Consecutive reactions

Not every reaction happens in one clean step. In a consecutive (or sequential) reaction, a reactant $A$ first converts into an intermediate $B$ , which then converts into the final product $C$ : $A \xrightarrow{k_1} B \xrightarrow{k_2} C$ . This two-step scheme behaves very differently from a single first-order decay — $[A]$ falls monotonically as usual, but the intermediate $[B]$ rises from zero, passes through a maximum, and then falls back to zero as $[C]$ takes over. That rise-peak-fall shape is the signature of a consecutive mechanism, and it shows up everywhere from radioactive decay chains (parent to daughter to granddaughter nuclide) to organic reaction mechanisms with a reactive intermediate. Consecutive reactions are the second major topic in the CY11001 Chemical Kinetics unit, and they introduce the rate-determining step — the idea that the slowest step in a sequence controls the overall observed rate, a concept you will meet again and again in mechanistic chemistry.

The core insight

In $A \xrightarrow{k_1} B \xrightarrow{k_2} C$ the intermediate $[B]$ rises to a maximum at $t_{max}=\ln(k_2/k_1)/(k_2-k_1)$ and then decays, and the slower of $k_1,k_2$ is the rate-determining step that controls how fast $C$ forms overall.

The core idea

Setting up the coupled rate equations. Take only $A$ present initially, at concentration $a_0$ , and let both steps be first order. The three species obey coupled differential equations that simply track what is made and destroyed at each stage: $A$ is destroyed by step 1; $B$ is made by step 1 and destroyed by step 2; $C$ is made by step 2 and never destroyed.

\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]

where: $[A],[B],[C]$ are the time-dependent concentrations, $k_1,k_2$ are the first-order rate constants (units $\text{time}^{-1}$ ) of the two steps, and $a_0$ is the initial concentration of $A$ .

Solving the chain. The first equation is the ordinary first-order decay you already know, $[A]=a_0e^{-k_1t}$ . Substituting this into the second equation gives a linear first-order ODE in $[B]$ whose solution (valid whenever $k_1\neq k_2$ ) is

[B] = \frac{k_1a_0}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right)

and mass balance, $[A]+[B]+[C]=a_0$ (nothing is lost, everything that starts as $A$ ends up as one of the three species), then fixes $[C]$ :

[C] = a_0\left[1+\frac{k_1e^{-k_2t}-k_2e^{-k_1t}}{k_2-k_1}\right]

Why $B$ has a maximum. Early on, $A$ is abundant and $B$ is scarce, so $B$ is made faster than it is destroyed and $[B]$ climbs. As $A$ depletes, the production term $k_1[A]$ shrinks while the destruction term $k_2[B]$ grows with $[B]$ itself; the two rates cross, $[B]$ peaks, and beyond that point destruction wins and $[B]$ falls back toward zero. Setting $d[B]/dt=0$ locates that peak.

Which step is rate-determining. The slower of the two constants controls the overall appearance of $C$ . If $k_1\ll k_2$ (step 1 slow), $B$ is mopped up almost as fast as it forms, $[B]$ stays small throughout, and $[C]\approx a_0(1-e^{-k_1t})$ — the whole two-step sequence behaves like a single slow step with constant $k_1$ . If $k_1\gg k_2$ (step 2 slow), $A$ converts to $B$ almost instantly, $[B]$ builds up toward $a_0$ and then drains slowly, and $[C]\approx a_0(1-e^{-k_2t})$ . Either way, the sequence is only as fast as its slowest link — the same physical idea as a chain being no stronger than its weakest link.

Figure

Concentration versus time curves for A falling smoothly B rising to a peak then falling and C rising in an S shape to the total starting concentration

Derivation ATime of maximum intermediate concentration

The goal is to derive the time $t_{max}$ at which the intermediate concentration $[B]$ reaches its peak in the consecutive scheme $A \xrightarrow{k_1} B \xrightarrow{k_2} C$ , by differentiating the closed-form expression for $[B](t)$ and finding where the rate of change is exactly zero, then simplifying the resulting exponential equation to an explicit formula in $k_1$ and $k_2$ .

Step 1. Start from the derived expression $[B] = \dfrac{k_1a_0}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right)$ . At the maximum, the slope $d[B]/dt$ must vanish, since a smooth curve is momentarily flat at its peak.

Step 2. Differentiate term by term. Since $\dfrac{k_1a_0}{k_2-k_1}$ is just a constant multiplier, $\dfrac{d[B]}{dt} = \dfrac{k_1a_0}{k_2-k_1}\left(-k_1e^{-k_1t}+k_2e^{-k_2t}\right)$ .

Step 3. Set this equal to zero. The constant prefactor is nonzero (for $k_1\neq k_2$ , $a_0\neq0$ ), so the bracket itself must vanish: $-k_1e^{-k_1t}+k_2e^{-k_2t}=0$ , i.e. $k_2e^{-k_2t}=k_1e^{-k_1t}$ .

Step 4. Rearrange into a ratio of exponentials. Dividing both sides by $k_1e^{-k_2t}$ gives $\dfrac{k_2}{k_1}=e^{-k_1t}/e^{-k_2t}=e^{(k_2-k_1)t}$ .

Step 5. Take the natural log of both sides to bring the exponent down: $\ln(k_2/k_1) = (k_2-k_1)t$ .

Step 6. Solve for $t$ , giving the compact result

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

This is symmetric in the sense that swapping $k_1\leftrightarrow k_2$ just flips the sign of both numerator and denominator, leaving $t_{max}$ unchanged — the peak time only depends on the pair of rate constants, not on which one is labelled "first."

Worked example

A reaction proceeds as $A\xrightarrow{k_1}B\xrightarrow{k_2}C$ with $k_1=0.20\ \text{min}^{-1}$ and $k_2=0.050\ \text{min}^{-1}$ , starting from pure $A$ at $a_0=2.0\ \text{mol L}^{-1}$ . Find the time of maximum $[B]$ and the peak value $[B]_{max}$ .

Step 1. Apply the derived formula for the peak time: $t_{max} = \dfrac{\ln(k_2/k_1)}{k_2-k_1} = \dfrac{\ln(0.050/0.20)}{0.050-0.20} = \dfrac{\ln(0.25)}{-0.15} = \dfrac{-1.386}{-0.15} = 9.24\ \text{min}$ .

Step 2. Evaluate the two exponentials at $t=9.24$ min: $e^{-k_1t}=e^{-0.20\times9.24}=e^{-1.848}=0.1576$ and $e^{-k_2t}=e^{-0.050\times9.24}=e^{-0.462}=0.630$ .

Step 3. Substitute into $[B]=\dfrac{k_1a_0}{k_2-k_1}\left(e^{-k_1t}-e^{-k_2t}\right)$ : $[B]_{max} = \dfrac{0.20\times2.0}{0.050-0.20}(0.1576-0.630) = \dfrac{0.40}{-0.15}\times(-0.472) = 1.259\ \text{mol L}^{-1}$ .

Step 4. Sanity check. Here $k_1>k_2$ , so step 2 is slower and rate-determining; $B$ should accumulate substantially before slowly draining into $C$ . A peak of $1.26\ \text{mol L}^{-1}$ out of a starting $2.0\ \text{mol L}^{-1}$ — well over half the initial material sitting as intermediate — is exactly consistent with that picture. The intermediate peaks at $[B]_{max}=1.26\ \text{mol L}^{-1}$ at $t_{max}=9.24\ \text{min}$ .

Competitive stretch

For this same system, check the rate-determining-step shortcut at long time. Since $k_2\ll k_1$ , the approximation $[C]\approx a_0(1-e^{-k_2t})$ should hold once $A$ has essentially all converted to $B$ . At $t=40$ min the exact formula gives $[C]=2.0\left[1+\dfrac{0.20e^{-2.0}-0.050e^{-8.0}}{-0.15}\right]=2.0(0.8197)=1.64\ \text{mol L}^{-1}$ , while the approximation gives $[C]\approx2.0(1-e^{-2.0})=2.0(0.8647)=1.73\ \text{mol L}^{-1}$ . The two agree to within about 5%, confirming that the shortcut is reasonable — though not exact, since it ignores the short but finite time needed to build up $B$ in the first place.

Common traps

Treating $[B]$ as simply $a_0-[A]$ . That is only true if $B$ never reacted further. With $k_2\neq0$ , $B$ is a genuinely separate function that rises and falls; you must use the exponential-difference formula, not a static mass-balance shortcut, at intermediate times.
Picking the "first" step as rate-determining out of habit. The slower rate constant controls the overall rate, regardless of whether it belongs to step 1 or step 2. Compare $k_1$ and $k_2$ numerically before deciding.
Plugging $k_1=k_2$ straight into the $[B]$ formula. The expression has $k_2-k_1$ in the denominator and blows up there; the correct degenerate-case result, obtained by taking the limit, is $[B]=k_1a_0t\,e^{-k_1t}$ .
Applying the rate-determining-step approximation when $k_1,k_2$ are comparable. The shortcut $[C]\approx a_0(1-e^{-k_{\text{slow}}t})$ is only good when the two constants differ by roughly an order of magnitude or more; for similar $k_1,k_2$ the full exponential expressions are needed.

Ask the tutor: want to work through the $k_1=k_2$ limiting-case derivation step by step, or see how this maps onto a three-member radioactive decay chain?

Why does the intermediate concentration have a maximum instead of just building up steadily?

Because $B$ is being made and destroyed at the same time. Early on there is plenty of $A$ , so $B$ forms faster than it can react away, and $[B]$ climbs. As $A$ runs out, the production rate $k_1[A]$ drops while the destruction rate $k_2[B]$ keeps growing along with $[B]$ itself. The two rates become equal at the peak, and past that point destruction outpaces production, so $[B]$ falls back toward zero as it all converts to $C$ .

How is this different from a reversible reaction, which was the previous topic?

A reversible reaction is $A \rightleftharpoons B$ — the same two species can convert into each other in both directions, and the system settles into an equilibrium mixture. A consecutive reaction is $A \to B \to C$ — three distinct species, one-way steps only, and the system does not settle into an equilibrium; it runs to completion with everything eventually ending up as $C$ . The intermediate's rise-and-fall behaviour is unique to the consecutive case and has no equilibrium analogue.

Does the identity of the intermediate matter, or is this just abstract algebra?

It matters a great deal in practice. If $B$ is a reactive, possibly hazardous or unstable species (a common situation in organic synthesis and industrial chemistry), knowing $t_{max}$ and $[B]_{max}$ tells you exactly when the intermediate is most concentrated — the point of greatest risk or greatest opportunity to isolate it if $B$ itself is the desired product. The same mathematics also describes a radioactive decay chain, where $B$ is a short-lived daughter nuclide whose activity peaks and then falls as it decays into a stable granddaughter $C$ .

Consecutive reactions

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads