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Reaction in Series

Kinetics of Consecutive First-Order Reactions ($A \rightarrow B \rightarrow C$).

By chemca Team • Updated Jan 2026

Series or sequential reactions involve the formation of an intermediate which further reacts to form the final product. The simplest case is two consecutive first-order reactions.

1. The Reaction Scheme

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

Where:
A is the Reactant.
B is the Intermediate.
C is the Final Product.
$k_1, k_2$ are first-order rate constants.

2. Differential Rate Equations

We can write the rate of change for each species:

1. For A (Consumption only):
$$ \frac{d[A]}{dt} = -k_1[A] $$
2. For B (Formation from A, Consumption to C):
$$ \frac{d[B]}{dt} = k_1[A] - k_2[B] $$
3. For C (Formation from B):
$$ \frac{d[C]}{dt} = k_2[B] $$

3. Concentration Profiles (Integrated Laws)

Concentration at time 't'

Assuming at $t=0$, $[A]=[A]_0$ and $[B]_0 = [C]_0 = 0$.

1. Concentration of A: Decreases exponentially.

$$ [A]_t = [A]_0 e^{-k_1 t} $$

2. Concentration of B: Increases to a max, then decreases.

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

3. Concentration of C: Increases continuously (Mass Balance).

$$ [C]_t = [A]_0 - [A]_t - [B]_t $$ $$ [C]_t = [A]_0 \left( 1 + \frac{k_1 e^{-k_2 t} - k_2 e^{-k_1 t}}{k_2 - k_1} \right) $$

4. Kinetics of the Intermediate (B)

Time for Maximum Concentration ($t_{max}$)

To find when concentration of B is maximum, we set $\frac{d[B]}{dt} = 0$.

$$ t_{max} = \frac{1}{k_1 - k_2} \ln \left( \frac{k_1}{k_2} \right) = \frac{\ln k_2 - \ln k_1}{k_2 - k_1} $$

Maximum Concentration ($[B]_{max}$)

Substituting $t_{max}$ into the equation for $[B]_t$:

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

5. Rate Determining Step (RDS)

The overall rate depends on the relative magnitude of $k_1$ and $k_2$.

Case 1: $k_1 \ll k_2$ (Step 1 is Slow/RDS)
B is consumed as fast as it forms. $[B]$ remains very low (Steady State).
Reaction behaves like $A \rightarrow C$ with rate $k_1$.

Case 2: $k_1 \gg k_2$ (Step 2 is Slow/RDS)
A converts to B rapidly, then B slowly converts to C.
Reaction behaves like $B \rightarrow C$ with rate $k_2$.

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