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

Definition Symbols used on this page (defined once, here)

$E_{a}$ = activation energy — the height of the energy barrier between reactants and products. The bigger $E_{a}$ , the fewer collisions carry enough energy to react, so the slower the reaction. A catalyst lowers $E_{a}$ .
$Δ G$ = Gibbs free-energy change of the reaction — the net "downhill drop" from reactants to products. It fixes the direction and extent of reaction. A catalyst never changes $Δ G$ .
$Δ H$ = enthalpy change of the reaction — the net heat released (if negative, exothermic) or absorbed (if positive) as reactants turn into products. On an energy diagram $Δ H$ is simply the height difference between the reactant level and the product level. Like $Δ G$ , a catalyst never changes $Δ H$ — it only alters the barrier between those two levels.
$K_{e q}$ = equilibrium constant — the ratio of products to reactants once the reaction settles. Set by $Δ G$ , so a catalyst never changes $K_{e q}$ either.
$k$ = a rate constant (a proportionality factor in a rate law); in Sabatier's law below it is just the overall speed scale.
$R = 8.314 J mol^{- 1} K^{- 1}$ = the gas constant; $T$ = absolute temperature in kelvin. These appear together in the Arrhenius factor $e^{- E_{a} / R T}$ = the fraction of collisions energetic enough to clear the barrier.
The arrow shorthand $V^{+ 5} ⇌ V^{+ 4}$ means vanadium's oxidation state cycles back and forth across the two catalytic steps (reduced in one step, re-oxidised in the next) — it is a repeating catalytic cycle, not a single equilibrium.

What the vertical energy axis means on every diagram below: we plot enthalpy $H$ (heat content). Reactant and product levels are their enthalpies; the difference between them is $Δ H$ ; the hump between them is the activation energy $E_{a}$ . (Using $H$ rather than $G$ keeps the picture concrete — a "height you must climb, then a height you drop to" — and none of the catalysis conclusions change, since a catalyst alters neither $H$ nor $G$ of the endpoints.)

Level 1 — Recognition

Problem 1.1

Name the catalyst used in each industrial process, and state whether the mechanism is oxidation-state relay or surface adsorption: (a) Contact process, (b) Haber process, (c) hydrogenation of vegetable oil, (d) Ostwald process.

Recall Solution

(a) Contact process — V₂O₅, oxidation-state relay ( $V^{+ 5} ⇌ V^{+ 4}$ , oxygen-transfer cycle). (b) Haber process — Fe (with Mo promoter), surface adsorption. (c) Hydrogenation of oils — Ni, surface adsorption. (d) Ostwald process — Pt (gauze), surface adsorption.

A note on "homogeneous vs heterogeneous": those words strictly describe phase (whether catalyst and reactants share one phase). V₂O₅ is a solid, so by phase it is a heterogeneous catalyst. But its mechanism is the oxidation-state relay, which is the mechanism usually met in homogeneous catalysis. On this page we classify by mechanism, not phase, to keep the chemistry clear — never say "V₂O₅ is homogeneous" in an exam; say "V₂O₅ is a solid (heterogeneous) catalyst that works by an oxidation-state (redox-cycle) mechanism."

Problem 1.2

State the two structural features of transition metals that make them good catalysts, and match each feature to the mechanism it powers.

Recall Solution

Variable oxidation states → powers the oxidation-state relay, e.g. V₂O₅.
Partly filled and empty d-orbitals → powers surface adsorption / chemisorption, e.g. Fe, Ni, Pt.

Level 2 — Application

Problem 2.1

Write the two half-cycle equations for V₂O₅ in the Contact process, label the oxidation state of vanadium in each step, and confirm the net reaction.

Recall Solution

$S O_{2} + V_{2} O_{5} \to S O_{3} + V_{2} O_{4} (V^{+ 5} \to V^{+ 4})$ $V_{2} O_{4} + \frac{1}{2} O_{2} \to V_{2} O_{5} (V^{+ 4} \to V^{+ 5})$ Add the two steps (the $V_{2} O_{5}$ and $V_{2} O_{4}$ cancel): $S O_{2} + \frac{1}{2} O_{2} \to S O_{3}$ Multiply by 2 for the standard form: $2 S O_{2} + O_{2} \to 2 S O_{3}$ . Vanadium is regenerated → catalytic. ✓ The two half-cycles together form one turn of the repeating cycle written compactly as $V^{+ 5} ⇌ V^{+ 4}$ .

Problem 2.2

For Ni-catalysed hydrogenation, explain in three steps how a single $H_{2}$ molecule ends up adding across a $C = C$ bond. Why can't $H_{2}$ do this alone?

Recall Solution

$H_{2}$ lands on the Ni surface and undergoes dissociative chemisorption: $H_{2} \to 2 H_{(a d s)}$ — the strong H–H bond (436 kJ/mol) is broken by overlap with Ni d-orbitals.
The oil's $C = C$ double bond also adsorbs onto neighbouring Ni sites.
Surface H atoms migrate and add across the C=C, giving a saturated C–C. Why not alone? An isolated $H_{2}$ molecule is unreactive — its bond is too strong to break at ordinary temperature, so a direct $H_{2} + C = C$ collision has a huge $E_{a}$ . Ni pre-splits $H_{2}$ into reactive atoms, slashing $E_{a}$ .

Level 3 — Analysis

Problem 3.1

The N≡N bond energy is 945 kJ/mol — one of the strongest known. A student says: "So Fe must supply at least 945 kJ/mol of energy to break it." Explain precisely what is wrong, and what Fe actually does. Refer to an energy diagram.

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

Recall Solution

A catalyst never supplies energy. $Δ G$ (reactants → products) is fixed by thermodynamics; the catalyst leaves it untouched. What Fe changes is the activation energy $E_{a}$ — the height of the barrier between reactants and products (look at the amber peak dropping to the cyan peak in Figure 1). Fe achieves this by chemisorbing $N_{2}$ : its d-orbitals draw electron density out of the N≡N bond, so the molecule is split while bound to the surface, one weakened increment at a time, rather than in a single 945 kJ gulp. The reaction still needs the same net enthalpy change $Δ H$ (defined in the symbol box: the height difference between reactant and product levels) — but the peak it must climb over is lowered. Uncatalysed and catalysed paths start and end at the same enthalpies; only the pass between them is lower.

Problem 3.2

$S c^{3 +}$ is $d^{0}$ and $Z n^{2 +}$ is $d^{10}$ , and both are poor catalysts. Explain, for each of the two catalytic mechanisms, why these specific configurations fail. Why is this the strongest evidence for the whole two-feature theory?

Recall Solution

Oxidation-state relay:

$S c^{3 +}$ ( $d^{0}$ ) has no d-electrons left to lose and does not readily go higher than +3 in accessible chemistry → no second, easily reached oxidation state → cannot relay electrons.
$Z n^{2 +}$ ( $d^{10}$ ) has a full, stable d-shell; it essentially only exists as +2 → again no variable oxidation state. Surface adsorption:
Adsorption needs partly filled d-orbitals to overlap with incoming molecules. $d^{0}$ (nothing to donate) and $d^{10}$ (nothing available — the shell is closed) both lack the partly-filled orbitals needed to chemisorb. Why this is the killer evidence: the theory predicts "the two features cause catalysis." The elements that happen to lack both features ( $S c$ , $Z n$ ) are exactly the transition metals that fail. A prediction that correctly names the exceptions is far stronger than one that only explains the successes.

Level 4 — Synthesis

Problem 4.1

The heterogeneous rate follows Sabatier's principle: $rate \propto θ (1 - θ) k$ where $θ$ is the fraction of surface sites covered by reactant ( $0 \leq θ \leq 1$ ) and $k$ is the overall rate constant (a fixed speed scale). (a) Using calculus, find the value of $θ$ that maximises the rate. (b) Interpret both limits $θ \to 0$ and $θ \to 1$ physically. (c) Explain how catalyst poisoning and promoters shift a real catalyst along this curve.

Recall Solution

(a) Let $f (θ) = θ (1 - θ) = θ - θ^{2}$ (drop the constant $k$ ; it doesn't move the peak). Why calculus? The maximum is where the curve stops rising and starts falling — i.e. where the slope is zero. The slope is the derivative: $\frac{df}{d θ} = 1 - 2 θ .$ Set it to zero: $1 - 2 θ = 0 \Rightarrow θ = \frac{1}{2}$ . Check it's a maximum: the second derivative $\frac{d ^{2} f}{d θ ^{2}} = - 2 < 0$ → curve is concave-down → yes, a peak. Peak rate value $= \frac{1}{2} (1 - \frac{1}{2}) = \frac{1}{4}$ (in units of $k$ ). (b)

$θ \to 0$ : almost nothing sticks → binding too weak → few reacting sites → rate $\to 0$ .
$θ \to 1$ : surface saturated, products can't desorb and no bare sites for fresh reactant → surface "poisoned by its own coverage" → rate $\to 0$ . This is exactly why Ni, Pt, Pd (moderate binders) sit near the peak. (c) See Figure 3 for the visual of the argument below.
Poisoning (As, S, CO bind irreversibly) removes usable sites: the poison occupies active sites so the reactant's usable coverage $θ$ drops and part of the surface is permanently blocked. Fewer working sites → you slide down the left (low- $θ$ ) side of the volcano toward lower rate (amber dot in Figure 3, pushed left of the peak).
Promoters (e.g. Mo in Haber, alkali on Fe) don't catalyse themselves but tune the surface's binding strength so a catalyst that was sitting off-peak is nudged toward $θ = \frac{1}{2}$ , i.e. horizontally toward the top of the volcano → higher rate (cyan dot in Figure 3, moved toward the peak). See Catalyst poisoning and promoters.

Problem 4.2

Both the Contact process ( $2 S O_{2} + O_{2} \to 2 S O_{3}$ , V₂O₅) and the Haber process ( $N_{2} + 3 H_{2} \to 2 N H_{3}$ , Fe) are exothermic and use a catalyst. Explain why raising temperature helps the rate but hurts the yield, and why the catalyst does not resolve this tension. Which quantity does the catalyst actually help with?

Recall Solution

Rate vs yield are two different questions.

Yield is governed by equilibrium (Le Chatelier). Both reactions are exothermic ( $Δ H < 0$ ), so higher $T$ shifts equilibrium backward → lower equilibrium yield.
Rate is governed by kinetics. Higher $T$ gives more molecules over the barrier → faster. So $T$ is a compromise (~450 °C for both, chosen for a workable rate at acceptable yield). The catalyst does not touch $K_{e q}$ — it can't fix the yield problem; it only lowers $E_{a}$ , letting you run at a lower temperature (better yield) while still reacting fast enough. That is its true contribution: it buys back rate that a yield-friendly low temperature would otherwise cost. See Haber process — Le Chatelier conditions and Contact process — manufacture of H2SO4.

Level 5 — Mastery

Problem 5.1 (quantitative)

An uncatalysed reaction has $E_{a} = 75 kJ/mol$ . Adding Pt lowers it to $E_{a} = 45 kJ/mol$ at $T = 500 K$ . Using the Arrhenius idea that rate $\propto e^{- E_{a} / R T}$ (with the same pre-exponential factor), by what factor does the rate increase? Take the gas constant $R = 8.314 J mol^{- 1} K^{- 1}$ (defined in the symbol box above).

Recall Solution

Why the exponential? The Arrhenius factor $e^{- E_{a} / R T}$ is the fraction of collisions with enough energy to clear the barrier. Lowering $E_{a}$ raises that fraction — that is the mechanism of catalysis, expressed numerically. The ratio of catalysed to uncatalysed rate: $\frac{r _{cat}}{r _{uncat}} = \frac{e ^{- E_{a, cat} / R T}}{e ^{- E_{a, uncat} / R T}} = exp (\frac{E _{a, uncat} - E _{a, cat}}{R T}) .$ Plug in $Δ E_{a} = 75 - 45 = 30 kJ/mol = 30000 J/mol$ : $\frac{30000}{8.314 \times 500} = \frac{30000}{4157} = 7.217.$ $factor = e^{7.217} \approx 1.36 \times 1 0^{3} .$ The reaction runs about 1360× faster — from one 30 kJ/mol cut in the barrier. This is why tiny amounts of Pt transform industrial rates.

Problem 5.2 (transfer)

A student proposes using Mn (which shows oxidation states $+ 2, + 3, + 4, + 6, + 7$ ) as an oxidation-state-relay catalyst for a reaction, and Zn as a comparison. Predict which is promising and which is hopeless, and give the single deepest reason each way. Then explain why "many oxidation states" alone is not a guarantee of a good catalyst.

Recall Solution

Mn: promising. It has many easily accessible oxidation states ( $+ 2 ⇌ + 3$ , $+ 6 ⇌ + 7$ , etc.), so it can genuinely act as an electron relay — the core requirement for oxidation-state catalysis. (Indeed $M n O_{2}$ catalyses $H_{2} O_{2}$ decomposition.)
Zn: hopeless. $Z n^{2 +}$ is $d^{10}$ with essentially only the +2 state → no accessible second oxidation state and no partly filled d-orbitals → fails both mechanisms.
Caveat: having many oxidation states is necessary but not sufficient. The relay only works if the two states are (i) close in energy (small step $E_{a}$ ) and (ii) kinetically fast to interconvert, and if the intermediates it forms are the right ones for the target reaction. A metal can have a rich redox table yet bind the substrate too strongly (Sabatier) or form dead-end intermediates. Real catalyst choice is oxidation-state access plus favourable binding energetics.

Recap

Recall One-line takeaways (cover and test)

Catalyst's real gift to an exothermic industrial process ::: lets you run cooler (protecting yield) while keeping rate high, by lowering $E_{a}$ — never by changing $K_{e q}$ . Sabatier-optimal coverage ::: $θ = \frac{1}{2}$ , where $θ (1 - θ)$ peaks. V₂O₅ mechanism in one phrase ::: oxygen-transfer redox cycle $V^{+ 5} ⇌ V^{+ 4}$ (a repeating catalytic cycle, not a single equilibrium). V₂O₅ phase vs mechanism ::: it is a solid (heterogeneous by phase) catalyst that works by an oxidation-state relay mechanism. Why Zn fails both mechanisms ::: $d^{10}$ → no variable OS and no partly filled d-orbitals. What the vertical axis of the energy diagrams is ::: enthalpy $H$ ; the hump is $E_{a}$ , the reactant–product level gap is $Δ H$ ; a catalyst changes neither endpoint. Rate boost from a 30 kJ/mol barrier cut at 500 K ::: about $1.4 \times 1 0^{3}$ (i.e. ~1360×).