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As the intermolecular attractive forces between molecules increase in magnitude, do you expect each of the following to increase or decrease in magnitude? (a) vapor pressure, (b) heat of vaporization, (c) boiling point, (d) freezing point, (e) viscosity, (f) surface tension, (g) critical temperature.

Short Answer

Expert verified
As the intermolecular attractive forces between molecules increase, we can expect: (a) vapor pressure to decrease, (b) heat of vaporization to increase, (c) boiling point to increase, (d) freezing point to increase, (e) viscosity to increase, (f) surface tension to increase, and (g) critical temperature to increase. This is because stronger intermolecular forces lead to more stable structures, a higher resistance to flow, and higher energy required to overcome the attractive forces between molecules.

Step by step solution

01

(a) Vapor Pressure

As the intermolecular attractive forces between molecules increase, the molecules will be more attracted to each other, making it less likely for them to escape into the gas phase. As a result, vapor pressure will decrease with an increase in intermolecular forces.
02

(b) Heat of Vaporization

Heat of vaporization is the amount of energy required to convert a substance from its liquid phase to its vapor phase. When the intermolecular forces are stronger, it requires more energy to overcome the attractive forces between the molecules in order to vaporize the substance. Therefore, the heat of vaporization will increase with an increase in intermolecular forces.
03

(c) Boiling Point

The boiling point is the temperature at which a substance changes from a liquid to a gas. As intermolecular forces increase, more energy is needed to separate the molecules, so the boiling point will also increase.
04

(d) Freezing Point

The freezing point is the temperature at which a substance changes from a liquid to a solid. As the intermolecular forces increase, the molecules will be more attracted to each other and will form a more stable structure at higher temperatures. Therefore, the freezing point will increase with an increase in intermolecular forces.
05

(e) Viscosity

Viscosity is the measure of a liquid's resistance to flow. When the intermolecular forces are stronger, the molecules are more attracted to each other and resist flowing past one another. So, the viscosity will increase with an increase in intermolecular forces.
06

(f) Surface Tension

Surface tension is the force that acts on the surface of a liquid and tends to minimize the surface area. Stronger intermolecular forces lead to a stronger tendency of molecules to remain close to their neighbors, and thus the surface tension increases with an increase in intermolecular forces.
07

(g) Critical Temperature

The critical temperature is the temperature above which a substance cannot be liquefied by applying pressure, regardless of how high the pressure is. Substances with stronger intermolecular forces require a higher temperature to overcome their attraction, so the critical temperature will increase with an increase in intermolecular forces.

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Key Concepts

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

Vapor Pressure
Vapor pressure is a key concept in understanding how a liquid behaves when exposed to air. It refers to the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature. When intermolecular forces (the forces that hold molecules together) are strong, less liquid will evaporate into vapor because molecules hold tightly together. As a result, vapor pressure is inversely related to the strength of these forces.

A liquid with high vapor pressure at a given temperature tends to evaporate more quickly and is therefore more volatile. In contrast, substances with strong intermolecular attractions, such as hydrogen bonds, display lower vapor pressure because the molecules do not escape easily. This concept is crucial in understanding boiling and evaporation processes in different liquids.
Heat of Vaporization
The heat of vaporization is the energy required to transform a given quantity of a substance from a liquid into a vapor at a constant temperature and pressure. It reflects how much energy is necessary to overcome the intermolecular forces holding the molecules in the liquid phase.

For substances with strong intermolecular forces, a higher heat of vaporization is needed. This happens because more energy must be input to break the strong attractions between molecules. For instance, water has a high heat of vaporization due to its hydrogen bonding. This property is essential in many natural processes and technologies, such as climate control and refrigeration.
Boiling Point
The boiling point of a substance is the temperature at which its liquid phase changes to vapor. This temperature depends on the atmospheric pressure as well as the intermolecular forces present in the liquid.

Stronger intermolecular forces lead to higher boiling points. This is because more energy is needed to separate the molecules from each other and transition them into the gas state.
  • Water boils at 100°C under standard atmospheric pressure.
  • Alcohols typically have lower boiling points than water, reflecting different strengths in intermolecular attractions.
Understanding boiling points is crucial in many real-world applications, such as cooking and industrial distillation processes.
Critical Temperature
Critical temperature is a unique characteristic representing the highest temperature at which a substance can exist as a liquid, no matter how much pressure is applied. It is significantly influenced by the intermolecular forces present in a substance.

The stronger the intermolecular attractions, the higher the critical temperature. This is because stronger forces require more thermal energy to allow the substance to remain in a liquid state under higher pressures.
  • Water has a critical temperature of around 374°C due to its strong hydrogen bonding.
  • Non-polar substances have lower critical temperatures due to weaker intermolecular forces.
Critical temperature is an essential concept in supercritical fluid applications and for understanding the limitations of phases in substances.

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Most popular questions from this chapter

The following quote about ammonia \(\left(\mathrm{NH}_{3}\right)\) is from a textbook of inorganic chemistry: "It is estimated that \(26 \%\) of the hydrogen bonding in \(\mathrm{NH}_{3}\) breaks down on melting, \(7 \%\) on warming from the melting to the boiling point, and the final \(67 \%\) on transfer to the gas phase at the boiling point." From the standpoint of the kinetic energy of the molecules, explain (a) why there is a decrease of hydrogen-bonding energy on melting and (b) why most of the loss in hydrogen bonding occurs in the transition from the liquid to the vapor state.

The boiling points, surface tensions, and viscosities of water and several alchohols are as follows: $$ \begin{array}{lrcc} & \begin{array}{l} \text { Boiling } \\ \text { Point }\left({ }^{\circ} \mathbf{C}\right) \end{array} & \begin{array}{l} \text { Surface } \\ \text { Tension }\left(\mathbf{J} / \mathbf{m}^{2}\right) \end{array} & \begin{array}{l} \text { Viscosity } \\ (\mathbf{k g} / \mathbf{m}-\mathbf{s}) \end{array} \\ \hline \text { Water, } \mathrm{H}_{2} \mathrm{O} & 100 & 7.3 \times 10^{-2} & 0.9 \times 10^{-3} \\ \text {Ethanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH} & 78 & 2.3 \times 10^{-2} & 1.1 \times 10^{-3} \\ \text {Propanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 97 & 2.4 \times 10^{-2} & 2.2 \times 10^{-3} \\ n \text { -Butanol, } \mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 117 & 2.6 \times 10^{-2} & 2.6 \times 10^{-3} \\\ \text {Ethylene glycol, } \mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH} & 197 & 4.8 \times 10^{-2} & 26 \times 10^{-3} \end{array} $$ (a) For ethanol, propanol, and \(n\) -butanol the boiling points, surface tensions, and viscosities all increase. What is the reason for this increase? (b) How do you explain the fact that propanol and ethylene glycol have similar molecular weights \((60\) versus \(62 \mathrm{amu}),\) yet the viscosity of ethylene glycol is more than 10 times larger than propanol? (c) How do you explain the fact that water has the highest surface tension but the lowest viscosity?

Look up and compare the normal boiling points and normal melting points of \(\mathrm{H}_{2} \mathrm{O}\) and \(\mathrm{H}_{2} \mathrm{~S}\). Based on these physical properties, which substance has stronger intermolecular forces? What kinds of intermolecular forces exist for each molecule?

At standard temperature and pressure the molar volume of \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) gases are \(22.06 \mathrm{~L}\) and \(22.40 \mathrm{~L},\) respectively (a) Given the different molecular weights, dipole moments, and molecular shapes, why are their molar volumes nearly the same? (b) \(\mathrm{On}\) cooling to \(160 \mathrm{~K}\), both substances form crystalline solids. Do you expect the molar volumes to decrease or increase on cooling to \(160 \mathrm{~K} ?\) (c) The densities of crystalline \(\mathrm{Cl}_{2}\) and \(\mathrm{NH}_{3}\) at \(160 \mathrm{~K}\) are \(2.02 \mathrm{~g} / \mathrm{cm}^{3}\) and \(0.84 \mathrm{~g} / \mathrm{cm}^{3}\), respectively. Calculate their molar volumes. (d) Are the molar volumes in the solid state as similar as they are in the gaseous state? Explain. (e) Would you expect the molar volumes in the liquid state to be closer to those in the solid or gaseous state?

The generic structural formula for a 1 -alkyl-3-methylimidazolium cation is where \(\mathrm{R}\) is a \(-\mathrm{CH}_{2}\left(\mathrm{CH}_{2}\right)_{n} \mathrm{CH}_{3}\) alkyl group. The melting points of the salts that form between the 1 -alkyl-3-methylimidazolium cation and the \(\mathrm{PF}_{6}^{-}\) anion are as follows: \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=60^{\circ} \mathrm{C}\right), \mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{m.p.}=40^{\circ} \mathrm{C}\right)\) \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=10^{\circ} \mathrm{C}\right)\) and \(\mathrm{R}=\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{CH}_{3}\left(\mathrm{~m} . \mathrm{p} .=-61^{\circ} \mathrm{C}\right) . \mathrm{Why}\) does the melting point decrease as the length of alkyl group increases?

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