Parallel Reaction Networks

Introduction

A reactant often has more than one way to react. An alkyl halide may undergo substitution or elimination; a radical may add to two different sites; a fuel may burn fully or partially. When pathways compete, chemists care not only about how fast the reactant disappears but also about which products form and in what proportion. Parallel-reaction kinetics provides the tools to predict and control that selectivity.

Core explanation

The scheme. A reactant A forms B with rate constant k₁ and C with rate constant k₂, both first order and irreversible:

A → B (k₁); A → C (k₂)

The rate equations are

−d[A]/dt = (k₁ + k₂)[A] d[B]/dt = k₁[A] d[C]/dt = k₂[A]

Decay of A. Because both paths depend on [A] to the same power, A decays exponentially with the sum of the constants:

[A] = [A]₀ exp[−(k₁ + k₂)t]

The measured half-life of A is ln 2/(k₁ + k₂); on its own, it cannot reveal the individual constants.

Product ratio. Dividing the two product rates gives d[B]/d[C] = k₁/k₂ at all times. Integrating from zero:

[B]/[C] = k₁/k₂

throughout the reaction. This constant ratio is the hallmark of competing parallel paths of equal order. The branching ratio for B is k₁/(k₁ + k₂).

Separating the constants. Measuring the overall decay gives k₁ + k₂; measuring the product ratio gives k₁/k₂. Together they identify both constants, a good illustration that different observables are needed to identify all the parameters of a network.

Unequal orders. If A → B is first order but A + R → C is second order, then [B]/[C] is no longer constant: the ratio depends on [R], so selectivity can be steered by concentration. Diluting favours the lower-order path; concentrating favours the higher-order path.

Temperature and selectivity. Each constant follows Arrhenius behaviour, so

k₁/k₂ = (A₁/A₂) exp[−(Eₐ₁ − Eₐ₂)/RT]

Raising temperature always favours the pathway with the higher activation energy, because its rate constant grows faster. Lowering temperature favours the lower-barrier path. This is the principle behind running selective reactions cold.

Kinetic versus thermodynamic control. The analysis above assumes irreversible steps, so products do not interconvert and the ratio reflects rates. If the steps become reversible, for example at higher temperature, the mixture drifts towards the most stable product, and selectivity is determined by equilibrium constants instead.

Formulae

[A] = [A]₀e^(−(k₁+k₂)t); [B]/[C] = k₁/k₂; branching ratio = k₁/(k₁ + k₂); k₁/k₂ = (A₁/A₂)exp[−(Eₐ₁ − Eₐ₂)/RT].

Step-by-step reasoning

1. Write the rate of loss of A as the sum of all pathway rates. 2. Integrate to obtain the decay of A with the summed rate constant. 3. Divide the product rates to show the ratio equals k₁/k₂. 4. Combine the overall decay constant with the product ratio to find each k. 5. Use the difference in activation energies to predict how temperature shifts selectivity.

Visual explanation

Draw A decaying exponentially, with B and C rising towards two different plateaus. At every moment the heights of B and C stand in the same ratio. A branching diagram with A splitting into two arrows, each thickness proportional to its rate constant, conveys the same idea.

Real-world analogy

Water from a tank draining through two pipes of different widths empties at a rate set by both pipes together, but the two buckets beneath fill in a fixed ratio set by the pipe sizes.

Real-world example

In industrial ethylene oxide production, ethene can be partially oxidised to the desired epoxide or fully oxidised to carbon dioxide and water. Silver catalysts, additives and carefully controlled temperature are used to favour the epoxide path; too high a temperature tips selectivity towards the more strongly activated combustion route.

Why?

Why does the higher-activation-energy path gain when heated? The fractional increase in k per kelvin is Eₐ/RT². A pathway with a larger barrier has a larger temperature coefficient, so its share of the total rate rises with temperature.

Common misconception

"The faster pathway always dominates at every temperature." The pathway that is faster at low temperature may be overtaken at high temperature if its competitor has a higher activation energy and a larger pre-exponential factor.

Worked example

Question: A decays by two first-order paths. The overall half-life is 50 s, and the final product ratio [B]/[C] = 3.0. Find k₁ and k₂.

Reasoning: k₁ + k₂ = 0.693/50 = 0.01386 s⁻¹. k₁ = 3.0k₂, so 4.0k₂ = 0.01386, giving k₂ = 3.47 × 10⁻³ s⁻¹ and k₁ = 1.04 × 10⁻² s⁻¹.

Answer: k₁ ≈ 1.0 × 10⁻² s⁻¹; k₂ ≈ 3.5 × 10⁻³ s⁻¹.

Quick check

1. For two competing first-order paths from A, what rate constant governs the overall disappearance of A? Answer: The sum of the two rate constants, k₁ + k₂, governs the exponential decay of A.

Exam focus

Show that the product ratio equals the ratio of rate constants when both paths share the same order. Explain temperature effects using activation energies explicitly. Distinguish kinetic control (irreversible, rate-based ratio) from thermodynamic control (equilibrium-based ratio).

Advanced insight

Enantioselective catalysis is a parallel-reaction problem: two diastereomeric transition states compete. A difference in ΔG‡ of just 11.4 kJ mol⁻¹ at 298 K gives a rate ratio near 100:1, or about 98% enantiomeric excess. Small energy differences in competing transition states therefore translate into large practical selectivities.

Summary

In parallel first-order reactions, the reactant decays with the summed rate constant, while products form in the constant ratio k₁/k₂. Combining overall decay with product ratios identifies each constant. Unequal orders let concentration steer selectivity, and temperature favours the path with the higher activation energy. Irreversibility is required for this kinetic control.

Practice questions

1. For A → B (k₁ = 0.020 s⁻¹) and A → C (k₂ = 0.005 s⁻¹), find the branching ratio for B. Answer: 0.020/0.025 = 0.80, so 80% of A forms B. 2. Pathway 1 has Eₐ = 60 kJ mol⁻¹ and pathway 2 has Eₐ = 90 kJ mol⁻¹. Which is favoured by heating? Answer: Pathway 2, because the higher activation energy gives the larger increase in rate constant with temperature. 3. Why does measuring only the decay of A fail to give k₁ and k₂ separately? Answer: The decay depends only on the sum k₁ + k₂, so infinitely many pairs of constants fit the same curve. 4. How can dilution change selectivity between a first-order and a second-order pathway? Answer: Dilution lowers the second-order rate more than the first-order rate, so it favours the first-order product.