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Ethylene glycol \(\left(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right)\) is the major component of antifreeze. It is a slightly viscous liquid, not very volatile at room temperature, with a boiling point of \(198^{\circ} \mathrm{C}\). Pentane \(\left(\mathrm{C}_{5} \mathrm{H}_{12}\right),\) which has about the same molecular weight, is a nonviscous liquid that is highly volatile at room temperature and whose boiling point is \(36.1^{\circ} \mathrm{C}\). Explain the differences in the physical properties of the two substances.

Short Answer

Expert verified
The differences in physical properties of ethylene glycol (HOCH2CH2OH) and pentane (C5H12) stem from their molecular structures and the resulting intermolecular forces. Ethylene glycol's polar hydroxyl groups lead to strong hydrogen bonding, making it more viscous, less volatile, and resulting in a higher boiling point of 198°C. Pentane, on the other hand, interacts via weaker dispersion forces and has no functional groups, leading to a less viscous liquid, higher volatility, and a boiling point of 36.1°C.

Step by step solution

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1. Molecular structure and intermolecular forces for ethylene glycol

Ethylene glycol is an organic compound with the molecular formula HOCH2CH2OH. The molecule contains two hydroxyl groups (-OH), which make it an alcohol. The presence of these polar hydroxyl groups gives rise to strong intermolecular hydrogen bonding between ethylene glycol molecules.
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2. Molecular structure and intermolecular forces for pentane

Pentane, on the other hand, is an alkane with the molecular formula C5H12. It is a simple, non-polar hydrocarbon with no functional groups, which results in weaker dispersion forces (also known as London dispersion forces or van der Waals forces) acting between pentane molecules.
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3. Viscosity comparison

Viscosity depends mainly on the strength of intermolecular forces in a substance. Ethylene glycol has strong hydrogen bonding between its molecules due to the presence of hydroxyl groups, which makes it more viscous than pentane. In contrast, pentane molecules experience weaker dispersion forces, making it a less viscous liquid.
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4. Volatility comparison

Volatility is directly related to the strength of intermolecular forces in a substance. A more volatile substance has weaker intermolecular forces, allowing molecules to escape from the liquid phase more easily. Since ethylene glycol experiences strong intermolecular hydrogen bonding, it is less volatile at room temperature compared to pentane, which has weaker dispersion forces.
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5. Boiling point comparison

Boiling point is also influenced by the strength of intermolecular forces. Substances with stronger intermolecular forces require more energy to overcome these forces and transition from the liquid to the gas phase. This is why ethylene glycol, with its strong hydrogen bonding, has a higher boiling point at 198°C, while pentane, with its weaker dispersion forces, has a boiling point of only 36.1°C.
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Conclusion

The physical properties of ethylene glycol and pentane can be explained by the differences in their molecular structures and the resulting intermolecular forces. Ethylene glycol, with its polar hydroxyl groups, experiences strong hydrogen bonding, leading to a more viscous, less volatile liquid with a higher boiling point than pentane, which interacts via weaker dispersion forces and has no functional groups.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ethylene Glycol
Ethylene glycol, with the chemical formula \( ext{HOCH}_2 ext{CH}_2 ext{OH}\), is an organic compound commonly used in antifreeze mixtures. Its significance lies in its molecular structure, which includes two hydroxyl groups—\(- ext{OH}\).
These groups are essential because they contribute to forming strong hydrogen bonds between ethylene glycol molecules. This bonding is a type of intermolecular force that plays a crucial role in determining the physical properties of the substance.

Ethylene glycol's strong intermolecular hydrogen bonding leads to higher boiling points and viscosity levels compared to molecules of similar size without such bonding capability. This is why ethylene glycol is less volatile and more viscous compared to compounds like pentane.
Pentane
Pentane, with the chemical formula \( ext{C}_5 ext{H}_{12}\), is a hydrocarbon belonging to the alkane family. It is a simple, non-polar molecule without any hydroxyl groups or other polar functional groups.
This simplicity results in pentane primarily experiencing weak intermolecular forces known as dispersion forces or London dispersion forces. Although weak, these forces are enough to hold the pentane molecules together in the liquid state at room temperature.

The lack of strong bonding like hydrogen bonds explains pentane's characteristics: it is nonviscous, highly volatile, and has a low boiling point. Its properties are mainly influenced by its molecular structure, which encourages weak interactions between its molecules.
Hydrogen Bonding
Hydrogen bonding is a special type of dipole-dipole attraction, involving hydrogen atoms. This bonding occurs when hydrogen is covalently bonded to highly electronegative atoms like oxygen, nitrogen, or fluorine.
In ethylene glycol, hydrogen bonding is prevalent due to its hydroxyl groups \(- ext{OH}\). These bonds are relatively strong intermolecular forces that require more energy to break compared to dispersion forces.

Hydrogen bonds profoundly impact the physical properties of substances, like ethylene glycol, causing them to have higher boiling points, increased viscosity, and reduced volatility compared to non-polar substances such as pentane.
Dispersion Forces
Dispersion forces, also known as London dispersion forces or van der Waals forces, are the weakest type of intermolecular forces. They are present in all molecular substances but are the only forces acting in nonpolar molecules like pentane.
These forces arise from temporary dipoles created when the electrons within a molecule temporarily shift, resulting in a weak attraction.
  • They do not require any permanent polarity within the molecule.
  • They are generally weaker than hydrogen bonds.
In pentane, dispersion forces support the liquid state at room temperature but do not provide the significant bonding seen with substances that can hydrogen bond.
Boiling Point
The boiling point of a substance is the temperature at which it transitions from a liquid to a gas. This property is heavily influenced by the strength of intermolecular forces.
Stronger intermolecular forces imply a higher boiling point as more energy is necessary to overcome these forces. Ethylene glycol, with robust hydrogen bonding, has a boiling point of \(198^{\circ} \text{C}\), significantly higher than pentane's \(36.1^{\circ} \text{C}\).

Pentane requires less heat to transition to a gaseous state due to its weaker dispersion forces. This difference illustrates how pivotal molecular structure and bonding are in determining the physical characteristics of substances.
Viscosity
Viscosity is a measure of a fluid's resistance to flow. It is influenced by the strength of intermolecular forces between the molecules of a substance.
Ethylene glycol displays higher viscosity due to the presence of hydrogen bonds, which increase molecular interaction, causing the liquid to flow less freely.

Pentane, exhibiting weaker dispersion forces and lacking polar functional groups, has lower viscosity. This allows pentane to flow more easily and rapidly compared to ethylene glycol. Understanding viscosity helps in explaining why substances behave differently when poured or pumped.
Volatility
Volatility refers to how readily a substance vaporizes. Like other physical properties, it is directly influenced by intermolecular forces.
Substances with weaker intermolecular forces, like pentane with its dispersion forces, are more volatile. This means pentane evaporates easily at room temperature.

Conversely, ethylene glycol is less volatile due to its strong hydrogen bonding. Understanding volatility is crucial for applications like antifreeze, where low volatility ensures minimal evaporation and effective performance over time.

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Most popular questions from this chapter

Describe the intermolecular forces that must be overcome to convert these substances from a liquid to a gas: (a) \(\mathrm{SO}_{2}\), (b) \(\mathrm{CH}_{3} \mathrm{COOH},\) (c) \(\mathrm{H}_{2} \mathrm{~S}\).

Do you expect the viscosity of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3},\) to be larger or smaller than that of 1 -propanol, \(\mathrm{C}_{3} \mathrm{H}_{7} \mathrm{OH}\) ? Explain. [Section 11.3\(]\)

(a) Two pans of water are on different burners of a stove. One pan of water is boiling vigorously, while the other is boiling gently. What can be said about the temperature of the water in the two pans? (b) A large container of water and a small one are at the same temperature. What can be said about the relative vapor pressures of the water in the containers?

Identify the type or types of intermolecular forces present in each substance and then select the substance in each pair that has the higher boiling point: (a) propane \(\mathrm{C}_{3} \mathrm{H}_{8}\) or \(n\) -butane \(\mathrm{C}_{4} \mathrm{H}_{10},\) (b) diethyl ether \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}\) or 1 -butanol \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH},\) (c) sulfur dioxide \(\mathrm{SO}_{2}\) or sulfur trioxide (d) phosgene \(\mathrm{Cl}_{2} \mathrm{CO}\) or formaldehyde \(\mathrm{H}_{2} \mathrm{CO}\). \(\mathrm{SO}_{3},\)

The following table gives the vapor pressure of hexafluorobenzene \(\left(\mathrm{C}_{6} \mathrm{~F}_{6}\right)\) as a function of temperature: $$ \begin{array}{lc} \text { Temperature (K) } & \text { Vapor Pressure (torr) } \\ \hline 280.0 & 32.42 \\ 300.0 & 92.47 \\ 320.0 & 225.1 \\ 330.0 & 334.4 \\ 340.0 & 482.9 \end{array} $$ (a) By plotting these data in a suitable fashion, determine whether the Clausius-Clapeyron equation (Equation 11.1 ) is obeyed. If it is obeyed, use your plot to determine \(\Delta H_{\text {vap }}\) for \(\mathrm{C}_{6} \mathrm{~F}_{6}\) (b) Use these data to determine the boiling point of the compound.

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