Parallel reactions

Not every reactant has just one fate. Split a beam of molecules $A$ and some of them react one way, some another — a substrate giving a mixture of substitution and elimination products, or an excited atom relaxing by two competing routes. Whenever a single species is consumed by two or more independent pathways running at the same time, chemists call it a parallel (or competitive) reaction, and it is the third of the three canonical multi-pathway kinetic schemes in CY11001, alongside reversible and consecutive reactions. The good news is that the mathematics stays completely within first-order kinetics you already know — no new integration technique is needed, only the bookkeeping of adding up every pathway that drains $A$ . What falls out is a strikingly clean pair of results: the reactant vanishes exactly as if a single, larger rate constant were acting on it, while the ratio of the products it leaves behind is frozen for all time, fixed the instant the reaction starts. That frozen ratio is the kinetic seed of "selectivity" — the idea, used constantly in organic and inorganic mechanism, that whichever pathway is intrinsically faster simply wins, in a proportion you can read off from a single product analysis taken at any moment.

The core insight

In parallel first-order reactions from a common reactant, the reactant decays with rate constant $k_1+k_2$ while the product ratio stays fixed forever at $k_1/k_2$ .

The core idea

Setting up the scheme. Consider a reactant $A$ that can react by two independent, simultaneous first-order pathways, one to $B$ and one to $C$ :

B \xleftarrow{k_1} A \xrightarrow{k_2} C

Both pathways draw from the same pool of $A$ at the same time, so the total rate at which $A$ disappears is the sum of the two individual rates:

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

where: $[A]$ is the instantaneous concentration of the reactant and $k_1,k_2$ are the rate constants of the two competing first-order steps.

The reactant decays as one combined process. This differential equation is identical in form to ordinary first-order decay, so it integrates the same way, but with an effective (observed) rate constant that is the sum of the individual ones:

[A] = a_0\,e^{-(k_1+k_2)t}, \qquad k_{obs}=k_1+k_2

where $a_0=[A]_0$ is the initial concentration. A kinetics experiment that only tracks how much $A$ is left — a UV absorbance trace, say — cannot by itself tell you $k_1$ and $k_2$ separately; it only ever hands you their sum.

The products grow, but in fixed proportion. Because $d[B]/dt=k_1[A]$ and $d[C]/dt=k_2[A]$ , and both share the same exponentially decaying $[A]$ , integrating from $[B]=[C]=0$ gives

[B] = \frac{k_1}{k_1+k_2}\,a_0\left(1-e^{-(k_1+k_2)t}\right), \qquad [C] = \frac{k_2}{k_1+k_2}\,a_0\left(1-e^{-(k_1+k_2)t}\right)

Dividing one by the other, the entire time-dependent bracket cancels — leaving a ratio that never changes. The figure below makes the shape of this concrete: $A$ falls away on a single combined exponential, while $B$ and $C$ both rise as mirror-image "growth" curves that always stay in the same $3:1$ proportion to each other at every instant, even though each is individually still climbing.

Figure

Concentration versus time for a parallel reaction showing A decaying on one combined exponential while B and C both rise with B staying three times C at every instant

Common traps

Using only $k_1$ (or $k_2$ ) in the decay of $[A]$ . The reactant always decays with the sum $k_1+k_2$ because every pathway drains the same pool simultaneously.
Thinking the product ratio drifts as the reaction proceeds. For same-order parallel first-order steps, $[B]/[C]=k_1/k_2$ is a true constant of the motion, valid from $t=0$ all the way to completion.
Mixing up parallel with consecutive schemes. In a parallel scheme both products form directly and independently from $A$ ; nothing here converts $B$ into $C$ or vice versa. Consecutive reactions need a completely different pair of differential equations.
Applying the simple ratio to mismatched orders. If the two branches are not both first order in $A$ (say one is second order), $[B]/[C]$ will depend on $[A]$ and hence change with time — the clean constant-ratio result is specific to same-order parallel steps.

Derivation AWhy the product ratio never changes with time

The goal is to show, starting only from the two coupled first-order rate laws for a reactant $A$ splitting into products $B$ and $C$ by independent first-order pathways with rate constants $k_1$ and $k_2$ , that the ratio $[B]/[C]$ is exactly $k_1/k_2$ at every instant of the reaction, not merely once it has gone to completion — a result that underlies every selectivity argument built on parallel kinetics.

Step 1. Write the rate of formation of each product. Since $B$ forms only from $A$ via the $k_1$ pathway and $C$ forms only from $A$ via the $k_2$ pathway, $\dfrac{d[B]}{dt}=k_1[A]$ and $\dfrac{d[C]}{dt}=k_2[A]$ , with the same $[A](t)$ appearing in both, since it is the same reactant feeding both branches.

Step 2. Substitute the already-known decay of $A$ . From the combined first-order decay derived above, $[A]=a_0e^{-(k_1+k_2)t}$ . Plugging this into both product rate laws gives $\dfrac{d[B]}{dt}=k_1a_0e^{-(k_1+k_2)t}$ and $\dfrac{d[C]}{dt}=k_2a_0e^{-(k_1+k_2)t}$ — two expressions that differ only by the constant multiplying them.

Step 3. Integrate both from $t=0$ , where $[B]=[C]=0$ . Each integral is a standard exponential integral, giving $[B]=\dfrac{k_1}{k_1+k_2}a_0\left(1-e^{-(k_1+k_2)t}\right)$ and $[C]=\dfrac{k_2}{k_1+k_2}a_0\left(1-e^{-(k_1+k_2)t}\right)$ ; both share the identical bracket $\left(1-e^{-(k_1+k_2)t}\right)$ , only the prefactor differs.

Step 4. Divide $[B]$ by $[C]$ . The shared factors $a_0$ and $\left(1-e^{-(k_1+k_2)t}\right)$ cancel completely, leaving

\boxed{\dfrac{[B]}{[C]} = \dfrac{k_1}{k_2}}

Because the time-dependent part cancelled exactly, this ratio holds at $t=1$ second just as much as at $t=\infty$ — it is a genuine constant, not a limiting value approached only late in the reaction. This is the mathematical basis for reading branching ratios or selectivities off a single product-mixture snapshot taken at any convenient conversion.

Worked example

A compound $A$ decomposes by two parallel first-order pathways to give $B$ and $C$ . Overall, $A$ decays with an observed half-life of $34.7\ \text{min}$ , and analysis of the product mixture at any stage shows $[B]:[C] = 3:1$ . Find $k_1$ (for $A\to B$ ) and $k_2$ (for $A\to C$ ), and then predict how much $B$ and $C$ have formed after one half-life if $a_0 = 0.60\ \text{mol L}^{-1}$ .

Step 1. Get the overall (observed) rate constant from the half-life, since the reactant still obeys ordinary first-order decay with $k_{obs}=k_1+k_2$ :

k_{obs}=k_1+k_2=\frac{\ln 2}{t_{1/2}}=\frac{0.693}{34.7\ \text{min}}=0.01997\ \text{min}^{-1}\approx0.0200\ \text{min}^{-1}

Step 2. Bring in the product-ratio result derived above: $[B]/[C]=k_1/k_2=3$ , so $k_1=3k_2$ .

Step 3. Solve the two equations simultaneously. Substituting $k_1=3k_2$ into $k_1+k_2=0.0200$ gives $4k_2=0.0200$ , so $k_2=0.00500\ \text{min}^{-1}$ and $k_1=3(0.00500)=0.0150\ \text{min}^{-1}$ . Check: $k_1+k_2=0.0200\ \text{min}^{-1}$ ✓ and $k_1/k_2=3$ ✓.

Step 4. Predict $[B]$ and $[C]$ after one half-life ( $t=34.7\ \text{min}$ , so $1-e^{-k_{obs}t}=1-\tfrac12=0.500$ ):

[B]=\frac{k_1}{k_{obs}}a_0(0.500)=\frac{0.0150}{0.0200}\times0.60\times0.500=0.225\ \text{mol L}^{-1}

[C]=\frac{k_2}{k_{obs}}a_0(0.500)=\frac{0.00500}{0.0200}\times0.60\times0.500=0.0750\ \text{mol L}^{-1}

Final answer: $k_1=0.0150\ \text{min}^{-1}$ , $k_2=0.00500\ \text{min}^{-1}$ , and after one half-life $[B]=0.225\ \text{mol L}^{-1}$ , $[C]=0.0750\ \text{mol L}^{-1}$ , with $[B]/[C]=3$ exactly, as it must be at every time.

Competitive stretch

The same logic runs the other way too. For an aryl halide $A$ at $a_0=0.40\ \text{mol L}^{-1}$ with $k_1=4.0\times10^{-3}\ \text{s}^{-1}$ (substitution, $B$ ) and $k_2=1.0\times10^{-3}\ \text{s}^{-1}$ (elimination, $C$ ), after $t=200\ \text{s}$ : $k_{obs}=5.0\times10^{-3}\ \text{s}^{-1}$ , fraction reacted $=1-e^{-1.0}=0.6321$ , so $[B]=\tfrac{4.0}{5.0}\times0.40\times0.6321=0.2023\ \text{mol L}^{-1}$ and $[C]=\tfrac{1.0}{5.0}\times0.40\times0.6321=0.05056\ \text{mol L}^{-1}$ , giving $[B]/[C]=4.00=k_1/k_2$ — the same $4:1$ ratio you would find at $t=50\ \text{s}$ or $t=1000\ \text{s}$ .

Ask the tutor: want to see how this framework extends to three or more competing pathways, or connects to selectivity arguments in organic mechanisms?

Why does the product ratio not depend on how far the reaction has gone?

Because both $[B]$ and $[C]$ grow through the identical bracket $\left(1-e^{-(k_1+k_2)t}\right)$ , which comes from the shared, time-dependent decay of the one reactant feeding both of them. That bracket is a common factor in both product expressions, so when you take the ratio $[B]/[C]$ it cancels exactly, leaving only the constant $k_1/k_2$ behind. This is a genuine algebraic cancellation, not an approximation valid at large $t$ — it holds equally well one second into the reaction and at practical completion, which is precisely why chemists can measure a branching ratio from a single snapshot of the product mixture taken at any convenient time.

Why can't I get k1 and k2 separately just from watching A disappear?

Because the disappearance of $A$ only ever reports the sum $k_1+k_2$ , since every molecule of $A$ that reacts — regardless of which product it heads toward — is removed from the pool being tracked. A UV trace, a titration, or any measurement that only tells you how much $A$ remains cannot distinguish which pathway removed it. To split the sum into its two parts you need a second, independent piece of information: typically the product ratio $[B]/[C]$ (or the individual amounts of $B$ and $C$ ), measured by analysing the products themselves rather than the disappearing reactant.

How is this different from a reversible or a consecutive reaction?

In a parallel scheme, $B$ and $C$ each form directly and independently from $A$ , and neither product feeds back into $A$ or converts into the other — that is what makes the ratio $[B]/[C]$ a true constant. A consecutive scheme instead has $A\to B\to C$ , where $B$ is an intermediate that itself decays into $C$ , so $[B]/[C]$ changes with time as $B$ builds up and is then consumed. A reversible scheme has $A\rightleftharpoons B$ , where the back reaction matters and the system approaches an equilibrium ratio rather than a ratio fixed from the start. All three use first-order rate laws, but the differential equations you set up, and what stays constant, are genuinely different.

Parallel reactions

The core idea

Common traps

Worked example

Competitive stretch

Before this

Where this leads