Multiple reactions: parallel and series, selectivity and yield

Product distribution in parallel and series reactions: fractional yield, selectivity, the effect of concentration, temperature and reactor type, and optimum residence time for intermediates.

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Why it matters

Most industrial reactions are accompanied by side reactions, and the value of a process is set by how much desired product you make per mole of feed, not by conversion alone. For multiple reactions, reactor type, feed concentration, contacting pattern and residence time are chosen to steer product distribution — often a bigger economic lever than reactor size.

Key ideas

Parallel (competing) reactions. A → R (desired), A → S (undesired), with rR = k1·CA^a1 and rS = k2·CA^a2. The ratio rR/rS = (k1/k2)·CA^(a1−a2) shows that:

  • If a1 > a2 (desired reaction has the higher order), keep CA high: use a batch reactor or PFR, avoid dilution, use high pressure for gases.
  • If a1 < a2, keep CA low: use a CSTR, dilute with inerts or recycle, run at low pressure.
  • If a1 = a2, concentration and reactor type do not matter; only k1/k2 (temperature or catalyst) can help. With two reactants, A + B → R and A + B → S, the same reasoning applies to each reactant separately and leads to contacting schemes (both high, both low, one high and one low by staged feed).

Temperature. Since k1/k2 = (k10/k20)·exp[−(E1 − E2)/RT], a high temperature favours the reaction with the higher activation energy and a low temperature the one with the lower activation energy.

Measures of product distribution.

  • Instantaneous fractional yield φ(R/A) = rR/(−rA) = moles R formed per mole A reacted at that point.
  • Overall fractional yield Φ(R/A) = CRf/(CA0 − CAf) — the average of φ over the reactor.
  • Selectivity S(R/S) = moles R formed / moles S formed (some texts define it on rates).
  • Overall yield based on feed: CRf/CA0 = Φ·XA. For a CSTR the whole tank is at exit concentration, so Φ = φ at exit. For a PFR, Φ is the average of φ over the concentration range.

Series (consecutive) reactions. A → R → S, both first order (k1, k2). R rises, passes through a maximum and then falls. To get the most R, stop at the right time: in a PFR or batch reactor at τ_opt; in a CSTR at its own τ_opt. Because back-mixing mixes fresh A with product R that is being destroyed, a PFR always gives a higher maximum of R than a CSTR. Never mix fluids of different compositions (different stages of reaction) when the intermediate is wanted.

Series-parallel reactions. A + B → R, R + B → S: with respect to B they are parallel, with respect to A, R, S they are series. Adding B slowly to A (semi-batch) or using a PFR favours R.

Formulas

φ(R/A) = rR/(−rA) = 1/[1 + (k2/k1)·CA^(a2 − a1)] (parallel A → R, A → S)

Φ(R/A) = φ(CAf) (CSTR); Φ(R/A) = [1/(CA0 − CAf)]·∫(CAf→CA0) φ dCA (PFR)

CRf = Φ·(CA0 − CAf)

Series A → R → S (first order, CR0 = CS0 = 0):

CR/CA0 = [k1/(k2 − k1)]·[exp(−k1·τ) − exp(−k2·τ)] (PFR/batch, k1 ≠ k2)

τ_opt = ln(k2/k1)/(k2 − k1), CR,max/CA0 = (k1/k2)^[k2/(k2 − k1)] (PFR/batch)

CR/CA0 = k1·τ/[(1 + k1·τ)·(1 + k2·τ)] (CSTR)

τ_opt = 1/√(k1·k2), CR,max/CA0 = 1/[(k2/k1)^0.5 + 1]² (CSTR)

  • k1, k2 in s⁻¹ (or min⁻¹); τ in matching time units; concentrations in kmol/m³.

Worked examples

Example 1 (standard, parallel). A → R, rR = k1·CA with k1 = 1 min⁻¹; A → S, rS = k2·CA² with k2 = 1 m³/kmol·min. CA0 = 2 kmol/m³, CAf = 0.5 kmol/m³. Find CRf in a CSTR and in a PFR.

  1. φ = k1·CA/(k1·CA + k2·CA²) = 1/(1 + CA).
  2. CSTR: Φ = φ(0.5) = 1/1.5 = 0.667; CRf = 0.667 × (2 − 0.5) = 1.00 kmol/m³.
  3. PFR: Φ = [1/1.5]·∫(0.5→2) dCA/(1 + CA) = (1/1.5)·ln(3/1.5) = 0.693/1.5 = 0.462; CRf = 0.462 × 1.5 = 0.693 kmol/m³.

CSTR: 1.00 kmol/m³; PFR: 0.693 kmol/m³ — the desired reaction is of lower order, so low CA (mixed flow) wins.

Example 2 (GATE level, series). A → R → S, k1 = 0.2 min⁻¹, k2 = 0.1 min⁻¹, CA0 = 1 kmol/m³. Find the maximum CR and the corresponding τ for a PFR and a CSTR.

  1. PFR: τ_opt = ln(k2/k1)/(k2 − k1) = ln 0.5/(−0.1) = 6.93 min.
  2. CR,max/CA0 = (k1/k2)^[k2/(k2 − k1)] = 2^(−1) = 0.50 → CR,max = 0.50 kmol/m³.
  3. CSTR: τ_opt = 1/√(0.2 × 0.1) = 1/0.1414 = 7.07 min.
  4. CR,max/CA0 = 1/[√0.5 + 1]² = 1/(1.7071)² = 0.343 → CR,max = 0.343 kmol/m³.

PFR: 0.50 kmol/m³ at 6.93 min; CSTR: 0.343 kmol/m³ at 7.07 min.

Common mistakes

  • Confusing yield based on A fed with fractional yield based on A reacted.
  • Using φ at the exit for a PFR (that is only correct for a CSTR).
  • Thinking a higher temperature always improves selectivity — it favours the reaction with the higher E.
  • Running a series reaction past τ_opt in pursuit of conversion and destroying the intermediate.
  • For equal orders, trying to improve selectivity by changing reactor type.

For GATE CH

Typical questions: φ or Φ for parallel reactions in a CSTR or PFR, which reactor or concentration level favours the desired product, the effect of temperature from activation energies, and τ_opt and CR,max for first-order series reactions in a PFR or CSTR. Practise the series-reaction formulas and the integral of φ for simple orders.

Quick check

  1. Desired reaction order 2, undesired order 1: high or low CA?
  2. E1 = 80 kJ/mol (desired), E2 = 50 kJ/mol: high or low temperature?
  3. For A → R → S with k1 = k2 = k in a PFR, what is τ_opt?
  4. Which gives a higher CR,max for series reactions: PFR or CSTR?

Answers: 1. high CA (PFR or batch, no dilution); 2. high temperature; 3. τ_opt = 1/k (limit of the formula), with CR,max/CA0 = 1/e = 0.368; 4. PFR.

Interview questions

All Chemical Reaction Engineering interview questions

Try answering each one aloud before you open it.

  1. 1.What is the difference between parallel and series reactions in chemical engineering?Concept

    In chemical engineering, parallel reactions occur when multiple reactions happen simultaneously from the same reactants, leading to different products. Series reactions, on the other hand, occur sequentially, where the product of one reaction becomes the reactant for the next. Understanding these differences is crucial for optimizing reaction conditions and maximizing desired product yield.

  2. 2.Explain the concept of selectivity in chemical reactions.Concept

    Selectivity measures how much of the desired product is formed relative to the undesired product, commonly S(R/S) = moles of R formed / moles of S formed (or the ratio of their rates at a point). A related measure is the instantaneous fractional yield φ = rR/(−rA), moles of R formed per mole of A reacted. For parallel reactions rR/rS = (k1/k2)·CA^(a1 − a2), so selectivity is controlled by concentration level when the orders differ and by temperature or catalyst through k1/k2. High selectivity cuts raw-material use and separation cost.

  3. 3.What is yield in the context of chemical reactions, and how is it different from selectivity?Concept

    Yield in chemical reactions is the amount of desired product obtained from a reaction relative to the theoretical maximum amount possible. It is usually expressed as a percentage. Unlike selectivity, which focuses on the preference for a particular product, yield measures the efficiency of converting reactants into the desired product. Both are important for evaluating the performance of a chemical process.

  4. 4.Why is it important to consider both selectivity and yield in industrial chemical processes?Application

    Considering both selectivity and yield is crucial in industrial chemical processes because they impact the efficiency and cost-effectiveness of production. High selectivity ensures that the desired product is predominantly formed, reducing waste and by-products. High yield ensures that the maximum possible amount of the desired product is obtained from the reactants, optimizing resource use and minimizing costs.

  5. 5.What happens if a parallel reaction has low selectivity towards the desired product?Application

    If a parallel reaction has low selectivity towards the desired product, it means that a significant portion of the reactants is converted into undesired by-products. This can lead to increased waste, higher purification costs, and reduced overall efficiency of the process. It may also necessitate additional steps to separate and recycle reactants, increasing operational complexity and costs.

  6. 6.How can reaction conditions be optimized to improve selectivity in a series reaction?Application

    To improve selectivity in a series reaction, reaction conditions such as temperature, pressure, and catalyst choice can be optimized. For instance, controlling the temperature can favor the formation of the desired intermediate product over further reactions. Using a selective catalyst can also enhance the rate of the desired reaction pathway. Additionally, adjusting reactant concentrations and residence time can help achieve better selectivity.

  7. 7.In a series reaction A → R → S, how would you maximize the yield of intermediate R?Application

    Use a reactor without back-mixing — a batch reactor or PFR — because mixing fresh A with R that is already being destroyed lowers the maximum attainable R; a CSTR always gives a lower maximum. Then stop at the optimum time: for first-order steps in a PFR, τ_opt = ln(k2/k1)/(k2 − k1) and CR,max/CA0 = (k1/k2)^[k2/(k2 − k1)]. Beyond that time, conversion of A rises but R falls. Temperature or catalyst can shift k1/k2: a higher temperature favours whichever step has the larger activation energy.

  8. 8.Calculate the selectivity of a reaction where 80 moles of desired product D and 20 moles of undesired product U are formed.Numerical

    Selectivity is calculated as the ratio of the moles of desired product to the moles of undesired product. In this case, selectivity = 80 moles of D / 20 moles of U = 4. This means the reaction is four times more selective towards the desired product D compared to the undesired product U.

  9. 9.A reaction has a theoretical yield of 100 moles of product but only produces 75 moles. What is the percentage yield?Numerical

    Percentage yield is calculated as (actual yield / theoretical yield) × 100. Here, percentage yield = (75 moles / 100 moles) × 100 = 75%. This indicates that 75% of the theoretical maximum product was obtained in the reaction.

  10. 10.Explain how catalysts can affect the selectivity and yield of a chemical reaction.Application

    Catalysts can significantly affect both the selectivity and yield of a chemical reaction by providing an alternative reaction pathway with a lower activation energy. This can increase the rate of the desired reaction relative to undesired side reactions, improving selectivity. Catalysts can also enhance yield by increasing the overall reaction rate, allowing more reactants to be converted into products within a given time. The choice of catalyst is crucial for optimizing both selectivity and yield.

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