Visual walkthrough — Collision theory — frequency factor, steric factor

This page rebuilds the central result of collision theory — the equation

$k=\rho Z{e}^{-E_a/RT}$

— from absolutely nothing. We will not assume you know what any of those letters mean. By the end you will have watched every one of them being born from a picture of two molecules approaching each other. Follow the red object in each figure: it is always the thing the step is about.

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Step 1 — When does a "collision" actually happen?

WHAT. Two molecules are little balls. Ball A has radius $r_A$; ball B has radius $r_B$. A radius is just "how far the surface sticks out from the centre." The two balls touch exactly when the distance between their centres shrinks to $r_A+r_B$.

WHY. We can't track surfaces — they're messy. But centres are single points, easy to track. So we translate the whole question into: how close do two centre-points get? The answer "they collide when centres are $r_A+r_B$ apart" turns geometry into one clean number.

PICTURE. Figure s01 freezes molecule A and draws a circle around A's centre of radius $r_A+r_B$ (red). If B's centre ever crosses into that red circle, the surfaces have already touched — a collision. So A effectively drags around a target disc.

Figure s01 — the red dashed disc of radius $r_A+r_B$ around molecule A; a B-centre entering it means a collision.

The area of that red target disc is called the collision cross-section:

$\underset{\text{target area}}{\underbrace{{\sigma }_{AB}}}=\underset{\text{circle}}{\underbrace{\pi }}\underset{{\text{radius}}^2}{\underbrace{(r_A+r_B{)}^2}}$

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Step 2 — How much of that target gets swept per second?

WHAT. Give molecule A a speed. In one second A moves forward a distance equal to its speed. Its target disc sweeps out a cylinder: base ${\sigma }_{AB}$, length = distance travelled = speed $\times$ 1 second.

WHY. A collision is "a B-centre lands inside the disc." So we don't want the disc's area — we want the volume it sweeps, because every B sitting inside that swept tube gets hit. Volume is the natural bookkeeping unit: how much space did A patrol?

PICTURE. Figure s02 shows the red tube A cleans out in one second. Any B-centre (black dots) caught inside the tube is a collision.

Figure s02 — the swept cylinder: base area ${\sigma }_{AB}$, length $v\times 1\text{s}$; enclosed B-centres are hits.

$\underset{\text{volume patrolled / second}}{\underbrace{V_{\text{swept}}}}=\underset{\text{disc area}}{\underbrace{{\sigma }_{AB}}}\times \underset{\text{how far A goes / second}}{\underbrace{v}}$

The faster A moves ($v$ up) the longer the tube, the more B's it can sweep up. Hold this thought — we now have to decide which speed $v$ to use.

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Step 3 — Which speed? Not A's, not B's — the relative speed

WHAT. B is not standing still. What decides whether they collide is how fast they approach each other — the relative speed ${\bar{v}}_{rel}$.

WHY. Imagine two cars on a motorway. If both drive at 100 km/h in the same direction, they never crash — relative speed zero. If they drive toward each other, closing speed is huge. Collisions care only about closing speed, so individual speeds are the wrong tool; we need ${\bar{v}}_{rel}$.

PICTURE. Figure s03, left panel: same direction → red "approach" arrow is tiny. Right panel: head-on → red arrow is long. The red arrow is the relative velocity.

Figure s03 — same-direction motion gives a tiny approach speed; head-on motion gives a large one. Red = relative velocity.

Statistical mechanics (the Maxwell-Boltzmann Distribution of speeds) gives the average closing speed of a whole gas as:

$\underset{\text{avg approach speed}}{\underbrace{{\bar{v}}_{rel}}}=\sqrt{\frac{8\underset{\text{Boltzmann const (per particle)}}{\underbrace{k_B}}\underset{\text{temperature}}{\underbrace{T}}}{\pi \underset{\text{reduced mass}}{\underbrace{\mu }}}}$

where the reduced mass packages both masses into one effective mass:

$\underset{\text{effective mass}}{\underbrace{\mu }}=\frac{\underset{\text{product of masses}}{\underbrace{m_A{m}_B}}}{\underset{\text{sum of masses}}{\underbrace{m_A+m_B}}}$

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Step 4 — Count all collisions: the collision frequency $Z_{AB}$

WHAT. Multiply the swept volume per second by how many B's sit in each unit of volume, then by how many A's are doing the sweeping. That total count is the collision frequency $Z_{AB}$. Crucially, $Z_{AB}$ already contains the concentrations $[A]$ and $[B]$ — it is a full count of collisions per unit volume per second, not a per-pair rate.

WHY. We built a tube of volume ${\sigma }_{AB}{\bar{v}}_{rel}$ per molecule per second (Steps 2–3). Number of hits = (tube volume) $\times$ (density of B's inside). Then every A does this, so scale by the density of A's. Densities are just the concentrations $[A]$ and $[B]$.

PICTURE. Figure s04 shows a box full of A's (each dragging a red tube) and B's. We tally every B-centre swept up, across the whole box, in one second.

Figure s04 — a box of A's (red target discs) and B's; every enclosed B is a collision, summed over the box gives $Z_{AB}$.

$\underset{{\text{total collisions / m}}^3\text{ / sec}}{\underbrace{Z_{AB}}}=\underset{\text{target area}}{\underbrace{{\sigma }_{AB}}}\underset{\text{approach speed}}{\underbrace{{\bar{v}}_{rel}}}\underset{{\text{A per m}}^3}{\underbrace{[A]}}\underset{{\text{B per m}}^3}{\underbrace{[B]}}$

Read it as knobs: bigger molecules (bigger $\sigma$) → more collisions; hotter gas (bigger ${\bar{v}}_{rel}$) → more collisions; more crowded ($[A],[B]$ up) → more collisions. Every arrow points the intuitive way. Because the $[A][B]$ is already inside $Z_{AB}$, we will never multiply by $[A][B]$ again later.

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Step 5 — Every corner case: does the count behave at its edges?

WHAT. Test the formula at all its extremes — both the "nothing happens" corners and the "runaway" corner $T\to \infty$ — so no reader ever hits a case we skipped.

WHY. A formula you trust must survive its own boundaries. Check every corner, high and low.

PICTURE. Figure s05 stacks four cases: (a) $T\to 0$, molecules freeze; (b) $[B]\to 0$, no partners; (c) $\sigma \to 0$, point-like particles that never touch; (d) $T\to \infty$, the runaway corner where ${\bar{v}}_{rel}\to \infty$.

Figure s05 — three collapse-to-zero limits plus the $T\to \infty$ runaway limit.

Corner case	What happens to $Z_{AB}$	Physical reading
$T\to 0$	${\bar{v}}_{rel}\to 0\Rightarrow Z_{AB}\to 0$	frozen gas, no bumps
$[A]$ or $[B]\to 0$	$Z_{AB}\to 0$	no partner to hit
$\sigma \to 0$ (point particles)	$Z_{AB}\to 0$	zero-size targets never meet
$T\to \infty$	${\bar{v}}_{rel}\propto \sqrt{T}\to \infty \Rightarrow Z_{AB}\to \infty$	see caution below

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Step 6 — Not every collision is hard enough: the energy filter

WHAT. Even a real collision does nothing unless it carries at least the Activation Energy $E_a$ (a molar energy, J/mol — recall the convention box). The fraction of collisions with energy $\ge E_a$ is the Boltzmann factor $e^{-E_a/RT}$, using the molar gas constant $R$.

WHY. Bonds are strong. A gentle tap bounces off. Only collisions above the energy hill $E_a$ can rearrange bonds. The Maxwell-Boltzmann Distribution tells us exactly what fraction of molecules sit in that high-energy tail: it is $e^{-E_a/RT}$ — a number between 0 and 1.

PICTURE. Figure s06 shows the energy distribution curve with the tail beyond $E_a$ shaded red. That red slice is the surviving fraction. Raise $T$ and the red tail fattens.

Figure s06 — Maxwell–Boltzmann energy spread; red tail beyond $E_a$ is the fraction $e^{-E_a/RT}$ that clears the barrier.

$\underset{\text{fraction hard enough}}{\underbrace{f_{\text{energy}}}}=\underset{-\frac{\stackrel{\text{molar energy hill}}{\overbrace{E_a}}}{\underset{\text{molar gas const}}{\underbrace{R}}\underset{\text{temperature}}{\underbrace{T}}}}{\underbrace{e}}$

• $E_a$ big → exponent very negative → tiny fraction survives (hard reaction).
• $T$ big → exponent closer to 0 → fraction rises toward 1 (heat helps a lot).
• $E_a=0$ → $e^0=1$ → every collision is energetic enough.

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Step 7 — Even hard enough can miss: the orientation filter $\rho$

WHAT. Of the collisions that are energetic, only some strike with the reactive parts lined up. That surviving fraction is the steric factor $\rho$ (a number, $0<\rho \le 1$).

WHY. Consider $NO+O_3$. The nitrogen end of $NO$ must meet the ozone; an oxygen-first, backwards, or sideways hit — however violent — just bounces. So after energy, we apply one more probability: is the geometry right? Simple atoms (noble gases) are spheres, every angle works, $\rho \approx 1$. Big floppy molecules have a tiny reactive patch, $\rho$ as low as $10^{-6}$.

PICTURE. Figure s07 shows the same energetic collision in three orientations: reactive patch (red) facing in → success; facing away → bounce; sideways → bounce. Only the red-aligned hit reacts.

Figure s07 — three approach orientations; only the one with the red reactive patch aligned reacts.

$\underset{\text{fraction correctly aimed}}{\underbrace{\rho }}=\frac{\text{reactive-patch area}}{\text{total surface}}\Rightarrow 0<\rho \le 1$

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Step 8 — Multiply the survivors → the rate constant

WHAT. First define the target. The reaction rate is the number of successful reaction events per unit volume per unit time. We build it by starting from $Z_{AB}$ (which already counts all collisions per volume per second, concentrations included) and applying the two filters, then we isolate the plain rate constant $k$.

WHY. Each filter is an independent probability (Step 7's box) slicing the previous survivors. Independent probabilities multiply. This is the whole logic of the derivation compressed to a few lines.

PICTURE. Figure s08 is a funnel: everything pours in at the top ($Z_{AB}$), the energy sieve removes the gentle bumps, the orientation sieve removes the mis-aimed ones, and the trickle out the bottom (red) is the reaction rate.

Figure s08 — the funnel: all collisions $Z_{AB}$ → energy filter → orientation filter → surviving rate (red).

Line 1 — apply both filters to the full collision count. Take every collision ($Z_{AB}$), keep the fraction that is energetic ($\times e^{-E_a/RT}$), then keep the fraction of those that are aimed right ($\times \rho$):

$\underset{{\text{reactions / m}}^3\text{ / sec}}{\underbrace{\text{Rate}}}=\underset{\text{all collisions (already has }[A][B])}{\underbrace{Z_{AB}}}\times \underset{\text{energy filter}}{\underbrace{e^{-E_a/RT}}}\times \underset{\text{orientation filter}}{\underbrace{\rho }}$

Line 2 — expand $Z_{AB}$. Substitute its full form $Z_{AB}={\sigma }_{AB}{\bar{v}}_{rel}[A][B]$ from Step 4, so the concentrations become visible again:

$\text{Rate}=\underset{\text{everything except }[A][B]}{\underbrace{\rho {\sigma }_{AB}{\bar{v}}_{rel}{e}^{-E_a/RT}}}\underset{\text{the two concentrations}}{\underbrace{[A][B]}}$

Line 3 — compare with the rate law and divide out $[A][B]$. A rate law defines the constant $k$ by $\text{Rate}=k[A][B]$. Setting Line 2 equal to $k[A][B]$ and dividing both sides by $[A][B]$ cancels the concentrations and leaves $k$ standing alone:

$k[A][B]=\rho {\sigma }_{AB}{\bar{v}}_{rel}{e}^{-E_a/RT}[A][B]⟹\boxed{k=\underset{\text{orientation}}{\underbrace{\rho }}\underset{\text{collision rate per unit }[A][B]}{\underbrace{{\sigma }_{AB}{\bar{v}}_{rel}}}\underset{\text{energy}}{\underbrace{e^{-E_a/RT}}}}$

Line 4 — read off the frequency factor. Comparing with the empirical Arrhenius equation $k=A{e}^{-E_a/RT}$, everything in front of the exponential is the pre-exponential $A$:

$A=\rho {\sigma }_{AB}{\bar{v}}_{rel}⟺k=\rho Z{e}^{-E_a/RT}$

where $Z={\sigma }_{AB}{\bar{v}}_{rel}$ is the collision rate per unit concentration. So the opening equation $k=\rho Z{e}^{-E_a/RT}$ is now fully earned. The frequency factor was never "just collision frequency" — it is the collision rate per unit concentration times the orientation fraction.

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The one-picture summary

Everything above, in a single flow: molecules bump ($Z_{AB}$, concentrations already inside) → keep only the hard hits ($e^{-E_a/RT}$) → keep only the well-aimed hits ($\rho$) → what survives is the rate, and dividing out $[A][B]$ leaves $k$. Figure s09 compresses the whole derivation: many collisions enter, two filters bite, a red trickle of successful reactions leaves.

Figure s09 — the entire derivation in one frame: collisions in (black), energy $\times$ orientation filters, reactions out (red) — the rate, hence $k=\rho {\sigma }_{AB}{\bar{v}}_{rel}{e}^{-E_a/RT}$.

Recall Feynman retelling — say it back in plain words

Two molecules are balls. A ball only counts a hit when the centres get within the sum of the radii — so each ball drags around a target disc of area $\sigma =\pi (r_A+r_B{)}^2$. In one second the disc sweeps a tube; how long the tube is depends on how fast the two approach, which is the relative speed ${\bar{v}}_{rel}\propto \sqrt{T/\mu }$ — square root of temperature, and reduced mass so a boulder-partner doesn't fool us. (That funny $\sqrt{8/\pi }$ out front is just what you get when you average a lopsided speed bell curve.) Multiply the swept volume by how many partners are around — counted as molecules per cubic metre, so remember the Avogadro conversion — and you've counted every collision: that's $Z_{AB}$, concentrations already baked in, so we never multiply by them again. But most bumps are wasted. First, only the fraction $e^{-E_a/RT}$ hit hard enough to climb the energy hill (that $E_a$ is per mole, paired with $R$) — hotter gas fattens that fraction, but it can never exceed 1. Second, of those, only the fraction $\rho$ point the reactive end the right way — spheres react from any side ($\rho \approx 1$), fussy big molecules almost never line up ($\rho$ tiny). Speed and aim are unrelated, so their chances just multiply. Multiply all three — total collisions, energy survivors, orientation survivors — that's the reaction rate, $\text{Rate}=Z_{AB}{e}^{-E_a/RT}\rho =\rho {\sigma }_{AB}{\bar{v}}_{rel}{e}^{-E_a/RT}[A][B]$. Match it to $\text{Rate}=k[A][B]$, cancel the concentrations, and the number staring back is the plain rate constant $k=\rho {\sigma }_{AB}{\bar{v}}_{rel}{e}^{-E_a/RT}$. And $A$, the "frequency factor"? It's just $\rho {\sigma }_{AB}{\bar{v}}_{rel}$ — the collisions and the aiming, bundled together, drifting gently as $\sqrt{T}$ but swamped by the exponential.

One-line-each self-test: Why $r_A+r_B$? ::: Centres touch when separation equals the sum of the two radii — that's the collision distance. Why relative speed, not individual speed? ::: Collisions depend on how fast molecules close in; same-direction motion gives zero approach. What are $[A],[B]$ inside $Z_{AB}$, in what units? ::: Number densities — molecules per cubic metre — obtained from mol/L via $\times N_A\times 1000$. How do we get $k$ from the Rate expression? ::: Write $\text{Rate}=Z_{AB}{e}^{-E_a/RT}\rho$, compare with $\text{Rate}=k[A][B]$, and divide out $[A][B]$. What happens to $k$ as $T\to \infty$? ::: The energy filter saturates at 1, so $k\to \rho {\sigma }_{AB}{\bar{v}}_{rel}$, growing only as $\sqrt{T}$ — no runaway (and the hard-sphere model breaks first anyway). Why do the energy and orientation fractions multiply? ::: Speed and orientation are statistically independent, so both-happening = product of the two chances. How is plain $k$ different from $k_B$? ::: $k$ is the reaction's rate constant (the answer); $k_B$ is the fixed Boltzmann constant of nature inside the speed formula.

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See also: Transition State Theory (the next model, which explains $\rho$ via the shape of the reaction path), Molecularity, and the Hinglish version.