What is the steady-state approximation and when is it valid?

The steady-state approximation assumes that a reactive intermediate (a radical, an enzyme–substrate complex) stays at a low, nearly constant concentration, so d[I]/dt ≈ 0. That lets you express the intermediate in terms of reactants and derive an overall rate law. It requires the intermediate to be consumed much faster than it is formed — k₂ ≫ k₁ in A → B → C, or [E]₀ ≪ [S]₀ + K_M for Michaelis–Menten. The simulator compares the SSA prediction with the full numerical solution and rates deviations below 1% as excellent.

How does the pre-equilibrium approximation differ from the steady-state approximation?

The pre-equilibrium approximation assumes a fast reversible step reaches equilibrium (forward and reverse rates nearly equal) before a slow step controls the overall rate — for example 2NO ⇌ N₂O₂ (fast) followed by N₂O₂ + O₂ → 2NO₂ (slow), giving d[NO₂]/dt = 2k₂K[NO]²[O₂]. It needs k₋₁ ≫ k₂[O₂]. The steady-state approximation only requires a constant intermediate concentration; when k₋₁ ≫ k₂ both give the same result. The rate table marks steps whose forward and reverse rates differ by less than 5% as "≈ equilibrium".

How do you find the maximum concentration of B in A → B → C?

[B] = k₁[A]₀/(k₂ − k₁)·(e^(−k₁t) − e^(−k₂t)). Setting the derivative to zero gives t_max = ln(k₁/k₂)/(k₁ − k₂), and substituting back gives [B]max = [A]₀(k₂/k₁)^(k₂/(k₁−k₂)). With k₁ = 1, k₂ = 0.5 and [A]₀ = 1, t_max = 1.386 and [B]max = 0.5. If k₁ = k₂, t_max = 1/k₁ and [B]max = [A]₀/e.

Why is the H₂ + Br₂ reaction three-halves order?

It is a chain reaction: Br₂ → 2Br initiates, Br + H₂ → HBr + H and H + Br₂ → HBr + Br propagate, H + HBr → H₂ + Br inhibits, and 2Br → Br₂ terminates. Applying the steady-state approximation to Br and H gives [Br] = (k₁[Br₂]/k₅)^½ and d[HBr]/dt = 2k₂(k₁/k₅)^½[H₂][Br₂]^½/(1 + (k₄/k₃)[HBr]/[Br₂]). Early on [HBr] ≈ 0, so the rate is proportional to [H₂][Br₂]^½ — order 3/2 — and accumulating HBr slows the reaction. In the simulator you can see the radicals sit at only 10⁻⁴–10⁻⁷ M and reach steady state almost immediately.

⛓️

⛓️Reaction Mechanism Simulator - Elementary Steps, Steady-State & Pre-Equilibrium

A complex reaction is a sequence of elementary steps, and each step obeys the law of mass action: its order equals the stoichiometric coefficients of its reactants. This simulator lets you write the steps line by line — "A -> B ; k1 = 1", "B -> C ; k2 = 0.5" or a reversible "A <=> B ; kf = 1, kr = 0.25" — enter initial concentrations, and it builds the rate equations d[X]/dt and integrates them with an adaptive Dormand–Prince RK45 solver. Stiff systems, where rate constants differ by a factor of a million, are detected automatically and finished with an implicit Rosenbrock method, so the curves stay stable and accurate. Take consecutive first-order reactions A → B → C with k₁ = 1 s⁻¹, k₂ = 0.5 s⁻¹ and [A]₀ = 1 M: the intermediate B peaks at t_max = ln(k₁/k₂)/(k₁ − k₂) = 1.386 s with [B]max = 0.5 M, and the simulation matches the analytic solution to better than 10⁻⁶. Ten classic textbook mechanisms are built in — parallel reactions (product ratio k₁/k₂), a reversible reaction approaching equilibrium (K = k_f/k_r), Lindemann–Hinshelwood unimolecular fall-off, the Michaelis–Menten enzyme mechanism, the H₂ + Br₂ chain reaction, 2NO + O₂ pre-equilibrium and ozone decomposition — each with the rate law derived by the steady-state or pre-equilibrium approximation and a point-by-point comparison with the full simulation that shows when the approximation holds. You also get a rate table that flags equilibrated and slowest steps, maxima and half-lives, conserved quantities from the stoichiometric matrix with an atom-balance check, optional Arrhenius temperature dependence, log axes, hover readout and CSV export.

Free · No sign-up · Runs in your browserLast updated: 2026-10-29

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How to use Reaction Mechanism Simulator

1Pick a classic mechanism or type your own elementary steps, one per line, as "A + B -> C ; k = 0.5"
2Enter initial concentrations and the time span; optionally turn on Arrhenius temperature dependence with Ea values
3Read the concentration–time plot, rate table, derived rate law and approximation test, and export CSV or SVG

FAQ

What is the steady-state approximation and when is it valid?: The steady-state approximation assumes that a reactive intermediate (a radical, an enzyme–substrate complex) stays at a low, nearly constant concentration, so d[I]/dt ≈ 0. That lets you express the intermediate in terms of reactants and derive an overall rate law. It requires the intermediate to be consumed much faster than it is formed — k₂ ≫ k₁ in A → B → C, or [E]₀ ≪ [S]₀ + K_M for Michaelis–Menten. The simulator compares the SSA prediction with the full numerical solution and rates deviations below 1% as excellent.
How does the pre-equilibrium approximation differ from the steady-state approximation?: The pre-equilibrium approximation assumes a fast reversible step reaches equilibrium (forward and reverse rates nearly equal) before a slow step controls the overall rate — for example 2NO ⇌ N₂O₂ (fast) followed by N₂O₂ + O₂ → 2NO₂ (slow), giving d[NO₂]/dt = 2k₂K[NO]²[O₂]. It needs k₋₁ ≫ k₂[O₂]. The steady-state approximation only requires a constant intermediate concentration; when k₋₁ ≫ k₂ both give the same result. The rate table marks steps whose forward and reverse rates differ by less than 5% as "≈ equilibrium".
How do you find the maximum concentration of B in A → B → C?: [B] = k₁[A]₀/(k₂ − k₁)·(e^(−k₁t) − e^(−k₂t)). Setting the derivative to zero gives t_max = ln(k₁/k₂)/(k₁ − k₂), and substituting back gives [B]max = [A]₀(k₂/k₁)^(k₂/(k₁−k₂)). With k₁ = 1, k₂ = 0.5 and [A]₀ = 1, t_max = 1.386 and [B]max = 0.5. If k₁ = k₂, t_max = 1/k₁ and [B]max = [A]₀/e.
Why is the H₂ + Br₂ reaction three-halves order?: It is a chain reaction: Br₂ → 2Br initiates, Br + H₂ → HBr + H and H + Br₂ → HBr + Br propagate, H + HBr → H₂ + Br inhibits, and 2Br → Br₂ terminates. Applying the steady-state approximation to Br and H gives [Br] = (k₁[Br₂]/k₅)^½ and d[HBr]/dt = 2k₂(k₁/k₅)^½[H₂][Br₂]^½/(1 + (k₄/k₃)[HBr]/[Br₂]). Early on [HBr] ≈ 0, so the rate is proportional to [H₂][Br₂]^½ — order 3/2 — and accumulating HBr slows the reaction. In the simulator you can see the radicals sit at only 10⁻⁴–10⁻⁷ M and reach steady state almost immediately.

Features

Line-by-line elementary steps with reversible steps, termolecular steps and third bodies (M)
Rate equations generated automatically from the law of mass action
Adaptive RK45 with stiffness detection and automatic switch to an implicit Rosenbrock solver — stable for k ratios of 10⁶
10 presets: consecutive, parallel, reversible, second order, Lindemann, Michaelis–Menten, H₂–Br₂ chain, 2NO + O₂, ozone and Arrhenius
Steady-state and pre-equilibrium rate laws compared with the full simulation (maximum deviation)
Forward and reverse rates of every step at a chosen time, with equilibrium and slowest steps highlighted
Intermediate maxima, half-lives, conserved quantities and per-step atom balance
Log time / concentration axes, hover readout, CSV and SVG export

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