^{\alpha_1} C_B^{\beta_1} \\ A + B &\xrightarrow{k_2} U \quad (\text{Undesired}), \quad &r_U &= k_2 C_A^{\alpha_2} C_B^{\beta_2} \end{aligned}$$ The instantaneous selectivity becomes: $$S_{D/U} = \frac{k_1}{k_2} C_A^{\alpha_1 - \alpha_2} C_B^{\beta_1 - \beta_2}$$ The optimal contacting scheme must adjust $C_A$ and $C_B$ independently: | Kinetic Orders | Target Concentrations | Recommended Continuous Reactor | Recommended Batch / Semi-Batch Scheme | | :--- | :--- | :--- | :--- | | $\alpha_1 > \alpha_2$ and $\beta_1 > \beta_2$ | High $C_A$, High $C_B$ | **Tubular PFR** with co-fed pure $A$ and $B$ | **Batch Reactor** with all $A$ and $B$ charged initially | | $\alpha_1 < \alpha_2$ and $\beta_1 < \beta_2$ | Low $C_A$, Low $C_B$ | **CSTR** or PFR with high recycle ratio | **Semi-batch Reactor** slowly co-feeding dilute $A$ and $B$ | | $\alpha_1 > \alpha_2$ and $\beta_1 < \beta_2$ | High $C_A$, Low $C_B$ | **PFR with side-stream feed** of $B$ along tube length | **Semi-batch Reactor** charged with $A$; $B$ added slowly | | $\alpha_1 < \alpha_2$ and $\beta_1 > \beta_2$ | Low $C_A$, High $C_B$ | **PFR with side-stream feed** of $A$ along tube length | **Semi-batch Reactor** charged with $B$; $A$ added slowly | ``` CONTACTING SCHEME: HIGH C_A, LOW C_B (PFR WITH SIDE FEEDS) Pure A Feed | v +=========================================================+ | Stage 1 | Stage 2 | Stage 3 | Stage 4 | ---> Effluent +=========================================================+ ^ ^ ^ ^ | | | | Feed B Feed B Feed B Feed B (Staged) (Staged) (Staged) (Staged) ``` > [!TIP] > **Membrane Reactors for Selectivity Control:** > When kinetic rules require keeping reactant $B$ at trace levels while maintaining high $C_A$, an advanced industrial alternative to discrete side-stream injection is a **membrane reactor**. Species $A$ flows down a central ceramic or polymeric tube while species $B$ slowly permeates radially through the membrane wall along the entire reactor axis. --- ## 4. Consecutive (Series) Reactions: PFR vs. CSTR Performance In consecutive (series) reactions, the desired product is an **intermediate species** ($R$) that can undergo further reaction to form an over-oxidized or degraded waste product ($S$): $$A \xrightarrow{k_1} R \xrightarrow{k_2} S$$ Assume irreversible, liquid-phase, elementary first-order kinetics ($r_1 = k_1 C_A$ and $r_2 = k_2 C_R$). The initial concentrations are $C_{A0}$, $C_{R0} = 0$, and $C_{S0} = 0$. ### Analytical Derivations for a Plug Flow Reactor (PFR) In a PFR (or batch reactor where $t = \tau$): $$\frac{dC_A}{d\tau} = -k_1 C_A \implies C_A(\tau) = C_{A0} e^{-k_1 \tau}$$ The net rate of intermediate $R$ production is the rate of step 1 minus step 2: $$\frac{dC_R}{d\tau} = k_1 C_A - k_2 C_R = k_1 C_{A0} e^{-k_1 \tau} - k_2 C_R$$ Multiplying by the integrating factor $e^{k_2 \tau}$ and integrating subject to $C_R(0) = 0$: $$C_R(\tau) = C_{A0} \frac{k_1}{k_2 - k_1} \left( e^{-k_1 \tau} - e^{-k_2 \tau} \right) \quad (k_1 \neq k_2)$$ As space time $\tau$ increases, $C_R$ rises to a peak and then decays to zero as all mass is converted to $S$. Setting $\frac{dC_R}{d\tau} = 0$ identifies the **optimum space time** ($\tau_{opt}$): $$k_1 C_{A0} e^{-k_1 \tau_{opt}} = k_2 C_R = k_2 \left[ C_{A0} \frac{k_1}{k_2 - k_1} \left( e^{-k_1 \tau_{opt}} - e^{-k_2 \tau_{opt}} \right) \right]$$ Simplifying yields the classical PFR optimization formula: $$\tau_{opt, PFR} = \frac{\ln\left( \frac{k_2}{k_1} \right)}{k_2 - k_1} = \frac{\ln\left( \frac{k_1}{k_2} \right)}{k_1 - k_2}$$ Substituting $\tau_{opt}$ back into the concentration profile gives the **maximum intermediate concentration** ($C_{R,max}$): $$C_{R,max, PFR} = C_{A0} \left( \frac{k_1}{k_2} \right)^{\frac{k_2}{k_2 - k_1}}$$ ``` Concentration vs. Space Time in a PFR (A -> R -> S) Conc ^ C_A0|---\ | \ C_A(t) | \ C_R,max | \ /\ | \ / \ C_R(t) [Intermediate] | \/ \ | /\ \--------- | / \ C_S(t) [Waste] | / \----------- +-----------------------------------> Space Time (tau) tau_opt ``` ### Analytical Derivations for a CSTR In an ideal CSTR, steady-state component material balances equate feed, effluent, and generation: $$C_{A0} - C_A - k_1 C_A \tau = 0 \implies C_A = \frac{C_{A0}}{1 + k_1 \tau}$$ For intermediate $R$ (with $C_{R0} = 0$): $$0 - C_R + (k_1 C_A - k_2 C_R)\tau = 0 \implies C_R (1 + k_2 \tau) = k_1 C_A \tau$$ Substituting the expression for $C_A$: $$C_R = \frac{k_1 \tau C_{A0}}{(1 + k_1 \tau)(1 + k_2 \tau)}$$ To find the space time that maximizes $C_R$, set $\frac{dC_R}{d\tau} = 0$: $$\frac{d}{d\tau} \left[ \frac{\tau}{1 + (k_1 + k_2)\tau + k_1 k_2 \tau^2} \right] = 0 \implies 1 - k_1 k_2 \tau^2 = 0$$ $$\tau_{opt, CSTR} = \frac{1}{\sqrt{k_1 k_2}}$$ Substituting $\tau_{opt, CSTR}$ back into the CSTR expression yields: $$C_{R,max, CSTR} = C_{A0} \frac{1}{\left( 1 + \sqrt{\frac{k_2}{k_1}} \right)^2}$$ > [!IMPORTANT] > **The Severe Backmixing Penalty of CSTRs in Series Networks:** > In an ideal PFR, fluid elements travel with uniform residence time. Fluid leaving the reactor at $\tau_{opt}$ is discharged before $R$ can degrade to $S$. In contrast, an ideal CSTR possesses an exponential residence time distribution: some fluid elements exit almost immediately (unreacted $A$), while others remain for prolonged durations where $R$ degrades into waste $S$. As a mathematical consequence, **$C_{R,max, PFR} > C_{R,max, CSTR}$ for all non-zero, finite rate constants**. --- ## 5. Summary Comparison Table: Multiple Reaction Optimization | Reaction Network | Key Metric / Target | Optimal Reactor Configuration | Governing Analytical Formula | | :--- | :--- | :--- | :--- | | **Parallel ($a_1 > a_2$)** | Maximize $S_{D/U} \propto C_A^{a_1 - a_2}$ | **PFR** or **Batch** (Keep $C_A$ high) | $\hat{S}_{D/U} = \frac{\int_0^V k_1 C_A^{a_1} dV}{\int_0^V k_2 C_A^{a_2} dV}$ | | **Parallel ($a_1 < a_2$)** | Maximize $S_{D/U} \propto C_A^{-(a_2 - a_1)}$ | **CSTR** (Keep $C_A$ low) | $\hat{S}_{D/U} = \frac{k_1}{k_2} C_{A,exit}^{a_1 - a_2}$ | | **Parallel ($E_1 > E_2$)** | Maximize $k_1 / k_2$ | **High Temperature** operation | $\frac{d \ln(k_1/k_2)}{dT} = \frac{E_1 - E_2}{R T^2} > 0$ | | **Parallel ($E_1 < E_2$)** | Maximize $k_1 / k_2$ | **Low Temperature** operation | $\frac{d \ln(k_1/k_2)}{dT} = \frac{E_1 - E_2}{R T^2} < 0$ | | **Series ($A \to R \to S$) PFR** | Maximize Intermediate $R$ | **PFR** at $\tau_{opt, PFR}$ | $\tau_{opt} = \frac{\ln(k_2/k_1)}{k_2 - k_1}$; $C_{R,max} = C_{A0} \left( \frac{k_1}{k_2} \right)^{\frac{k_2}{k_2 - k_1}}$ | | **Series ($A \to R \to S$) CSTR** | Maximize Intermediate $R$ | **CSTR** at $\tau_{opt, CSTR}$ | $\tau_{opt} = \frac{1}{\sqrt{k_1 k_2}}$; $C_{R,max} = \frac{C_{A0}}{(1 + \sqrt{k_2/k_1})^2}$ | --- ## 6. Comprehensive Worked Numerical Example: Series Reaction Optimization ### Problem Statement A pharmaceutical intermediate $R$ is synthesized via the liquid-phase consecutive reaction network: $$A \xrightarrow{k_1} R \xrightarrow{k_2} S$$ The reaction is first-order in both steps with rate constants $k_1 = 0.60 \text{ min}^{-1}$ and $k_2 = 0.15 \text{ min}^{-1}$ at the process temperature of $65^\circ\text{C}$. The feed enters at a volumetric flow rate of $v_0 = 50.0 \text{ L/min}$ ($0.050 \text{ m}^3/\text{min}$) containing reactant $A$ at initial concentration $C_{A0} = 2.50 \text{ mol/L}$ ($2,500 \text{ mol/m}^3$) with no $R$ or $S$ present. **Calculate:** 1. For an ideal **Plug Flow Reactor (PFR)**: the optimum space time ($\tau_{opt}$), required reactor volume ($V_{PFR}$), maximum intermediate concentration ($C_{R,max}$), reactant conversion ($X_A$), and overall yield ($\hat{Y}_R$). 2. For an ideal **Continuous Stirred-Tank Reactor (CSTR)**: the optimum space time ($\tau_{opt}$), required reactor volume ($V_{CSTR}$), maximum intermediate concentration ($C_{R,max}$), reactant conversion ($X_A$), and overall yield ($\hat{Y}_R$). 3. The percentage reduction in intermediate production rate incurred by choosing a CSTR instead of a PFR. --- ### Step 1: PFR Design Calculations First, evaluate the rate constant difference and ratio: $$k_2 - k_1 = 0.15 - 0.60 = -0.45 \text{ min}^{-1}$$ $$\frac{k_2}{k_1} = \frac{0.15}{0.60} = 0.250 \implies \ln(0.250) = -1.3863$$ Calculate optimum space time: $$\tau_{opt, PFR} = \frac{\ln(k_2 / k_1)}{k_2 - k_1} = \frac{-1.3863}{-0.45 \text{ min}^{-1}} = \mathbf{3.081 \text{ min}}$$ Determine required PFR volume: $$V_{PFR} = v_0 \cdot \tau_{opt, PFR} = 50.0 \text{ L/min} \times 3.081 \text{ min} = \mathbf{154.0 \text{ L}} \quad (0.154 \text{ m}^3)$$ Calculate reactant concentration and conversion at exit: $$C_{A} = C_{A0} e^{-k_1 \tau_{opt}} = 2.50 \times e^{-(0.60 \times 3.081)} = 2.50 \times e^{-1.8486} = 2.50 \times 0.15746 = \mathbf{0.3937 \text{ mol/L}}$$ $$X_A = 1 - \frac{C_A}{C_{A0}} = 1 - \frac{0.3937}{2.50} = 1 - 0.1575 = \mathbf{0.8425} \quad (84.25\%$$ Calculate maximum intermediate concentration: $$\frac{k_1}{k_2} = \frac{0.60}{0.15} = 4.00$$ $$\frac{k_2}{k_2 - k_1} = \frac{0.15}{-0.45} = -\frac{1}{3}$$ $$C_{R,max, PFR} = C_{A0} \left( \frac{k_1}{k_2} \right)^{\frac{k_2}{k_2 - k_1}} = 2.50 \times (4.00)^{-1/3} = \frac{2.50}{1.5874} = \mathbf{1.575 \text{ mol/L}}$$ Calculate overall yield of $R$: $$\hat{Y}_{R, PFR} = \frac{C_{R,max}}{C_{A0} - C_A} = \frac{1.575 \text{ mol/L}}{(2.50 - 0.3937) \text{ mol/L}} = \frac{1.575}{2.1063} = \mathbf{0.7478} \quad (74.78\%)$$ --- ### Step 2: CSTR Design Calculations Calculate optimum space time for the CSTR: $$\tau_{opt, CSTR} = \frac{1}{\sqrt{k_1 k_2}} = \frac{1}{\sqrt{0.60 \times 0.15}} = \frac{1}{\sqrt{0.090}} = \frac{1}{0.30 \text{ min}^{-1}} = \mathbf{3.333 \text{ min}}$$ Determine required CSTR volume: $$V_{CSTR} = v_0 \cdot \tau_{opt, CSTR} = 50.0 \text{ L/min} \times 3.333 \text{ min} = \mathbf{166.7 \text{ L}} \quad (0.1667 \text{ m}^3)$$ Calculate exit reactant concentration and conversion: $$C_{A} = \frac{C_{A0}}{1 + k_1 \tau_{opt}} = \frac{2.50}{1 + (0.60 \times 3.333)} = \frac{2.50}{1 + 2.00} = \frac{2.50}{3.00} = \mathbf{0.8333 \text{ mol/L}}$$ $$X_A = 1 - \frac{0.8333}{2.50} = 1 - 0.3333 = \mathbf{0.6667} \quad (66.67\%)$$ Calculate maximum intermediate concentration: $$\sqrt{\frac{k_2}{k_1}} = \sqrt{\frac{0.15}{0.60}} = \sqrt{0.25} = 0.500$$ $$C_{R,max, CSTR} = C_{A0} \frac{1}{\left( 1 + \sqrt{\frac{k_2}{k_1}} \right)^2} = \frac{2.50}{(1 + 0.500)^2} = \frac{2.50}{(1.50)^2} = \frac{2.50}{2.25} = \mathbf{1.111 \text{ mol/L}}$$ Calculate overall yield of $R$ in the CSTR: $$\hat{Y}_{R, CSTR} = \frac{C_{R,max}}{C_{A0} - C_A} = \frac{1.111 \text{ mol/L}}{(2.50 - 0.8333) \text{ mol/L}} = \frac{1.111}{1.6667} = \mathbf{0.6667} \quad (66.67\%)$$ --- ### Step 3: Performance Comparison Evaluate molar intermediate output rates ($\dot{n}_R = v_0 C_R$): - **PFR Output:** $\dot{n}_{R, PFR} = 50.0 \text{ L/min} \times 1.575 \text{ mol/L} = \mathbf{78.75 \text{ mol/min}}$ - **CSTR Output:** $\dot{n}_{R, CSTR} = 50.0 \text{ L/min} \times 1.111 \text{ mol/L} = \mathbf{55.56 \text{ mol/min}}$ $$\text{Production Penalty} = \frac{78.75 - 55.56}{78.75} \times 100\% = \mathbf{29.45\%}$$ Selecting a CSTR results in a **$29.45\%$ loss in intermediate production** while requiring a slightly larger vessel ($166.7 \text{ L}$ vs. $154.0 \text{ L}$), underscoring the decisive superiority of plug flow for consecutive networks. --- ## 7. Critical PE Exam Traps & Pitfalls > **Trap 1: Confusing Instantaneous Selectivity ($S_{D/U}$) with Overall Selectivity ($\hat{S}_{D/U}$)** > Instantaneous selectivity $r_D / r_U$ only equals effluent molar ratio $F_D / F_U$ in a single ideal CSTR! In a PFR, $r_D / r_U$ changes continuously from inlet to outlet. When asked for total product amounts in a PFR, you must integrate the rates along the reactor volume ($F_D = \int r_D dV$), not simply multiply inlet selectivity by total feed. > **Trap 2: Applying PFR Series Optimization Formulas to a CSTR** > The equation $\tau_{opt} = \ln(k_2/k_1) / (k_2 - k_1)$ is strictly valid **only for plug flow and batch reactors**. Applying this formula to a CSTR produces gross design errors. For a CSTR, the geometric mean relation $\tau_{opt} = 1 / \sqrt{k_1 k_2}$ must be used. > **Trap 3: Overlooking Feed Dilution Effects in Parallel Reactions** > When $a_1 > a_2$, adding inert solvent or operating at reduced pressure lowers $C_A$, which unintentionally shifts selectivity in favor of the undesired reaction $U$. Verify whether problem statements specify pure versus diluted feeds when evaluating parallel reactor choices. > **Trap 4: Misinterpreting Semi-Batch Feed Policies** > If $A + B \to D$ is second-order in $A$ and first-order in $B$, while $A + B \to U$ is first-order in $A$ and second-order in $B$, selectivity requires **high $C_A$ and low $C_B$**. Students often guess "charge $B$ and drip in $A$." Dripping in $A$ creates low $C_A$ and high $C_B$, maximizing the undesired product $U$! The vessel must be charged with $A$, with $B$ metered in slowly.