Is the shaded fraction exactly equal to exp(−E_a/RT)?

No. The Arrhenius factor is a convenient approximation related to the high-energy tail, not a literal integral identity for every form of f(E). The simulator shows both so you can see they track similarly with T and E_a without being identical curves.

Why does a small temperature increase change the rate so much?

Only the far tail of f(E) lies above a typical E_a ≫ RT. That tail grows roughly exponentially with T, so rates often change by large factors for a few tens of kelvin. A misconception is that all molecules suddenly gain energy E_a; most remain below threshold.

Does this prove a reaction mechanism?

It only illustrates the kinetic idea that barrier crossing correlates with the energetic fraction of collisions. Real mechanisms need elementary steps, transition-state theory, and experimental rate laws; the plot is pedagogical gas-kinetic motivation for Arrhenius temperature dependence.

Maxwell–Boltzmann vs Eₐ

Translational energy density f(E); shaded fraction above activation energy; compare Arrhenius exp(−Eₐ/RT).

About this model

This simulator plots the Maxwell–Boltzmann translational energy distribution f(E) for an ideal gas and shades the high-energy fraction of molecules with E > E_a. That tail is compared with the Arrhenius factor exp(−E_a/RT) that appears in many simple rate laws. Temperature T and activation energy E_a are the main controls: raising T fattens the tail so more molecules clear E_a even though the Arrhenius factor is only an approximate integral of the distribution. The model is classical ideal-gas kinetics with a sharp energy threshold; it ignores steric factors, tunneling, and detailed collision theory. You vary T and E_a to see why modest heating can accelerate reactions dramatically.

Who it's for: Introductory physical chemistry and chemical kinetics courses linking molecular speeds to Arrhenius behavior.

Key terms

Maxwell–Boltzmann distribution
Activation energy
Arrhenius equation
Energy distribution
Reaction rate
Thermal energy

How it works

Translational energy distribution for a classical ideal gas in 3D: f(E) ∝ √E exp(−E/kT). The shaded tail is the fraction of collisions with kinetic energy exceeding Eₐ per molecule (compare with the Arrhenius factor exp(−Eₐ/RT), which uses molar energies in reaction rate theory).

Key equations

f(E) = (2/√π)(kT)^(−3/2) √E exp(−E/kT)

P(E > Eₐ) = ∫_{Eₐ}^∞ f(E) dE

Frequently asked questions

Is the shaded fraction exactly equal to exp(−E_a/RT)?: No. The Arrhenius factor is a convenient approximation related to the high-energy tail, not a literal integral identity for every form of f(E). The simulator shows both so you can see they track similarly with T and E_a without being identical curves.
Why does a small temperature increase change the rate so much?: Only the far tail of f(E) lies above a typical E_a ≫ RT. That tail grows roughly exponentially with T, so rates often change by large factors for a few tens of kelvin. A misconception is that all molecules suddenly gain energy E_a; most remain below threshold.
Does this prove a reaction mechanism?: It only illustrates the kinetic idea that barrier crossing correlates with the energetic fraction of collisions. Real mechanisms need elementary steps, transition-state theory, and experimental rate laws; the plot is pedagogical gas-kinetic motivation for Arrhenius temperature dependence.

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About this model

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Key equations

Frequently asked questions

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How it works

Key equations

Frequently asked questions