9.1 Consecutive reactions and their kinetics

Consecutive Reactions

Consecutive reactions are a series of chemical transformations where one reaction's product becomes the next reaction's reactant. Understanding how these multi-step processes unfold over time is central to fields ranging from nuclear chemistry to biochemistry. The tools here (rate equations, concentration-time profiles, and approximations like the steady-state assumption) let you predict how reactants, intermediates, and products behave and identify which step controls the overall rate.

Consecutive reactions in chemical systems

In a consecutive reaction, each step depends on the product of the previous step. The intermediate species produced along the way are both a product of one reaction and a reactant for the next. This sequential dependence is what makes the kinetics more complex than a single-step reaction.

You'll encounter consecutive reactions across many areas of chemistry:

• Radioactive decay series: Uranium-238 decays through a chain of alpha and beta emissions, passing through intermediates like thorium-234 and radium-226, before ultimately reaching stable lead-206.
• Metabolic pathways: In the citric acid cycle, each enzyme-catalyzed step converts one intermediate into the next. Citrate is converted to isocitrate, then to $\alpha$-ketoglutarate, and so on.
• Multistep organic synthesis: Complex target molecules are built from simpler precursors through a planned sequence of reactions, where each step's product feeds into the next.

One thing to note: each step doesn't need to wait for the previous step to fully complete. Once some intermediate forms, the next step can begin consuming it. All steps run simultaneously once their reactants are available.

Rate equations for consecutive reactions

Consider the simplest consecutive reaction, where both steps are first-order:

$A \stackrel{k_1}{\longrightarrow} B \stackrel{k_2}{\longrightarrow} C$

Here $A$ is the reactant, $B$ is the intermediate, $C$ is the final product, and $k_1$ and $k_2$ are the rate constants for the first and second steps.

You can write a differential rate equation for each species by tracking what creates it and what destroys it:

• Reactant $A$: Only consumed in step 1.

$\frac{d[A]}{dt} = -k_1[A]$

• Intermediate $B$: Produced by step 1, consumed by step 2.

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

• Product $C$: Only produced by step 2.

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

The equation for $A$ is straightforward first-order decay, giving the familiar solution $[A] = [A]_0 e^{-k_1 t}$. The equation for $B$ is trickier because it depends on both $[A]$ (which is changing) and $[B]$ itself. Solving it by substituting the expression for $[A]$ yields:

$[B] = \frac{k_1[A]_0}{k_2 - k_1}\left(e^{-k_1 t} - e^{-k_2 t}\right)$

This expression is valid when $k_1 \neq k_2$. The concentration of $C$ at any time follows from mass balance: $[C] = [A]_0 - [A] - [B]$.

How each species behaves over time:

• $[A]$ decreases exponentially from its initial value.
• $[B]$ starts at zero, rises to a maximum, then falls back toward zero as it gets consumed in step 2.
• $[C]$ starts at zero and increases, eventually approaching $[A]_0$ as the reaction goes to completion.

The time at which $[B]$ reaches its maximum can be found by setting $\frac{d[B]}{dt} = 0$ and solving for $t$:

$t_{max} = \frac{\ln(k_1/k_2)}{k_1 - k_2}$

Concentration-time profiles analysis

The rate-determining step (RDS) is the slowest step in the sequence, and it controls the overall rate of product formation. You can figure out which step is rate-determining by looking at the relative magnitudes of $k_1$ and $k_2$ and at the shape of the concentration-time profiles.

Case 1: $k_1 \ll k_2$ (first step is rate-determining)

• $A$ is consumed slowly because $k_1$ is small.
• As soon as $B$ forms, it's rapidly converted to $C$ (since $k_2$ is large), so $[B]$ stays very low throughout the reaction.
• $[C]$ increases slowly, at a rate governed by $k_1$.
• The overall rate of product formation is approximately $k_1[A]$.

Case 2: $k_1 \gg k_2$ (second step is rate-determining)

• $A$ is consumed quickly because $k_1$ is large.
• $B$ accumulates to a significant concentration because it forms fast but is consumed slowly.
• $[C]$ increases slowly, at a rate governed by $k_2$.
• The concentration-time profile for $B$ shows a pronounced peak before it gradually declines.

Being able to read these profiles is a practical skill. If you see a large buildup of intermediate in experimental data, that's a strong signal that the step consuming the intermediate is rate-determining.

Kinetics problem-solving for consecutive reactions

The steady-state approximation (SSA) is a powerful simplification for consecutive reactions. It assumes that after a brief initial period, the concentration of the intermediate $B$ changes so slowly that you can treat $\frac{d[B]}{dt}$ as approximately zero.

This approximation works best when the intermediate is highly reactive (consumed almost as fast as it forms), which keeps $[B]$ small relative to $[A]$ and $[C]$.

Applying the steady-state approximation step by step:

1. Write the rate equation for the intermediate: $\frac{d[B]}{dt} = k_1[A] - k_2[B]$

2. Set it equal to zero (the steady-state condition): $k_1[A] - k_2[B] = 0$

3. Solve for the steady-state concentration of $B$: $[B]_{ss} = \frac{k_1[A]}{k_2}$

4. Substitute this into the rate equation for product $C$: $\frac{d[C]}{dt} = k_2[B]_{ss} = k_2 \cdot \frac{k_1[A]}{k_2} = k_1[A]$

The result tells you that under steady-state conditions, the rate of product formation depends only on $[A]$ and $k_1$. This makes physical sense: if $B$ is consumed as fast as it's made, the bottleneck is the rate at which $A$ produces $B$.

A word of caution: the SSA is not valid at the very beginning of the reaction (when $[B]$ is still building up from zero) or when the intermediate accumulates to high concentrations. It's most reliable when $k_2 \gg k_1$, which keeps $[B]$ small throughout.