Consecutive reactions

Change the rate constants and starting concentrations and watch the time course update. The open circles show the steady-state approximation (SSA) for B. Hover or drag across the plot to see the concentrations at any time in the panel below it.

A k1k2 B k3k4 C

A ⇌ B ⇌ C Kinetics

ABCSSA for B
How this is calculated

The rate laws are first order in each species:

d[A]/dt = −k1[A] + k2[B] d[B]/dt = k1[A] − (k2 + k3)[B] + k4[C] d[C]/dt = k3[B] − k4[C]

Because the system is linear, the page solves it exactly with the matrix exponential, [X](t) = eKt[X]₀, so no numerical integrator is involved. This shortcut works only because every step here is first order. Add a bimolecular step, such as A + B ⇌ C, and the rate laws are no longer linear. That model has to be solved by numerical integration, as MATLAB's ode15s does.

The two observable relaxation rates are the nonzero eigenvalues of the rate matrix K. With S = k1+k2+k3+k4 and P = k1k3 + k1k4 + k2k4:

1/τ± = ½ [ S ± √(S² − 4P) ]

The time course is a sum of two exponentials with these rates. It never oscillates. When the two rates differ by more than about fivefold, the reaction looks biphasic.

The steady-state approximation sets d[B]/dt = 0, which gives [B]SSA = (k1[A] + k4[C]) / (k2 + k3). It does not require [B] to be small. It requires B to relax (at rate k2 + k3) much faster than [A] and [C] change, which happens at roughly the slow relaxation rate 1/τslow. After the initial transient, the relative error in [B]SSA is about τB/τslow, where τB = 1/(k2 + k3). Michaelis–Menten kinetics is the same case: the SSA on ES holds even when nearly all the enzyme is in the ES form, as long as ES turns over fast compared with how quickly substrate is used up.

Same model and automatic end-time rule as the MATLAB function AtoBtoCkin (J. Kahn, UMD).