Question bank — Heavy water D₂O, hydrogen peroxide H₂O₂ — structure, preparation, reactions

Before we start, one word we lean on: an isotope is an atom with the same number of protons (so the same element, same chemistry-of-electrons) but a different number of neutrons (so different mass). Deuterium (D or ²H) is hydrogen with one extra neutron. Keep that picture — same electron cloud, heavier core — because half the traps hinge on it.

The picture above shows the two molecules side by side. Same bent shape, same lone pairs (the two grey lobes), same ~104.5° angle — but the D nuclei carry an extra neutron each. Every physical difference traces back to that heavier core, not to the electrons.

Building the tools these traps use

Several traps below quote three ideas. Let us earn each one with a picture before we test you on it, so no symbol is a mystery.

Tool 1 — reduced mass $\mu$ (why a vibrating bond is a two-body problem)

When an O and an H are joined by a bond, both atoms wobble — the light H swings a lot, the heavy O barely twitches. Physics has a trick: a two-mass "spring" behaves exactly like a single effective mass $\mu$ tied to a wall, where

$\mu =\frac{m_1{m}_2}{m_1+m_2}$

Here $m_1,m_2$ are the two atomic masses (in amu). Notice it is the product over the sum, never the plain sum — the sum would describe the whole molecule sliding, not the internal stretch.

Tool 2 — zero-point energy and the 5–7× kinetic isotope effect

A bond is a spring in a valley (the potential-energy well). Quantum rules forbid it from sitting perfectly still at the bottom; it always keeps a minimum wobble whose energy is the zero-point energy (ZPE):

$E_{ZPE}=\frac{1}{2}h\nu ,\nu =\frac{1}{2\pi }\sqrt{\frac{k}{\mu }}$

$h$ is Planck's constant, $k$ the bond stiffness (identical for O–H and O–D because electrons set it), and $\nu$ the vibration frequency. Because $\nu \propto 1/\sqrt{\mu }$ and ${\mu }_{OD}\approx 2{\mu }_{OH}$:

$\frac{{\nu }_{OD}}{{\nu }_{OH}}=\sqrt{\frac{{\mu }_{OH}}{{\mu }_{OD}}}=\sqrt{\frac{0.94}{1.78}}\approx 0.73$

The figure shows one well with two floors: the O–D floor sits lower (smaller ZPE) than the O–H floor. To break either bond you climb from your floor to the same rim, so the deeper D floor means a taller climb — extra activation energy $\Delta E\approx \frac{1}{2}h({\nu }_{OH}-{\nu }_{OD})$. Feeding that into the rate expression $k\propto e^{-\Delta E/RT}$ gives roughly a 5–7× slowdown for O–D bond breaking at room temperature. That is the kinetic isotope effect — and why heavy water reacts more sluggishly.

Tool 3 — the Rayleigh law $R_{final}/R_{initial}=L^{(1/\alpha )-1}$

Now define every symbol before the traps use them:

Why this exact form? Picture the liquid as a huge pool of H atoms with a rare sprinkle of D. Remove a tiny slice of liquid; H leaves $\alpha$ times more eagerly than D, so the slice that departs is poor in D and the pool is left slightly enriched. Track this repeatedly:

$\frac{d(\text{D in pool})}{d(\text{H in pool})}=\frac{1}{\alpha }R\Rightarrow \frac{dR}{R}=\left(\frac{1}{\alpha }-1\right)\frac{dL}{L}$

Integrating from the start ($L=1$) to a residue fraction $L$ gives the clean power law

$\boxed{\frac{R_{final}}{R_{initial}}=L^{(1/\alpha )-1}}$

The exponent $(1/\alpha )-1$ is negative whenever $\alpha >1$, so shrinking $L$ pushes the ratio up. The curve in the figure climbs steeply as $L\to 0$ — that upward sweep is the enrichment.

Tool 4 — the Girdler two-tower exchange cycle

The exchange reaction that shuttles deuterium between gas and liquid is

${\text{H}}_2{\text{S}}_{(g)}+{\text{HDO}}_{(l)}\rightleftharpoons {\text{HDS}}_{(g)}+{\text{H}}_2{\text{O}}_{(l)}$

Its equilibrium constant shifts with temperature: cold ⇒ D prefers the water; hot ⇒ D prefers the H₂S gas. The flowchart below shows how two towers exploit that.

Circulating H₂S grabs D where D likes water least and releases it where D likes water most, pumping deuterium steadily into the cold-tower bottoms.

Tool 5 — pD, pH and the self-ionisation of D₂O

$\text{pH}=-{\log }_{10}[{\text{H}}^{+}]$; by exact analogy $\text{pD}=-{\log }_{10}[{\text{D}}^{+}]$. Water self-ionises, and so does heavy water, but less, because splitting an O–D bond is harder (Tool 2):

${\text{D}}_2\text{O}\rightleftharpoons {\text{D}}^{+}+{\text{OD}}^{-},K_w({\text{D}}_2\text{O})\approx 1.35\times 10^{-15}(25^{\circ }\text{C})$

Compared with $K_w({\text{H}}_2\text{O})=1.0\times 10^{-14}$, this is smaller, so neutral pD is higher: ${\text{pD}}_{neutral}=\frac{1}{2}\text{p}{K}_w\approx \frac{1}{2}(14.87)\approx 7.4$, versus neutral pH $=7.0$.

True or false — justify

Spot the error

Why questions

Edge cases