Problem 42
Question
The value of \(\ldots \ldots \ldots\) is less for \(\mathrm{D}_{2} \mathrm{O}\) compared to that of \(\mathrm{H}_{2} \mathrm{O}_{2}\) (a) boiling point (b) latent heat of vaporization (J \(\mathrm{g}^{-1}\) ) (c) density \(\left(\mathrm{g} / \mathrm{mL}^{-1}\right)\) at \(20^{\circ} \mathrm{C}\) (d) dielectric constant at \(20^{\circ} \mathrm{C}\)
Step-by-Step Solution
Verified Answer
The dielectric constant of \(\mathrm{D}_{2}\mathrm{O}\) is less than that of \(\mathrm{H}_{2}\mathrm{O}_{2}\).
1Step 1: Understanding the Property
We need to compare different properties of two substances: \(\mathrm{D}_{2}\mathrm{O}\) (heavy water) and \(\mathrm{H}_{2}\mathrm{O}_{2}\) (hydrogen peroxide). The properties considered are boiling point, latent heat of vaporization, density, and dielectric constant. We must identify which property of \(\mathrm{D}_{2}\mathrm{O}\) is less than that of \(\mathrm{H}_{2}\mathrm{O}_{2}\).
2Step 2: Boiling Point Comparison
Boiling point is a measure of the temperature at which a substance changes from liquid to gas. Typically, hydrogen peroxide has a higher boiling point than water and, consequently, higher than heavy water. This is due to stronger hydrogen bonding in \(\mathrm{H}_{2}\mathrm{O}_{2}\).
3Step 3: Latent Heat of Vaporization Analysis
The latent heat of vaporization is the heat required to turn a substance from a liquid into a vapor at constant temperature. \(\mathrm{D}_{2}\mathrm{O}\) has a significant latent heat since it is an isotope of water, and heavier molecules require more energy to vaporize. \(\mathrm{H}_{2}\mathrm{O}_{2}\) typically has lower latent heat compared to \(\mathrm{D}_{2}\mathrm{O}\).
4Step 4: Density Comparison
Density indicates how much mass a substance has in a given volume. \(\mathrm{D}_{2}\mathrm{O}\) is known as heavy water because it is denser than ordinary water, and \(\mathrm{H}_{2}\mathrm{O}_{2}\) generally has less density than \(\mathrm{D}_{2}\mathrm{O}\).
5Step 5: Dielectric Constant Investigation
The dielectric constant is a measure of a substance's ability to store electrical energy in an electric field. \(\mathrm{H}_{2}\mathrm{O}_{2}\) has a higher dielectric constant than \(\mathrm{D}_{2}\mathrm{O}\) because it can store more energy, owing to the presence of an extra oxygen atom which increases molecular polarity and shape.
6Step 6: Conclusion on Property with Lesser Value
Among the evaluated properties, the property that has less value for \(\mathrm{D}_{2}\mathrm{O}\) compared to \(\mathrm{H}_{2}\mathrm{O}_{2}\) is the dielectric constant, as \(\mathrm{H}_{2}\mathrm{O}_{2}\)'s molecular structure allows it to store more electrical energy.
Key Concepts
Boiling PointLatent Heat of VaporizationDensityDielectric Constant
Boiling Point
The boiling point of a substance is the temperature at which its liquid form transitions to a gas. This temperature remains constant during the phase change. Super easy fact: you know when water boils, right? It's at 100°C at sea level.
- Hydrogen peroxide (H_2O_2) boasts a higher boiling point than regular water (H_2O) and heavy water (D_2O), thanks to its stronger hydrogen bonds.
- These bonds require more energy (i.e., higher temperature) to break, which is why H_2O_2 has a higher boiling point.
- As a result, heavy water boils at a lower temperature compared to hydrogen peroxide.
Latent Heat of Vaporization
The latent heat of vaporization is the energy needed to transform a liquid at its boiling point into a gas without changing the temperature. It's like the invisible magic that turns a pot of boiling water into steam!
For D_2O, or heavy water, the larger, heavier molecules require even more energy to make this change. Imagine trying to blow up a heavier balloon. It takes a little more air, right? That’s because:
For D_2O, or heavy water, the larger, heavier molecules require even more energy to make this change. Imagine trying to blow up a heavier balloon. It takes a little more air, right? That’s because:
- Heavy water molecules, due to their mass, demand more energy input.
- Thus, D_2O has a larger latent heat of vaporization than H_2O_2.
Density
Density tells us how tightly packed the mass of a substance is within a unit volume. Think about it as how "heavy" something feels for its size.
Heavy water, noted as D_2O, is literally heavier per given volume because:
This means that if you were to measure out equal volumes of D_2O and H_2O_2 at the same temperature, D_2O would weigh more.
Heavy water, noted as D_2O, is literally heavier per given volume because:
- It contains deuterium, a heavier isotope of hydrogen.
- This makes D_2O denser than regular water (H_2O) and hydrogen peroxide.
This means that if you were to measure out equal volumes of D_2O and H_2O_2 at the same temperature, D_2O would weigh more.
Dielectric Constant
The dielectric constant is a measure of a material's ability to store electrical energy. It’s like a trick tool for holding electric power.
Hydrogen peroxide (H_2O_2) features a higher dielectric constant compared to heavy water (D_2O). Here’s why:
So, if you're looking for a compound with a stronger ability to handle electric fields, H_2O_2 would be the go-to choice.
Hydrogen peroxide (H_2O_2) features a higher dielectric constant compared to heavy water (D_2O). Here’s why:
- The presence of an extra oxygen atom in H_2O_2 increases its molecular polarity.
- This molecular structure allows H_2O_2 to store more electrical energy.
So, if you're looking for a compound with a stronger ability to handle electric fields, H_2O_2 would be the go-to choice.
Other exercises in this chapter
Problem 39
A commercial sample of hydrogen peroxide is labelled as 10 volume, its percentage strength is (a) \(1 \%\) (b) \(3 \%\) (c) \(10 \%\) (d) \(90 \%\)
View solution Problem 40
Which of the properties of interstitial hydrides is correct? (a) They give rise to metals fit for fabrication (b) They generally form non-stoichiometric species
View solution Problem 43
Which of the following is not correct regarding the electrolytic preparation of \(\mathrm{H}_{2} \mathrm{O}_{2}\) ? (a) sulphuric acid undergoes oxidation (b) h
View solution Problem 44
The reaction of \(\mathrm{H}_{2} \mathrm{O}\) with \(\mathrm{X}\) does not liberate gaseous product. Which of the following is \(X\) ? (a) \(\mathrm{PbO}_{2}\)
View solution