Visual walkthrough — Sulfur — allotropes (rhombic, monoclinic); SO₂, SO₃; H₂SO₄ (Contact process); oxoacids of S
Step 1 — Start from zero: what is "oxidation state" and why does it rise?
WHAT. Before any reaction, we have neutral sulfur atoms. We label each atom with a number called the oxidation state — a bookkeeping charge that answers the question "if I gave every shared electron to the greedier atom, what charge would this atom carry?"
WHY this tool. Oxygen is greedier than sulfur (it pulls bonding electrons harder). So every time we attach an oxygen to sulfur, sulfur loses claim to two electrons and its number goes up by 2. This single rule predicts the whole chain: . We use oxidation state — not real charge — because the bonds are covalent, so no atom truly carries a whole ; it's a scoreboard, not a physical charge.
PICTURE. A number line for sulfur from to . The red dot marks where sulfur sits at each stage. It only ever moves right in the Contact process.
Step 2 — Burn the sulfur: making SO₂ (the easy oxidation, 0 → +4)
WHAT. We set fire to sulfur in air. Each S atom grabs two oxygen atoms:
WHY this is the first step and why it's easy. Sulfur burns spontaneously once lit — the reaction is strongly exothermic and needs no catalyst, no pressure, nothing clever. It's the "free" jump on the oxidation-state line. Nature does most of the work; we just supply air. (See Oxygen and its oxides for why O₂ is such an eager oxidiser.)
PICTURE. A single S atom (red) with a lone pair and two S=O bonds, bent at ~119°. The lone pair on top is why the molecule is bent, not straight — the pair pushes the two oxygens down.
Recall Why is SO₂ bent and not linear like CO₂?
Sulfur in SO₂ keeps one lone pair (2 bond pairs + 1 lone pair). ::: The lone pair repels the two bond pairs, folding the molecule to ~119°. CO₂ has no lone pair on carbon, so it stays straight at 180°.
Step 3 — The hard jump: SO₂ → SO₃ is a tug-of-war (+4 → +6)
WHAT. Now we must add one more half-molecule of oxygen to reach :
WHY this step is the whole difficulty. Look at the harpoon arrows — this reaction is reversible. It does not go to completion; it settles at a balance point (equilibrium). And is negative, meaning the forward reaction releases heat. That negative sign is the villain of this entire story, and Step 4 explains why.
PICTURE. A see-saw (balance beam). Left pan = reactants (), right pan = product (). Heat is drawn leaking out of the right pan (exothermic). The red label marks the tilt we want: toward product.
Step 4 — The temperature trap (why "hotter = better" is wrong here)
WHAT. We now decide the temperature. Two instincts fight:
- Faster: higher temperature always speeds molecules up.
- More product: by Le Chateliers Principle, adding heat to an exothermic reaction is like adding product — the balance shifts backward, destroying SO₃.
WHY these clash. For an exothermic reaction, heat is effectively a product. Pour in heat (raise ) and Le Chatelier's principle says the system pushes back by consuming SO₃ to soak up that heat. So high gives you a fast reaction that reaches a bad equilibrium. The resolution is a compromise temperature ≈ 720 K — hot enough to be quick, cool enough to keep a good yield.
PICTURE. Two curves against temperature: yield of SO₃ (falls as rises) and reaction rate (rises as rises). They cross in a usable window. The red vertical line marks the chosen compromise, 720 K.
Step 5 — The catalyst breaks the deadlock (V₂O₅ gives speed for free)
WHAT. We drop in ==vanadium pentoxide, ==, as a catalyst. It speeds the reaction without changing the equilibrium position at all.
WHY a catalyst and not more heat. A catalyst lowers the energy hill (activation energy) both ways equally — forward and backward. So it lets us reach equilibrium quickly at our chosen low-ish 720 K, where the equilibrium already favours SO₃. It buys us the speed that high temperature would have cost us in yield. This is the key escape from the Step 4 trap. (More on how cycles between and in Catalysis and V2O5.)
PICTURE. An energy-versus-progress hill. Black curve = uncatalysed (tall hill). Red curve = catalysed (lower hill). Both start and end at the same heights — the catalyst only shrinks the barrier, never the drop.
Step 6 — The absorption trick: never feed SO₃ to water directly
WHAT. We now have SO₃ and want . The naive equation is balanced and true — but industrially forbidden. Instead we take a detour through oleum:
WHY the detour. SO₃ + water is so violently exothermic it creates a fine, floating acid mist that refuses to condense — you lose product to the air and it's dangerous. Dissolving SO₃ into existing concentrated H₂SO₄ (which already "wants" it) is smooth and controllable; you then add water to the liquid oleum at your own pace. Equation-correct ≠ factory-correct.
PICTURE. Two paths side by side. Left (black, crossed out): SO₃ + H₂O → uncontrolled red mist cloud. Right (chosen): SO₃ into liquid H₂SO₄ → clear oleum → controlled dilution.
Step 7 — Read the product: where does S sit now, and what's inside H₂SO₄?
WHAT. Apply the Step 1 scoreboard to the final acid: Sulfur has reached the top of its number line — the same it hit as SO₃. Building the acid added no more oxidation; it only wrapped water around the sulfur.
WHY this closes the loop. The whole Contact process was one long climb . Once at , sulfur can climb no higher (its outer shell is fully committed to oxygen). is just SO₃ dressed with two –OH groups. (For the family of acids that live at other rungs of this ladder, see Oxidation states and oxoacids.)
PICTURE. The structure of : a central S (red) with two terminal =O (double bonds, drawn boldly) and two –OH single bonds. The oxidation-state number floats beside the sulfur.
The one-picture summary
Everything above, compressed into a single flow: sulfur climbs the oxidation ladder , meets one reversible exothermic bottleneck (solved by catalyst + moderate T, not high T), and finishes with a safety detour through oleum instead of straight into water.
Recall The whole walkthrough retold in plain words (Feynman style)
We start with cheap yellow sulfur, oxidation number . Burning it in air is the easy jump — it grabs two oxygens on its own and becomes SO₂ at , bent because sulfur keeps a lone pair. The hard jump is adding the last oxygen to reach SO₃ at , because that reaction is a reversible see-saw that gives off heat. That "gives off heat" is the whole trouble: if we heat it to go faster, Le Chatelier punishes us by shifting the balance backward and destroying our SO₃. So we cheat — a catalyst lowers the energy hill and gives us speed for free, letting us keep the temperature at a modest 720 K where the yield is good. We skip high pressure because we're already at ~96% conversion. Finally we refuse to pour SO₃ into water (it makes an uncontrollable acid fog) and instead dissolve it into existing sulfuric acid to make oleum, then dilute that calmly into — where sulfur sits at the top of its ladder, , having climbed exactly as far as oxygen allows.