Question

All of the following physical properties of \(\mathrm{D}_{2} \mathrm{O}\) are considered to be related to isotope effects except \(\mathrm{D}_{2} \mathrm{O}\), which has a: (a) Higher melting point than \(\mathrm{H}_{2} \mathrm{O}\) (b) Smaller equilibrium dissociation constant than \(\mathrm{H}_{2} \mathrm{O}\) (c) Higher boiling point than \(\mathrm{H}_{2} \mathrm{O}\) (d) Smaller density than \(\mathrm{H}_{2} \mathrm{O}\)

Short answer

Option (d) is not related to isotope effects, as \(\text{D}_2\text{O}\) has a higher density than \(\text{H}_2\text{O}\), due to deuterium's greater mass.

Step-by-step solution

Understanding Isotope Effects

Isotope effects occur due to the replacement of an atom in a molecule with its isotope, causing changes in the molecule's physical and chemical properties. For \(\text{D}_2\text{O}\), the heavier deuterium atoms replace the hydrogen atoms in \(\text{H}_2\text{O}\), affecting properties like melting and boiling points, equilibrium dissociation constant, and density.

Considering Melting and Boiling Points

Deuterium (1D) is heavier than hydrogen (1H), which can affect intermolecular forces. \(\text{D}_2\text{O}\), with stronger hydrogen bonds due to deuterium's higher mass, generally exhibits a higher melting point and boiling point than \(\text{H}_2\text{O}\). This reflects isotope effects.

Analyzing the Equilibrium Dissociation Constant

The equilibrium dissociation constant, \(K_d\), for \(\text{D}_2\text{O}\), is smaller than for \(\text{H}_2\text{O}\) because of stronger O-D bonds compared to O-H bonds. This results in \(\text{D}_2\text{O}\) dissociating less easily, another effect of isotopic substitution.

Evaluating Density Difference

The density of a substance is its mass per unit volume. \(\text{D}_2\text{O}\) is denser than \(\text{H}_2\text{O}\) because deuterium is heavier than hydrogen, leading to a greater mass per unit volume. Therefore, this property is not affected inversely by the isotope effect, making this option the exception.

Key concepts

Deuterium Oxide (D2O)

Deuterium Oxide, commonly known as heavy water, is denoted as \( \text{D}_2 \text{O} \). It is composed of two deuterium atoms, which are isotopes of hydrogen, and one oxygen atom. This substitution of deuterium for hydrogen in water gives rise to what is known as isotope effects.
These isotope effects result in changes in both chemical and physical properties compared to ordinary water \( \text{H}_2 \text{O} \). Since deuterium has nearly twice the mass of hydrogen, these differences can be significant. Deuterium oxide is utilized in various nuclear reactors as a neutron moderator and is found naturally in small quantities in normal water. Its role in scientific research and nuclear technology highlights its importance beyond being just a heavier form of water.

Physical Properties

When discussing the physical properties of Deuterium Oxide (\( \text{D}_2 \text{O} \)), it is important to consider the influences of isotope effects. Deuterium, being heavier, alters the physical characteristics of water.

• Melting and Boiling Points: \( \text{D}_2 \text{O} \) has both higher melting and boiling points than \( \text{H}_2 \text{O} \). The increased mass and stronger hydrogen bonds due to deuterium make it more energetically demanding to break these bonds, resulting in a higher temperature required for phase changes.


• Density: Contrary to the exception stated in the textbook, \( \text{D}_2 \text{O} \) is actually denser than \( \text{H}_2 \text{O} \) because of the greater atomic mass of deuterium compared to hydrogen, contributing to an overall higher mass for the same volume.

In summary, the physical properties of \( \text{D}_2 \text{O} \) are significantly influenced by the mass difference in isotopic substitution.

Equilibrium Dissociation Constant

The equilibrium dissociation constant \( (K_d) \) is an important concept in understanding the behavior of chemical reactions in equilibrium, specifically how substances dissociate into ions. For deuterium oxide, the dissociation into ions is represented by the equilibrium:\[ \text{D}_2 \text{O} \rightleftharpoons \text{D}^+ + \text{OD}^- \]Due to the stronger O-D bonds, \( \text{D}_2 \text{O} \) has a smaller equilibrium dissociation constant compared to \( \text{H}_2 \text{O} \). This means that heavy water dissociates less readily than regular water.
The stronger bonds result from deuterium's larger mass, which makes the O-D bond more stable and less likely to break. Understanding the lower \( K_d \) for \( \text{D}_2 \text{O} \) is essential when considering its role in biochemical reactions and other applications where reactivity with acids and bases might differ from regular water.