Parallel and Sequential Pathways

Introduction

Two products can arise by parallel routes from one reactant or by a sequence in which the first product reacts again. Both arrangements can yield the same final products but suggest different catalyst and reactor improvements. Time profiles, conversion dependence and product-addition experiments help identify which network is operating. Selectivity must be defined on a conserved-atom basis before comparing the alternatives.

Core explanation

For parallel first-order paths A → P and A → Q with rate constants kP and kQ, the initial rates are kP[A] and kQ[A]. If products do not react further and stoichiometry is one-to-one, instantaneous selectivity to P is kP/(kP + kQ), independent of [A] under this simple model. Changing temperature can alter the ratio if the two steps have different activation energies. A catalyst modification that lowers one branch's barrier can change selectivity without necessarily raising total conversion dramatically.

For sequential A → P → Q, P is an intermediate as well as a product. Early in a batch run, P forms and Q may lag; later P can fall as Q rises. Its maximum occurs when formation from A equals consumption to Q. At high conversion or long residence time, Q can dominate even if A initially makes only P. A catalyst that speeds P → Q may appear poorly selective at high conversion but not at low conversion. An ACS educational treatment of multistep kinetics uses time-resolved profiles to examine complex mechanisms.

Many real networks combine the two: A can form P and Q directly, while P can also convert to Q. Product ratio at one time cannot uniquely distinguish them. Varying residence time, initial product concentration and feed composition creates different predictions. If added P immediately makes Q in the absence of A, a sequential route is plausible. Isotopic labelling of P can test whether its atoms enter Q. Analytical calibration and closed mass balance are essential; an apparent lag may simply reflect detection limits.

Selectivity can depend on conversion for other reasons too. Product inhibition, surface coverage and temperature gradients may shift branch rates. Thus a falling P fraction with time suggests but does not prove a sequence. Use a kinetic model, independent product-addition tests and matched-conversion comparisons to evaluate the mechanism.

Step-by-step reasoning

1. Draw parallel and sequential candidate networks that explain the products. 2. Write simple flux expressions and predict early-time behaviour. 3. Measure A, P and Q over time or residence time with a conserved-atom balance. 4. Add P as a starting material or label it to test P → Q directly. 5. Fit and challenge each network under changed conditions.

Visual explanation

Draw two side-by-side reaction diagrams. In the parallel one, A forks to P and Q. In the sequential one, A leads to P then Q. Beneath each draw P and Q versus time: parallel products rise together from the start, while sequential Q lags and P may peak. Include error bars or a detection threshold line to caution against overinterpreting an apparent delay.

Real-world analogy

Two buses departing one station toward different towns are parallel routes. A bus going through Town P on its way to Town Q is sequential. Counting passengers at Q after a day does not reveal which arrangement was used; observations at intermediate times and starting a bus at P help. Chemical paths differ because reactant concentrations and rate constants change continuously.

Real-world example

In partial oxidation, desired aldehyde P may form from alcohol A and then oxidise further to acid Q. If Q is mainly sequential, removing P quickly or reducing contact time may improve aldehyde yield. If Q forms directly by a competing route, changing catalyst site chemistry may be more important. The network diagnosis guides a different intervention for each case.

Why?

Why compare catalysts at matched conversion? A sequential pathway automatically makes more Q at longer residence time, even if the underlying elementary selectivity is unchanged. Matching conversion reduces the confounding effect of reaction extent when asking whether catalysts differ in branching.

Common misconception

“Two final products mean two parallel paths” ignores consecutive chemistry. “A rising Q curve proves direct A → Q” is false. “A P peak proves only a sequence” can be complicated by reversible or changing conditions. “A selectivity ratio from one time point reveals every rate constant” ignores structural non-identifiability.

Worked example

For parallel A → P and A → Q, let kP = 0.3 min⁻¹ and kQ = 0.1 min⁻¹. Instantaneous branch selectivity to P is 0.3/(0.3 + 0.1) = 75%. In a different model A → P → Q with k₁ = 0.4 min⁻¹ and k₂ = 0.1 min⁻¹, at an instant [A] = 0.5 M and [P] = 0.2 M, P forms at 0.4 × 0.5 = 0.20 M/min and is consumed at 0.1 × 0.2 = 0.02 M/min; it rises at 0.18 M/min. Later, when A is nearly depleted, P can fall and Q continue to grow. These two models can both initially favour P but predict distinct later profiles.

Quick check

1. What time-profile feature often suggests a sequential A → P → Q pathway? Answer: P rises and may peak before declining, while Q appears with a lag and continues rising.

Exam focus

Sketch parallel and sequential networks and their qualitative product curves. Calculate a branch ratio for simple first-order parallel paths. Propose a product-addition or isotope experiment to distinguish direct from secondary Q formation.

Advanced insight

At low conversion, initial-rate selectivity can isolate branching before secondary product consumption becomes substantial. Yet if intermediate lifetimes are extremely short, even early sampling may miss them. Rapid-mixing or transient spectroscopy may then be required; absence of a detected intermediate does not prove a direct route.

Summary

Parallel paths compete from a common species, while sequential paths transform an earlier product into a later one. Time, conversion and perturbation data distinguish them more reliably than a single final product ratio.

Practice questions

1. For kP = 2 s⁻¹ and kQ = 3 s⁻¹ in a simple parallel model, what is P branch selectivity? Answer: 2/(2 + 3) = 40%. 2. Why can Q lag in A → P → Q? Answer: P must first accumulate before the second step can make much Q. 3. What does adding pure P test? Answer: Whether P can directly convert to Q under the reaction conditions. 4. Does selectivity falling with conversion prove a sequential pathway by itself? Answer: No. Coverage, inhibition or transport changes can also alter selectivity.