9.2 Parallel reactions and product distribution

Parallel Reactions and Product Distribution

Parallel reactions occur when the same reactant can transform into two or more different products at the same time, each through its own pathway. Understanding how these pathways compete is essential for predicting (and controlling) which product dominates in a mixture.

Significance of Parallel Reactions

In a parallel reaction system, a single reactant follows multiple pathways simultaneously. Each pathway has its own rate constant ($k$) and activation energy ($E_a$), and the relative speeds of these pathways determine how much of each product you get.

• The faster pathway produces more of its product. This is called kinetic control of product distribution.
• By manipulating reaction conditions like temperature, chemists can shift selectivity toward a desired product by making one pathway faster relative to the others.

This matters in real-world chemistry. In pharmaceutical synthesis, for instance, a reactant might form both the desired drug molecule and an unwanted byproduct through parallel pathways. Controlling which pathway wins is the whole challenge.

Rate Equations for Parallel Reactions

Consider a simple system where reactant A can form either product B or product C:

• $A \rightarrow B$ with rate constant $k_1$
• $A \rightarrow C$ with rate constant $k_2$

Assuming both are first-order in A, the rate equations are:

• $\frac{d[B]}{dt} = k_1[A]$
• $\frac{d[C]}{dt} = k_2[A]$

The overall rate of consumption of A is the sum of both pathways:

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

What controls which product dominates:

• Relative magnitudes of $k_1$ and $k_2$: Whichever rate constant is larger produces more of its corresponding product.
• Activation energies: The pathway with the lower $E_a$ has a larger rate constant at a given temperature, so its product is favored.
• Temperature adjustments can change the ratio of $k_1$ to $k_2$, shifting selectivity (more on this below).

Temperature and Concentration Effects

Temperature affects each rate constant through the Arrhenius equation:

$k = Ae^{-E_a/RT}$

where $A$ is the pre-exponential factor, $R$ is the gas constant, and $T$ is the absolute temperature.

The key insight here: reactions with higher $E_a$ are more sensitive to temperature changes. If you raise the temperature, the rate constant for the high-$E_a$ pathway increases by a larger factor than the low-$E_a$ pathway. So increasing temperature shifts selectivity toward the product of the higher-activation-energy reaction.

• At low temperatures, the pathway with the lower $E_a$ dominates because it has the larger rate constant.
• At high temperatures, the pathway with the higher $E_a$ catches up and can even become dominant.

Concentration has a different effect. Since both pathways depend on the same reactant A, increasing $[A]$ speeds up both reactions proportionally. The ratio $k_1/k_2$ stays the same, so the product distribution doesn't change. Concentration alone cannot shift selectivity in simple parallel reactions where both pathways have the same order in A.

Product Distribution Calculations

For parallel first-order reactions $A \rightarrow B$ and $A \rightarrow C$, the product ratio depends only on the rate constants:

1. Product ratio: $\frac{[B]}{[C]} = \frac{k_1}{k_2}$
2. Fraction of B: $\frac{[B]}{[B]+[C]} = \frac{k_1}{k_1+k_2}$
3. Fraction of C: $\frac{[C]}{[B]+[C]} = \frac{k_2}{k_1+k_2}$

These hold because both rates draw from the same pool of A, so the integrated amounts are always in the ratio $k_1 : k_2$.

Worked example: Suppose $k_1 = 0.6 \, \text{s}^{-1}$ and $k_2 = 0.3 \, \text{s}^{-1}$.

• Product ratio: $\frac{[B]}{[C]} = \frac{0.6}{0.3} = 2$, so B forms at twice the rate of C.
• Fraction of B: $\frac{0.6}{0.6 + 0.3} = \frac{0.6}{0.9} = 0.667$, or about 66.7% of the product mixture is B.
• Fraction of C: $\frac{0.3}{0.9} = 0.333$, or about 33.3%.

If you then raised the temperature and the new rate constants became $k_1 = 1.2 \, \text{s}^{-1}$ and $k_2 = 0.9 \, \text{s}^{-1}$, the ratio would shift to $1.2/0.9 = 1.33$. Product B still dominates, but less so, because pathway 2 (with its higher $E_a$) gained more from the temperature increase.