Chapter 11: Liquids and Solids

11.3 Phase Transitions

Learning Outcomes

  • Define phase transitions and phase transition temperatures
  • Explain the relation between phase transition temperatures and intermolecular attractive forces
    • Use molecular-level representations to describe what happens to a pure substance during a phase change
  • Use the Clausius-Clapeyron equation to determine the enthalpy of vaporization or to predict the vapor pressure of a pure substance using experimental data

We witness and utilize changes of physical state, or phase transitions, in a great number of ways. As one example of global significance, consider the evaporation, condensation, freezing, and melting of water. These changes of state are essential aspects of our earth’s water cycle as well as many other natural phenomena and technological processes of central importance to our lives. In this section, the essential aspects of phase transitions are explored.

Vaporization and Condensation

When a liquid vaporizes in a closed container, gas molecules cannot escape. As these gas phase molecules move randomly about, they will occasionally collide with the surface of the condensed phase, and in some cases, these collisions will result in the molecules re-entering the condensed phase. The change from the gas phase to the liquid is called condensation. When the rate of condensation becomes equal to the rate of vaporization, neither the amount of the liquid nor the amount of the vapor in the container changes. The vapor in the container is then said to be in equilibrium with the liquid. Keep in mind that this is not a static situation, as molecules are continually exchanged between the condensed and gaseous phases. Such is an example of a dynamic equilibrium, the status of a system in which reciprocal processes (for example, vaporization and condensation) occur at equal rates. The pressure exerted by the vapor in equilibrium with a liquid in a closed container at a given temperature is called the liquid’s vapor pressure (or equilibrium vapor pressure). The area of the surface of the liquid in contact with a vapor and the size of the vessel have no effect on the vapor pressure, although they do affect the time required for the equilibrium to be reached. We can measure the vapor pressure of a liquid by placing a sample in a closed container, like that illustrated in Figure 11.3.1, and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase.

Three images are shown and labeled “a,” “b,” and “c.” Each image shows a round bulb connected on the right to a tube that is horizontal, then is bent vertically, curves, and then is vertical again to make a u-shape. A valve is located in the horizontal portion of the tube. Image a depicts a liquid in the bulb, labeled, “Liquid,” and upward-facing arrows leading away from the surface of the liquid. The phrase, “Molecules escape surface and form vapor” is written below the bulb, and a gray liquid in the u-shaped portion of the tube is shown at equal heights on the right and left sides. Image b depicts a liquid in the bulb, labeled, “Liquid,” and upward-facing arrows leading away from the surface of the liquid to molecules drawn in the upper portion of the bulb. A gray liquid in the u-shaped portion of the tube is shown slightly higher on the right side than on the left side. Image c depicts a liquid in the bulb, labeled, “Liquid,” and upward-facing arrows leading away from the surface of the liquid to molecules drawn in the upper portion of the bulb. There are more molecules present in c than in b. The phrase “Equilibrium reached, vapor pressure determined,” is written below the bulb and a gray liquid in the u-shaped portion of the tube is shown higher on the right side. A horizontal line is drawn level with each of these liquid levels and the distance between the lines is labeled with a double-headed arrow. This section is labeled with the phrase, “Vapor pressure.”
Figure 11.3.1. In a closed container, dynamic equilibrium is reached when (a) the rate of molecules escaping from the liquid to become the gas (b) increases and eventually (c) equals the rate of gas molecules entering the liquid. When this equilibrium is reached, the vapor pressure of the gas is constant, although the vaporization and condensation processes continue.

The chemical identities of the molecules in a liquid determine the types (and strengths) of intermolecular attractions possible; consequently, different substances will exhibit different equilibrium vapor pressures. Relatively strong intermolecular attractive forces will serve to impede vaporization as well as favoring “recapture” of gas-phase molecules when they collide with the liquid surface, resulting in a relatively low vapor pressure. Weak intermolecular attractions present less of a barrier to vaporization, and a reduced likelihood of gas recapture, yielding relatively high vapor pressures. The following example illustrates this dependence of vapor pressure on intermolecular attractive forces.

Example 11.3.1: Explaining Vapor Pressure in Terms of IMFs

Given the shown structural formulas for these four compounds, explain their relative vapor pressures in terms of types and extents of IMFs:

Four Lewis structures are shown. The first structure, labeled “ethanol,” shows a carbon bonded to three hydrogen atoms that is single bonded to a second carbon that is bonded to two hydrogen atoms and a hydroxyl group. The second structure, labeled “ethylene glycol, shows two carbon atoms, single bonded to one another, single bonded each to two hydrogen atoms, and each single bonded to a hydroxyl group. The third image, labeled “diethyl ether,” shows an oxygen atom single bonded on both sides to a carbon that is bonded to two hydrogens, and a second carbon, that is itself bonded to three hydrogen atoms. The fourth image, labeled “water,” shows an oxygen atom that is single bonded on both sides to hydrogen atoms.

Show Solution

Diethyl ether has a very small dipole and most of its intermolecular attractions are London forces. Although this molecule is the largest of the four under consideration, its IMFs are the weakest and, as a result, its molecules most readily escape from the liquid. It also has the highest vapor pressure. Due to its smaller size, ethanol exhibits weaker dispersion forces than diethyl ether. However, ethanol is capable of hydrogen bonding and, therefore, exhibits stronger overall IMFs, which means that fewer molecules escape from the liquid at any given temperature, and so ethanol has a lower vapor pressure than diethyl ether. Water is much smaller than either of the previous substances and exhibits weaker dispersion forces, but its extensive hydrogen bonding provides stronger intermolecular attractions, fewer molecules escaping the liquid, and a lower vapor pressure than for either diethyl ether or ethanol. Ethylene glycol has two [latex]\ce{-OH}[/latex] groups, so, like water, it exhibits extensive hydrogen bonding. It is much larger than water and thus experiences larger London forces. Its overall IMFs are the largest of these four substances, which means its vaporization rate will be the slowest and, consequently, its vapor pressure the lowest.

Check Your Learning

As temperature increases, the vapor pressure of a liquid also increases due to the increased average KE of its molecules. Recall that at any given temperature, the molecules of a substance experience a range of kinetic energies, with a certain fraction of molecules having a sufficient energy to overcome IMF and escape the liquid (vaporize). At a higher temperature, a greater fraction of molecules have enough energy to escape from the liquid, as shown in Figure 11.3.2. The escape of more molecules per unit of time and the greater average speed of the molecules that escape both contribute to the higher vapor pressure.

A graph is shown, with kinetic energy increasing in height and kinetic energy increasing along the bottom.
Figure 11.3.2. Temperature affects the distribution of kinetic energies for the molecules in a liquid. At the higher temperature, more molecules have the necessary kinetic energy, KE, to escape from the liquid into the gas phase.

Boiling Points

When the vapor pressure increases enough to equal the external atmospheric pressure, the liquid reaches its boiling point. The boiling point of a liquid is the temperature at which its equilibrium vapor pressure is equal to the pressure exerted on the liquid by its gaseous surroundings. For liquids in open containers, this pressure is that due to the earth’s atmosphere. The normal boiling point of a liquid is defined as its boiling point when surrounding pressure is equal to 1 atm (101.3 kPa). Figure 11.3.3 shows the variation in vapor pressure with temperature for several different substances. Considering the definition of boiling point, these curves may be seen as depicting the dependence of a liquid’s boiling point on surrounding pressure.

A graph is shown where the x-axis is labeled “Temperature ( degree sign, C )” and has values of 200 to 1000 in increments of 200 and the y-axis is labeled “Pressure ( k P a )” and has values of 20 to 120 in increments of 20. A horizontal dotted line extends across the graph at point 780 on the y-axis while three vertical dotted lines extend from points 35, 78, and 100 to meet the horizontal dotted line. Four lines are graphed. The first line, labeled “ethyl ether,” begins at the point “0 , 200” and extends in a slight curve to point “45, 1000” while the second line, labeled “ethanol”, extends from point “0, 20” to point “88, 1000” in a more extreme curve. The third line, labeled “water,” begins at the point “0, 0” and extends in a curve to point “108, 1000” while the fourth line, labeled “ethylene glycol,” extends from point “80, 0” to point “140, 100” in a very shallow curve.
Figure 11.3.3. The boiling points of liquids are the temperatures at which their equilibrium vapor pressures equal the pressure of the surrounding atmosphere. Normal boiling points are those corresponding to a pressure of 1 atm (101.3 kPa.)

Example 11.3.2: A Boiling Point at Reduced Pressure

A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in Figure 11.3.3 to determine the boiling point of water at this elevation.

Show Solution

The graph of the vapor pressure of water versus temperature in Figure 11.3.3 indicates that the vapor pressure of water is 68 kPa at about 90 °C. Thus, at about 90 °C, the vapor pressure of water will equal the atmospheric pressure in Leadville, and water will boil.

Check Your Learning

The quantitative relation between a substance’s vapor pressure and its temperature is described by the Clausius-Clapeyron equation:

[latex]P=A{e}^{-\Delta {H}_{\text{vap}}\text{/}RT}[/latex]

where [latex]\Delta[/latex]Hvap is the enthalpy of vaporization for the liquid, R is the gas constant, and ln A is a constant whose value depends on the chemical identity of the substance. This equation is often rearranged into logarithmic form to yield the linear equation:

[latex]\text{ln}P= -\dfrac{\Delta {H}_{\text{vap}}}{RT}+\text{ln}A[/latex]

This linear equation may be expressed in a two-point format that is convenient for use in various computations, as demonstrated in the example exercises that follow. If at temperature T1, the vapor pressure is P1, and at temperature T2, the vapor pressure is T2, the corresponding linear equations are:

[latex]\text{ln}{P}_{1}=-\dfrac{\Delta {H}_{\text{vap}}}{R{T}_{1}}+\text{ln}A[/latex]


[latex]\text{ln}{P}_{2}=-\dfrac{\Delta {H}_{\text{vap}}}{R{T}_{2}}+\text{ln}A[/latex]

Since the constant, ln A, is the same, these two equations may be rearranged to isolate ln A and then set them equal to one another:

[latex]\text{ln}{P}_{1}+\dfrac{\Delta {H}_{\text{vap}}}{R{T}_{1}}=\text{ln}{P}_{2}+\dfrac{\Delta {H}_{\text{vap}}}{R{T}_{2}}[/latex]

which can be combined into:

[latex]\text{ln}\left(\dfrac{{P}_{2}}{{P}_{1}}\right)=\dfrac{\Delta {H}_{\text{vap}}}{R}\left(\dfrac{1}{{T}_{1}}-\dfrac{1}{{T}_{2}}\right)[/latex]

Example 11.3.3: Estimating Enthalpy of Vaporization

Isooctane (2,2,4-trimethylpentane) has an octane rating of 100. It is used as one of the standards for the octane-rating system for gasoline. At 34.0 °C, the vapor pressure of isooctane is 10.0 kPa, and at 98.8 °C, its vapor pressure is 100.0 kPa. Use this information to estimate the enthalpy of vaporization for isooctane.

Show Solution

The enthalpy of vaporization, ΔHvap, can be determined by using the Clausius-Clapeyron equation:

[latex]\text{ln}\left(\dfrac{{P}_{2}}{{P}_{1}}\right)=\dfrac{\Delta {H}_{\text{vap}}}{R}\left(\dfrac{1}{{T}_{1}}-\dfrac{1}{{T}_{2}}\right)[/latex]

Since we have two vapor pressure-temperature values (T1 = 34.0 °C = 307.2 K, P1 = 10.0 kPa and T2 = 98.8 °C = 372.0 K, P2 = 100 kPa), we can substitute them into this equation and solve for ΔHvap. Rearranging the Clausius-Clapeyron equation and solving for ΔHvap yields:

[latex]\displaystyle\Delta {H}_{\text{vap}}=\frac{R\cdot \text{ln}\left(\frac{{P}_{2}}{{P}_{1}}\right)}{\left(\frac{1}{{T}_{1}}-\frac{1}{{T}_{2}}\right)}=\frac{\left(-8.3145\text{J/mol}\cdot \text{K}\right)\cdot \text{ln}\left(\frac{\text{100 kPa}}{\text{10.0 kPa}}\right)}{\left(\frac{1}{307.2\text{K}}-\frac{1}{372.0\text{K}}\right)}=\text{33,800 J/mol}=\text{33.8 kJ/mol}[/latex]

Note that the pressure can be in any units, so long as they agree for both P values, but the temperature must be in kelvin for the Clausius-Clapeyron equation to be valid.

Check Your Learning

Example 11.3.4: Estimating Temperature (or Vapor Pressure)

For benzene (C6H6), the normal boiling point is 80.1 °C and the enthalpy of vaporization is 30.8 kJ/mol. What is the boiling point of benzene in Denver, where atmospheric pressure = 83.4 kPa?

Show Solution

If the temperature and vapor pressure are known at one point, along with the enthalpy of vaporization, ΔHvap, then the temperature that corresponds to a different vapor pressure (or the vapor pressure that corresponds to a different temperature) can be determined by using the Clausius-Clapeyron equation:

[latex]\text{ln}\left(\dfrac{{P}_{2}}{{P}_{1}}\right)=\dfrac{\Delta {H}_{\text{vap}}}{R}\left(\dfrac{1}{{T}_{1}}-\dfrac{1}{{T}_{2}}\right)[/latex]

Since the normal boiling point is the temperature at which the vapor pressure equals atmospheric pressure at sea level, we know one vapor pressure-temperature value (T1 = 80.1 °C = 353.3 K, P1 = 101.3 kPa, [latex]\Delta[/latex]Hvap = 30.8 kJ/mol) and want to find the temperature (T2) that corresponds to vapor pressure P2 = 83.4 kPa. We can substitute these values into the Clausius-Clapeyron equation and then solve for T2. Rearranging the Clausius-Clapeyron equation and solving for T2 yields:

[latex]\displaystyle{T}_{2}={\left(\frac{-R\cdot \text{ln}\left(\frac{{P}_{2}}{{P}_{1}}\right)}{\Delta {H}_{\text{vap}}}+\frac{1}{{T}_{1}}\right)}^{-1}={\left(\frac{-\left(8.3145\text{J/mol}\cdot \text{K}\right)\cdot \text{ln}\left(\frac{83.4\text{kPa}}{101.3\text{kPa}}\right)}{\text{30,800 J/mol}}+\frac{1}{353.3\text{K}}\right)}^{-1}=\text{346.9 K or}{73.8}^{\circ }\text{C}[/latex]

Check Your Learning

Enthalpy of Vaporization

Vaporization is an endothermic process. The cooling effect can be evident when you leave a swimming pool or a shower. When the water on your skin evaporates, it removes heat from your skin and causes you to feel cold. The energy change associated with the vaporization process is the enthalpy of vaporization, [latex]\Delta[/latex]Hvap. For example, the vaporization of water at standard temperature is represented by:

[latex]\ce{H2O}(l)\longrightarrow \ce{H2}\ce{O}(g)\qquad\Delta {H}_{\text{vap}}=\text{44.01 kJ/mol}[/latex]

As described in the chapter on thermochemistry, the reverse of an endothermic process is exothermic. And so, the condensation of a gas releases heat:

[latex]\ce{H2O}(g)\longrightarrow \ce{H2O}(l)\qquad\Delta {H}_{\text{con}}=-\Delta {H}_{\text{vap}}=-44.01\text{kJ/mol}[/latex]

Example 11.3.5: Using Enthalpy of Vaporization

A photograph of a person's back beaded in sweat is shown.
Figure 11.3.4. Evaporation of sweat helps cool the body. (credit: “Kullez”/Flickr)

One way our body is cooled is by evaporation of the water in sweat (Figure 11.3.4). In very hot climates, we can lose as much as 1.5 L of sweat per day. Although sweat is not pure water, we can get an approximate value of the amount of heat removed by evaporation by assuming that it is. How much heat is required to evaporate 1.5 L of water (1.5 kg) at T = 37 °C (normal body temperature); [latex]\Delta[/latex]Hvap = 43.46 kJ/mol at 37 °C.

Show Solution

We start with the known volume of sweat (approximated as just water) and use the given information to convert to the amount of heat needed:

[latex]1.5\text{L}\times \dfrac{1000\cancel{\text{g}}}{\text{1 L}}\times \dfrac{1\cancel{\text{mol}}}{18\cancel{\text{g}}}\times \dfrac{43.46\text{kJ}}{1\cancel{\text{mol}}}=3.6\times {10}^{3}\text{kJ}[/latex]

Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water.

Check Your Learning

Melting and Freezing

When we heat a crystalline solid, we increase the average energy of its atoms, molecules, or ions and the solid gets hotter. At some point, the added energy becomes large enough to partially overcome the forces holding the molecules or ions of the solid in their fixed positions, and the solid begins the process of transitioning to the liquid state, or melting. At this point, the temperature of the solid stops rising, despite the continual input of heat, and it remains constant until all of the solid is melted. Only after all of the solid has melted will continued heating increase the temperature of the liquid (Figure 11.3.5).

This figure shows four photos each labeled, “a,” “b,” “c,” and, “d.” Each photo shows a beaker with ice and a digital thermometer. The first photo shows ice cubes in the beaker, and the thermometer reads negative 12.0 degrees C. The second photo shows slightly melted ice, and the thermometer reads 0.0 degrees C. The third photo shows more water than ice in the beaker. The thermometer reads 0.0 degrees C. The fourth photo shows the ice completely melted, and the thermometer reads 22.2 degrees C.
Figure 11.3.5. (a) This beaker of ice has a temperature of −12.0 °C. (b) After 10 minutes the ice has absorbed enough heat from the air to warm to 0 °C. A small amount has melted. (c) Thirty minutes later, the ice has absorbed more heat, but its temperature is still 0 °C. The ice melts without changing its temperature. (d) Only after all the ice has melted does the heat absorbed cause the temperature to increase to 22.2 °C. (credit: modification of work by Mark Ott)

If we stop heating during melting and place the mixture of solid and liquid in a perfectly insulated container so no heat can enter or escape, the solid and liquid phases remain in equilibrium. This is almost the situation with a mixture of ice and water in a very good thermos bottle; almost no heat gets in or out, and the mixture of solid ice and liquid water remains for hours. In a mixture of solid and liquid at equilibrium, the reciprocal process of melting and freezing occur at equal rates, and the quantities of solid and liquid therefore remain constant. The temperature at which the solid and liquid phases of a given substance are in equilibrium is called the melting point of the solid or the freezing point of the liquid. Use of one term or the other is normally dictated by the direction of the phase transition being considered, for example, solid to liquid (melting) or liquid to solid (freezing).

The enthalpy of fusion and the melting point of a crystalline solid depend on the strength of the attractive forces between the units present in the crystal. Molecules with weak attractive forces form crystals with low melting points. Crystals consisting of particles with stronger attractive forces melt at higher temperatures.

The amount of heat required to change one mole of a substance from the solid state to the liquid state is the enthalpy of fusion, [latex]\Delta[/latex]Hfus of the substance. The enthalpy of fusion of ice is 6.0 kJ/mol at 0 °C. Fusion (melting) is an endothermic process:

[latex]\ce{H2O}(s)\longrightarrow \ce{H2O}(l)\qquad\Delta {H}_{\text{fus}}=\text{6.01 kJ/mol}[/latex]

The reciprocal process, freezing, is an exothermic process whose enthalpy change is −6.0 kJ/mol at 0 °C:

[latex]\ce{H2O}(l)\longrightarrow \ce{H2O}(s)\qquad\Delta {H}_{\text{frz}}=-\Delta {H}_{\text{fus}}=-6.01\text{kJ/mol}[/latex]

Sublimation and Deposition

This figure shows a test tube. In the bottom is a dark substance which breaks up into a purple gas at the top.
Figure 11.3.6. Sublimation of solid iodine in the bottom of the tube produces a purple gas that subsequently deposits as solid iodine on the colder part of the tube above. (credit: modification of work by Mark Ott)

Some solids can transition directly into the gaseous state, bypassing the liquid state, via a process known as sublimation. At room temperature and standard pressure, a piece of dry ice (solid [latex]\ce{CO2}[/latex]) sublimes, appearing to gradually disappear without ever forming any liquid. Snow and ice sublime at temperatures below the melting point of water, a slow process that may be accelerated by winds and the reduced atmospheric pressures at high altitudes. When solid iodine is warmed, the solid sublimes and a vivid purple vapor forms (Figure 11.3.6). The reverse of sublimation is called deposition, a process in which gaseous substances condense directly into the solid state, bypassing the liquid state. The formation of frost is an example of deposition.

Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, [latex]\Delta[/latex]Hsub, is the energy required to convert one mole of a substance from the solid to the gaseous state. For example, the sublimation of carbon dioxide is represented by:

[latex]{\ce{CO}}_{2}\left(s\right)\longrightarrow {\ce{CO}}_{2}\text{(}g\text{)}\qquad\Delta {H}_{\text{sub}}=\text{26.1 kJ/mol}[/latex]

Likewise, the enthalpy change for the reverse process of deposition is equal in magnitude but opposite in sign to that for sublimation:

[latex]{\ce{CO}}_{2}\left(g\right)\longrightarrow {\ce{CO}}_{2}\text{(}s\text{)}\qquad\Delta {H}_{\text{dep}}=-\Delta {H}_{\text{sub}}=-26.1\text{kJ/mol}[/latex]

Consider the extent to which intermolecular attractions must be overcome to achieve a given phase transition. Converting a solid into a liquid requires that these attractions be only partially overcome; transition to the gaseous state requires that they be completely overcome. As a result, the enthalpy of fusion for a substance is less than its enthalpy of vaporization. This same logic can be used to derive an approximate relation between the enthalpies of all phase changes for a given substance. Though not an entirely accurate description, sublimation may be conveniently modeled as a sequential two-step process of melting followed by vaporization in order to apply Hess’s Law. Viewed in this manner, the enthalpy of sublimation for a substance may be estimated as the sum of its enthalpies of fusion and vaporization, as illustrated in Figure 11.3.7. For example:

[latex]\begin{array}{l}\text{solid}\longrightarrow \text{liquid}\qquad\Delta {H}_{\text{fus}}\\ \underline{\text{liquid}\longrightarrow \text{gas}\qquad\Delta {H}_{\text{vap}}}\\ \text{solid}\longrightarrow \text{gas}\qquad\Delta {H}_{\text{sub}}=\Delta {H}_{\text{fus}}+\Delta {H}_{\text{vap}}\end{array}[/latex]

A graph showing energy increasing from bottom to top is shown. Solid is the lowest bar, with liquid in the middle and gas at the top.
Figure 11.3.7. For a given substance, the sum of its enthalpy of fusion and enthalpy of vaporization is approximately equal to its enthalpy of sublimation.

Key Concepts and Summary

Phase transitions are processes that convert matter from one physical state into another. There are six phase transitions between the three phases of matter. Melting, vaporization, and sublimation are all endothermic processes, requiring an input of heat to overcome intermolecular attractions. The reciprocal transitions of freezing, condensation, and deposition are all exothermic processes, involving heat as intermolecular attractive forces are established or strengthened. The temperatures at which phase transitions occur are determined by the relative strengths of intermolecular attractions and are, therefore, dependent on the chemical identity of the substance.

Key Equations

  • [latex]P=A{e}^{-\Delta {H}_{\text{vap}}\text{/}RT}[/latex]
  • [latex]\text{ln}P=-\dfrac{\Delta {H}_{\text{vap}}}{RT}+\text{ln}A[/latex]
  • [latex]\text{ln}\left(\dfrac{{P}_{2}}{{P}_{1}}\right)=\dfrac{\Delta {H}_{\text{vap}}}{R}\left(\dfrac{1}{{T}_{1}}-\dfrac{1}{{T}_{2}}\right)[/latex]

Try It

  1. Carbon tetrachloride, [latex]\ce{CCl4}[/latex], was once used as a dry cleaning solvent, but is no longer used because it is carcinogenic. At 57.8 °C, the vapor pressure of [latex]\ce{CCl4}[/latex] is 54.0 kPa, and its enthalpy of vaporization is 33.05 kJ/mol. Use this information to estimate the normal boiling point for [latex]\ce{CCl4}[/latex].
  2. Titanium tetrachloride, [latex]\ce{TiCl4}[/latex], has a melting point of −23.2 °C and has a [latex]\Delta[/latex]H fusion = 9.37 kJ/mol.
    1. How much energy is required to melt 263.1 g [latex]\ce{TiCl4}[/latex]?
    2. For [latex]\ce{TiCl4}[/latex], which will likely have the larger magnitude: [latex]\Delta[/latex]H fusion or [latex]\Delta[/latex]H vaporization? Explain your reasoning.
Show Selected Solutions

2. (a) [latex]263.1{\ce{g TiCl}}_{4}\times \frac{\text{1 mol}}{189.9\text{g}}={\text{1.385 mol} \ce{TiCl}}_{4}[/latex]. Heat required to melt this amount of TiCl4 is nΔHfusion = 1.385 mol [latex]\times[/latex] 9.37 kJ/mol = 13.0kJ.

(b) It is likely that the heat of vaporization will have a larger magnitude since in the case of vaporization the intermolecular interactions have to be completely overcome, while melting weakens or destroys only some of them.


boiling point: temperature at which the vapor pressure of a liquid equals the pressure of the gas above it

Clausius-Clapeyron equation: mathematical relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance

condensation: change from a gaseous to a liquid state

deposition: change from a gaseous state directly to a solid state

dynamic equilibrium: state of a system in which reciprocal processes are occurring at equal rates

freezing: change from a liquid state to a solid state

freezing point: temperature at which the solid and liquid phases of a substance are in equilibrium; see also melting point

melting: change from a solid state to a liquid state

melting point: temperature at which the solid and liquid phases of a substance are in equilibrium; see also freezing point

normal boiling point: temperature at which a liquid’s vapor pressure equals 1 atm (760 torr)

sublimation: change from solid state directly to gaseous state

vapor pressure: (also, equilibrium vapor pressure) pressure exerted by a vapor in equilibrium with a solid or a liquid at a given temperature

vaporization: change from liquid state to gaseous state

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