Catalytic properties — examples (V₂O₅, Fe, Ni, Pt): Visual walkthrough

We only need one new idea from outside: coverage. Everything else we picture.

Step 1 — WHAT is "coverage" $θ$ ? (build the symbol before we use it)

WHY we need this one number. A reaction on a surface can only happen where reactant molecules are actually sitting. So the first thing we must count is how full the surface is. That single fraction $θ$ is what the whole derivation will turn on.

PICTURE. Three snapshots of the same surface: nearly empty ( $θ$ small), half-full, and jammed ( $θ \approx 1$ ). The blue dots are metal sites; orange discs are stuck gas molecules.

Figure — Catalytic properties — examples (V₂O₅, Fe, Ni, Pt)

Step 2 — WHAT must happen for a reaction? Two ingredients must MEET

For a surface reaction (say $H_{2} + C=C$ on Ni), you need two different things next to each other:

A molecule already stuck to a site (so its bond is weakened and ready).
A free neighbouring site for the partner to land on, or for the product to peel off into.

WHY multiply? Probabilities of two independent conditions both being true get multiplied (like needing two coins both to land heads: $\frac{1}{2} \times \frac{1}{2}$ ). Here:

Chance a given spot is occupied $= θ$ .
Chance its neighbour is empty $= 1 - θ$ (whatever isn't occupied is empty, and the two must add to the whole: $θ + (1 - θ) = 1$ ).

So the chance of finding the winning "occupied-next-to-empty" pair is: $this spot filled θ \times neighbour free (1 - θ)$

PICTURE. Two adjacent sites highlighted: left one filled (orange), right one empty (blue outline) — the only arrangement that lets the reaction fire. Beside it, the two failing arrangements crossed out in red.

Step 3 — WHAT is $k$ ? Fold in the intrinsic speed

Finding the right pair is necessary but not sufficient — once the pair exists, the atoms still have to actually rearrange. That intrinsic per-pair speed is a rate constant.

WHY this exact shape and not, say, $k θ$ ? A plain $k θ$ would say "more coverage is always better" — which is wrong, because a jammed surface has nowhere to react. The extra $(1 - θ)$ factor is the mathematics remembering that the surface also needs empty room.

PICTURE. The Arrhenius barrier: a hill of height $E_{a}$ ; the catalyst lowers the hill, so the green "molecules with enough energy" fraction grows.

Step 4 — WHERE does $θ (1 - θ)$ peak? Find the summit

We now have $rate = k θ (1 - θ)$ . Treat $k$ as fixed and ask: which coverage gives the biggest rate?

Expand first so we can see the shape: $θ (1 - θ) = θ - θ^{2}$

$θ$ — grows the rate as more sites fill.
$- θ^{2}$ — fights back harder and harder as the surface clogs.

This is an upside-down parabola (a hump). A hump has exactly one top. By symmetry — or by asking "where does the rise ( $θ$ ) exactly balance the choke ( $θ^{2}$ )?" — the top sits at the midpoint: $θ^{⋆} = \frac{1}{2}$

At the summit: $rate_{m a x} = k \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} k$

PICTURE. The full parabola of $θ (1 - θ)$ vs $θ$ : zero at both ends, a single peak at $θ = \frac{1}{2}$ marked in orange with value $0.25$ .

Step 5 — EDGE CASE ①: binding too WEAK ( $θ \to 0$ )

If the metal barely grips the gas (weak chemisorption), almost nothing sticks, so $θ \approx 0$ .

Plug in: $rate = k \cdot θ 0 \cdot = 1 (1 - 0) = 0$

The $θ$ factor collapses the rate: no reactant on the surface, no reaction.
Notice the free-site factor $(1 - θ) = 1$ is perfectly happy — plenty of room — but rooms full of nothing react with nothing.

Which metals? The right-hand side of a period, e.g. $Z n^{2 +}$ with a full $d^{10}$ configuration — no hungry half-empty d-orbital to grip the gas. This is exactly why Zn is a poor catalyst, matching §5 of the parent.

PICTURE. A nearly-empty surface; a single $H_{2}$ molecule bouncing straight back off (it never sticks). Rate meter reads ~0.

Step 6 — EDGE CASE ②: binding too STRONG ( $θ \to 1$ )

If the metal grips too hard, every site fills and products refuse to leave — the surface poisons itself, $θ \approx 1$ .

Plug in: $rate = k \cdot θ 1 \cdot = 0 (1 - 1) = 0$

Now the $θ$ factor is happy (surface loaded), but $(1 - θ) = 0$ kills it: no free site means the product can never desorb, and no fresh reactant can land.
This is the same maths behind catalyst poisoning: a strong-binding impurity (As, S, CO) drives its own local $θ \to 1$ and freezes the site.

Which metals? Far-left transition metals that bind adsorbates too tightly. So both ends of the series fail — for opposite reasons — and only the middle survives. That two-sided failure is the "diagnostic test" the parent note points to.

PICTURE. A jammed surface, every site orange, a finished product molecule glued down with a red "STUCK" tag; nothing can move. Rate meter reads ~0.

Step 7 — PUT IT TOGETHER: the volcano curve

Sweep binding strength from weak (left) to strong (right). Coverage $θ$ climbs $0 \to 1$ as we go, and the rate $k θ (1 - θ)$ traces a volcano: zero at the weak end, zero at the strong end, a peak in between.

WHY Ni, Pt, Pd sit on top. Their partly-filled d-orbitals grip $H_{2}$ , $N_{2}$ , C=C moderately — strong enough to weaken the bond (parent §3), weak enough to let product leave. They land near $θ \approx \frac{1}{2}$ , the summit. $V_{2} O_{5}$ works by a different engine — the oxidation-state relay $V^{+ 5} ⇌ V^{+ 4}$ (parent §4a) — but the same "regenerate unchanged" logic applies.

PICTURE. The volcano plot: x-axis "binding strength", y-axis "rate"; weak-end metals and strong-end metals both low, Ni/Pt marked at the peak.

The one-picture summary

Recall Feynman retelling of the whole walkthrough

Picture a metal surface as a car park of hungry parking spots. Fill a fraction $θ$ of them with gas molecules. To react you need one filled spot next to one empty spot — a molecule ready to go and room for its partner (and for the product to drive off). The chance of "filled next to empty" is $θ \times (1 - θ)$ . Multiply by how fast a ready pair actually reacts, $k$ , and you get $rate = k θ (1 - θ)$ . Now stress-test the ends. Empty park ( $θ = 0$ ): nothing stuck, nothing reacts. Jammed park ( $θ = 1$ ): no room, products can't leave, nothing reacts. Both ends give zero. So the rate must hump in the middle, peaking when the park is exactly half full, $θ = \frac{1}{2}$ . That hump is the volcano. Metals that grip too weakly (Zn, $d^{10}$ ) sit on the empty-park side; metals that grip too strongly (and any poison like CO or S) sit on the jammed-park side. Ni, Pt, Pd grip just right — half-full park, top of the volcano. That, and nothing more mysterious, is Sabatier's principle.

Recall Quick self-test

Why does rate $= 0$ when $θ = 0$ ? ::: The $θ$ factor is zero — no reactant on the surface. Why does rate $= 0$ when $θ = 1$ ? ::: The $(1 - θ)$ factor is zero — no free site, products can't desorb. At what coverage is the rate maximum? ::: $θ = \frac{1}{2}$ (top of the upside-down parabola $θ - θ^{2}$ ). What is $rate_{m a x}$ in terms of $k$ ? ::: $\frac{1}{4} k$ . Why is Zn a poor catalyst in this picture? ::: $d^{10}$ full shell binds too weakly, $θ \to 0$ (weak-binding end of the volcano).