Parallel and Consecutive Reactions

Introduction

Many reactions do not lead to one product by one path. A reactant can choose parallel pathways, or a product of an early step can react again in a consecutive sequence. These networks change how concentrations evolve and how product ratios should be interpreted. A balanced overall equation for one pathway may hide competition.

Core explanation

For simple parallel first-order paths A → P with constant k₁ and A → Q with k₂, disappearance of A follows −d[A]/dt=(k₁+k₂)[A]. Thus [A]=[A]0e^(−(k₁+k₂)t). The instantaneous P formation rate is k₁[A] and Q formation rate k₂[A]. If products do not react further, their eventual fractions are k₁/(k₁+k₂) and k₂/(k₁+k₂). Selectivity is controlled by competing rate constants, not by a coefficient in one isolated equation.

For consecutive first-order steps A → I → P, I is first produced and then consumed. At t=0 with [I]=0, its formation begins while consumption is initially zero; [I] rises. Later, as A is depleted, I production slows while its conversion to P continues, so [I] can peak and then fall. This rise-and-fall is evidence of an intermediate, though another mechanism could mimic a similar pattern.

The timing of product collection matters. If I is the desired product, stopping the reaction near its concentration maximum may improve yield. Waiting indefinitely can convert I onward to P. In parallel chemistry, changing temperature or catalyst may alter the ratio k₁/k₂ and favor one product. A catalyst's selectivity can matter more than simply increasing total rate.

The simple formulas assume irreversible first-order steps and constant conditions. If pathways are reversible, product inhibition occurs, or one route has different concentration order, the ratios can vary with time and concentration. A measured product ratio alone does not prove constant branching without checking the network.

Mass balance is a useful check. In the parallel example, if A converts only to P or Q in one-to-one stoichiometry, [A]+[P]+[Q]=[A]0 in a fixed-volume closed system. In the consecutive case, [A]+[I]+[P]=[A]0. Deviations can indicate missing species, measurement error or volume change.

Reaction networks can combine both patterns, such as A → I, then I → P or I → Q. The steady-state approximation from the previous page can estimate branching from I when it remains low, while full time-course equations reveal its transient rise. Choosing a model depends on what data are available and what question is asked.

Step-by-step reasoning

1. Draw arrows among reactants, intermediates and products. 2. Identify parallel branches from a shared reactant versus a consecutive chain. 3. Write production and loss terms for each species. 4. For simple parallel first-order paths, add k values for A disappearance and form branching ratios. 5. Check mass balance and whether products undergo further reaction.

Visual explanation

Draw A splitting into P and Q with arrows k₁ and k₂. Beside it draw A → I → P, and sketch I rising then falling against time. Place a bracket under the split showing total disappearance constant k₁+k₂.

Real-world analogy

A passenger can choose one of two direct trains, a parallel choice, or take a connecting train through a transfer station, a consecutive route. The number waiting at the transfer station first rises then falls as arrivals and departures change.

Real-world example

In organic synthesis, a desired intermediate may react further if left in the reactor too long. Monitoring its concentration and stopping at a suitable time can improve isolated yield without changing the starting stoichiometry.

Why?

Why does A disappear with k₁+k₂ in simple parallel first-order paths? Each pathway independently removes A at rate kᵢ[A], so the two loss rates add to (k₁+k₂)[A].

Common misconception

“An intermediate's concentration must always rise until the reaction finishes.” In consecutive reactions it can peak and decline when its consumption overtakes its production.

Worked example

Let A form P with k₁=0.30 min⁻¹ and Q with k₂=0.10 min⁻¹. Overall A disappearance is first order with constant 0.40 min⁻¹. If all A eventually follows only these paths, fraction to P is 0.30/0.40=0.75 and to Q is 0.10/0.40=0.25. The P:Q ratio is 3:1. This conclusion assumes no secondary reaction of either product.

Quick check

1. What time-course shape can a consecutive intermediate show? Answer: It can rise, reach a maximum and then decline.

Exam focus

Draw the network before writing equations, add parallel loss rates, and state assumptions behind constant branching fractions. Identify a consecutive intermediate from production and consumption terms.

Advanced insight

Product selectivity can change with temperature if parallel pathways have different activation energies. Even if total A disappearance accelerates, the desired fraction may rise or fall, so optimization requires more than one rate constant.

Summary

Parallel paths compete for one reactant; consecutive paths pass through intermediates. In the simplest parallel first-order model, disappearance constants add and product fractions follow their relative k values. Consecutive intermediates can peak and fall.

Practice questions

1. If k₁=2k₂ for parallel first-order products P and Q, what fraction becomes P? Answer: 2/3, assuming no further product reaction. 2. What is A's disappearance constant for k₁=0.2 and k₂=0.3 s⁻¹? Answer: 0.5 s⁻¹. 3. Why might stopping a consecutive reaction early improve intermediate yield? Answer: It can be collected before its later conversion to the final product consumes it.