Worked examples — Heavy water D₂O, hydrogen peroxide H₂O₂ — structure, preparation, reactions

This page is the drill ground for the parent topic. Before we work anything, let us build the map of every kind of question this topic can ask — so that when an exam throws a case at you, you have already seen its twin.

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The scenario matrix

Every worked example below is tagged with the cell it fills. If you can do all cells, you can do any variant.

Cell	Case class	What makes it tricky	Example
A	Reduced-mass / ZPE ratio	pure ratio, no units to trip on	Ex 1
B	Single Rayleigh stage (finite $L$)	negative exponent, $L\to$ small	Ex 2
C	Rayleigh limiting case ($L\to 0$ and $L\to 1$)	degenerate endpoints	Ex 3
D	Multi-stage cascade (solve for $n$)	invert an exponential (logs)	Ex 4
E	Single-phase equilibrium partition	conservation + ratio	Ex 5
F	H₂O₂ "volume strength" ↔ molarity	unit web, definition trap	Ex 6
G	H₂O₂ redox titration (KMnO₄)	mole ratio, n-factor	Ex 7
H	H₂O₂ as reducing and oxidising (both signs)	which way does it go?	Ex 8
I	Degenerate / zero input ($\alpha =1$, $n=0$, 0-volume)	formula must still behave	Ex 9
K	Rayleigh with $\alpha <1$ (heavy goes to vapour)	exponent flips sign	Ex 10
J	Real-world word twist (reactor + peroxide first-aid)	reading, not algebra	Ex 11

Cells A–E and K are the D₂O / isotope side; F–H the H₂O₂ / redox side; I is the degenerate guard; J is the word twist.

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First, build the Rayleigh law from scratch

Cells B, C and K all lean on one equation. Let us earn it before we use it, and draw the picture behind it.

Derivation (strict mass balance on the vanishing sip). Let $N$ = light atoms (H) and $n$ = heavy atoms (D) currently in the liquid; the ratio is $R=n/N$.

1. Remove a tiny sip. Say the sip carries away $-dN$ light atoms and $-dn$ heavy atoms (the minus signs because both counts fall as liquid leaves; $dN<0$, $dn<0$). Why this step? Nothing is created; we are only bookkeeping atoms that physically walk out of the pot with the vapour/gas.
2. The sip's composition is set by $\alpha$. By definition, per unit removed the light species leaves $\alpha$ times more readily than the heavy one, weighted by how much of each is present. So the removal rates are in proportion $\frac{-dn}{-dN}=\frac{1}{\alpha }\cdot \frac{n/(\text{liquid})}{N/(\text{liquid})}=\frac{1}{\alpha }\frac{n}{N}.$ Why this step? "How readily each leaves" must be scaled by "how much of each is sitting in the pot" — a species can only leave in proportion to its own abundance. The $1/\alpha$ is precisely the statement "heavy leaves $\alpha$ times slower than light." The minus signs cancel, giving the clean differential relation below.
3. The differential equation (minus signs cancelled): $\frac{dn}{dN}=\frac{1}{\alpha }\frac{n}{N}.$ Separate the variables and integrate: $\int \frac{dn}{n}=\frac{1}{\alpha }\int \frac{dN}{N}\Rightarrow \ln n=\frac{1}{\alpha }\ln N+\text{const}\Rightarrow n\propto N^{1/\alpha }.$
4. Form the ratio $R=n/N\propto N^{1/\alpha }/N=N^{(1/\alpha )-1}$.
5. Switch to $L$. From the definition $N=N_{0}L$ we have $N\propto L$, so $\frac{R_{final}}{R_{initial}}={\left(\frac{N}{N_0}\right)}^{(1/\alpha )-1}=L^{(1/\alpha )-1}.$ Why this step? $L$ was defined as $N/N_0$ in the definition box, so replacing $N$ by $L$ is a substitution, not an assumption — that closes the gap.

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Cell A — Reduced mass and vibration ratio

Forecast: Guess — will O–D vibrate faster or slower? (Heavier ball on a spring... slower.) Guess a rough ratio: between $0.7$ and $0.75$?

1. Write reduced mass $\mu =\frac{m_1{m}_2}{m_1+m_2}$. Why this step? The vibrating unit is not a single atom — it is two atoms tied by one bond. The spring "feels" an effective mass $\mu$, not $m_H$ alone. This is the only mass that enters the frequency.
2. Plug O–H: ${\mu }_{OH}=\frac{16\times 1}{16+1}=\frac{16}{17}=0.9412$ amu.
3. Plug O–D: ${\mu }_{OD}=\frac{16\times 2}{16+2}=\frac{32}{18}=1.7778$ amu. Why this step? Only the H→D swap changes; oxygen is untouched, so any difference is a pure mass effect — the isotope effect isolated.
4. Frequency ratio. Since $\nu \propto \frac{1}{\sqrt{\mu }}$, $\frac{{\nu }_{OD}}{{\nu }_{OH}}=\sqrt{\frac{{\mu }_{OH}}{{\mu }_{OD}}}=\sqrt{\frac{0.9412}{1.7778}}=\sqrt{0.5294}=\mathrm{0.7276.}$ Why this step? Frequency answers "how fast does the bond wobble?" A slower wobble = lower zero-point energy = a bond that sits deeper in its well = harder to break. That single number 0.73 is the reason D₂O melts and boils higher.

Verify: $0.7276<1$ ✓ (slower, as forecast). Matches the parent note's ${\nu }_{OD}\approx 0.73{\nu }_{OH}$. Units cancel (ratio) ✓.

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Cell B — One Rayleigh electrolysis stage

Forecast: Electrolysis eats H faster, so D piles up. Big enrichment or small? Guess ~40–50×.

1. Use the Rayleigh law $\frac{R_{final}}{R_{initial}}=L^{(1/\alpha )-1}$ (just derived; $R$ = heavy/light ratio in liquid, $L$ = fraction of liquid remaining, $\alpha$ = separation factor). Why this step? D is a trace species removed $\alpha$ times more slowly than H — exactly the situation the law was built for.
2. Exponent $=\frac{1}{6}-1=0.1667-1=-0.8333$. Why this step? The exponent is negative because $1/\alpha <1$; a negative power of a small $L$ blows up — that is enrichment.
3. Evaluate $\frac{R_{final}}{R_{initial}}=(0.01{)}^{-0.8333}=10^{2\times 0.8333}=10^{1.667}=46.4$. Why this step? ${\log }_{10}(0.01)=-2$, so $(0.01{)}^{-0.8333}=10^{1.667}$ — clean because $L$ is a power of ten.
4. Final D% $=0.0156\%\times 46.4=0.724\%$. Why this step? Enrichment multiplies the ratio $R$, i.e. the atom-fraction, directly.

Verify: $0.724\%\ll 99.8\%$ ✓ — one stage is nowhere near pure, so cascades are genuinely needed. Enrichment $46\times$ matches parent note ✓.

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Cell C — Rayleigh limiting cases (degenerate endpoints)

Forecast: If you electrolyse nothing, enrichment should be 1 (unchanged). If you electrolyse everything, enrichment should shoot to infinity (all the leftover D crammed into a vanishing drop).

The figure above (the $\alpha =6$ red curve) is exactly this scenario plotted — read it as you follow the three points below: the curve starts pinned at height 1 on the right edge ($L=1$) and rises without bound as you sweep left toward $L=0$.

1. Case $L=1$: $1^{-0.8333}=1$. Why this step? $L=1$ means fraction remaining is the whole thing → no separation has happened → ratio must equal 1. This is the right-hand anchor of the red curve.
2. Case $L\to 0^{+}$: $L^{-0.8333}\to +\infty$. Why this step? A negative power of a number shrinking to zero explodes — the red curve shoots up at the left edge. Physically: the last microlitre is almost pure heavy species. (In reality mixing and re-equilibration cap this — but the ideal law diverges, and knowing that is the exam point.)
3. Sample the middle at $L=0.5$: $0.5^{-0.8333}=1.78$. Why this step? The curve passes through height $1.78$ at $L=0.5$ — confirming smooth monotone growth from 1 (right) upward as $L$ falls, with no wiggles or sign flips.

Verify: $1^{-0.8333}=1$ ✓, $0.5^{-0.8333}=1.78$ ✓, and $0.01^{-0.8333}=46.4>1.78>1$ ✓ (monotone, matching the plotted marked points).

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Cell D — Cascade: solve for stage count

Forecast: Each stage multiplies by 2.3. Going up ~6300× — guess about 10 stages.

1. Total enrichment needed $=\frac{0.99}{0.000156}=6346$. Why this step? Convert both concentrations to the same fraction basis ($99\%=0.99$, $0.0156\%=0.000156$) so their ratio is the pure multiplication factor the cascade must supply.
2. Set $2.3^n=6346$. Why this step? Stages act in series and each multiplies by $\alpha$, so $n$ stages multiply by ${\alpha }^n$ — an exponential in $n$.
3. Take logs (the tool that undoes an exponential): $n=\frac{\ln 6346}{\ln 2.3}=\frac{8.756}{0.8329}=\mathrm{10.5.}$ Why logs and not division? Division would answer "how many times bigger"; we need "how many multiplications" — that is precisely what a logarithm counts.

Verify: $2.3^{10.5}=6353\approx 6346$ ✓. This is the ideal (thermodynamic) count; real towers use hundreds of trays — the parent's caveat holds ✓.

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Cell E — Single-phase equilibrium partition

Forecast: D flows toward whichever phase it "prefers" (water at 30 °C). So the water's D should rise a little above its starting $0.02$ mmol — guess ~0.024 mmol.

We work in absolute amounts of D (mmol), not percentages, precisely because a percent in the water phase and a percent in the H₂S phase are measured against different bases (moles of water vs. moles of H₂S) and therefore cannot be added. Amounts of D atoms can be added — atoms are atoms in either phase.

1. Conservation of D atoms: $d_{water}+d_{H_{2}S}=0.02+0.015=0.035$ mmol. Why this step? D atoms are neither created nor destroyed — they only migrate between phases. The total number of D atoms (an absolute count in mmol) is what is conserved, and counts from different phases are freely additive.
2. Partition ratio. Equal moles of each phase, so equilibrium pins the ratio of D amounts to $\alpha$: $\frac{d_{water}}{d_{H_{2}S}}=\alpha =\mathrm{2.3.}$ Why this step? At equilibrium the ratio of D in the two phases is fixed by $K=\alpha$; with equal phase sizes the ratio of amounts equals the ratio of concentrations, so no base-mismatch sneaks in.
3. Solve: $d_{water}=\frac{2.3}{3.3}\times 0.035=0.0244$ mmol, and $d_{H_{2}S}=0.035-0.0244=0.0106$ mmol. Why this step? Two equations, two unknowns; substitute $d_{H_{2}S}=d_{water}/2.3$ into the sum from step 1.
4. Report as mole-fraction if desired. Per mole of water, $0.0244$ mmol D $\Rightarrow$ D content $=0.0244\%$ of the water. Why this step? Only now, after conserving absolute atoms, do we convert back to a percent — and only for a single phase against its own base, which is legitimate.

Verify: $d_{water}+d_{H_{2}S}=0.0244+0.0106=0.035$ mmol ✓ (atoms conserved). Ratio $0.0244/0.0106=2.3$ ✓. Water D rose from $0.020$ to $0.0244$ mmol (tiny per stage) ✓.

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Cell F — H₂O₂ volume strength ↔ molarity

Forecast: "20 volume" means 1 L of solution releases 20 L of O₂ at STP. Guess molarity somewhere near 1.8 M.

1. Meaning of label. 1 L solution → 20 L O₂ at STP. Why this step? "Volume strength" is defined as litres of O₂ (STP) per litre of solution — a definition trap if you skip it.
2. Moles of O₂: $\frac{20}{22.4}=0.8929$ mol. Why this step? 1 mole of any gas occupies 22.4 L at STP — converts the labelled volume into chemistry-usable moles.
3. Moles of H₂O₂: stoichiometry says 2 H₂O₂ → 1 O₂, so H₂O₂ $=2\times 0.8929=1.7857$ mol per litre. Why this step? The label is about the gas produced; molarity is about the H₂O₂ present — the 2:1 ratio bridges them.
4. Molarity $=\frac{1.7857\text{ mol}}{1\text{ L}}=1.786$ M. Why this step? Molarity is by definition moles of solute per litre of solution; we found the moles in step 3 and the label fixed the volume at 1 L, so dividing gives the concentration directly.

Verify: Shortcut check — $\text{Molarity}=\frac{\text{Vol strength}}{11.2}=\frac{20}{11.2}=1.786$ M ✓ (the $11.2=22.4/2$ constant). Units mol/L ✓.

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Cell G — Redox titration with KMnO₄

Forecast: Here H₂O₂ is the reducing agent (it hands electrons to Mn, its O goes 0). Mole ratio 5:2 favours lots of H₂O₂. Guess ~0.04 M.

1. Moles KMnO₄: $0.020\text{ M}\times 0.0200\text{ L}=4.0\times 10^{-4}$ mol. Why this step? Titrant amount is known exactly; it anchors the whole calculation.
2. Establish the mole ratio by counting electrons. Each Mn drops from $+7$ (in ${\text{MnO}}_4^{-}$) to $+2$ (in ${\text{Mn}}^{2+}$), gaining $5$ electrons. Each H₂O₂ has O at $-1$ going to $0$ (in O₂), losing $1$ electron per O, i.e. $2$ electrons per H₂O₂. Electrons lost must equal electrons gained, so $\underset{{\text{MnO}}_4^{-}}{\underbrace{2}}\times 5=\underset{{\text{H}}_2{\text{O}}_2}{\underbrace{5}}\times 2=10\text{electrons},$ which fixes the ratio ${\text{MnO}}_4^{-}:{\text{H}}_2{\text{O}}_2=2:5$. Why this step? A titration is an electron trade; the coefficients are not arbitrary — they are forced by the demand that electrons gained by Mn exactly balance electrons lost by H₂O₂. Counting them yourself is the guard against copying a wrong ratio.
3. Moles H₂O₂: $4.0\times 10^{-4}\times \frac{5}{2}=1.0\times 10^{-3}$ mol. Why this step? Apply the $2:5$ ratio just derived to convert titrant moles into H₂O₂ moles.
4. Molarity: $\frac{1.0\times 10^{-3}\text{ mol}}{0.0250\text{ L}}=0.040$ M. Why this step? Molarity means moles of H₂O₂ per litre of the H₂O₂ solution; step 3 gave the moles reacted (which is all the H₂O₂ in the 25 mL sample), so dividing by that sample's volume in litres yields its concentration.

Verify: Equivalents: KMnO₄ $=4\times 10^{-4}\times 5=2\times 10^{-3}$ eq; H₂O₂ $=1\times 10^{-3}\times 2=2\times 10^{-3}$ eq — equal ✓. Answer $0.040$ M ✓.

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Cell H — H₂O₂ as BOTH reducing and oxidising (both signs)

Forecast: In H₂O₂ oxygen is at $-1$ — an in-between state, so it can climb to $0$ (act as reducer) or drop to $-2$ (act as oxidiser). Sign depends on the partner.

1. O state in H₂O₂ is $-1$. Why this step? H is $+1$, molecule neutral: $2(+1)+2x=0\Rightarrow x=-1$. Being mid-scale is why H₂O₂ is two-faced.
2. (a) With I⁻: ${\text{H}}_2{\text{O}}_2+2{\text{I}}^{-}+2{\text{H}}^{+}\to {\text{I}}_2+2{\text{H}}_2\text{O}$. O goes $-1\to -2$ (reduced) → H₂O₂ is the oxidising agent. Why this step? I⁻ has no lower state to give from; H₂O₂ must accept electrons, so its O falls to $-2$.
3. (b) With MnO₄⁻: (Ex 7 equation) O goes $-1\to 0$ (oxidised) → H₂O₂ is the reducing agent. Why this step? MnO₄⁻ is a stronger oxidiser; it pulls electrons out of H₂O₂, pushing O up to $0$ (O₂ gas bubbles off).

Verify: Both signs realised from the same $-1$ start: down to $-2$ (Ex a), up to $0$ (Ex b) ✓. The two half-directions bracket the $-1$ state — the definition of amphoteric-redox behaviour ✓.

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Cell I — Degenerate / zero inputs (the guard)

Forecast: All three "do nothing" cases must return the neutral value: enrichment 1, molarity 0.

1. (a) $\alpha =1$: exponent $=1/1-1=0$, so $L^0=1$ for any $L$ — no matter how far you boil down, the residue's D/H ratio never changes. Why this step? $\alpha =1$ means H and D leave equally fast — there is nothing to separate, and the formula correctly refuses to enrich (factor 1) regardless of $L$. A good formula must be inert to a null input; this is the $\alpha =1$ row promised in the formula box.
2. (b) $n=0$: ${\alpha }^0=1$. Why this step? Zero stages = untouched feed = factor 1. Confirms the cascade formula has the right "identity" behaviour.
3. (c) "0 volume": $M=0/11.2=0$ M — pure water, no peroxide. Why this step? Volume strength of 0 means no O₂ can ever be released, i.e. no H₂O₂ present. The definition degrades gracefully to zero.

Verify: $L^0=1$ for all $L$ ✓, $2.3^0=1$ ✓, $0/11.2=0$ ✓. Every "do nothing" input yields the neutral output — no formula breaks.

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Cell K — Rayleigh with $\alpha <1$ (heavy leaves first)

Forecast: Now the residue should lose its heavy species — enrichment factor below 1 (depletion), the mirror image of Cell B.

1. Exponent $=\frac{1}{\alpha }-1=\frac{1}{0.5}-1=2-1=+1$. Why this step? With $\alpha <1$ we have $1/\alpha >1$, so the exponent is now positive — the sign has flipped versus every earlier example. The formula itself tells us the physics reversed.
2. Evaluate $\frac{R_{final}}{R_{initial}}=(0.01{)}^{+1}=0.01$. Why this step? A positive power of a small $L$ shrinks toward zero — the opposite of the blow-up in Cell B.
3. Interpret: the residue's heavy/light ratio drops to $1\%$ of its start — heavy has been stripped out into the vapour. Why this step? This is the guard against blindly assuming "residue always enriches." Enrichment direction is dictated entirely by whether $\alpha \gtrless 1$.

Verify: exponent $+1>0$ ✓ (sign flipped from Cell B's $-0.833$). $(0.01{)}^1=0.01<1$ ✓ depletion, mirror of Cell B's $46>1$ enrichment. On the figure this would be a curve falling to the left instead of rising ✓.

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Cell J — Real-world word twist

Forecast: (a) Slightly more than the 10.5 stages we found for 99% — maybe 11. (b) 3% is weak; volume strength around 10.

1. (a) Ratio needed: $\frac{0.998}{0.000156}=6397$. Why this step? Same cascade logic as Ex 4, just a higher target; convert both to fractions and divide.
2. (a) Stages: take logs to invert $2.3^n=6397$: $n=\frac{\ln 6397}{\ln 2.3}=\frac{8.764}{0.8329}=10.52\Rightarrow \text{11 ideal stages}.$ Why round up? You cannot buy a fraction of a stage, and fewer stages would fall short of spec — always round up for a purity floor.
3. (b) 3% mass → g/L: $3\text{ g}$ H₂O₂ per $100$ g solution; with $\rho =1.0$ g/mL, $100$ g $=100$ mL, so $30$ g/L. Why this step? Volume strength needs mass per litre; density turns mass-% into a per-litre figure.
4. (b) Moles/L: $30/34=0.882$ M (molar mass H₂O₂ $=34$).
5. (b) Volume strength $=11.2\times M=11.2\times 0.882=9.88\approx 10$ volume. Why this step? Invert the Ex 6 constant: strength $=11.2\times$ molarity.

Verify: (a) $2.3^{11}=8113>6397$ ✓ (11 stages clear the bar); $2.3^{10}=3527<6397$ ✗ (10 fall short) — so 11 is correct. (b) $\approx 10$ volume matches the familiar "10 vol" antiseptic label ✓.

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Recall Self-test: name the cell before you solve

Which matrix cell is "given volume strength, find molarity"? ::: Cell F Which cell tests $\alpha =1$ and $n=0$? ::: Cell I (degenerate) Which cell flips the Rayleigh exponent positive? ::: Cell K ($\alpha <1$, heavy leaves first) In Ex 8, why can H₂O₂ act both ways? ::: Its oxygen sits at $-1$, an intermediate state, so it can rise to $0$ or fall to $-2$. What tool "undoes" the exponential ${\alpha }^n$ to find $n$? ::: The logarithm.