Sulfur — allotropes (rhombic, monoclinic); SO₂, SO₃; H₂SO₄ (Contact process); oxoacids of S: Question bank

Three foundations you need before the traps

The traps below lean on three ideas the parent note used quickly. Let's earn each symbol first, with a picture, so nothing catches you off guard.

Definition Gibbs free energy

$G = H - T S$ To decide which form of a substance is stable, chemists don't compare energy alone — they compare ==Gibbs free energy $G$ ==, defined by $G = H - T S$ where $H$ is the enthalpy (heat content, roughly "how low-energy the bonds/packing are", units $kJ/mol$ ), $T$ is the absolute temperature in kelvin ( $K$ ), and $S$ is the entropy (a measure of disorder / number of ways the atoms can arrange, units $J K^{- 1} mol^{- 1}$ ). The stable form is the one with the lower $G$ . Look at the crossing lines in the figure: at low $T$ the $- T S$ term is tiny, so $G$ tracks $H$ and the low- $H$ (denser, better-packed) form wins; as $T$ grows, the $- T S$ term pulls harder on the high- $S$ (looser, more disordered) form, dragging its $G$ below — so it takes over. Where the two lines cross is the transition temperature.

Definition The Contact-process Stage 2 reaction (and its

$Δ H$ ) The Contact process makes H₂SO₄ in three stages; Stage 2 is the catalytic oxidation that turns sulfur dioxide into sulfur trioxide: $2 S O_{2} + O_{2} ⇌ V_{2} O_{5}, 720 K 2 S O_{3} Δ H = - 196 kJ/mol$ (equivalently, per molecule, $S O_{2} + \frac{1}{2} O_{2} \to S O_{3}$ ). Here == $Δ H$ is the enthalpy change== — the heat released or absorbed by the reaction, in kilojoules per mole ( $kJ/mol$ ). The negative sign means heat is released, i.e. the forward reaction is exothermic. == $V_{2} O_{5}$ is vanadium(V) pentoxide==, the solid catalyst: it repeatedly hands an oxygen from itself to $S O_{2}$ and is then re-oxidised by $O_{2}$ , giving the reaction a faster route without being used up (more in Catalysis and V2O5). The figure shows how $Δ H$ sits between reactants and products, and how the catalyst lowers the hill in between.

True or false — justify

Recall Rhombic and monoclinic sulfur are different because their

molecules are different. False ::: Both are made of the same S₈ crown ring; only the way the rings pack in the crystal lattice differs. The molecule is identical — the difference is purely crystallographic. See Allotropy - carbon and phosphorus for the same "same atoms, different packing" idea.

Recall Above the transition temperature (369 K) rhombic sulfur is more stable.

False ::: Below 369 K rhombic wins; above 369 K (up to the melting point 393 K) monoclinic wins because its higher entropy makes the $- T S$ term in $G = H - T S$ dominate as $T$ rises, pulling monoclinic's $G$ below rhombic's (see the crossing lines in figure s01).

Recall SO₂ can only act as a reducing agent.

False ::: S is at +4 in SO₂, the middle of its −2 to +6 range, so it can go up (reducing agent) or down (oxidising agent) depending on the partner.

Recall SO₃ and SO₂ have the same molecular shape.

False ::: SO₂ is bent (2 bond pairs + 1 lone pair, ~119°); gaseous SO₃ is trigonal planar (3 bond pairs, no lone pair on S, 120°).

Recall In the Contact process, using a catalyst increases the equilibrium yield of SO₃.

False ::: A catalyst ( $V_{2} O_{5}$ , vanadium pentoxide) speeds up how fast equilibrium is reached but does not shift its position — see Catalysis and V2O5 and Le Chateliers Principle. Yield is set by temperature and pressure, not the catalyst.

Recall Plastic sulfur is a stable crystalline allotrope.

False ::: It is amorphous and metastable — long S–S chains frozen from the melt (figure s02) that slowly revert to rhombic sulfur. No regular lattice, so it is not crystalline.

Recall Bleaching by SO₂ is permanent, just like bleaching by Cl₂.

False ::: SO₂ bleaches by reduction and is temporary — air re-oxidises the dye and colour returns. Cl₂ bleaches by oxidation and is permanent.

Spot the error

Recall "SO₃ + H₂O → H₂SO₄ is the industrial absorption step." Find the flaw.

The equation is balanced but the process is wrong ::: Direct SO₃ + water is so exothermic it forms an uncontrollable acid mist/fog that won't condense. Industry absorbs SO₃ in conc. H₂SO₄ to make oleum, then dilutes. Equation-correct ≠ industrially correct.

Recall "Run Stage 2 (

$2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ ) at very high temperature so SO₃ forms faster and in greater amount." Find the flaw. Speed rises but yield falls ::: This reaction is exothermic ( $Δ H = - 196 kJ/mol$ , heat released), so high $T$ shifts equilibrium backward, giving less SO₃. Speed comes from the $V_{2} O_{5}$ catalyst, not brute heat; ~720 K is a compromise.

Recall "The average oxidation state of S in thiosulfate is +2, so both sulfur atoms are +2." Find the flaw.

Averages hide real structure ::: The two S atoms in $S_{2} O_{3}^{2 -}$ are not equivalent — one central S is ~+5, the terminal (thio) S is ~−1. Only the low-oxidation S is oxidised by iodine (see Redox titrations - thiosulfate and iodine).

Recall "S must use S=S double bonds to build S₈, just as O₂ uses O=O." Find the flaw.

S prefers single S–S bonds ::: Sulfur's larger, more diffuse orbitals give poor $π$ -overlap, so it avoids S=S double bonds and instead catenates through S–S single bonds into rings and chains — this is why S₈ exists but O₈ does not.

Recall "In H₂SO₅ and H₂S₂O₈ the sulfur is not +6 because they contain extra oxygen." Find the flaw.

The extra O is a peroxo link, not oxo ::: The extra oxygens sit in –O–O– (peroxo) bonds where each O is only $- 1$ , not $- 2$ . Accounting for that, S is still $+ 6$ in both Caro's and Marshall's acid. See Oxidation states and oxoacids.

Why questions

Recall Why does rhombic sulfur have a

higher density than monoclinic even though both are S₈? Tighter packing ::: The S₈ rings sit in a more compact, efficient lattice arrangement in rhombic ( $2.06 g/cm^{3}$ ) than in the looser monoclinic lattice ( $1.98 g/cm^{3}$ ). Same molecule, denser stacking — that denser (lower- $H$ ) form is why rhombic wins at low $T$ .

Recall Why is viscosity a

signature that plastic-sulfur chains have formed? Long chains tangle ::: When S₈ rings break open above ~160 °C and link into long polymeric chains (figure s02), the chains entangle and drag past each other slowly, so viscosity rises sharply — a hallmark of polymer formation.

Recall Why is high pressure

not used in Stage 2 ( $2 S O_{2} + O_{2} ⇌ 2 S O_{3}$ ) despite the mole count dropping (3 → 2)? Not cost-effective ::: Le Chatelier says high $P$ would help, but conversion is already ~96% at 1–2 atm. The tiny extra yield doesn't justify the expense and hazard of high-pressure plant.

Recall Why does sulfur have the most allotropes of any element?

Catenation flexibility ::: S–S single bonds are strong and stable, so sulfur can arrange atoms into many ring sizes and chain lengths — each stable enough to be a distinct allotrope (see Group 16 - Oxygen family - general trends).

Recall Why is SO₃ called the

anhydride of H₂SO₄? Acid minus water ::: $S O_{3} + H_{2} O \to H_{2} S O_{4}$ : SO₃ is what remains when you formally strip one water molecule from sulfuric acid. See Oxygen and its oxides for acidic-oxide behaviour.

Recall Why do both S–O bonds in SO₂ have the same length (~143 pm) if we draw one single and one double?

Resonance averaging ::: The two resonance structures make the real molecule an average, giving bond order 1.5 and two identical S–O bonds — intermediate between a pure single and pure double bond.

Edge cases

Recall What exactly is true

at 369 K (the transition temperature)? Both coexist ::: At exactly 369 K rhombic and monoclinic have equal Gibbs free energy ( $G = H - T S$ is the same for both), so both forms are stable simultaneously and interconvert. It's the boundary — where the two $G$ lines cross in figure s01 — not a "one wins" point.

Recall What happens to sulfur

at and above its melting point 393 K but before ~433 K? Mobile S₈ liquid ::: It melts to a straw-coloured, thin (low-viscosity) liquid still made of intact S₈ rings that slide easily — chain-breaking (and the viscosity spike) only begins nearer ~160 °C/433 K.

Recall If a sulfur oxoacid contains an S–S bond, does the "H=+1, O=−2, solve" shortcut still work?

Only for the average ::: The formula gives a correct average oxidation state, but with S–S (or O–O) bonds the individual sulfurs differ (e.g. dithionous $H_{2} S_{2} O_{4}$ , avg +3). Use the shortcut only when no S–S or O–O bonds are present.

Recall Is SO₂ acidic, basic, or neutral toward water — and what's the boundary case with dry vs moist SO₂?

Acidic oxide ::: SO₂ is an acidic oxide giving sulfurous acid with water. Note the edge case: dry SO₂ does not bleach — bleaching needs moist SO₂ so the reducing $H_{2} S O_{3}$ species can form.

Recall In the limit of zero catalyst, does the Contact-process equilibrium change?

Position unchanged, rate → tiny ::: Without $V_{2} O_{5}$ (vanadium pentoxide) the equilibrium position is identical, but the reaction is impractically slow. Removing the catalyst kills the rate, never the yield — a clean demonstration that catalysts and equilibrium are separate.