Collision theory — frequency factor, steric factor: Question bank

This bank drills the ideas from the parent note: the frequency factor and steric factor, and it leans on Activation Energy, the Maxwell-Boltzmann Distribution, Temperature Dependence of Reaction Rates, and Transition State Theory.

The whole toolkit on one page (read before quizzing)

Before the traps, here is every symbol used below, defined in plain words and pinned to a picture. Nothing on this page uses a symbol you have not met here first.

Definition The core equation and its letters

The reaction speed is captured by one master formula (the Arrhenius equation): $k = A e^{- E_{a} / R T}$

$k$ = the rate constant: the "speed dial" of the reaction. Bigger $k$ , faster reaction. Its units match the rate law: $s^{- 1}$ (unimolecular) or $M^{- 1} s^{- 1}$ (bimolecular).
$A$ = the frequency factor: how fast the reaction would go if energy were never a problem. It is built from two pieces (see below).
$E_{a}$ = the activation energy: the minimum collision energy a pair needs to react — the height of the hill in Figure s01.
$R$ = the gas constant $= 8.314 J mol^{- 1} K^{- 1}$ : the fixed conversion factor between temperature and molecular energy.
$T$ = the absolute temperature in kelvin: hotter means faster-moving molecules.
$e^{- E_{a} / R T}$ = the Boltzmann factor: the fraction of collisions energetic enough to clear the hill (the shaded tail in Figure s02).

Formula Why

$\overset{v}{ˉ}_{r e l}$ carries $μ$ , and why $A \propto T$ $\overset{v}{ˉ}_{r e l}$ is the mean relative speed — how fast two molecules approach each other on average. From the Maxwell-Boltzmann Distribution applied to the relative motion: $\overset{v}{ˉ}_{r e l} = \frac{8 k _{B} T}{π μ}$ Here $μ$ is the reduced mass. Why reduced mass? A two-body approach (both molecules moving) can be re-written as one effective particle of mass $μ$ moving toward a fixed point — this is the standard two-body-to-one-body trick, so the two masses collapse into a single $μ$ : $μ = \frac{m _{A} m _{B}}{m _{A} + m _{B}}$ Notice $μ$ is always smaller than either mass: dividing the product by the (larger) sum shrinks it — physically, the pair responds more nimbly than either lump alone.

The $T$ chain (this is the missing derivation): $Z_{unit} = σ \overset{v}{ˉ}_{r e l}, \overset{v}{ˉ}_{r e l} \propto \frac{T}{μ} \Rightarrow Z_{unit} \propto T$ Since $A = ρ Z_{unit}$ and $ρ, σ$ do not depend on $T$ , we get $A \propto T$ Figure s02 shows why this weak $T$ growth is drowned out by the exponential.

Every figure below is referenced by the quiz items that follow — glance at it before answering.

True or false — justify

Every claim is either subtly true or subtly false. State why, don't just verdict.

The frequency factor

A

equals the number of collisions per second.

False.

A = ρ Z_{unit}

— it is the collision frequency

Z_{unit}

scaled down by the steric factor

ρ

for orientation. Only when

ρ \approx 1

(colliding bare atoms) does

A \approx Z_{unit}

The Boltzmann factor

e^{- E_{a} / R T}

already includes the orientation requirement.

False.

e^{- E_{a} / R T}

is purely the fraction with enough energy (the shaded tail in Figure s02). Orientation is a separate, multiplicative filter

ρ

— the two never overlap.

If a reaction has zero activation energy, every collision leads to reaction.

False.

E_{a} = 0

makes

e^{- E_{a} / R T} = 1

(energy filter passes all), but the orientation filter

ρ

still applies — molecules can still hit at the wrong angle unless

ρ = 1

too.

Larger molecules always react faster because they collide more often.

False. They do collide more (bigger cross-section

σ = π (r_{A} + r_{B})^{2}

), but they also have more ways to be mis-oriented, so

ρ

drops sharply — the net rate often falls.

The steric factor

ρ

can never exceed 1.

False in general. Simple collision theory expects

ρ \leq 1

, but a few real reactions (e.g. harpoon/long-range electron transfer, Figure s03 right) show

ρ > 1

— a sign that the hard-sphere model underestimates the true reactive cross-section.

For the identical-molecule reaction

A + A

, we divide

Z

by 2.

True. The factor

\frac{1}{2}

prevents double-counting: "molecule 1 hits molecule 2" and "molecule 2 hits molecule 1" are one and the same collision.

Doubling the absolute temperature roughly doubles the value of

A

False. Since

A \propto T

, doubling

T

raises

A

by only

2 \approx 1.41

, i.e. about 41%. The dramatic rate increase with

T

comes from the exponential term, not from

A

(Figure s02).

The units of

A

are always

s^{- 1}

False.

A

carries the same units as the rate constant

k

s^{- 1}

for unimolecular,

M^{- 1} s^{- 1}

for bimolecular (see Rate Laws).

Relative velocity, not individual velocity, governs collision frequency.

True. What matters is how fast two molecules approach; two molecules travelling side-by-side at equal speed have zero relative velocity and never collide, hence the reduced mass

μ

appears in

\overset{v}{ˉ}_{r e l}

Spot the error

Each statement contains one flaw. Name it and correct it.

k = A e^{+ E_{a} / R T}

, so higher

E_{a}

gives a bigger rate."

The sign is wrong: it is

e^{- E_{a} / R T}

. A larger barrier means a more negative exponent (a taller hill in Figure s01), so fewer collisions clear it and

k

decreases.

"To get the Boltzmann factor I plug

E_{a}

in kJ into

e^{- E_{a} / R T}

with

R = 8.314

Unit mismatch.

R = 8.314 J mol^{- 1} K^{- 1}

, so

E_{a}

must be in J/mol (multiply kJ by 1000) before dividing by

R T

ρ = k / (Z e^{- E_{a} / R T})

came out to 2000, so orientation helps 2000-fold."

A value that large flags a modelling error, not a real orientation boost — likely

Z

was computed too small or

E_{a}

too high. Genuine steric factors for hard spheres sit at or below 1.

"Collision cross-section uses

σ = π r_{A} r_{B}

Wrong combination. It is

σ = π (r_{A} + r_{B})^{2}

— collision happens when centres approach within the sum of radii (Figure s04), so that sum (squared) is the target-disc radius.

"Because

A

contains

ρ

, and

ρ \leq 1

A

must be smaller than

Z_{unit}

always."

Usually true, but not "always": when

ρ > 1

(long-range reactive encounters)

A

exceeds

Z_{unit}

, exposing the limit of the hard-sphere picture.

"Reduced mass

μ = m_{A} + m_{B}

That's the total mass. Reduced mass is

μ = \frac{m _{A} m _{B}}{m _{A} + m _{B}}

, always smaller than either individual mass.

"In an Arrhenius plot of

ln k

1/ T

, the slope gives

A

The slope gives

- E_{a} / R

; the intercept (

1/ T \to 0

) gives

ln A

(Figure s05). Slope is energy, intercept is frequency factor.

Why questions

Answer the mechanism, not just the fact.

Why is the steric factor for reactions between complex molecules (like proteins) as small as

1 0^{- 6}

Only a tiny patch of each large molecule is the reactive site (Figure s03 left); the vast majority of collisions strike inert regions, so the fraction of correctly aligned hits collapses.

Why do we use the reduced mass

μ

instead of the mass of one molecule in

\overset{v}{ˉ}_{r e l}

Because both partners move; the two-body approach is re-cast as one effective particle of mass

μ

moving toward a fixed point, so the two masses collapse into

μ

and the Maxwell-Boltzmann Distribution of the relative speed uses it.

Why can we treat

A

as temperature-independent in ordinary Arrhenius analysis?

Its

T

variation is trivial next to the exponential, which can swing by factors of

1 0^{4}

–

1 0^{6}

over the same range (Figure s02) — so

A

's change is lost in the noise.

Why does collision theory systematically over-predict rates for molecular reactions?

It counts every energetic collision as reactive (

ρ = 1

), ignoring geometry; real molecules need specific alignment, so measured rates fall below the prediction — the gap is the steric factor. Transition State Theory later explained

ρ

from first principles.

Why is orientation irrelevant for a reaction between two noble-gas atoms?

A sphere has no "wrong end" — any approach direction is equivalent, so

ρ \approx 1

and

A \approx Z_{unit}

; this is exactly why early collision theory worked for atomic reactions.

Why does increasing concentration raise the collision frequency

Z

but not the rate constant

k

Z = Z_{unit} [A] [B]

scales with the concentrations, but

k = ρ Z_{unit} e^{- E_{a} / R T}

is defined per unit concentration; concentration effects live in the rate law

Rate = k [A] [B]

, not in

k

itself.

Why must the approach angle, not just which end, sometimes be constrained for a low

ρ

Beyond hitting the right atom, the colliding species must let the reacting orbitals overlap; a glancing hit at the correct site with poor orbital alignment still fails, further shrinking

ρ

Edge cases

The limits and degenerate scenarios the formulas quietly assume.

What happens to

e^{- E_{a} / R T}

T \to \infty

It tends to

1

: every collision now clears the energy barrier, so

k \to ρ Z_{unit} = A

. This is why

A

is called the "maximum possible rate constant" (Figure s02, curve flattens to

A

What happens to

e^{- E_{a} / R T}

T \to 0

It tends to

0

: essentially no molecule has the needed energy, so

k \to 0

and the reaction freezes out — consistent with Temperature Dependence of Reaction Rates.

E_{a} = 0

and

ρ = 1

, what is

k

k = Z_{unit}

— the collision-limited maximum, since both filters pass everything and

A = ρ Z_{unit} = Z_{unit}

. This is the fastest a bimolecular reaction can go (diffusion/collision limit).

What does

ρ \to 1

physically demand about the colliding species?

They must react from any mutual orientation — true only for structureless or highly symmetric partners like single atoms; molecules with a specific reactive site cannot reach

ρ = 1

If a computed

ρ

comes out negative, what went wrong?

Impossible physically — a negative fraction signals a sign or unit error (often the

E_{a}

What happens to collision frequency for a very heavy pair (large

μ

\overset{v}{ˉ}_{r e l} = 8 k_{B} T / π μ

falls as

μ

grows, so heavier partners move sluggishly and collide less often, lowering

Z_{unit}

even before orientation is considered.

Recall One-line self-test

Cover every answer above and re-derive the reasoning (not the verdict) for three items you found hardest. If you can explain the because, you own the concept.