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

Level 1 — Recognition

Recall Solution 1.1

WHAT we recall: The two crystalline forms are rhombic (α) and monoclinic (β). Both are built from the same S₈ crown ring — they differ only in how those rings pack.

The transition temperature is $369 K$ ( $= 96^{\circ} C$ ). $S_{(α)} cool ⇌ 369 K S_{(β)}$ Below 369 K: rhombic (α) is stable — it is the denser form ( $2.06 g/cm^{3}$ ), and dense = lower enthalpy, which wins when temperature (and hence the entropy term) is small.

Recall Solution 1.2

Sulfur in $S O_{2}$ has 2 bond pairs + 1 lone pair. The lone pair pushes the bonds together, so the molecule is bent (V-shaped) with a bond angle of about $11 9^{\circ}$ . Oxidation state: let it be $x$ . Oxygen is $- 2$ each, molecule neutral: $x + 2 (- 2) = 0 \Rightarrow x = + 4.$ So sulfur is $+ 4$ — the middle of its range ( $- 2$ to $+ 6$ ), which is why $S O_{2}$ can act as both oxidising and reducing agent.

Level 2 — Application

Recall Solution 2.1

The rule and WHY it works: in a neutral molecule the oxidation states of all atoms sum to $0$ . We know H behaves as $+ 1$ and O as $- 2$ here (no funny peroxo or S–S links), so the only unknown is S.

$H_{2} S O_{4}$ : $2 (+ 1) + x + 4 (- 2) = 0 \Rightarrow 2 + x - 8 = 0 \Rightarrow x = + 6.$ $H_{2} S O_{3}$ : $2 (+ 1) + x + 3 (- 2) = 0 \Rightarrow 2 + x - 6 = 0 \Rightarrow x = + 4.$ So sulfuric acid is $+ 6$ (maximum) and sulfurous acid is $+ 4$ . The difference is literally one oxygen — sulfurous is "sulfuric with one S=O removed."

Recall Solution 2.2

WHAT we do: balance Fe, then S, then O last (O appears in two products, so fix it last).

Fe: put a 2 on $F e_{2} O_{3}$ 's iron source → 4 $F e S_{2}$ gives 4 Fe, needs 2 $F e_{2} O_{3}$ .
S: 4 $F e S_{2}$ has $4 \times 2 = 8$ S → 8 $S O_{2}$ .
O: right side has $2 \times 3 + 8 \times 2 = 6 + 16 = 22$ O atoms → 11 $O_{2}$ . $4 F e S_{2} + 11 O_{2} \to 2 F e_{2} O_{3} + 8 S O_{2}$ Check: left O $= 22$ , right O $= 22$ . ✓

Level 3 — Analysis

Recall Solution 3.1

See the mole/energy bookkeeping in the figure below.

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

(a) Raise temperature. The forward reaction is exothermic ( $Δ H < 0$ ) — it releases heat. Treat heat as a product. Adding heat (raising $T$ ) shifts the equilibrium backward, so yield of SO₃ falls.

(b) Raise pressure. Count gas moles: left $= 2 + 1 = 3$ , right $= 2$ . Higher pressure pushes the system toward the side with fewer moles = the product side, so yield of SO₃ rises.

(c) Add $V_{2} O_{5}$ . A catalyst speeds both forward and backward reactions equally. It changes only how fast equilibrium is reached, not the position — so yield is unchanged.

Why 720 K and not lower? Low $T$ gives more SO₃ (part a) but the reaction is then too slow. $720 K$ is a compromise: fast enough (helped by the catalyst) while still keeping conversion high (~96%).

Why only 1–2 atm and not very high? Part (b) says high $P$ helps, but conversion is already ~96% at 1–2 atm, so squeezing out the last few percent isn't worth the expensive high-pressure equipment. Economics, not chemistry, sets this. See Le Chateliers Principle and Catalysis and V2O5.

Recall Solution 3.2

$S O_{2}$ bleaches by reduction. Sulfur is $+ 4$ and wants to climb to $+ 6$ , so it grabs oxygen from the coloured dye, reducing the dye to a colourless form. But this reduced dye is unstable: the oxygen in the air slowly re-oxidises it, and the colour returns — hence temporary.

$C l_{2}$ bleaches by oxidation. In water $C l_{2}$ forms nascent oxygen / hypochlorite that oxidises the dye, destroying the coloured structure irreversibly — hence permanent.

The analysis punchline: same visible result (colour vanishes), opposite mechanism (reduction vs oxidation), opposite permanence. See Oxidation states and oxoacids.

Level 4 — Synthesis

Recall Solution 4.1

(a) Oxidation states. Thiosulfate $S_{2} O_{3}^{2 -}$ : charge $= - 2$ , oxygens $- 2$ each. $2 x + 3 (- 2) = - 2 \Rightarrow 2 x = 4 \Rightarrow x = + 2 (average) .$ Tetrathionate $S_{4} O_{6}^{2 -}$ : $4 y + 6 (- 2) = - 2 \Rightarrow 4 y = 10 \Rightarrow y = + 2.5 (average) .$ So sulfur is oxidised from an average $+ 2$ to $+ 2.5$ — a small average change because only one sulfur per thiosulfate actually reacts (see the L4 mistake box).

(b) Titration. WHY moles first: the balanced equation tells us the ratio in which the two react, and moles is the currency that ratio speaks in. $n (S_{2} O_{3}^{2 -}) = M \times V = 0.100 mol/L \times 0.0200 L = 2.00 \times 1 0^{- 3} mol .$ From the equation, $2$ thiosulfate react with $1$ iodine, so $n (I_{2}) = \frac{1}{2} n (S_{2} O_{3}^{2 -}) = \frac{1}{2} (2.00 \times 1 0^{- 3}) = 1.00 \times 1 0^{- 3} mol .$ This is dissolved in $25.0 mL = 0.0250 L$ : $M (I_{2}) = \frac{1.00 \times 1 0 ^{- 3} mol}{0.0250 L} = 0.0400 M .$ $M (I_{2}) = 0.0400 M$ See Redox titrations - thiosulfate and iodine.

Recall Solution 4.2

WHAT the label means: oleum is $H_{2} S O_{4}$ with extra dissolved $S O_{3}$ (as $H_{2} S_{2} O_{7}$ ). If you add water, that free $S O_{3}$ becomes more $H_{2} S O_{4}$ : $S O_{3} + H_{2} O \to H_{2} S O_{4} .$ "109%" means: $100 g$ of oleum, on full dilution, yields $109 g$ of pure $H_{2} S O_{4}$ . The extra $9 g$ of mass comes from the water taken up by the free $S O_{3}$ .

Find the water absorbed: the diluted acid mass ( $109$ ) minus original oleum ( $100$ ) = $9 g$ of $H_{2} O$ added. Moles of water $= 9/18 = 0.5 mol$ . Each $S O_{3}$ consumes one $H_{2} O$ , so moles of free $S O_{3}$ $= 0.5 mol$ . Mass of free $S O_{3}$ $= 0.5 \times 80 = 40 g$ . $40 g of free S O_{3} per 100 g oleum$

Level 5 — Mastery

Recall Solution 5.1

WHY this needs the whole chapter: we chain three stages — $S O_{2} S O_{2} V_{2} O_{5} 96% S O_{3} oleum, H_{2} O H_{2} S O_{4} .$ Each stage carries 1 sulfur atom through, so the mole ratio $S : H_{2} S O_{4}$ is $1 : 1$ , except for the 96% conversion loss at Stage 2.

Step 1 — moles of S burned: $n (S) = \frac{1 000 000 g}{32 g/mol} = 31250 mol .$ Step 2 — moles reaching SO₃ (only 96% convert): $n (S O_{3}) = 0.96 \times 31250 = 30000 mol .$ Step 3 — moles of $H_{2} S O_{4}$ (1:1 from SO₃): $n (H_{2} S O_{4}) = 30000 mol .$ Step 4 — mass: $m = 30000 \times 98 = 2940000 g = 2940 kg .$ $2940 kg of H_{2} S O_{4} per hour$

Recall Solution 5.2

Use the "read the linkages" method from the parent note. See the linkage figure.

Acid	Feature	Naïve (all O = −2)	True S state
$H_{2} S_{2} O_{7}$ (pyrosulfuric/oleum)	–O– bridge between two S	$+ 6$	$+ 6$ ✓
$H_{2} S_{2} O_{8}$ (Marshall's)	–O–O– peroxo bridge between two S	$+ 7$ ✗	$+ 6$
$H_{2} S O_{5}$ (Caro's)	one –O–O– peroxo link	$+ 8$ ✗	$+ 6$
$H_{2} S_{2} O_{4}$ (dithionous)	S–S bond	$+ 3$	$+ 3$

WHY the peroxo acids stay at +6: sulfur can lose at most 6 valence electrons (it has 6), so $+ 7$ or $+ 8$ is chemically impossible. The naïve calculation over-counts because it assumes every O is $- 2$ . In a peroxo –O–O– link the two oxygens are $- 1$ each, not $- 2$ . Re-doing $H_{2} S O_{5}$ with one $- 1, - 1$ pair and three $- 2$ oxygens: $2 (+ 1) + x + 3 (- 2) + 2 (- 1) = 0 \Rightarrow 2 + x - 6 - 2 = 0 \Rightarrow x = + 6. ✓$ Structure keeps sulfur honest at its true maximum, $+ 6$ . See Oxidation states and oxoacids.

Recall Self-test checklist (reveal to grade yourself)

Rhombic stable below 369 K, denser ::: yes — 2.06 g/cm³ SO₂ is bent, ~119°, S = +4 ::: yes Catalyst changes rate not yield ::: yes Iodine molarity in 4.1(b) ::: 0.0400 M Free SO₃ in 109% oleum ::: 40 g per 100 g H₂SO₄ per hour in 5.1 ::: 2940 kg Peroxo acids: S is +6 not +7/+8 ::: yes (peroxo O is −1)