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

This page is a problem gym for the parent topic. We will not learn new theory — instead we take every kind of question this topic can throw and solve one of each, out loud, guessing before we compute. If a symbol shows up you have not met, we build it here first.

The scenario matrix

Before solving, let us list every case class this topic can produce. Each later example is tagged with the cell it fills. The two symbols encode the engine each cell uses, and the check/cross tells you whether catalysis works in that cell — a quick visual anchor before you read any maths.

🔁 = oxidation-state relay engine · 🧲 = surface adsorption engine · ⚙️ = general kinetics / thermodynamics (engine-neutral)
✅ = catalysis works in this cell · ❌ = catalysis fails (a degenerate / limiting case)

Cell	Engine	Works?	Case class	What makes it tricky	Example
C1	🔁	✅	Oxidation-state relay, normal	balance the two half-steps	Ex 1 (V₂O₅)
C2	🧲	✅	Surface (heterogeneous), normal	identify chemisorption + bond weakening	Ex 2 (Fe/Haber)
C3	⚙️	✅	Energy/kinetics numeric	compute rate speed-up from $E_{a}$ drop	Ex 3
C4	🔁🧲	❌	Degenerate metal (fails)	empty ( $d^{0}$ ) or full ( $d^{10}$ ) d-subshell — engine won't turn	Ex 4 (Sc/Zn)
C5	🧲	❌ at ends	Limiting value of surface coverage	fraction covered $θ \to 0$ and $θ \to 1$ endpoints	Ex 5 (Sabatier)
C6	⚙️	✅ (rate only)	Sign/direction twist (equilibrium)	catalyst does NOT move $K_{e q}$	Ex 6
C7	🧲	✅	Real-world word problem	pick catalyst + defend choice	Ex 7 (vanaspati)
C8	🔁	✅	Exam trap (wrong mechanism)	steel-man then correct	Ex 8 (V₂O₅ "sponge")
C9	🧲	❌ (poisoned)	Poisoning / promoter numeric	fraction of sites lost	Ex 9 (Pt + CO)

We now hit each cell.

Worked example Ex 1 — Balance the V₂O₅ relay (Cell C1 · 🔁 ✅)

Statement: In the Contact process $2 S O_{2} + O_{2} \to 2 S O_{3}$ , write the two catalytic half-steps and show vanadium is regenerated. State the oxidation number of V in each species.

Forecast: Guess — does V go up or down when it hands SO₂ an oxygen?

Assign oxidation numbers. In $V_{2} O_{5}$ : oxygen is $- 2$ , five O's give $- 10$ , two V share $+ 10$ , so each V is $+ 5$ . In $V_{2} O_{4}$ : four O's give $- 8$ , two V share $+ 8$ , each V is $+ 4$ . Why this step? We must name the two states of the relay before we can claim it cycles — see Variable oxidation states of transition metals.
Step A (V gives oxygen to SO₂): $S O_{2} + V_{2} O_{5} \to S O_{3} + V_{2} O_{4} (V^{+ 5} \to V^{+ 4})$ Why this step? SO₂ needs one more oxygen to become SO₃; V₂O₅ donates it and is reduced (loses that O, drops to +4).
Step B (O₂ re-oxidises the vanadium): $V_{2} O_{4} + \frac{1}{2} O_{2} \to V_{2} O_{5} (V^{+ 4} \to V^{+ 5})$ Why this step? The relay is only catalytic if the metal returns to start; molecular O₂ supplies the oxygen back.
Add the two steps. The $V_{2} O_{5}$ and $V_{2} O_{4}$ cancel: $S O_{2} + \frac{1}{2} O_{2} \to S O_{3} .$ Double it: $2 S O_{2} + O_{2} \to 2 S O_{3}$ . ✔

Verify: Vanadium appears on both sides and cancels → truly a catalyst. Oxidation change per V-atom: $+ 5 \to + 4 \to + 5$ , net zero — regenerated. Atom count: left $2 S, 6 O$ ; right $2 S, 6 O$ . Balanced.

Worked example Ex 2 — Which bond does Fe actually attack? (Cell C2 · 🧲 ✅)

Statement: In the Haber process, $N_{2} + 3 H_{2} F e 2 N H_{3}$ , identify (i) the mechanism type, (ii) the rate-limiting bond, (iii) what the Fe surface physically does to it.

Forecast: Guess which single bond is the "boss level" of ammonia synthesis.

Type. Fe is a solid, reactants are gases → heterogeneous (surface) catalysis, not an oxidation-state relay. Why this step? Choosing the wrong engine is the #1 error; solid + gas ⇒ surface. See Adsorption — physisorption vs chemisorption.
Rate-limiting bond. The N≡N triple bond (bond enthalpy $945 kJ mol^{- 1}$ ) is far stronger than H–H ( $436 kJ mol^{- 1}$ ). Breaking N≡N is the bottleneck. Why this step? The catalyst is judged on whether it lowers the biggest barrier.
What Fe does. Fe's partly-filled d-orbitals overlap N₂'s orbitals (chemisorption), draining electron density out of the N≡N bond so it stretches and weakens. Weakened, it splits into surface N atoms; adsorbed H atoms then add stepwise → NH₃, which desorbs. Why this step? This is the whole reason a catalyst helps — it lowers the barrier to split N≡N, it does not add energy.

Verify: Bond enthalpy sanity check: $945 > 436$ , so N≡N is indeed the harder split — consistent with calling it rate-limiting.

Worked example Ex 3 — How much faster? Numeric kinetics (Cell C3 · ⚙️ ✅)

Statement: A catalyst lowers activation energy from $E_{a} = 75 kJ mol^{- 1}$ to $50 kJ mol^{- 1}$ at $T = 500 K$ . By what factor does the rate increase? (Use $R = 8.314 J mol^{- 1} K^{- 1}$ .)

Forecast: Guess an order of magnitude — 10×? 1000×? more?

Pick the tool: the Arrhenius factor. Rate $\propto e^{- E_{a} / R T}$ . So the ratio of rates is $\frac{k _{cat}}{k _{uncat}} = \frac{e ^{- E_{a, cat} / R T}}{e ^{- E_{a, uncat} / R T}} = e^{(E_{a, uncat} - E_{a, cat}) / R T} .$ Why this tool and not another? The exponential $e^{- E_{a} / R T}$ is exactly "the fraction of molecules with enough energy to cross the barrier." Lowering $E_{a}$ raises that fraction — see Activation energy and reaction kinetics. No other single number captures this.
Plug in numbers. $Δ E_{a} = (75 - 50) \times 1 0^{3} = 25000 J mol^{- 1}$ . $\frac{25000}{8.314 \times 500} = \frac{25000}{4157} = 6.014.$ Why this step? (Why is the exponent a pure number?) The exponent is $E_{a} / (R T)$ . Write the units of each piece: $E_{a}$ is $J mol^{- 1}$ ; $R$ is $J mol^{- 1} K^{- 1}$ ; $T$ is $K$ . Multiply $R \times T$ : the kelvins cancel, leaving $J mol^{- 1}$ . Now divide $E_{a}$ by $R T$ : $\frac{J mol ^{- 1}}{J mol ^{- 1}} = 1$ — every unit cancels, so the exponent is dimensionless. That must be true, because you can only raise $e$ to a plain number.
Exponentiate. $e^{6.014} \approx 409$ . Why this step? The exponent is how many "e-folds" the rate jumps; converting to a plain number gives the speed-up.

Verify: Answer ≈ 409×. Sanity: a 25 kJ drop at 500 K should give a few hundred-fold — plausible (real catalysts give $1 0^{2}$ – $1 0^{10}$ ). Units of exponent: $J mol^{- 1} \div (J mol^{- 1} K^{- 1} \cdot K)$ = dimensionless ✔.

Worked example Ex 4 — The degenerate metals that FAIL (Cell C4 · 🔁🧲 ❌)

Statement: Explain, using electron configuration, why $S c^{3 +}$ and $Z n^{2 +}$ are poor catalysts even though both are technically "transition-adjacent."

Forecast: Guess which of the two engines (relay / adsorption) each one lacks.

Write the d-configurations. Sc is $[A r] 3 d^{1} 4 s^{2}$ ; removing 3 electrons gives $S c^{3 +} = [A r] 3 d^{0}$ (an empty d-subshell). Zn is $[A r] 3 d^{10} 4 s^{2}$ ; removing 2 gives $Z n^{2 +} = [A r] 3 d^{10}$ (a completely full d-subshell). Why this step? The whole diagnosis lives in the $d^{n}$ count we defined at the top — see d-orbital electronic configuration.
Test the relay engine. A relay needs an accessible neighbouring oxidation state. $S c^{3 +}$ ( $d^{0}$ ) cannot easily go lower without breaking the noble-gas core; $Z n^{2 +}$ ( $d^{10}$ ) is fully filled and very stable — neither cycles. Relay: dead. Why this step? No spare, easily-swapped electron ⇒ no electron relay.
Test the adsorption engine. Adsorption needs partly filled d-orbitals to overlap gas orbitals. $d^{0}$ has no d-electrons to donate; $d^{10}$ has no vacancies to accept into. Adsorption: dead. Why this step? Both a giver and a taker are needed; each ion is missing one.

Verify: This is the "diagnostic test" of the parent theory: the elements lacking both features are exactly the ones that fail — so the two-engine theory is falsifiable and passes. Configurations check: $S c (Z = 21)$ : $2, 8, 9, 2$ ; $Z n (Z = 30)$ : $2, 8, 18, 2$ ✔.

Worked example Ex 5 — Limiting behaviour of surface coverage θ (Cell C5 · 🧲 ❌ at ends)

Statement: The Sabatier rate law is $rate \propto θ (1 - θ)$ , where $θ$ is the fraction of surface covered (the number from $0$ to $1$ we defined at the top). Find the two limiting endpoints and the coverage that maximises the rate.

Forecast: Guess the best coverage — 0? 1? somewhere between?

Read the two factors. $θ$ = "reactant is present on the surface"; $(1 - θ)$ = "there is still bare surface to receive/release." You need both > 0 to react. Why this step? Interpreting each factor tells us the endpoints without any calculus.
Endpoint $θ \to 0$ (too-weak binding). rate $\propto 0 \times 1 = 0$ . Nothing sticks → no reaction. Why this step? This is the "binds too weakly" failure of Sabatier's principle mentioned in the parent.
Endpoint $θ \to 1$ (too-strong binding / poisoned). rate $\propto 1 \times 0 = 0$ . Surface saturated, products can't leave. Why this step? This is the "binds too strongly / poisoned" failure — connects to Catalyst poisoning and promoters.
Find the maximum. Use the derivative: the tool that finds where a smooth curve turns flat. $\frac{d}{d θ} [θ - θ^{2}] = 1 - 2 θ = 0 ⟹ θ = \frac{1}{2} .$ Why this tool? Setting the slope $= 0$ locates the peak; the second derivative $- 2 < 0$ confirms it is a maximum, not a dip.

Verify: At $θ = \frac{1}{2}$ : rate $\propto \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$ , the largest value of $θ (1 - θ)$ on $[0, 1]$ . Endpoints both give $0$ ✔.

The figure below is this whole example in one picture. The magenta curve is $θ (1 - θ)$ ; read it left-to-right as "coverage rising from empty to full." The violet dot at the top is the maximum we just found at $θ = \frac{1}{2}$ (rate $= \frac{1}{4}$ ). The two orange dots at the ends are our failure endpoints: $θ \to 0$ (nothing sticks) and $θ \to 1$ (poisoned, nothing leaves) — both sit on the floor at rate $0$ . The single hump shape is why this plot is nicknamed the "volcano," and why moderate binders like Ni and Pt (which live near the peak) beat both weak and strong binders.

Worked example Ex 6 — The equilibrium sign trap (Cell C6 · ⚙️ ✅ rate only)

Statement: For $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ ( $Δ G^{\circ} < 0$ ), a student claims "adding Fe catalyst raises the equilibrium yield of NH₃." True or false? Justify with what a catalyst changes and what it does not.

Forecast: Guess — does more Fe give more ammonia at equilibrium?

List what a catalyst touches. It lowers $E_{a}$ → speeds up both forward and reverse rates by the same factor (it lowers the same barrier both ways). Why this step? The barrier sits between the same two valleys; lowering it helps traffic in both directions equally.
List what it does NOT touch. $K_{e q}$ depends only on $Δ G^{\circ}$ (energies of reactants vs products), which the catalyst never alters. So $K_{e q}$ and the equilibrium position are unchanged. Why this step? Yield is fixed by $K_{e q}$ ; if $K_{e q}$ is untouched, yield is untouched.
Conclusion. The claim is False. Fe makes the mixture reach the same equilibrium faster, not further. To raise yield you change $T$ / $P$ (Le Chatelier) — see Haber process — Le Chatelier conditions.

Verify: Consistency check: if a catalyst could raise $K_{e q}$ , you could build a perpetual concentration machine by cycling catalyst in and out — violates thermodynamics. So $K_{e q}$ must be catalyst-independent ✔.

Worked example Ex 7 — Real-world word problem: making vanaspati (Cell C7 · 🧲 ✅)

Statement: A factory hydrogenates liquid vegetable oil (many C=C bonds) into semi-solid fat. H₂ gas alone barely reacts. Choose a catalyst, state the mechanism, and explain each design choice.

Forecast: Guess the metal — and guess why plain H₂ is so lazy.

Diagnose the sluggishness. The H–H bond ( $436 kJ mol^{- 1}$ ) is strong; molecular H₂ rarely splits on its own, so it can't add across C=C. Why this step? The bottleneck is splitting H₂ — the catalyst must target that.
Choose the engine. Solid metal + gas + we want to split H₂ on a surface ⇒ heterogeneous adsorption, not an oxidation relay. Why this step? No oxidation-state change is needed here; it is pure surface chemistry.
Choose the metal: Ni. Ni dissociatively adsorbs H₂ into two surface H atoms; the C=C also adsorbs; surface H atoms then add across it → C–C. Ni binds moderately (Sabatier "Goldilocks"), so products desorb. Why this step? We need a moderate binder — too weak won't split H₂, too strong won't release the fat.
Why not Pt? Pt works too but is far costlier; Ni is the cheap industrial choice for oils. Why this step? Real engineering weighs cost, not just activity.

Verify: Product test: C=C → C–C means the oil gains H atoms and becomes saturated (higher melting point → semi-solid), exactly what "vanaspati" is. Mechanism and Sabatier logic consistent with parent §4(c) ✔.

Worked example Ex 8 — Exam trap: the wrong mechanism (Cell C8 · 🔁 ✅)

Statement: "V₂O₅ is a solid, so it must work by adsorbing SO₂ on its surface like a sponge." Steel-man this, then correct it.

Forecast: Guess the one word that separates "sticking" from "reacting."

Steel-man (why it feels right). V₂O₅ genuinely is a solid, and many solid catalysts (Fe, Ni, Pt) do work by surface adsorption. So "solid ⇒ heterogeneous" sounds automatic. Why this step? Honestly stating the tempting logic makes the correction stick.
Correct it. The accepted V₂O₅ mechanism is the oxidation-state cycle $V^{+ 5} ⇌ V^{+ 4}$ : it donates an oxygen atom to SO₂, then grabs O₂ back. The driver is oxygen (electron) transfer, not mere physical sticking. Why this step? "Adsorbs like a sponge" implies no chemistry happens; here real O-atom transfer and redox happen.
General lesson. Bind each catalyst to its correct engine: V₂O₅ → relay, Fe/Ni/Pt → surface. Being a solid is necessary for heterogeneous, not sufficient to prove it. Why this step? Exams test whether you match mechanism to catalyst, not phase to mechanism.

Verify: Cross-check with Ex 1: the balanced half-steps used redox ( $+ 5 \leftrightarrow + 4$ ), never a "physisorption" step — internally consistent. The correct answer is False (it is the oxidation-state relay, not sponge adsorption) ✔.

Worked example Ex 9 — Poisoning numeric (Cell C9 · 🧲 ❌ poisoned)

Statement: A Pt catalyst has $2.0 \times 1 0^{20}$ active sites. Trace CO impurity irreversibly binds and blocks $3.0 \times 1 0^{19}$ of them. What fraction of activity is lost, and what fraction remains?

Forecast: Guess — is losing $3 \times 1 0^{19}$ out of $2 \times 1 0^{20}$ "a little" or "a lot"?

Fraction poisoned. Divide blocked by total (activity here ∝ number of free sites): $f_{lost} = \frac{3.0 \times 1 0 ^{19}}{2.0 \times 1 0 ^{20}} = 0.15.$ Why this step? Poisoning removes sites; the fractional loss is just the ratio of blocked to total — see Catalyst poisoning and promoters.
Fraction remaining. $f_{remain} = 1 - 0.15 = 0.85$ . Why this step? Every site is either blocked or free; the two fractions sum to 1.
Interpret. $15%$ lost is significant but not catastrophic; because CO binds too strongly to Pt, it never desorbs — a $θ \to 1$ "stuck" situation from Ex 5. Contrast this with a promoter (Ex 2's Mo), which does the opposite — it raises activity rather than blocking it. Why this step? Links the number back to the Sabatier failure mode and ties poisons vs promoters together.

Verify: Answer: 15% lost, 85% remaining. Check: $0.15 + 0.85 = 1.00$ ✔. Magnitude sanity: $3 \times 1 0^{19}$ is $\frac{3}{20}$ of $2 \times 1 0^{20}$ , which is indeed $0.15$ ✔.

Recall

Recall Cover the answers

Which engine does V₂O₅ use? ::: Oxidation-state relay $V^{+ 5} ⇌ V^{+ 4}$ (homogeneous-style), NOT surface adsorption. Rate speed-up from a 25 kJ/mol drop in $E_{a}$ at 500 K? ::: $e^{25000/ (8.314 \cdot 500)} = e^{6.01} \approx 409 \times$ . Coverage $θ$ that maximises Sabatier rate $θ (1 - θ)$ ? ::: $θ = \frac{1}{2}$ . Does a catalyst raise equilibrium yield? ::: No — it speeds both directions equally; $K_{e q}$ unchanged. Why do $S c^{3 +} (d^{0})$ and $Z n^{2 +} (d^{10})$ fail? ::: Neither has an accessible variable oxidation state nor partly-filled d-orbitals — both engines dead. What is Mo in the Haber process? ::: Molybdenum — a promoter (boosts the Fe catalyst, not a catalyst itself). 15% of Pt sites poisoned leaves what activity? ::: 85%.