Collision theory — frequency factor, steric factor: Exercises

This page is a self-test ladder for the parent topic. Each problem is graded from L1 (just recognise the idea) up to L5 (mastery — combine everything). Every solution is hidden inside a collapsible callout so you can try first, then reveal.

Before we start, let us pin down every symbol so nobody is lost from line one.

Definition The symbols you will use on this page

$k$ ::: the rate constant — the number that tells you how fast a reaction goes at a given temperature.
$A$ ::: the frequency factor (pre-exponential factor) — the value $k$ would reach if every collision had enough energy.
$Z$ ::: the collision frequency — how many collisions happen per unit volume per unit time.
$ρ$ ::: the steric factor — the fraction of energetic collisions that are also pointed the right way.
$E_{a}$ ::: the activation energy — the minimum collision energy needed to react (see Activation Energy).
$R = 8.314 J mol^{- 1} K^{- 1}$ ::: the gas constant.
$T$ ::: absolute temperature in kelvin (K).
$e^{- E_{a} / R T}$ ::: the Boltzmann factor — the fraction of collisions that clear the energy bar (from the Maxwell-Boltzmann Distribution).

The one master equation everything below flows from:

Figure — Collision theory — frequency factor, steric factor

The three filters above are independent gates — a collision must pass all three. That picture is the key to every problem here.

L1 — Recognition

Problem 1.1

Which of these has units of $s^{- 1}$ (never $M^{- 1} s^{- 1}$ ) no matter what: the steric factor $ρ$ , the collision frequency $Z$ , or the frequency factor $A$ ?

Recall Solution

The steric factor $ρ$ is a pure fraction — it is a ratio of "aimed-right collisions" to "energetic collisions", so all units cancel. It is dimensionless (which we can loosely write as $s^{0}$ , i.e. no units at all). Both $A$ and $Z$ carry the same units as $k$ (for a bimolecular step, $M^{- 1} s^{- 1}$ ). So the answer is $ρ$ — it is unitless.

Problem 1.2

State, in one sentence each, what the three factors $Z$ , $e^{- E_{a} / R T}$ , and $ρ$ physically filter for.

Recall Solution

$Z$ counts how often the molecules meet.
$e^{- E_{a} / R T}$ is the fraction hitting hard enough (energy $\geq E_{a}$ ).
$ρ$ is the fraction pointed correctly among those energetic hits.

L2 — Application

Problem 2.1

A reaction has $Z = 4.0 \times 1 0^{10} M^{- 1} s^{- 1}$ and a measured frequency factor $A = 8.0 \times 1 0^{8} M^{- 1} s^{- 1}$ . Find the steric factor $ρ$ .

Recall Solution

WHAT we do: invert $A = ρZ$ . WHY: $A$ and $Z$ are given, $ρ$ is the only unknown. $ρ = \frac{A}{Z} = \frac{8.0 \times 1 0 ^{8}}{4.0 \times 1 0 ^{10}} = 0.020$ Meaning: only 2% of energetic collisions are oriented well enough to react.

Problem 2.2

At $T = 300 K$ , a reaction has $E_{a} = 50 kJ mol^{- 1}$ . What fraction of collisions clear the energy bar?

Recall Solution

WHAT/WHY: we want the Boltzmann factor, so we compute $e^{- E_{a} / R T}$ — the single tool that answers "what fraction has energy $\geq E_{a}$ ". Keep units consistent: $E_{a} = 50000 J mol^{- 1}$ , $R = 8.314$ , $T = 300$ . $\frac{E _{a}}{R T} = \frac{50000}{8.314 \times 300} = \frac{50000}{2494.2} = 20.05$ $e^{- 20.05} = 1.96 \times 1 0^{- 9}$ Meaning: about 2 in a billion collisions are energetic enough.

L3 — Analysis

Problem 3.1

For $NO + O_{3} \to NO_{2} + O_{2}$ at $298 K$ : observed $k = 1.8 \times 1 0^{7} M^{- 1} s^{- 1}$ , $E_{a} = 10 kJ mol^{- 1}$ , collision frequency $Z = 5.0 \times 1 0^{10} M^{- 1} s^{- 1}$ . Find $ρ$ and interpret it.

Recall Solution

Step 1 — energy gate. Compute the Boltzmann factor: $\frac{E _{a}}{R T} = \frac{10000}{8.314 \times 298} = 4.036, e^{- 4.036} = 0.0177$ Step 2 — invert the master equation. From $k = ρZ e^{- E_{a} / R T}$ : $ρ = \frac{k}{Z e ^{- E_{a} / R T}} = \frac{1.8 \times 1 0 ^{7}}{( 5.0 \times 1 0 ^{10} ) ( 0.0177 )} = \frac{1.8 \times 1 0 ^{7}}{8.85 \times 1 0 ^{8}} = 0.0203$ Interpretation: roughly $2%$ of the energetic collisions have the N-end of NO aimed at $O_{3}$ correctly for electron transfer. The other $98%$ are energetic but mis-aimed glancing blows.

Problem 3.2

Two reactions have the same $Z$ and same $E_{a}$ . Reaction P (small atoms) has $ρ = 1$ ; reaction Q (bulky molecules) has $ρ = 1 0^{- 5}$ . By what factor is P faster than Q?

Recall Solution

Since $Z$ , $E_{a}$ , $T$ are shared, they cancel in the ratio: $\frac{k _{P}}{k _{Q}} = \frac{ρ _{P} Z e ^{- E_{a} / R T}}{ρ _{Q} Z e ^{- E_{a} / R T}} = \frac{ρ _{P}}{ρ _{Q}} = \frac{1}{1 0 ^{- 5}} = 1 0^{5}$ Meaning: orientation alone makes P one hundred thousand times faster. This is why complex-molecule reactions (see Transition State Theory) are so much slower than atom recombinations.

L4 — Synthesis

Problem 4.1

A bimolecular gas reaction is studied at $300 K$ then at $600 K$ (doubling $T$ ). Its $E_{a} = 100 kJ mol^{- 1}$ , and $ρ$ is temperature-independent. Because $Z \propto T$ , by what factor does (a) $A$ change, and (b) the Boltzmann factor change? Which dominates $k$ ?

Recall Solution

(a) Frequency factor. Since $A = ρZ$ and $Z \propto T$ : $\frac{A ( 600 )}{A ( 300 )} = \frac{600}{300} = 2 = 1.41$ (b) Boltzmann factor. Compute both exponents. $\frac{E _{a}}{R ( 300 )} = \frac{100000}{8.314 \times 300} = 40.09, \frac{E _{a}}{R ( 600 )} = 20.05$ $\frac{e ^{- 20.05}}{e ^{- 40.09}} = e^{40.09 - 20.05} = e^{20.04} = 5.0 \times 1 0^{8}$ Which dominates? The Boltzmann factor grows by $\sim 5 \times 1 0^{8}$ ; $A$ grows by only $1.41$ . The exponential utterly dominates — this is why we treat $A$ as constant in Arrhenius plots.

Problem 4.2

Combine the collision-frequency formula with the master equation. Given $σ = 2.83 \times 1 0^{- 19} m^{2}$ , $\overset{v}{ˉ}_{r e l} = 631 m s^{- 1}$ , number densities $[A] = [B] = 6.02 \times 1 0^{25} m^{- 3}$ , find the raw collision frequency $Z_{A B}$ (collisions per m³ per s).

Recall Solution

WHY this formula: $Z_{A B} = σ \overset{v}{ˉ}_{r e l} [A] [B]$ multiplies cross-section (how big a target), approach speed (how fast they meet), and how many of each are present. $Z_{A B} = (2.83 \times 1 0^{- 19}) (631) (6.02 \times 1 0^{25})^{2}$ First the density product: $(6.02 \times 1 0^{25})^{2} = 3.624 \times 1 0^{51}$ . Then $2.83 \times 1 0^{- 19} \times 631 = 1.786 \times 1 0^{- 16}$ . $Z_{A B} = 1.786 \times 1 0^{- 16} \times 3.624 \times 1 0^{51} = 6.47 \times 1 0^{35} collisions m^{- 3} s^{- 1}$

L5 — Mastery

Problem 5.1

A reaction at $500 K$ has $E_{a} = 80 kJ mol^{- 1}$ , $Z = 1.0 \times 1 0^{11} M^{- 1} s^{- 1}$ , and a measured $k = 3.0 \times 1 0^{- 2} M^{- 1} s^{- 1}$ . (a) Predict $k$ assuming $ρ = 1$ (perfect orientation). (b) Compare to the measured $k$ and extract the true $ρ$ . (c) What does $ρ$ tell you about the geometry of the reactive site?

Recall Solution

(a) Ideal prediction, $ρ = 1$ . $\frac{E _{a}}{R T} = \frac{80000}{8.314 \times 500} = 19.245, e^{- 19.245} = 4.39 \times 1 0^{- 9}$ $k_{ρ = 1} = Z e^{- E_{a} / R T} = (1.0 \times 1 0^{11}) (4.39 \times 1 0^{- 9}) = 4.39 \times 1 0^{2} M^{- 1} s^{- 1}$ (b) Extract $ρ$ . $ρ = \frac{k _{meas}}{k _{ρ = 1}} = \frac{3.0 \times 1 0 ^{- 2}}{4.39 \times 1 0 ^{2}} = 6.83 \times 1 0^{- 5}$ (c) Geometry. $ρ \approx 7 \times 1 0^{- 5}$ means only about 7 in 100 000 energetic collisions are aimed correctly. This is the signature of a bulky reactant where the reactive site covers a tiny patch of the surface — consistent with the molecularity involving large, orientation-sensitive species.

Problem 5.2

Show algebraically that the ratio of two rate constants at the same temperature depends only on the two frequency factors when the activation energies are equal — i.e. derive $k_{1} / k_{2} = A_{1} / A_{2}$ , and state when this fails.

Recall Solution

WHAT/WHY: write each $k$ in Arrhenius form and divide, so the shared machinery cancels. $\frac{k _{1}}{k _{2}} = \frac{A _{1} e ^{- E_{a, 1} / R T}}{A _{2} e ^{- E_{a, 2} / R T}} = \frac{A _{1}}{A _{2}} e^{- (E_{a, 1} - E_{a, 2}) / R T}$ If $E_{a, 1} = E_{a, 2}$ , the exponential is $e^{0} = 1$ , leaving $\frac{k _{1}}{k _{2}} = \frac{A _{1}}{A _{2}}$ Since $A = ρZ$ , and $Z$ is nearly the same for similar-sized molecules, this ratio is really a ratio of steric factors. It fails whenever the activation energies differ: then the exponential term does not cancel and dominates, exactly as in Problem 4.1.

Recall Quick self-check clozes

The frequency factor equals == $A = ρZ$ . The three independent gates a reactive collision must pass are collision frequency $Z$ , energy $e^{- E_{a} / R T}$ , and orientation $ρ$ ==. Doubling $T$ (300→600 K, $E_{a} = 100$ kJ) boosts $A$ by == $2 \approx 1.41$ but the Boltzmann factor by $\sim 5 \times 1 0^{8}$ ==. A computed $ρ > 1$ means the hard-sphere collision model has broken down; use transition-state theory.