Consecutive (Sequential) Reactions

Radioactive decay chains, drug metabolism, and industrial nitration all share a pattern. A reactant becomes an intermediate before it becomes the final product. That intermediate is not a spectator. It builds up, peaks, and decays, and its concentration profile is exactly what a plant operator or pharmacologist needs to predict. Consecutive reactions matter because most real mechanisms are not single steps. Even a "one-step" reaction, viewed closely, hides a sequence of elementary events, and the algebra of A→B→C is the template for all of them. Understanding when B accumulates and when it vanishes almost immediately separates slow-intermediate kinetics from the steady-state world. This article derives the full time-dependent concentrations for A→B→C, finds the time and magnitude of the intermediate's maximum, and works two numeric examples. It builds directly on the integrated first-order rate law, and it sets up the steady-state approximation used in later kinetics topics.

Key takeaways

A→B→C with rate constants $k_1$ and $k_2$ gives exact closed-form expressions for A, B, and C versus time.
A decays exponentially exactly as in a simple first-order reaction, unaffected by what happens to B.
The intermediate B rises, passes through a maximum concentration $t_{max}$ , then decays back toward zero.
$t_{max}$ depends only on $k_1$ and $k_2$ ; it is the time when B's formation rate equals its consumption rate.
When $k_2 \gg k_1$ , B stays low throughout, foreshadowing the steady-state approximation.
Key idea: The intermediate's concentration is a competition between how fast it forms and how fast it decays, and calculus locates the single moment of balance.

The core insight

In a consecutive reaction A→B→C, the intermediate B's concentration is set by a race between its formation from A (rate $k_1[A]$ ) and its consumption to C (rate $k_2[B]$ ); the maximum in $[B]$ occurs exactly when these two rates become equal, giving a single algebraic condition that pins down both the peak time and peak height.

The physical picture

Consider three species linked by two irreversible first-order (or pseudo-first-order) steps:

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

At $t=0$ only A is present, at concentration $[A]_0$ . As the reaction proceeds, A disappears with its own characteristic exponential decay — this part is identical to the simple first-order case already familiar from the integrated rate law. But B does not simply appear; it is created from A and destroyed by converting to C. Early on, when little B exists, formation dominates and $[B]$ climbs. Later, as A is depleted, formation slows while consumption continues, and $[B]$ falls. Somewhere in between lies a maximum. The shape of the B curve depends critically on the ratio $k_2/k_1$ . If B is destroyed much faster than it forms ( $k_2 \gg k_1$ ), the intermediate never accumulates appreciably — it is consumed almost as fast as it appears, a scenario that motivates the steady-state approximation used for reactive intermediates like radicals. If instead B is destroyed much more slowly than it forms ( $k_2 \ll k_1$ ), B piles up substantially before slowly bleeding into C, and the first step becomes the practical bottleneck — foreshadowing the idea of a rate-determining step. C, the final product, must make up the difference: mass balance requires $[A]+[B]+[C] = [A]_0$ at all times, so C's curve is an S-shaped (sigmoidal) rise with an inflection point related to the timing of B's maximum.

Figure

Concentration profiles for A→B→C with k1 = 0.20 per minute and k2 = 0.05 per minute

The core relations

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

where $[A]_0$ is the initial concentration of A and $k_1$ is the first-step rate constant.

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

where $k_2$ is the second-step rate constant, valid for $k_1 \neq k_2$ .

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

where $[C](t)$ follows from mass balance, $[C] = [A]_0 - [A] - [B]$ .

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

where $t_{max}$ is the time at which $[B]$ is maximal.

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

where $[B]_{max}$ is the peak intermediate concentration, a function of the ratio $k_2/k_1$ only.

Derivation ASolving the consecutive first-order kinetic scheme

Goal. Starting from the rate laws for A→B→C, obtain closed-form $[A](t)$ , $[B](t)$ , and then $t_{max}$ , $[B]_{max}$ .

Step 1. Write the rate equations. Each species obeys mass-action kinetics for elementary first-order steps:

\frac{d[A]}{dt} = -k_1[A], \qquad \frac{d[B]}{dt} = k_1[A] - k_2[B], \qquad \frac{d[C]}{dt} = k_2[B]

A is consumed only by step 1; B is fed by step 1 and drained by step 2; C only accumulates.

Step 2. Solve for A. The A equation is the ordinary first-order decay already known from the integrated rate law:

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

This does not depend on $k_2$ at all — A cannot "feel" what happens to B.

Step 3. Substitute into the B equation. This turns the B equation into a linear, first-order, inhomogeneous ODE:

\frac{d[B]}{dt} + k_2[B] = k_1[A]_0 e^{-k_1 t}

Use the integrating factor $e^{k_2 t}$ , multiplying both sides:

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

Step 4. Integrate both sides from 0 to $t$ . Using $[B](0)=0$ :

[B]\,e^{k_2 t} - 0 = \frac{k_1[A]_0}{k_2-k_1}\left(e^{(k_2-k_1)t} - 1\right)

Multiplying through by $e^{-k_2 t}$ gives the boxed intermediate law:

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

Step 5. Get C by mass balance. Since no species is destroyed except by conversion to the next, total concentration is conserved:

[C](t) = [A]_0 - [A](t) - [B](t)

Substituting Steps 2 and 4 and simplifying yields the C expression given above. Check: at $t=0$ , $[C]=0$ ; as $t\to\infty$ , $[C]\to[A]_0$ , as required.

Step 6. Locate the maximum of B. Differentiate $[B](t)$ and set the derivative to zero:

\frac{d[B]}{dt} = \frac{k_1[A]_0}{k_2-k_1}\left(-k_1 e^{-k_1 t} + k_2 e^{-k_2 t}\right) = 0

This requires

k_1 e^{-k_1 t_{max}} = k_2 e^{-k_2 t_{max}}

which is exactly the statement that B's formation rate ( $k_1[A]$ ) equals its consumption rate ( $k_2[B]$ ) at the peak — the physical condition $d[B]/dt=0$ always implies.

Step 7. Solve for $t_{max}$ . Taking logarithms of the balance condition:

\ln k_1 - k_1 t_{max} = \ln k_2 - k_2 t_{max}

(k_2-k_1)\,t_{max} = \ln k_2 - \ln k_1

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

Step 8. Get $[B]_{max}$ . Substitute $t_{max}$ back into $[B](t)$ ; after simplification (using $e^{-k_1 t_{max}} = (k_1/k_2)^{k_1/(k_2-k_1)}\cdot(1/k_1)\cdot k_1$ , carried through algebra) one obtains the compact result:

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

This depends only on the dimensionless ratio $k_2/k_1$ , not on $[A]_0$ separately, so doubling the initial concentration simply doubles $[B]_{max}$ at the same relative time.

Worked examples

Example 1 — Foundational: locating the intermediate maximum. A drug A is absorbed and converted in the body to an active metabolite B, which is then eliminated to an inactive product C. Suppose $k_1 = 0.20\ \text{min}^{-1}$ (absorption/first conversion) and $k_2 = 0.050\ \text{min}^{-1}$ (elimination), with $[A]_0 = 1.00\times10^{-3}\ \text{M}$ . Step 1 — time of maximum:

t_{max} = \frac{\ln(k_2/k_1)}{k_2-k_1} = \frac{\ln(0.050/0.20)}{0.050 - 0.20\ \text{min}^{-1}} = \frac{\ln(0.25)}{-0.150\ \text{min}^{-1}}

t_{max} = \frac{-1.3863}{-0.150\ \text{min}^{-1}} = 9.24\ \text{min}

Step 2 — peak concentration:

\frac{[B]_{max}}{[A]_0} = \left(\frac{k_1}{k_2}\right)^{k_2/(k_2-k_1)} = (4.0)^{0.050/(-0.150)} = (4.0)^{-0.3333}

(4.0)^{-0.3333} = e^{-0.3333\ln4.0} = e^{-0.3333(1.3863)} = e^{-0.4621} = 0.630

[B]_{max} = 0.630 \times 1.00\times10^{-3}\ \text{M} = 6.30\times10^{-4}\ \text{M}

Step 3 — check with the direct formula at $t=9.24$ min:

[A] = [A]_0 e^{-k_1 t} = 1.00\times10^{-3}\, e^{-0.20(9.24)} = 1.00\times10^{-3}(0.1580) = 1.58\times10^{-4}\ \text{M}

[B] = \frac{k_1[A]_0}{k_2-k_1}\left(e^{-k_1 t}-e^{-k_2 t}\right) = \frac{(0.20)(1.00\times10^{-3})}{-0.150}\left(0.1580 - 0.6321\right)

[B] = (-1.333\times10^{-3})(-0.4741) = 6.32\times10^{-4}\ \text{M}

This matches $6.30\times10^{-4}$ M within rounding — consistent with Step 2. At the peak, $[C] = 1.00\times10^{-3} - 1.58\times10^{-4} - 6.32\times10^{-4} = 2.10\times10^{-4}\ \text{M}$ , and all three sum back to $1.00\times10^{-3}$ M as required by mass balance.

Example 2 — Applied: comparing a fast-intermediate and slow-intermediate regime. An industrial reactor runs a two-step conversion A→B→C. Engineers want to know if switching catalysts (which changes $k_2$ from $0.10\ \text{s}^{-1}$ to $5.0\ \text{s}^{-1}$ , while $k_1 = 0.50\ \text{s}^{-1}$ stays fixed) meaningfully changes the maximum buildup of the potentially hazardous intermediate B. Take $[A]_0 = 2.00\ \text{M}$ . Case (i): slow second step, $k_2 = 0.10\ \text{s}^{-1}$ .

t_{max} = \frac{\ln(0.10/0.50)}{0.10-0.50\ \text{s}^{-1}} = \frac{\ln(0.20)}{-0.40\ \text{s}^{-1}} = \frac{-1.609}{-0.40\ \text{s}^{-1}} = 4.02\ \text{s}

\frac{[B]_{max}}{[A]_0} = (5.0)^{0.10/(-0.40)} = (5.0)^{-0.25} = e^{-0.25\ln5.0} = e^{-0.25(1.609)} = e^{-0.4024} = 0.669

[B]_{max} = 0.669 \times 2.00\ \text{M} = 1.34\ \text{M}

Here B accumulates to 67% of the starting concentration — a serious buildup, taking about 4 seconds to peak. Case (ii): fast second step, $k_2 = 5.0\ \text{s}^{-1}$ .

t_{max} = \frac{\ln(5.0/0.50)}{5.0-0.50\ \text{s}^{-1}} = \frac{\ln(10)}{4.50\ \text{s}^{-1}} = \frac{2.303}{4.50\ \text{s}^{-1}} = 0.512\ \text{s}

\frac{[B]_{max}}{[A]_0} = \left(\frac{0.50}{5.0}\right)^{5.0/4.50} = (0.10)^{1.111} = e^{1.111\ln0.10} = e^{1.111(-2.303)} = e^{-2.559} = 0.0774

[B]_{max} = 0.0774 \times 2.00\ \text{M} = 0.155\ \text{M}

Switching to the faster catalyst cuts the peak intermediate concentration from 1.34 M to 0.155 M — roughly a factor of 8.6 — and moves the peak nearly eightfold earlier. This is exactly the qualitative behavior expected: making $k_2 \gg k_1$ suppresses B's buildup, approaching steady-state-like behavior where B stays low throughout the reaction.

Limiting cases & common traps

Regime	Behavior of B	Physical picture
$k_2 \gg k_1$	$[B]_{max}$ small, $t_{max}$ small; B tracks A closely at low level	Approaches the steady-state approximation; first step is rate-determining
$k_2 \ll k_1$	$[B]_{max} \to [A]_0$ , $t_{max}$ large	B accumulates almost completely before slowly converting to C; second step is rate-determining
$k_1 = k_2 = k$	Formula must be replaced by $[B]=k[A]_0 t\, e^{-kt}$ (limit case, avoid 0/0)	Peak at $t_{max}=1/k$ , with $[B]_{max}=[A]_0/e$
$k_1 \gg k_2$ but both finite	Same as $k_2 \ll k_1$ above	—

The chemist behind it

The mathematics of consecutive first-order reactions was formalized in the early twentieth century, but its most famous incarnation is a chain of nuclear decays worked out by Ernest Rutherford and, algebraically, by chemist and physicist Frederick Soddy, whose work on radioactive decay series required exactly this A→B→C (→D→…) formalism to predict how much of each daughter nuclide would be present at a given time. Soddy's insight — that a decaying parent produces an intermediate that itself decays, generating characteristic buildup-and-decay curves — is mathematically identical to the chemical kinetics case, and it directly inspired the general treatment of sequential first-order kinetics used throughout physical chemistry today.

Origins · 1 minFrederick Soddy and how this was discovered

Frederick Soddy

1877-1956 · British radiochemist who formulated the radioactive series decay laws and coined the term "isotope"

Ask the tutor: Want to see how these formulas change if the first step is reversible, or how the steady-state approximation quantitatively approximates this exact solution?

Why doesn't A's decay depend on k2 at all

A is only ever consumed by the first step, A→B, so its differential rate law is simply $d[A]/dt = -k_1[A]$ with no term involving B, C, or $k_2$ . This is a standalone first-order ODE, solved exactly as in the integrated rate law topic, giving $[A]=[A]_0 e^{-k_1 t}$ . Physically, A cannot "know" what B eventually becomes; it only feels the rate at which it converts away from itself. Any coupling to $k_2$ enters only downstream, through B and C.

What happens at t_max physically, in terms of rates rather than concentrations

At the maximum, $[B]$ momentarily stops changing, so $d[B]/dt=0$ . Since $d[B]/dt = k_1[A]-k_2[B]$ , this means the instantaneous rate of B's formation from A exactly equals the instantaneous rate of B's consumption into C. Before $t_{max}$ , formation outpaces consumption and B grows; afterward, consumption wins and B shrinks. It is a dynamic balance point, not a state of rest for either underlying reaction.

Can this two-step treatment handle reversible steps or three or more consecutive steps

The specific closed-form equations here assume both steps are irreversible and first-order. Reversible first steps or longer chains (A→B→C→D…) can still be solved analytically using the same integrating-factor technique, but the algebra grows quickly and the maxima of later intermediates require solving transcendental equations. For three or more steps chemists often switch to numerical integration or invoke the steady-state approximation for short-lived intermediates to avoid the full algebra.

How does this connect to the steady-state approximation

The steady-state approximation assumes $d[B]/dt\approx 0$ for the entire reaction, not just at one instant. That assumption is good precisely when $k_2 \gg k_1$ , because then $[B]_{max}$ is small and $t_{max}$ is early, so B spends most of the reaction near this quasi-equilibrium low value. The exact consecutive-reaction solution derived here is the benchmark against which the steady-state approximation's accuracy is judged.

Is there a simple way to estimate k1 and k2 from an experimental B-versus-time curve

Yes, in outline. Measuring $t_{max}$ and $[B]_{max}$ from data gives two equations in two unknowns, $t_{max}=\ln(k_2/k_1)/(k_2-k_1)$ and the peak-height formula, which can be solved numerically (or graphically) for $k_1$ and $k_2$ individually. Alternatively, fitting the full $[A](t)$ decay gives $k_1$ directly, and substituting that into the $t_{max}$ equation isolates $k_2$ .