Worked examples — Collision theory — frequency factor, steric factor

This page is the practice arena for the parent topic. The parent built the ideas: collision frequency $Z$, the Boltzmann energy filter $e^{-E_a/RT}$, the orientation filter $\rho$, and how they multiply into the rate constant $k=\rho Z{e}^{-E_a/RT}$. Here we use them — and we deliberately hunt down every kind of question the topic can throw at you.

Before any symbol appears in a solution, remember what each one is, in plain words:

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The scenario matrix

Every problem in collision theory is really one of these cells. We will make sure each cell gets a worked example.

Cell	What varies / what's degenerate	Example
C1 Forward count	Compute $Z$ from radii, $T$, masses	Ex 1
C2 Identical partners	$A+A$ case with the $\frac{1}{2}$ double-count fix	Ex 2
C3 Solve backward	Given $k$, $Z$, $E_a$ → find $\rho$	Ex 3
C4 Limiting $\rho$	$\rho \to 1$ (noble-gas / atomic) vs $\rho \ll 1$ (big molecule)	Ex 4
C5 Degenerate speed	Same velocity → relative speed $\to 0$; and $T\to 0$ / $T\to \infty$	Ex 5
C6 Temperature scaling	How $A$ and $e^{-E_a/RT}$ each move when $T$ changes	Ex 6
C7 Real-world word problem	Atmospheric ozone destruction, full chain	Ex 7
C8 Exam twist	"Why is measured $A$ bigger than $Z$?" trap	Ex 8

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[!example] Ex 1 (C1) — Count the collisions: H₂ + I₂

Statement. At $T=500\text{K}$, take $r_{{\text{H}}_2}=1.0\text{A˚}$, $r_{{\text{I}}_2}=2.0\text{A˚}$, $[{\text{H}}_2]=[{\text{I}}_2]=0.1\text{M}$. Find the collision frequency $Z$.

Forecast: guess the order of magnitude — is $Z$ nearer $10^5$ or $10^{36}$ collisions per m³ per second? Write your guess before reading on.

Step 1 — Collision cross-section. $\sigma =\pi (r_{{\text{H}}_2}+r_{{\text{I}}_2}{)}^2=\pi (3.0\times 10^{-10}\text{m}{)}^2=2.83\times 10^{-19}{\text{m}}^2$ Why this step? A hit happens when the two centres come within $r_A+r_B$; that distance is the radius of the "target disc" one molecule sweeps out. In the figure below, molecule A (blue) sits at the centre and the yellow dashed circle is that target disc — any B-centre (red) entering it counts as a collision. Watch how the disc radius is $r_A+r_B$, the sum of both radii, not either one alone.

Step 2 — Reduced mass. $\mu =\frac{m_{{\text{H}}_2}{m}_{{\text{I}}_2}}{m_{{\text{H}}_2}+m_{{\text{I}}_2}}=\frac{(2)(254)}{256}\frac{\text{g/mol}}{N_A}=3.30\times 10^{-27}\text{kg}$ Why this step? What matters is how fast the pair approaches, and that approach behaves like a single body of mass $\mu$. (Convert grams to kg: divide $\text{g/mol}$ by $N_A$ to get grams per molecule, then by $10^3$ for kg.)

Step 3 — Mean relative speed. ${\bar{v}}_{\text{rel}}=\sqrt{\frac{8k_{B}T}{\pi \mu }}=\sqrt{\frac{8(1.38\times 10^{-23})(500)}{\pi (3.30\times 10^{-27})}}\approx 2.31\times 10^3\text{m/s}$ Why this step? Collisions per second scale with how quickly molecules close the gap; the $\sqrt{T}$ comes straight out of the speed distribution. (The light, low-mass reduced pair moves fast — a few km/s is exactly what kinetic theory predicts.)

Step 4 — Convert concentration, then multiply. $0.1\text{M}=0.1\frac{\text{mol}}{\text{L}}\times 10^3\frac{\text{L}}{{\text{m}}^3}\times N_A=6.02\times 10^{25}{\text{molecules/m}}^3$ $Z=\sigma {\bar{v}}_{\text{rel}}[{\text{H}}_2][{\text{I}}_2]=(2.83\times 10^{-19})(2.31\times 10^3)(6.02\times 10^{25}{)}^2\approx 2.37\times 10^{36}\frac{\text{coll}}{{\text{m}}^3\text{s}}$ Why this step? Each factor is one ingredient: how big a target, how fast, how many of each.

Verify. Units: ${\text{m}}^2\cdot \frac{\text{m}}{\text{s}}\cdot \frac{1}{{\text{m}}^3}\cdot \frac{1}{{\text{m}}^3}=\frac{1}{{\text{m}}^3\text{s}}$ ✓. The huge number ($\sim 10^{36}$) is expected — that is why reactions need the energy and orientation filters to slow them down.

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[!example] Ex 2 (C2) — Identical partners: the ½ trap

Statement. For ${\text{NO}}_2+{\text{NO}}_2\to$ products at $500\text{K}$, $\sigma =4.0\times 10^{-19}{\text{m}}^2$, $m=46\text{g/mol}$, $[{\text{NO}}_2]=6.02\times 10^{24}{\text{m}}^{-3}$. Find $Z_{AA}$.

Forecast: will $Z_{AA}$ be bigger or smaller than the "naive" $\sigma \bar{v}[A{]}^2$? Guess the factor.

Step 1 — Reduced mass of identical partners. $\mu =\frac{m\cdot m}{m+m}=\frac{m}{2}=\frac{46}{2}\frac{\text{g/mol}}{N_A}=3.82\times 10^{-26}\text{kg}$ Why this step? Two equal masses give $\mu =m/2$ — the effective inertia is half a molecule's.

Step 2 — Relative speed. ${\bar{v}}_{\text{rel}}=\sqrt{\frac{8k_{B}T}{\pi \mu }}=\sqrt{\frac{8(1.38\times 10^{-23})(500)}{\pi (3.82\times 10^{-26})}}\approx 6.78\times 10^2\text{m/s}$ Why this step? ${\text{NO}}_2$ is far heavier than the H₂/I₂ reduced pair, so its thermal speed is lower — a few hundred m/s.

Step 3 — Apply the ½. $Z_{AA}=\frac{1}{2}\sigma {\bar{v}}_{\text{rel}}[A{]}^2=\frac{1}{2}(4.0\times 10^{-19})(678)(6.02\times 10^{24}{)}^2\approx 4.92\times 10^{33}\frac{\text{coll}}{{\text{m}}^3\text{s}}$ Why this step? Molecule 1 hitting molecule 2 is the same event as 2 hitting 1. Counting the pair once means dividing by two.

Verify. Drop the $\frac{1}{2}$ and you'd get $9.83\times 10^{33}$ — exactly double, i.e. every collision counted twice. The physical (correct) answer is the smaller one, $4.92\times 10^{33}$.

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[!example] Ex 3 (C3) — Solve backward for ρ: NO + O₃

Statement. For $\text{NO}+{\text{O}}_3\to {\text{NO}}_2+{\text{O}}_2$ at $298\text{K}$: measured $k=1.8\times 10^7{\text{M}}^{-1}{\text{s}}^{-1}$, $E_a=10\text{kJ/mol}$, computed $Z=5.0\times 10^{10}{\text{M}}^{-1}{\text{s}}^{-1}$. Find $\rho$.

Forecast: is $\rho$ closer to $1$ or to $0.01$? A three-atom molecule has to be aimed carefully...

Step 1 — Boltzmann fraction. $e^{-E_a/RT}=\exp \left(\frac{-10000}{(8.314)(298)}\right)=e^{-4.036}=0.0177$ Why this step? Only this fraction of collisions clears the energy hill; we must strip that out before we can see the orientation effect alone. Use joules, so $E_a=10000\text{J/mol}$ matches $R=8.314{\text{J mol}}^{-1}{\text{K}}^{-1}$.

Step 2 — Rearrange $k=\rho Z{e}^{-E_a/RT}$. $\rho =\frac{k}{Z{e}^{-E_a/RT}}=\frac{1.8\times 10^7}{(5.0\times 10^{10})(0.0177)}=\frac{1.8\times 10^7}{8.85\times 10^8}\approx 0.020$ Why this step? Everything except orientation is now on the right; what's left, $\rho$, is the pure geometry fraction. Both $k$ and $Z$ are in ${\text{M}}^{-1}{\text{s}}^{-1}$, so $\rho$ comes out dimensionless — as a fraction must.

Verify. $\rho =0.02<1$ ✓ — sensible for a directional molecule. Rebuild $k$: $(0.02)(5.0\times 10^{10})(0.0177)=1.77\times 10^7\approx k$ ✓. Interpretation: only 2 % of energetic hits are aimed right (N-end toward ${\text{O}}_3$).

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[!example] Ex 4 (C4) — The two limits of ρ

Statement. Compare (a) $\text{Ar}+\text{Ar}$ (spherical atoms, no preferred direction) and (b) a protein-substrate collision where the active site covers $10^{-4}$ of the surface. What $\rho$ does collision theory predict for each?

Forecast: for a sphere, does orientation even mean anything?

Step 1 — Spherical atom (limit $\rho \to 1$). A single atom looks identical from every direction, so every energetic hit is "correctly oriented." ${\rho }_{\text{Ar}}\approx 1$ Why this step? Orientation is only a penalty when the molecule has a special reactive spot. A sphere has none.

Step 2 — Big molecule (limit $\rho \ll 1$). If the reactive patch is a fraction $f=10^{-4}$ of the surface, the geometric chance of a right approach is roughly that fraction: ${\rho }_{\text{protein}}\approx 10^{-4}$ Why this step? The reader arrives with random aim; only aim landing on the tiny patch reacts.

Verify. These bracket the whole real range: small molecules like ${\text{H}}_2+{\text{I}}_2$ sit between at $\rho \sim 0.1$–$0.5$. So $10^{-6}\le \rho \le 1$ covers everything — no fourth case exists. (See Transition State Theory for a deeper reason $\rho$ can even exceed $1$ — Ex 8.)

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[!example] Ex 5 (C5) — Degenerate & limiting speeds

Statement. (a) Two molecules travel side-by-side at the same velocity. What is $Z$? (b) What happens to ${\bar{v}}_{\text{rel}}$ as $T\to 0$ and as $T\to \infty$?

Forecast: can two molecules moving fast still never collide?

Step 1 — Equal velocities → zero relative speed. ${\vec{v}}_{\text{rel}}={\vec{v}}_A-{\vec{v}}_B=\vec{0}\Rightarrow Z\propto {\bar{v}}_{\text{rel}}=0$ Why this step? Collisions depend on closing the gap. If both move identically, the gap never shrinks — no matter how fast they go. In the figure, the top pair (green arrows, equal length and direction) keeps a fixed separation forever; the bottom pair (blue and red arrows pointing toward each other) closes the gap and collides. Look at the arrow directions, not their lengths — that is what decides a collision.

Step 2 — Cold limit $T\to 0$. ${\bar{v}}_{\text{rel}}=\sqrt{\frac{8k_{B}T}{\pi \mu }}\to 0$ Why this step? No thermal energy means no motion, so molecules never find each other — $Z\to 0$ and reactions freeze.

Step 3 — Hot limit $T\to \infty$. ${\bar{v}}_{\text{rel}}\to \infty ,e^{-E_a/RT}\to e^0=1$ so $k\to \rho Z\to A$ (times a slow $\sqrt{T}$ growth). Why this step? At infinite temperature every collision has enough energy; the energy filter disappears and the rate constant reaches its ceiling $A$ — exactly why $A$ is called the "maximum" rate constant.

Verify. All three are consistent with ${\bar{v}}_{\text{rel}}\propto \sqrt{T}$: it is $0$ at $T=0$, grows without bound, and the ceiling $k\to A$ matches ${\lim }_{T\to \infty }{e}^{-E_a/RT}=1$. ✓

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[!example] Ex 6 (C6) — Which factor moves more with T?

Statement. A reaction has $E_a=100\text{kJ/mol}$. Compare how the frequency factor $A$ and the Boltzmann term $e^{-E_a/RT}$ each change when $T$ goes $300\to 600\text{K}$.

Forecast: the parent says "treat $A$ as constant." By how many powers of ten is that a good approximation?

Step 1 — Change in $A$ (via $Z\propto \sqrt{T}$). $\frac{A(600)}{A(300)}=\sqrt{\frac{600}{300}}=\sqrt{2}=1.41$ Why this step? Only ${\bar{v}}_{\text{rel}}\propto \sqrt{T}$ inside $A$ depends on $T$; $\rho$ and $\sigma$ don't.

Step 2 — Change in the Boltzmann term. $\frac{e^{-E_a/(R\cdot 600)}}{e^{-E_a/(R\cdot 300)}}=\exp \left[\frac{E_a}{R}\left(\frac{1}{300}-\frac{1}{600}\right)\right]=\exp \left[\frac{100000}{8.314}\cdot \frac{1}{600}\right]=e^{20.05}\approx 5.1\times 10^8$ Why this step? This isolates the exponential's response; it moves by eight orders of magnitude.

Verify. Ratio of the two effects: $\frac{5.1\times 10^8}{1.41}\approx 3.6\times 10^8$. The exponential outruns $A$ by hundreds of millions to one — that is the rigorous justification for holding $A$ fixed in an Arrhenius plot.

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[!example] Ex 7 (C7) — Word problem: ozone in the stratosphere

Statement. Stratospheric ozone is destroyed by $\text{NO}+{\text{O}}_3\to {\text{NO}}_2+{\text{O}}_2$. Using Ex 3's numbers ($k=1.8\times 10^7{\text{M}}^{-1}{\text{s}}^{-1}$), with $[\text{NO}]=1.0\times 10^{-9}\text{M}$ and $[{\text{O}}_3]=5.0\times 10^{-8}\text{M}$, find the destruction rate of ozone.

Forecast: with a fast $k$ but tiny concentrations, is the rate large or vanishing?

Step 1 — Write the rate law. $\text{Rate}=k[\text{NO}][{\text{O}}_3]$ Why this step? Collision theory built a bimolecular (one-step, two-body) mechanism, so the rate is first order in each reactant.

Step 2 — Plug numbers. $\text{Rate}=(1.8\times 10^7)(1.0\times 10^{-9})(5.0\times 10^{-8})=9.0\times 10^{-10}\text{M/s}$ Why this step? This turns molecular collision counting into a lab-measurable rate.

Verify. Units: ${\text{M}}^{-1}{\text{s}}^{-1}\cdot \text{M}\cdot \text{M}=\text{M/s}$ ✓. Small concentrations dominate over the fast $k$, giving a slow but relentless rate — physically why a little NO catalytically damages a lot of ozone over time.

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[!example] Ex 8 (C8) — Exam twist: measured A larger than Z

Statement. A student measures $A=3\times 10^{13}{\text{M}}^{-1}{\text{s}}^{-1}$ but computes $Z=1\times 10^{13}{\text{M}}^{-1}{\text{s}}^{-1}$. This gives $\rho =A/Z=3>1$. "A fraction can't exceed 1 — is the data wrong?"

Forecast: must $\rho$ always be less than $1$?

Step 1 — Compute the apparent $\rho$. $\rho =\frac{A}{Z}=\frac{3\times 10^{13}}{1\times 10^{13}}=3$ Why this step? It exposes the paradox honestly before resolving it.

Step 2 — Diagnose. Simple collision theory treats molecules as hard spheres. Real molecules have long-range attractions (dipoles, ion pairs) that funnel partners together, so more reactive encounters happen than the hard-sphere $Z$ predicts. Why this step? It identifies which assumption broke — not the data, the model.

Step 3 — The fix. Interpret $\rho >1$ as a signal to upgrade to Transition State Theory, which handles attraction and loose transition states naturally. Why this step? $\rho$ is a correction factor, not a literal probability; values slightly above $1$ are physically meaningful.

Verify. $\rho =3$ is not an error: reactions between ions or strongly polar species routinely show $\rho >1$. The lesson — a "fraction over one" flags the limits of the sphere model, not a mistake. ✓

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Recall Self-test

Which matrix cell forces the $\frac{1}{2}$ factor? ::: C2 — identical collision partners ($A+A$), to avoid double counting. When $T\to \infty$, what does $k$ approach? ::: $k\to A=\rho Z$, because $e^{-E_a/RT}\to 1$. If side-by-side molecules share one velocity, what is $Z$? ::: Zero — relative speed is zero, so the gap never closes. A measured $\rho$ of $3$ means the data is wrong. True/False? ::: False — it flags long-range attraction; upgrade to Transition State Theory. In Ex 3, what fraction of energetic collisions were correctly aimed? ::: About $2\%$ ($\rho \approx 0.02$). In Ex 1, roughly what order of magnitude is the raw collision frequency $Z$? ::: About $10^{36}$ collisions per m³ per second.