Heavy water D₂O, hydrogen peroxide H₂O₂ — structure, preparation, reactions: Exercises

This is your self-test lab for the parent topic. Work each problem on paper before opening the solution. Levels climb from "do you recognise the idea?" up to "can you combine several ideas from scratch?"

Every symbol used below is defined the first time it appears, so you can start from line one even if the numbers look scary.

The figure below is your map for the whole page. Look at it before Level 3: the three curves show how a bigger $α$ enriches deuterium faster, and the dashed line marks the $\approx 46 \times$ answer you will derive in Exercise 3.1. Come back to the mint curve when you reach the $α \to 1$ discussion.

Figure — Heavy water D₂O, hydrogen peroxide H₂O₂ — structure, preparation, reactions

Figure 1. Enrichment $R_{f ina l} / R_{ini t ia l}$ versus fraction of liquid remaining $L$ , for three separation factors. Higher $α$ (lavender, $α = 6$ ) climbs steeply; the near-flat mint curve ( $α = 1.05$ ) shows why distillation needs thousands of stages. As $α \to 1$ the curve would flatten to a horizontal line at enrichment $= 1$ (no separation at all).

Level 1 — Recognition

Exercise 1.1

State the molecular formula of heavy water, name the isotope it contains, and say how many neutrons that isotope has versus ordinary hydrogen.

Recall Solution

Formula: $D_{2} O$ (also written $^{2} H_{2} O$ ).
Isotope: deuterium (symbol D or $^{2} H$ ).
Neutron count: deuterium has 1 neutron; ordinary hydrogen ( $^{1} H$ , "protium") has 0 neutrons. Both have 1 proton, so they are the same element — only the nucleus mass differs.

Exercise 1.2

Which has the higher boiling point, $H_{2} O$ or $D_{2} O$ , and in one sentence why?

Recall Solution

$D_{2} O$ boils higher ( $101. 4^{\circ} C$ vs $10 0^{\circ} C$ ). The heavier D nucleus vibrates more slowly, so O–D···O hydrogen bonds are effectively stronger, and it takes slightly more energy to pull the molecules apart into vapour.

Exercise 1.3

Give the molecular formula of hydrogen peroxide, state its oxygen oxidation state, and say in one line what class of reagent it is.

Recall Solution

Formula: $H_{2} O_{2}$ .
Oxygen oxidation state: $- 1$ (each H is $+ 1$ ; two H give $+ 2$ ; two O must total $- 2$ , so each O is $- 1$ ).
Class: an oxidising agent primarily, but also a reducing agent — it is redox-amphoteric because $- 1$ is an in-between oxidation state.

Level 2 — Application

Exercise 2.1

Compute the reduced mass $μ$ (in amu) of an O–H bond and an O–D bond. Take $m_{O} = 16$ , $m_{H} = 1$ , $m_{D} = 2$ amu.

Recall Solution

$μ_{O H} = \frac{16 \times 1}{16 + 1} = \frac{16}{17} \approx 0.941 amu$ $μ_{O D} = \frac{16 \times 2}{16 + 2} = \frac{32}{18} \approx 1.778 amu$ The O–D reduced mass is nearly double the O–H one, even though the whole molecule only got 11% heavier — because $μ$ is dominated by the lighter partner.

Exercise 2.2

Using $ν \propto \frac{1}{μ}$ (vibration frequency $ν$ falls as reduced mass rises), find the ratio $ν_{O D} / ν_{O H}$ .

Recall Solution

With the same $k$ : $\frac{ν _{O D}}{ν _{O H}} = \frac{μ _{O H}}{μ _{O D}} = \frac{0.941}{1.778} = 0.529 \approx 0.727$ So O–D vibrates at about 73% of the O–H frequency, matching the parent note's $ν_{O D} \approx 0.73 ν_{O H}$ . Lower frequency → lower zero-point energy → the O–D bond sits deeper in its energy well → harder to break.

Level 3 — Analysis

Exercise 3.1 — Rayleigh electrolysis enrichment

You electrolyse water with separation factor $α = 6$ until only the fraction $L = 0.01$ of the liquid remains. Deuterium leaves $α$ times slower than hydrogen. Find the enrichment ratio $R_{f ina l} / R_{ini t ia l}$ of deuterium in the leftover liquid.

Intuition Where the exponent

$(1/ α) - 1$ comes from (continuous removal) Imagine removing the liquid a tiny sip at a time. Let $L$ be the moles of liquid left and $D$ the moles of deuterium in it, so the D fraction is $R = D / L$ (both symbols were introduced before Figure 1). In each sip, hydrogen leaves at "full speed" but deuterium leaves $α$ times slower, so the D lost per unit of total liquid lost is only $\frac{1}{α}$ of D's current share: $\frac{d D}{d L} = \frac{1}{α} \frac{D}{L} .$ This says "the rate D disappears, per drop of total liquid, is $1/ α$ times its own concentration." Now do the calculus explicitly. First separate the variables — put everything with $D$ on the left, everything with $L$ on the right: $\frac{d D}{D} = \frac{1}{α} \frac{d L}{L} .$ Then integrate both sides (each side is the standard integral $\int \frac{d x}{x} = ln x$ ): $\int \frac{d D}{D} = \frac{1}{α} \int \frac{d L}{L} ⟹ ln D = \frac{1}{α} ln L + C,$ where $C$ is the constant of integration. Fix the constant using the start ( $D = D_{0}$ when $L = 1$ ): since $ln 1 = 0$ , we get $C = ln D_{0}$ . Substitute back: $ln D - ln D_{0} = \frac{1}{α} ln L ⟹ ln \frac{D}{D _{0}} = ln L^{1/ α} ⟹ \frac{D}{D _{0}} = L^{1/ α} .$ So $D \propto L^{1/ α}$ . Divide by $L$ to turn moles of D into the D fraction $R$ : $R = \frac{D}{L} \propto L^{(1/ α) - 1} .$ That is why the exponent is $(1/ α) - 1$ : the $1/ α$ comes from D's slower removal, and the $- 1$ comes from dividing by the shrinking total liquid $L$ . Comparing start and finish: $\frac{R _{f ina l}}{R _{ini t ia l}} = L^{(1/ α) - 1} .$

Recall Solution

Exponent: $\frac{1}{6} - 1 = - \frac{5}{6} \approx - 0.833$ . $\frac{R _{f ina l}}{R _{ini t ia l}} = (0.01)^{- 0.833}$ Write $0.01 = 1 0^{- 2}$ , so $(1 0^{- 2})^{- 0.833} = 1 0^{1.667} \approx 46$ . Enrichment $\approx 46 \times$ . Starting from $0.0156%$ D, one deep stage gives $0.0156 \times 46 \approx 0.72%$ — enriched, but nowhere near pure $D_{2} O$ , which is why real plants cascade many stages.

Exercise 3.2 — How much liquid is left?

In the same run, if you started with $1000$ L of water, how many litres of liquid remain, and is that leftover mostly $D_{2} O$ ?

Recall Solution

Liquid remaining $= 1000 \times L = 1000 \times 0.01 = 10$ L. No — it is not mostly $D_{2} O$ . $L$ tracks the whole liquid, and deuterium is still only $\approx 0.72%$ of the hydrogen even after enrichment. The 10 L is almost entirely ordinary water; the D is merely enriched relative to H.

Level 4 — Synthesis

Exercise 4.1 — Ideal stage count (Girdler process)

The Girdler sulfide process has a per-stage separation factor $α = 2.3$ . Each ideal stage multiplies the deuterium fraction by $α$ , so after $n$ stages the total enrichment is $α^{n}$ . How many ideal stages are needed to go from $0.0156%$ to $99%$ D?

Intuition Why one ideal stage multiplies the D fraction by

$α$ An ideal stage is one place where the two phases (water and $H_{2} S$ gas) sit together long enough to reach equilibrium. Recall the meaning of $α$ : at equilibrium, deuterium sits in the enriched phase $α$ times more richly than it entered. So the phase leaving that stage carries a D fraction $α$ times the fraction that came in — one multiplication by $α$ . Now chain the stages: the output of stage 1 (already $\times α$ ) becomes the input of stage 2, which multiplies by $α$ again to give $\times α^{2}$ , and so on. After $n$ stages the D fraction has been multiplied by $α$ a total of $n$ times: $final fraction = α^{n} \times initial fraction ⟹ enrichment = α^{n} .$ (This is exact only for the trace-deuterium limit, where each stage's $α$ stays constant — true here because D is always a tiny fraction.)

Recall Solution

Ratio: $\frac{0.99}{0.000156} \approx 6346$ . $n = \frac{l n 6346}{l n 2.3} = \frac{8.756}{0.8329} \approx 10.5 ideal stages.$ This is a thermodynamic lower bound. Because real trays are inefficient and back-mix, actual towers need hundreds of physical stages to imitate these ~10.5 ideal ones — hence the giant industrial columns.

Exercise 4.2 — Single-stage phase partition

At $3 0^{\circ} C$ one Girdler stage has separation factor $α = 2.3$ . Equal moles of $H_{2} O$ and $H_{2} S$ start at $0.02%$ and $0.015%$ D. At equilibrium, D partitions so that $\frac{x _{w a t er}}{x _{H_{2} S}} = α$ , and total D is conserved: $x_{w a t er} + x_{H_{2} S} = 0.035%$ . Find $x_{w a t er}$ .

Recall Solution

From $x_{w a t er} = α x_{H_{2} S}$ and $x_{w a t er} + x_{H_{2} S} = 0.035$ : $α x_{H_{2} S} + x_{H_{2} S} = 0.035 ⟹ x_{H_{2} S} = \frac{0.035}{α + 1} = \frac{0.035}{3.3} = 0.01061%$ $x_{w a t er} = α x_{H_{2} S} = 2.3 \times 0.01061 = 0.0244%$ Water rises from $0.02%$ to $0.0244%$ in one stage — a small gain, which is exactly why the process needs many stages counter-current.

Level 5 — Mastery

Exercise 5.1 — Reverse-engineer a target purity

You must reach $15%$ D using the Girdler process ( $α = 2.3$ ) starting from natural $0.0156%$ . How many ideal stages does the first plant need before it hands off to distillation? Then explain in one line why the answer is smaller than Exercise 4.1's.

Recall Solution

Ratio: $\frac{15}{0.0156} = 961.5$ . $n = \frac{l n 961.5}{l n 2.3} = \frac{6.868}{0.8329} \approx 8.2 ideal stages.$ Rounding up, about 9 ideal stages. Why smaller than 10.5: reaching only $15%$ (not $99%$ ) is a smaller total enrichment ratio, so fewer factors of $α$ are needed. Real plants deliberately stop around $15 - 20%$ because the final push to $99.8%$ is far cheaper by distillation/electrolysis.

Exercise 5.2 — Combine the effects: predict a rate ratio

A reaction's rate depends on breaking an O–H (or O–D) bond. For this exercise use the parent note's stated result directly: O–D breaking is $5 - 7 \times$ slower. If a certain hydrolysis in $H_{2} O$ finishes in $2$ minutes, estimate the range of times in $D_{2} O$ , and name the effect.

Recall Solution

Slower by $5$ to $7 \times$ : $t_{min} = 2 \times 5 = 10 min, t_{ma x} = 2 \times 7 = 14 min .$ So the same reaction takes roughly 10–14 minutes in $D_{2} O$ . The effect is the kinetic isotope effect.

Recall One-line self-check

Separation factor larger means enrichment is easier or harder? ::: Easier — a bigger $α$ means fewer stages, since total enrichment is $α^{n}$ . In the Rayleigh law, why does shrinking $L$ enrich the residue? ::: Because for $α > 1$ the exponent $(1/ α) - 1$ is negative, so a smaller $L$ raised to a negative power is a larger number. What happens to enrichment when $α = 1$ ? ::: Nothing — the exponent is $0$ , so $L^{0} = 1$ and the ratio never changes (no separation). What single equation must accompany the equilibrium ratio in a phase-partition problem? ::: Conservation of total deuterium (mass balance). Why can H₂O₂ act as both oxidiser and reducer? ::: Its oxygen is at the intermediate oxidation state $- 1$ , so it can go down to $- 2$ (oxidising) or up to $0$ (reducing).