Parallel (Competing) Reactions

Mix aniline with a nitrating agent and you get three products at once: ortho, meta, and para-nitroaniline, all forming simultaneously from the same starting material. This is a parallel reaction network, and its outcome is decided the instant molecules choose which transition state to cross. Chemists rarely have the luxury of one reactant giving one product; branching pathways are the norm in synthesis, biochemistry, and atmospheric chemistry alike. This article builds directly on the integrated first-order rate law. It shows how competing unimolecular or pseudo-first-order channels share a common decay of starting material while splitting into distinct products. You will learn the branching-ratio formula. You will see how it connects to the Boltzmann distribution when products can interconvert. Finally, you will see why "kinetic control" and "thermodynamic control" describe genuinely different experimental regimes. The ideas here set up catalysis, where a catalyst's entire job is to open a new, faster parallel pathway to a desired product without touching the other channels.

Key takeaways

In parallel reactions, one reactant decays through two or more simultaneous first-order (or pseudo-first-order) pathways, each with its own rate constant.
The total decay of reactant is governed by the sum of rate constants, $k_{obs} = k_1 + k_2 + \cdots$ , so the reactant still disappears with clean first-order kinetics.
The branching ratio (product distribution) depends only on the ratio of rate constants, not on time or concentration.
Kinetic control gives the product of the fastest pathway; thermodynamic control gives the most stable product, reached only if the reaction is reversible and allowed to equilibrate.
Temperature, catalysts, and reaction time are the experimental levers that shift a system between kinetic and thermodynamic control.
Key idea: competing pathways share one exponential reactant decay but split into products in a fixed ratio set by relative rate constants, unless reversibility lets the system relax toward the thermodynamic minimum.

The core insight

When a single reactant can react by several independent first-order routes, all channels drain the same pool of starting material at once, so the reactant vanishes with an effective rate constant equal to the sum of the individual ones, while the fraction of product formed via each route is fixed forever by the ratio of that channel's rate constant to the total — unless the reaction is reversible and thermodynamics is given time to take over.

The physical picture

Picture a reactant $A$ sitting at the top of an energy landscape with two valleys below it, reached over two different transition states. Molecules of $A$ do not "decide" collectively; each individual molecule independently samples its surroundings and, when it reacts, crosses whichever barrier it happens to surmount first. Statistically, this means A is consumed by two simultaneous first-order processes:

A \xrightarrow{k_1} B, \qquad A \xrightarrow{k_2} C

Because both processes draw from the same reservoir of $A$ , the loss of $A$ is exponential with a rate constant equal to the sum $k_1+k_2$ — faster than either channel alone would give. Meanwhile $B$ and $C$ accumulate in a fixed ratio determined by how often each barrier is crossed relative to the other. This ratio is called the branching ratio, and it is the central experimental quantity of parallel-reaction kinetics. If the products can revert to $A$ (or interconvert with each other), the system is no longer locked into that early ratio. Given enough time, it will drift toward the Boltzmann-weighted equilibrium distribution set by the relative Gibbs energies of $B$ and $C$ — this is the border between kinetic control (fast, often irreversible, barrier-height-determined) and thermodynamic control (slow, reversible, stability-determined).

Figure

Energy diagram for two competing pathways from a common reactant A

Figure

Branching ratio versus rate-constant ratio, r = k2/k1

The core relations

-\frac{d[A]}{dt} = (k_1 + k_2)[A] = k_{obs}[A]

where: $k_1, k_2$ are the individual first-order (or pseudo-first-order) rate constants for the two channels, and $k_{obs}$ is the observed overall decay constant.

[A]_t = [A]_0 \, e^{-(k_1+k_2)t}

where: $[A]_0$ is the initial concentration, $t$ is elapsed time.

\frac{[B]_t}{[C]_t} = \frac{k_1}{k_2} \quad \text{(constant at all times, irreversible case)}

where: the product ratio depends only on the rate constants, never on time or on $[A]_0$ .

[B]_\infty = \frac{k_1}{k_1+k_2}\big([A]_0-[A]_t\big), \qquad [C]_\infty = \frac{k_2}{k_1+k_2}\big([A]_0-[A]_t\big)

where: each product's yield is its branching fraction times the total amount of $A$ consumed.

\frac{[C]_{eq}}{[B]_{eq}} = e^{-\Delta G^\circ/RT} = K_{eq}

where: under thermodynamic control (reversible network, long time), the product ratio is set by the standard Gibbs energy difference $\Delta G^\circ = G^\circ_C - G^\circ_B$ , not by $k_1/k_2$ .

Derivation ABranching ratio for two parallel first-order channels

Goal. Starting from the coupled rate laws for $A \xrightarrow{k_1} B$ and $A \xrightarrow{k_2} C$ , derive the time-independent branching ratio $[B]/[C]$ and the explicit product concentrations.

Step 1 — Write the loss law for A. Both channels consume $A$ independently, so their rates add:

-\frac{d[A]}{dt} = k_1[A] + k_2[A] = (k_1+k_2)[A]

This is a single first-order differential equation in $[A]$ alone, exactly the form treated in the integrated first-order rate law, with effective constant $k_{obs}=k_1+k_2$ .

Step 2 — Integrate. Separating variables and integrating from $0$ to $t$ ,

\int_{[A]_0}^{[A]_t} \frac{d[A]}{[A]} = -(k_1+k_2)\int_0^t dt'

gives

[A]_t = [A]_0 e^{-(k_1+k_2)t}

This confirms that even though two reactions are occurring, $A$ 's disappearance still looks like ordinary single-exponential first-order decay — only the rate constant is a sum.

Step 3 — Write the production laws for B and C. Each product forms in direct proportion to the instantaneous concentration of $A$ :

\frac{d[B]}{dt} = k_1[A]_0 e^{-(k_1+k_2)t}, \qquad \frac{d[C]}{dt} = k_2[A]_0 e^{-(k_1+k_2)t}

Step 4 — Integrate each production law. Using $\int_0^t e^{-kt'}dt' = (1-e^{-kt})/k$ with $k=k_1+k_2$ ,

[B]_t = \frac{k_1[A]_0}{k_1+k_2}\Big(1-e^{-(k_1+k_2)t}\Big), \qquad [C]_t = \frac{k_2[A]_0}{k_1+k_2}\Big(1-e^{-(k_1+k_2)t}\Big)

Both expressions share the identical time-dependent factor $\big(1-e^{-(k_1+k_2)t}\big)$ ; only the prefactor differs between the two products.

Step 5 — Take the ratio. Dividing the two expressions from Step 4, the shared time-dependent factor cancels exactly:

\frac{[B]_t}{[C]_t} = \frac{k_1}{k_2}

for every value of $t > 0$ , not just at completion.

Result.

\boxed{\dfrac{[B]}{[C]} = \dfrac{k_1}{k_2} \ \text{ at all times, and each product's yield is } \dfrac{k_i}{k_1+k_2} \text{ of the total } A \text{ consumed}}

Worked examples

Example 1 — Foundational: branching ratio from rate constants. A compound $A$ decomposes by two independent, irreversible first-order pathways at 298 K:

A \xrightarrow{k_1 = 2.4\times10^{-3}\,\text{s}^{-1}} B, \qquad A \xrightarrow{k_2 = 8.0\times10^{-4}\,\text{s}^{-1}} C

Starting from $[A]_0 = 0.500\ \text{mol L}^{-1}$ , find (a) the observed decay constant, (b) $[A]$ after 300 s, and (c) the concentrations of $B$ and $C$ at that time. (a) The reactant decays with the sum of the rate constants:

k_{obs} = k_1+k_2 = 2.4\times10^{-3} + 8.0\times10^{-4} = 3.2\times10^{-3}\ \text{s}^{-1}

(b) After $t = 300$ s:

[A]_t = 0.500\, e^{-(3.2\times10^{-3})(300)} = 0.500\, e^{-0.96} = 0.500 \times 0.3829 = 0.1915\ \text{mol L}^{-1}

So $[A]$ consumed is $0.500 - 0.1915 = 0.3085\ \text{mol L}^{-1}$ . (c) The branching fractions are:

f_B = \frac{k_1}{k_1+k_2} = \frac{2.4\times10^{-3}}{3.2\times10^{-3}} = 0.750, \qquad f_C = \frac{k_2}{k_1+k_2} = \frac{8.0\times10^{-4}}{3.2\times10^{-3}} = 0.250

[B]_t = 0.750 \times 0.3085 = 0.2314\ \text{mol L}^{-1}, \qquad [C]_t = 0.250 \times 0.3085 = 0.0771\ \text{mol L}^{-1}

Sanity check: $[A]_t+[B]_t+[C]_t = 0.1915+0.2314+0.0771 = 0.5000\ \text{mol L}^{-1}$ , matching $[A]_0$ exactly. The ratio $[B]/[C] = 0.2314/0.0771 = 3.00$ , matching $k_1/k_2 = 2.4\times10^{-3}/8.0\times10^{-4} = 3.00$ as required.

Example 2 — Applied: kinetic versus thermodynamic control with reversibility. Electrophilic addition of HBr to 1,3-butadiene at low temperature (0 °C, kinetic conditions, short reaction times) gives predominantly the 1,2-adduct, while prolonged heating at 40 °C allows equilibration to the more stable 1,4-adduct. Suppose separate rate studies give, for the two initial (irreversible-on-the-timescale-measured) addition steps at 0 °C:

k_{1,2} = 6.0\times10^{-2}\ \text{s}^{-1}, \qquad k_{1,4} = 1.5\times10^{-2}\ \text{s}^{-1}

(a) Compute the kinetic-control branching ratio. (b) Given that the 1,4-adduct is more stable than the 1,2-adduct by $\Delta G^\circ = -6.5\ \text{kJ mol}^{-1}$ (1,4 lower in energy), compute the thermodynamic-control ratio at 313 K (40 °C) and compare. (a) Kinetic branching ratio:

\frac{[1,2]}{[1,4]}_{kinetic} = \frac{k_{1,2}}{k_{1,4}} = \frac{6.0\times10^{-2}}{1.5\times10^{-2}} = 4.0

So under kinetic control the 1,2-adduct dominates, about 80% of product ( $f_{1,2} = 4.0/5.0 = 0.80$ ), consistent with its lower activation barrier. (b) Thermodynamic branching ratio uses $K_{eq} = e^{-\Delta G^\circ/RT}$ with $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ , $T = 313\ \text{K}$ :

\frac{\Delta G^\circ}{RT} = \frac{-6500\ \text{J mol}^{-1}}{8.314 \times 313\ \text{J mol}^{-1}} = \frac{-6500}{2602.3} = -2.498

K_{eq} = \frac{[1,4]}{[1,2]} = e^{-(-2.498)} = e^{2.498} = 12.2

So at equilibrium the 1,4-adduct outnumbers the 1,2-adduct roughly 12:1 — a complete reversal from the kinetic result. In terms of mole fraction, $f_{1,4} = 12.2/13.2 = 0.924$ , i.e. about 92% 1,4-adduct at thermodynamic equilibrium, versus only 20% under kinetic control. Sanity check: the signs make sense — the lower-barrier product (1,2) wins kinetically; the lower-energy product (1,4) wins thermodynamically, and since $|\Delta G^\circ|$ (6.5 kJ/mol) is comparable to $RT$ (2.6 kJ/mol at 313 K) the equilibrium ratio (12.2) is large but not absurd, as expected from $e^{2.5}\approx 12$ .

Limiting cases & common traps

Situation	Behavior	Common trap
$k_1 \gg k_2$	Product $B$ dominates almost completely; $C$ is a trace side-product	Forgetting $C$ still forms measurably if $[A]_0$ is large or reaction runs long
$k_1 \approx k_2$	Roughly 50:50 product mixture	Assuming equal activation energies give equal rates; only $k$ , not $E_a$ alone, sets the ratio (pre-exponential factors matter too)
Reversible network, short time	Kinetic product dominates	Reporting the early-time ratio as if it were the true equilibrium composition
Reversible network, long time / high T	Thermodynamic product dominates, set by $\Delta G^\circ$	Assuming a catalyst that speeds up one path also changes the equilibrium ratio — it does not, since catalysts affect $k$ 's, not $\Delta G^\circ$
Three or more competing channels	$k_{obs}=\sum_i k_i$ ; each branching fraction is $k_i/\sum_i k_i$	Treating channels as competing only pairwise and missing normalization by the full sum

The chemist behind it

The kinetic/thermodynamic control framework for competing pathways is most famously associated with studies of diene addition reactions in the 1920s-30s, where chemists noticed that the product distribution from HBr or maleic anhydride addition to dienes changed dramatically with temperature and reaction time. This observation — that the same starting materials could give different major products depending on conditions alone — forced a conceptual split between "which product forms fastest" and "which product is most stable," a distinction that is now foundational to how chemists plan and interpret every branching synthetic step.

Origins · 1 minKurt Alder and how this was discovered

Kurt Alder

1902-1958 · German chemist whose work with Otto Diels on diene addition reactions provided some of the clearest early evidence distinguishing kinetic from thermodynamic control of product distribution, later recognized with the 1950 Nobel Prize in Chemistry

:::quiz[In an irreversible two-channel parallel reaction, what is the ratio [B]/[C] at any time t greater than zero?]

It equals k1/k2 and is independent of t ✓
It equals (k1/k2) times e to the minus (k1+k2)t
It increases linearly with t
It equals k1t/k2t and depends on t

:::quiz[If a catalyst selectively increases k1 for pathway A to B, what happens to the thermodynamic equilibrium ratio [B]eq/[C]eq, assuming the reaction is reversible?]

It increases proportionally to the new k1
It stays exactly the same, since catalysts change rates, not standard Gibbs energies ✓
It decreases because faster reactions favor less stable products
It becomes undefined because the system is no longer at equilibrium

Ask the tutor: How would the branching-ratio formula change if the two competing pathways were second order instead of first order?

Why does the reactant in a parallel reaction still decay as a clean single exponential?

Because each channel independently removes molecules of $A$ at a rate proportional to $[A]$ , the total rate of loss is still first order overall — just with an effective constant equal to the sum $k_1+k_2+\cdots$ . Mathematically, adding several terms each proportional to $[A]$ gives another term proportional to $[A]$ . Physically, molecules of $A$ have no memory of which pathway they will eventually take; they all decay from the same pool, so the population as a whole shrinks exponentially, exactly as in a single-channel first-order process, but faster.

Does the branching ratio ever change as the reaction proceeds, for a truly irreversible parallel reaction?

No. For strictly irreversible parallel first-order channels, the ratio $[B]/[C] = k_1/k_2$ holds at every instant, from the very start of the reaction to completion. This is because both products are produced from the same decaying pool of $A$ with the same time-dependence, so that time-dependent factor cancels exactly when you take the ratio. This makes branching ratios convenient experimentally — you can measure the product ratio at almost any conversion and get the same rate-constant ratio.

How do you experimentally tell whether you have kinetic or thermodynamic control?

Run the reaction under two different sets of conditions: short time and low temperature versus long time and higher temperature (or add a catalyst that accelerates equilibration). If the product ratio is the same in both cases, thermodynamic and kinetic control happen to coincide, or the system was already at equilibrium. If the ratio shifts — especially if it inverts, as in the 1,2- versus 1,4-addition example — you have clear evidence that the short-time product is kinetic and the long-time product is thermodynamic. Quenching the reaction rapidly at low temperature is the classic way to "freeze in" the kinetic product.

Can branching ratios depend on temperature even without invoking thermodynamic control?

Yes. Each rate constant $k_i$ typically follows the Arrhenius law, $k_i = A_i e^{-E_{a,i}/RT}$ , so the ratio $k_1/k_2$ depends on temperature unless the two channels happen to have identical activation energies. Raising the temperature disproportionately speeds up the channel with the higher activation energy, since its rate constant is more temperature-sensitive. This can shift the kinetic branching ratio with temperature, quite separately from any approach toward thermodynamic equilibrium.

How does this connect to catalysis, covered next?

A catalyst is, in essence, a way of opening an entirely new parallel pathway — or dramatically lowering the barrier of an existing one — without altering the thermodynamics of the overall reaction. Everything derived here for branching ratios and rate-constant sums applies directly: a catalyst raises one $k_i$ selectively, changing the kinetic product distribution, while leaving $\Delta G^\circ$ (and hence the thermodynamic product ratio) completely untouched. Recognizing this distinction is essential to understanding selective catalysis.