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Chapter 7 – States of Matter and Changes in State

Introduction

The materials we deal with every day exist in one of three states of matter: solid, liquid, or gas. The molecules in the liquid and solid phase are much closer to one another than are the molecules in the gas state. Consequently, the solid and liquid states are often called the condensed states. Some substances, such as water, are found in all three states under normal conditions (ice, water, and steam). However, most other substances are found in only one state under normal conditions. Thus, most of us have only encountered elemental iron as a solid, helium as a gas, and table salt as a solid. In this chapter, we examine the properties of the states of matter.

7.1 Gases

Introduction

Gases are characterized by their volume (V), pressure (P), temperature (T), and the number of moles of particles (n). In this section, we discuss how the volume of a gas is affected by changes in its pressure, temperature, and number of moles. Finally, we consider mixtures of gases.

Objectives

Boyle's Law and Pressure

7.1-1. Boyle's Law and Pressure

Barometric pressure is usually given by the height of a mercury column that the atmosphere will support.
As shown in Figure 7.1, a barometer is a glass tube, which is closed at one end and open at the other, that is filled with mercury and immersed with the open end down into a pool of mercury. Only part of the mercury drains from the tube, and the resulting column has a height that depends upon the atmospheric pressure. The column, which is supported by the molecules in the atmosphere, has a height of 760 mm at sea level at normal conditions. Gas pressures can be determined by measuring the height of the mercury column that they support. The height is usually given in mm, and the pressure is given in mm of Hg or torr (1 torr = 1 mm Hg). The following are common ways of expressing the normal pressure at sea level.
( 7.1 )
1 atm = 14.7 lb/in2 = 760 mm Hg = 760 torr = 1.01 × 105 Pa
Pa is the pascal, the SI unit of pressure.
1 Pa = 1 kg · m−1 · s−2
Note that the barometric pressures given in weather reports are given in inches of Hg. The median barometric pressure at sea level is 760 mm = 29.9 in. Some extreme barometric pressures observed in the U.S. are given below.
Figure 7.1: A Barometer
The pressure exerted by the weight of the mercury column of height h equals the pressure exerted by the weight of the atmosphere.

7.1-2. Measuring Pressure with a Manometer

The pressure of a gas in the laboratory is often found as the difference in heights of two columns of a liquid (most often mercury) in a manometer.
The device used to measure the pressure of a gas in the laboratory is called a manometer. One side of the manometer shown in Figure 7.2 is open to the atmosphere, while the other is attached to the gas whose pressure is to be measured. Consequently, the atmospheric pressure (Patm) is pushing on the right side, and the unknown gas pressure (P) is pushing on the left. When the two pressures are the same, the two mercury columns in the manometer are at the same height, and h = 0. However, when the two pressures are not the same, the side with the greater pressure exerts the greater force, which causes the mercury column on that side to drop, and the column on the other side to rise. The difference in the pressures of the two sides is given by the difference in the two mercury levels, h, as described below.
Figure 7.2: Determining Pressure with a Manometer
A manometer, which is used to measure the pressure of a gas: In the above experiment, the pressure of the gas (P) is greater than barometric pressure (Patm) by h mm of Hg, i.e.,
P = Patm + h.

7.1-3. Barometer Exercise

Exercise 7.1:

pressure greater than atmospheric by 335 mm
What is the pressure of the yellow gas in torr (mm Hg) if barometric pressure is 762 torr?
1097_0__ The pressure of the gas is greater than barometric pressure by 335 mm Hg because the column is higher on the atmosphere side. Thus, the pressure of the gas is P = Pbar + h = 762 + 335 = 1097 mm Hg or 1097 torr. torr
What is the pressure of the yellow gas in atmospheres?
1.44_.04__ 1 atm = 760 torr, so the pressure in atm is 1097 torr (1 atm/760 torr) = 1.44 atm. atm

7.1-4. Boyle's Law

PV = k, where k is a constant that depends only upon the amount and temperature of the gas.
The volume of a given amount of gas at a constant temperature varies with its pressure. The graph in Figure 7.3 shows the volume of one mole of gas as a function of its pressure at 0 °C. The relationship between pressure and volume
( 7.2 )
PV = k(n, T)
Boyle's Law
is known as Boyle's Law. The constant, k(n, T), is a function of the number of moles of gas (n) and its temperature (T). Boyle found that the product of the pressure of a gas and its volume is constant as long as the number of moles of the gas and the temperature of the gas are held constant. The following equation is often used for Boyle's Law. It indicates that the PV product is unchanged in processes that involve no change in temperature or amount of gas. Thus, the following form of Boyle's Law is also commonly used.
PfVf = PiVi
where PfVf is the final PV product and PiVi is the initial product.
graph of volume versus pressure
Figure 7.3: Volume of Gas vs. Pressure of Gas
The volume of one mole of a gas at 0 °C vs. the pressure of the gas is shown.

7.1-5 Boyle's Law Exercise

Exercise 7.2:

The pressure on 5.2 L of a gas is decreased from 1.0 to 0.75 atm at constant temperature. What is the new volume?
initial volume in liters 5.2___ The initial volume is the given volume, which is 5.2 L in this problem. L
initial pressure in atmospheres 1___ The question indicates that the gas started at 1 atm, which is the initial pressure. atm
final pressure in atmospheres 0.75___ The question indicates that the final pressure is 0.75 atm. atm
final volume in liters 6.9___ The final volume is determined from the following:
Vf = Vi ×
Pf
Pi
= (5.2 L) ×
1.0 atm
0.75 atm
= 6.9 L.
Note that the final volume equals the initial volume times a pressure change correction factor, which is a ratio of the pressures. Problems like this can be done with common sense rather than algebra if we recognize that pressure and volume must move in opposite directions. In this example, the pressure decreased, so we reason that the volume must increase, which means that the larger pressure must be in the numerator of the factor.
L

Temperature and Charles's Law

7.1-6. Charles's Law

V = kT, where k is a constant that depends on the amount and pressure of the gas, and T is the temperature in kelvins.
Experiments with gases in the late 1700s and early 1800s determined the relationship between the volume of a gas and its temperature. As shown in Figure 7.4, the volume of a given amount of gas at constant pressure increases linearly with its temperature. Although the slope of the straight line varies with the number of moles and pressure of the gas, extrapolation of all lines to the temperature of zero volume yields the same temperature, –273.15 °C (degrees Celsius). At lower temperatures, the volume of the gas would be negative, which is not possible, so we conclude that temperatures can be no lower than –273.15 °C. The absolute temperature, or Kelvin, scale was defined based on this absolute minimum in temperature.
( 7.3 )
K = °C + 273.15
Note that the word "degrees" is not used with Kelvin temperatures, so 273 K is said "273 kelvins." Equation 7.3 shows that you add 273.15 to a temperature in degrees Celsius to convert it to the absolute scale. For example, water freezes at 0 °C, which is 273 K, and boils at 100 °C, which is 373 K. Room temperature is usually assumed to be 25 °C, which is 298 K. The absolute temperature must be used in all equations involving the temperature of a chemical system. However, either the Celsius or absolute scales can be used in equations involving a change in the temperature of a system because the unit size is the same in both scales. Once the new temperature scale was defined, the volume of a gas was found to be directly proportional to its temperature expressed in kelvins, which is known as Charles's Law.
( 7.4 )
V = k(n, P)T
Charles's Law
Charles's Law indicates that the V/T ratio of a gas is constant as long as n and P are unchanged, so Charles's Law can also be given as follows.
Vf
Tf
=
Vi
Ti
The above shows that the final V/T ratio equals the initial V/T ratio as long as the amount of gas and its pressure are unchanged.
line graph volume vs. temperature for different number of moles of helium
Figure 7.4: Charles's Law
The volume of the specified number of moles of gas as a funtion of temperature: The bottom scale is in degrees Celsius, and the top scale is the Kelvin scale.

7.1-7. Charles's Law Exercise

Exercise 7.3:

What is the volume of gas at 0 °C if the volume is 250 mL at 100 °C?
initial volume in milliliters 250_0__ The initial volume is the given volume, which is 250 mL in this problem. Note that the units of the answer will be the same as the units of the given volume, so it does not matter what those units are. mL
initial temperature in kelvins 373___ The question indicates that the gas started at 100 °C, which must be converted to kelvins: K = 100 + 273 = 373 K. K
final temperature in kelvins 273___ The question indicates that the gas started at 0 °C, which must be converted to kelvins: K = 0 + 273 = 273 K. K
final volume in milliliters 182.98___ The final volume is determined from the following:
Vf = Vi ×
Tf
Ti
= (250 mL) ×
273 K
373 K
= 183 mL.
Note that the final volume equals the initial volume times a temperature change correction factor, which is a ratio of the temperatures in kelvins. Problems like this can be done with common sense rather than algebra if we recognize that temperature and volume must move in the same direction. In this example, the temperature decreased, so we reason that the volume must also decrease, which means that the smaller temperature must be in the numerator of the factor.
mL

The Ideal Gas Law

7.1-8. Ideal Gas Law

PV = nRT is the ideal gas law.
Avogadro's Law states that equal volumes of gases measured under the same conditions of temperature and pressure contain equal numbers of molecules. The volume of a gas at constant temperature and pressure is directly proportional to the number of moles of gas,
( 7.5 )
V = k(P, T)n
Avogadro's Law
where n is the number of moles of gas, and k(P, T) is a proportionality constant that depends upon the pressure and the temperature of the gas. We now have three equations that relate the volume of a gas to its pressure, its temperature, and to the number of moles of the gas. These three relationships can be combined into the ideal gas law.
( 7.6 )
PV = nRT
Ideal Gas Law
P V = n R T
Pressure expressed in atmospheres: It can be expressed in other units, but that would change the value of R. Volume of the gas expressed in liters: It can be expressed in other units, but that would change the value of R. Number of moles of gas R = 0.0821 L · atm · K–1 · mol–1: It is the gas law constant. T is the absolute temperature in kelvins.
Table 7.1

7.1-9. Ideal Gas Law Exercise

Exercise 7.4:

The ideal gas law can be used to solve for any one of the variables (P, V, n, or T) if the others are known. Things to remember when using R = 0.0821 L · atm · K–1 · mol–1:
  • T is in kelvins.
  • P is in atmospheres.
  • V is in liters.
What is the pressure in a 280-mL container that contains 1.8 g He at 330 °C?
n = 0.45___ n = the number of moles, which is equal to the mass of He divided by the molar mass of He:
1.8 g He ×
1 mol He
4.0 g He
= 0.45 mol He.
mol
R = 0.0821___ R = 0.0821 only when T is in K, P is in atm, and V is in L. L · atm · K–1 · mol–1
T = 603___ Degrees Celsius must be converted to kelvins: T = 330 °C + 273.15 = 603.15 K. K
V = 0.28_0__ V must be in liters:
280 mL ×
1 L
1000 mL
= 0.28 L.
L
P = 79.66___ Solve PV = nRT for the pressure:
(0.45 mol)(0.0821 L · atm · K−1 · mol−1)(603 K)
0.28 L
= 80 atm.
atm

Mixtures of Gases

7.1-10. Molarity

The molar concentration, or molarity, of a gas is the number of moles of gas present in 1 L.
Mixtures of gases are the same throughout, i.e., they are homogeneous mixtures, and homogeneous mixtures are called solutions. The molecules in a mixture (solution) of ideal gases do not interact, so they can be treated independently. This means that we can apply the ideal gas law to each component of a mixture as if the other components were not present. Each gas in a mixture is at the same temperature, and each has the same volume (the volume of the container). Thus, only the number of moles and the pressure of each component differ. The pressure exerted by gas A in a mixture is called the partial pressure of A and given the symbol PA. If A is an ideal gas, we can express its pressure with the ideal gas law as follows.
( 7.7 )
PA =
nART
V
Partial Pressure
(nA/V) is the number of moles of A per unit volume. The amount of substance per unit volume is called the concentration of the substance. When the amount is expressed in moles and the volume in liters, it is called the molar concentration, or molarity, of the substance, so nA/V is the molarity of A in the solution. Molarity is the most common unit of concentration in a chemistry laboratory (and the only one used in this course). The molarity of A is represented as [A] and is defined as follows.
( 7.8 )
[A] =
moles of A
total volume in liters
Definition of Molarity
[A] = 1 M is read as "the concentration of A is one molar, or one mole per liter." Substituting [A] = n/V into the expression for the partial pressure of A, we obtain the relationship between the pressure of a gas and its concentration.
( 7.9 )
PA = [A]RT
Pressure and Molarity
Thus, the pressure of a gas is directly proportional to its molar concentration.
Example:
If a mixture of gases in a 9.6-L flask at 300 K contains 0.53 moles of A, we would determine that [A] = 0.53/9.6 = 0.055 M, i.e., the solution of gases is 0.055 molar in A. In addition, the partial pressure of A would be:
PA = (0.055 mol/L)(0.0821 L · atm · mol−1 K−1)(300 K)
= 1.36 atm

7.1-11. Molarity Exercise

Exercise 7.5:

What is the concentration of fluorine (F2) in a 4.0-L flask that contains 4.2 g of fluorine?
1. Determine the number of moles of F2 (Mm = 38.00 g/mol).
n = 0.11___ The number of moles of F2 in the container is
n = 4.2 g F2 ×
1 mol F2
38.0 g F2
= 0.11 mol F2.
mol F2
2. Calculate the molarity.
[F2]
=
0.028___ The concentration of fluorine in moles per liter (the molarity) is
[F2] =
0.11 mol F2
4.0 L
= 0.028 mol · L−1 = 0.028 M.
The fluorine concentration is 0.028 M (0.028 molar).
M

7.1-12. Partial Pressures

The pressure of a mixture of gases equals the sum of the partial pressures of the component gases.
The total pressure in a container filled with a mixture of gases is equal to the sum of the partial pressures of the component gases.
( 7.10 )
Ptot =
Pi
all gases
Dalton's Law of Partial Pressures
where Pi is the partial pressure of the ith component of the gas. Thus, the total pressure in a container that contains two gases, A and B, can be expressed as follows.
P = PA + PB
Using Equation 7.7
PA =
nART
V
to evaluate PA and PB, we obtain the following.
P =
nART
V
+
nBRT
V
=
(nA + nB)RT
V
The total pressure of a mixture of gases equals the total number of moles of gas times RT/V.

7.1-13. Partial Pressure Exercise

Exercise 7.6:

What are the partial pressures of Ne and He in a 6.0-L flask that contains 0.12 mol Ne and
0.19 mol He at 55 °C?
PNe = 0.54___ The volume of each gas is V = 6.0 L, and the temperature of each gas is T = 273 + 55 = 328 K. Apply the Ideal Gas Law to each to obtain the following:
PNe =
nNeRT
V
=
(0.12 mol)(0.0821 L · atm · K−1 · mol−1)(328 K)
6.0 L
= 0.54 atm.
atm
PHe = 0.85___ The volume of each gas is V = 6.0 L, and the temperature of each gas is T = 273 + 55 = 328 K. Apply the Ideal Gas Law to each to obtain the following:
PHe =
nHeRT
V
=
(0.19 mol)(0.0821 L · atm · K−1 · mol−1)(328 K)
6.0 L
= 0.85 atm.
atm
What is the total pressure in the flask?
1.39___ The total pressure is the sum of the partial pressures: Ptot = 0.54 atm + 0.85 atm = 1.39 atm. Alternatively, the total pressure can be obtained from the Ideal Gas Law by using the total number of moles of gas, ntot = nNe + nHe = 0.12 mol + 0.19 mol = 0.31 mol:
Ptot =
ntotRT
V
=
(0.31 mol)(0.0821 L · atm · K−1 · mol−1)(328 K)
6.0 L
= 1.4 atm.
atm

7.1-14. Gas Law Exercise

Exercise 7.7:

A mixture contains 15 g Ne (AW = 20) and 10 g Ar (AW = 40) in a 5.0-L container at 30 °C.
Neon:
number of moles 0.75___ (15.0 g)(1 mol/20.0 g) = 0.75 mol Ne mol Ne
molar concentration 0.15___ 0.75 mol/5.0 L = 0.15 M Ne M Ne
partial pressure in atm 3.7___ P = nRT/V = (0.75 mol)(0.0821 L · atm/K · mol)(303 K)/5.0 L = 3.7 atm Ne atm Ne
Argon:
number of moles 0.25___ (10 g)(1 mol/40 g) = 0.25 mol Ar mol Ar
molar concentration 0.050___ 0.25 mol/5.0 L = 0.050 M Ar M Ar
partial pressure in atm 1.244_0.044__ P = nRT/V = (0.25 mol)(0.0821 L · atm/K · mol)(303 K)/5.0 L = 1.2 atm Ar atm Ar
What is the total pressure in the container?
4.9___ P = PNe + PAr = 3.7 + 1.2 = 4.9 atm atm

7.2 Kinetic-Molecular Theory and Thermal Energy

Introduction

The ideal gas law applies only to gases that exhibit ideal gas behavior, which can deviate from real gas behavior. In this section, we examine the properties of ideal gases.

Objectives

7.2-1. Postulates

The behavior of gases is explained by kinetic-molecular theory, which is based on the four postulates provided below. Gases that obey all four postulates are called ideal gases.

7.2-2. Thermal Energy

Thermal energy, which is approximated as RT, is a measure of the average kinetic energy of the molecules.
The average kinetic energy of the molecules in a system is often referred to as the thermal energy of the system. Thermal energy is an average energy because the molecules in a system do not all move at the same speed—some are moving faster and have more energy than others. All of the energies related to molecules or ions that we have encountered to this point have been potential energies arising from Coulombic interactions. The thermal energy of the molecules is energy that can be used to overcome Coulombic interactions. Thus, thermal energy can be used to break a bond or to melt a solid. Thermal energy is also required to make reactions occur. As we will see in Chapter 9, rates of reaction increase when the temperature is increased. Thus, thermal energy plays an important role in chemistry, but the thermal energy of a solid or a liquid is not easily determined. However, thermal energy is usually approximated as follows:
( 7.11 )
Ethermal ~ RT
Thermal Energy Approximation
where is the ideal gas law constant in J rather than L · atm.

7.2-3. Thermal Energy Exercise

Exercise 7.8:

What is thermal energy at room temperature (25 °C)?
2.479___ First, convert temperature to kelvins: T = 273 + 25 = 298. Then,
Ethermal ~ RT = (0.008314 kJ · mol−1 · K−1)(298 K) = 2.5 kJ · mol−1.
kJ/mol
What is thermal energy at 100 °C?
3.103___ T = 273 + 100 = 373, so
Ethermal ~ RT = (0.008314 kJ · mol−1 · K−1)(373 K) = 3.1 kJ · mol−1.
kJ/mol
What is the approximate temperature at which Ethermal ~ 100 kJ/mol?
12027___ E ~ RT, so
T ~
E
R
=
100 × 103 J
8.31
= 1.2 × 104 K.
K

7.3 Intermolecular Interactions

Introduction

Why are some molecules gases, while others are liquids and solids at room conditions? Why does water expand upon freezing? The answers to both questions lie in the nature of the interaction between molecules, the topic of this section.

Prerequisites

Objectives

7.3-1. Intermolecular Forces

Particles in the gas phase are relatively far apart, so their attractive forces are negligible compared to their thermal energy. However, the attractive forces between particles in the condensed states are strong enough to hold the particles in contact with one another. Ionic compounds are always solids at room temperature because their cations and anions are held in the solid network by strong electrostatic attractions called ionic bonds. However, molecular substances can occur as gases (O2), liquids (H2O), or solids (sugar) at room conditions. This diversity in states arises from a diversity in the strengths of the forces between the molecules. Whereas the strength of ionic bonds determines the physical properties of a salt crystal, the physical properties of molecular substances are not related to the strength of their covalent bonds. In molecular substances, we must distinguish between two different kinds of forces: The N≡N bond is a very strong bond, so the intramolecular force between the nitrogen atoms in an N2 molecule is very strong. However, N2 is a gas at room conditions because the intermolecular forces between different N2 molecules are very weak. Intermolecular forces, like all interactions in chemistry, are electrostatic. Thus, condensed phases in molecular substances result because there is an attraction between regions of opposite charge on the molecules. However, these charges are smaller, more diffuse, and farther apart than those in bonds, so the force of attraction between different molecules is much less than that between the atoms in a bond. Bond energies lie between 100 and 1000 kJ/mol, but the strengths of most intermolecular interactions are less than 10 kJ/mol. Consider that the bond formed with Super Glue is an intramolecular (bonding) interaction, while static cling is an intermolecular interaction. Although molecules are electrically neutral, many have regions of nonzero charge due to asymmetric distributions of their valence electrons that produce regions with above normal electron density that are slightly negative (δ–) and regions of depleted electron density that are slightly positive (δ+). Such molecules contain two poles, one positive (δ+) and one negative (δ–). The presence of two poles in the molecule results in a molecular dipole, much like the bond dipole discussed in Chapter 5.2 Bond Polarity. In the remainder of this section, we discuss the origin of these dipoles and the forces they create.

Dispersion Forces

7.3-2. Dispersion Forces

Dispersion forces, which result from the interaction of induced dipoles, increase with molar mass.
Electrical charge in molecules is derived from asymmetrical electron distributions within the molecules, i.e., there are regions within the molecule that are electron rich and other regions that are electron deficient. These regions of excess and deficient electron density result in negatively and positively charged regions in the molecule. The center of the positive regions and the center of the negative regions form two poles. If the two poles do not coincide, the molecule is said to have a dipole. Dipoles can be permanent or temporary. Temporary dipoles result from random electron motion, which can create a dipole in a molecule. The resulting dipole then induces dipoles in surrounding molecules as the electrons in surrounding molecules move toward the positive pole and away from the negative pole. Thus, temporary dipoles are also called induced dipoles. The force of attraction of the opposite charges of the temporary dipoles is called a dispersion force.
noninteracting molecules
Figure 7.5a: The Dispersion Force
Three molecules with no regions of charge have no way to interact.
interacting molecules
Figure 7.5b: The Dispersion Force
Random electron movement in the central molecule causes more electrons to be on one side and fewer on the other, which results in two charged regions.
dispersion force
Figure 7.5c: The Dispersion Force
The charged regions on the central molecule induce charged regions on the adjacent molecules. The result is that the adjacent molecules have induced dipoles. The interaction of the opposite charges of the induced dipoles is called a dispersion force.
Dispersion forces are present in all molecular substances, but they are most important in large molecules where the electron clouds are more easily deformed. Thus, dispersion forces increase with molar mass.

Molecular Dipoles

7.3-3. Dipolar Interactions

Dipolar forces act in addition to dispersion forces to increase the strength of the intermolecular interaction.
Most molecules contain atoms with partial positive charge and atoms with partial negative charge as a result of the electronegativity differences that produce polar bonds. If the centers of positive and negative charge do not coincide, two poles are produced and the molecule has a permanent dipole. The magnitude of the dipole depends upon both the size and the separation of the two centers of charge; if the centers of charge coincide, their separation is zero and they have no permanent dipole. Molecules with permanent dipoles are said to be polar molecules, or simply polar, while those that do not have permanent dipoles are said to be nonpolar. Interactions between the centers of opposite charge of the permanent dipoles of different molecules are called dipolar forces or dipole-dipole forces. Dipolar forces act in addition to dispersion forces. For example, the dispersion forces in HCl and F2 are expected to be nearly the same because they have similar molar masses (38 and 37 g/mol, respectively). However, HCl is polar, and as shown in Figure 7.6, the partial positive charge on the hydrogen atom of one HCl molecule interacts with the partial negative charge on the chlorine atom of an adjacent molecule to produce a dipolar force between the molecules. This force is in addition to the dispersion forces, so the intermolecular forces are greater in HCl than in F2, which makes HCl gas much easier to liquefy.
Figure 7.6: Dipolar interactions in HCl
Dipolar interactions (dotted lines) are between the chlorine atom (negative end of dipole) of one molecule and the hydrogen atom (positive end of dipole) on an adjacent molecule.

7.3-4. Charge Distribution

Molecules with a permanent asymmetric charge distribution are said to be polar or to have a molecular dipole.
Recall from Chapter 5.2 Bond Polarity that bond dipoles arise in bonds due to electronegativity differences between the bound atoms. Although the total charge on a molecule is zero, some of the atoms in a molecule with polar bonds can carry some charge. The charge on an atom lies somewhere between its formal charge and its oxidation state, depending upon bond polarities. The centers of positive and negative charge in a molecule are called poles, and a molecular dipole is defined as the product of the charge on the two poles of a molecule and the distance between them. If the centers of positive and negative charge coincide, their separation is zero, and there is no molecular dipole. Molecules with permanent dipoles are said to be polar, while those without molecular dipoles are said to be nonpolar. We now compare the charge distributions in SO3 and H2O. In Figures 7.7a and 7.7b, regions of negative charge are shown in red and regions of positive charge are shown in blue, with a deeper color used to depict more charge.
charge distribution in SO3
Figure 7.7a: Charge Distribution in SO3
O is more electronegative than S, so the three S–O bonds in SO3 are polar. As a result, negative charge resides on the O atoms (red regions in Figure 7.7a), while positive charge is found around the S atom (blue regions in Figure 7.7a). The geometric centers of the negative and positive charge both lie on the sulfur atom. Although SO3 has regions of charge, the centers of positive and negative charge coincide, so SO3 does not contain a molecular dipole and is a nonpolar molecule.
charge distribution in water
Figure 7.7b: Charge Distribution in H2O
O is more electronegative than H, so the two O–H bonds in H2O are polar. Increased electron density on the oxygen produces a partial negative charge there (red region in Figure 7.7b), while decreased electron density on the hydrogens results in positive charge on both hydrogen atoms (blue regions in Figure 7.7b). The coloring is more intense in H2O than in SO3 because O–H bonds are more polar than S–O bonds. The center of negative charge (near the O atom) does not coincide with the center of positive charge (between H atoms), so water is a polar molecule (has a molecular dipole).

7.3-5. Representing Molecular Dipoles

A molecular dipole is represented by an arrow that points from the positive pole toward the negative pole. A line is drawn through the arrow at the positive end to make "+" sign.
dipole arrow
Figure 7.8
A molecule's permanent dipole is indicated by an arrow pointing from the center of positive charge toward the center of negative charge. A short line crosses the arrow forming a "+" at the positive end.

7.3-6. Effect of Geometry on Molecular Polarity

Molecular dipoles also depend upon molecular shape.
The strength of the dipole depends on both the charge on the poles and their separation. Consider the examples of CO2 and SO2.
CO2 with charges
Figure 7.9a: Charge Distribution in CO2
CO2
SO2 with charge
Figure 7.9b: Charge Distribution in SO2
SO2
SO2 with dipole
Figure 7.9c: Polarity in SO2

7.3-7. Molecular Dipole of H2O

H2O with charge
Figure 7.10a
H2O with dipole
Figure 7.10b

7.3-8. Molecular Dipoles from Bond Dipoles

Molecular dipoles are the vector sum of the bond dipoles in a molecule.
Recall that the bond dipole (5.2 Bond Polarity) points from the less electronegative atom to the more electronegative atom in the bond. It is a vector that indicates the charge distribution in the bond. The molecular dipole can be determined as the vector sum of the bond dipole vectors.
CO2 bond dipoles
Figure 7.11a: Molecular dipoles are the sums of bond dipoles.
The bond dipole of each CO bond points from the carbon toward the oxygen. Because they point in opposite directions, they cancel one another, and CO2 does not have a molecular dipole, i.e., it is a nonpolar molecule.
SO2 bond dipoles
Figure 7.11b: Molecular dipoles are the sums of bond dipoles.
Each SO bond has a bond dipole pointing from the sulfur toward the more electronegative oxygen. The sum of the two bond dipoles results in the molecular dipole represented by the red arrow.
H2O bond dipoles
Figure 7.11c: Molecular dipoles are the sums of bond dipoles.
The OH bond dipoles (blue arrows) are at 104° angles. The resultant molecular dipole (yellow arrow) points from the center of the line connecting the hydrogens toward the oxygen.

7.3-9. Predicting Dipoles Exercise

Exercise 7.9:

Predict whether each of the following molecules is polar or nonpolar.
    ammonia
  • polar Each bond dipole points from a hydrogen to the more electronegative nitrogen, so the molecular dipole points from the center of the three hydrogen atoms toward the nitrogen.
  • nonpolar
    CS2
  • polar
  • nonpolar CS2 is linear, so the bond dipoles, which point from the carbon to a sulfur, cancel. Consequently, CS2 is not polar.
    HCN
  • polar HCN is linear, but the bond dipoles do not cancel because a C≡N bond is more polar than a C–H bond. Consequently, HCN is a polar molecule.
  • nonpolar
    CH2F2
  • polar The bond dipoles do not cancel because the C–F bond is much more polar than the C–H bond. The molecular dipole points from the center of the hydrogen atoms toward the center of the fluorine atoms.
  • nonpolar
    CH2Cl2
  • polar This is a commonly missed problem because the drawing is misleading. There are four electron groups around the carbon, so it is tetrahedral. None of the atoms in a tetrahedral structure are "opposite" one another as shown in the planar representation. See the preceding example for a better representation. As in the preceding example, the molecular dipole points from the center of the hydrogen atoms toward the center of the halogen atoms.
  • nonpolar
    trans C2H2F2
  • polar
  • nonpolar We can neglect the C–H bonds as they are not polar. The bond dipoles of the two C–F bonds cancel, so the molecule is not polar.
    cis C2H2F2
  • polar We can neglect the C–H bonds as they are not polar. The bond dipoles of the two C–F bonds do not cancel, so the molecule is polar.
  • nonpolar
    ammonia
  • polar Two of the P–Br bonds cancel leaving only the one P–Br bond. The two P–F bond dipoles do not cancel, and their sum is opposite that of the P–Br bond. F is more electronegative than Br, so the molecular dipole points from the bromine toward the center of the F atoms.
  • nonpolar
    ammonia
  • polar
  • nonpolar The two P–F bond dipoles cancel, as do the three P–Br bonds, so the molecule is not polar.

Hydrogen Bonding

7.3-10. Hydrogen Bonding

Hydrogen bonding is a strong dipolar force that can occur between a hydrogen atom and either a nitrogen, oxygen, or fluorine atom.
The strength of a dipole-dipole interaction depends upon the distance between the two interacting poles (the positive pole of one molecule and the negative pole of a different molecule) and the magnitude of the charge on the poles. Consequently, dipole-dipole interactions can be quite strong in molecules that contain small atoms of very different electronegativities. Hydrogen is the smallest and one of the least electronegative nonmetals, so when it is bound to another small, highly electronegative element (N, O, or F), a very strong dipole-dipole interaction, the hydrogen bond, is established. However, a molecule must contain one of the following bonds to participate in a hydrogen bond.

7.3-11. H-bonding in Water

An oxygen atom in water can hydrogen bond to two hydrogen atoms on different water molecules.
Hydrogen bonding in water is especially strong because each oxygen atom can bond to two different water molecules through its two lone pairs. Note that the oxygen is tetrahedral with two O–H covalent bonds that are responsible for H2O molecules and two longer O–H hydrogen bonds that hold water molecules close to one another and are responsible for water being a liquid at room temperature.
H bonds in water
Figure 7.12

7.3-12. Ice

In order to maximize the hydrogen bonding interactions, water molecules in ice orient relative to one another so that the two lone pairs on an oxygen interact with hydrogen atoms of two different water molecules. As a result, water crystallizes in a structure containing hexagonal channels (see Figure 7.13). These channels, which are created when water freezes, result in void space. The presence of the large amount of void space makes the solid less dense than the liquid. The following properties of ice are due to these channels.
crystal structure of ice
Figure 7.13: Structure of Ice

Predicting Relative Forces

7.3-13. Summary

As shown in Table 7.2, the relative strengths of the various interactions we have discussed overlap, so it is difficult to predict whether the interactions in a large nonpolar molecule are stronger or weaker than in a small polar molecule that hydrogen bonds. For example, I2 is a solid at room condition, while H2O is a liquid. Remember that dispersion forces are always present, so the interactions in a large polar molecule are probably stronger than in a small one. In the examples in Exercise 7.10, you will be asked to select the molecule in each pair that is expected to have the stronger interactions and the force responsible for the difference. Only one is responsible for the difference because the molecules have been chosen so that the other forces are comparable, not because they do not exist.
Type of Interaction Energy of Interaction
(kJ/mol)
Covalent Bond 100–1000
Hydrogen Bond 10–50
Dipolar Force 5–25
Dispersion 1–50
Table 7.2: Relative Strengths of Interaction

7.3-14. Predicting Relative Strengths of Interaction Exercise

Exercise 7.10:

Predict the compound with the stronger interaction in each pair. Then indicate the force that is most responsible for the difference.
    Compound:
  • CH3–O–CH3 No. The melting point of C2H5–O–H is much higher. The two compounds have the same formula but have very different melting points. Examine them carefully to determine the structural difference that would make their melting points so different.
  • C2H5–O–H Correct. The melting point of C2H5–O–H is much higher. The two compounds have the same formula but have very different melting points. Next, identify the force responsible for the difference.
    Compound:
  • CH3–O–CH3 No. The compounds have similar structures, but different formulas. Use this fact to identify which of the three types of intermolecular forces is responsible for the difference.
  • C6H13–O–C6H13 Correct. The compounds have similar structures, but different formulas. Use this fact to identify which of the three types of intermolecular forces is responsible for the difference.
    Compound:
  • NH4Cl Correct. PCl5 is a liquid at room conditions, while NH4Cl is a solid. Why?
  • PCl5 No. PCl5 is a liquid at room conditions, while NH4Cl is a solid. What is the difference between them?
    Compound:
  • PH3 NH3 is a higher melting substance even though the compounds have similar structures and NH3 has a lower molar mass.
  • NH3 Correct. Now identify the force responsible.
    Force:
  • dispersion The compounds have the same formula C2H6O, so they are expected to have very similar dispersion forces.
  • dipole-dipole C2H5–O–H is the more polar molecule, but dipole-dipole interactions are not the forces that are most responsible for the difference.
  • H-bonding The main difference between the intermolecular interactions of the two compounds is the H-bond. Remember that the requirement for a good H-bond is that H must be attached to an F, an O, or an N. The presence of an O–H bond in C2H5–O–H means that H-bonding is important in this molecule.
  • ionic bonding Neither compound is ionic.
    Force:
  • dispersion Neither compound is very polar, and neither displays H-bonding, so the main difference between the two is their size. C6H13–O–C6H13 is the larger molecule, so it has the greater dispersion forces.
  • dipole-dipole Neither compound is very polar.
  • H-bonding There are no N–H, O–H, or F–H bonds.
  • ionic bonding Neither compound is ionic.
    Force:
  • dispersion Dispersion forces are the most important forces in PCl5, but not in NH4Cl.
  • dipole-dipole The 5 P-Cl bond dipoles cancel, soPCl5 is not polar. NH4Cl is not a molecule so it cannot have dipole-dipole interactions.
  • H-bonding Although NH4Cl does have N–H bonds, it cannot display H-bonding because the nitrogen already has four electron regions.
  • ionic bonding Ammonium chloride is an ionic compound (NH41+ + Cl1–). As a general rule, ionic compounds are higher melting than molecular compounds because ionic bonds must be broken to melt ionic substances.
    Force:
  • dispersion Dispersion forces are the most important forces in PCl5, but not in NH4Cl.
  • dipole-dipole Both compounds are polar, and ammonia is slightly more polar, but dipole-dipole interactions are not the main reason for the difference.
  • H-bonding Ammonia has N–H bonds, so it displays H-bonding.
  • ionic bonding Neither is ionic.

7.4 Solids

Introduction

Molecules in the solid and liquid states are much closer to one another than they are in the gas phase, so the solid and liquid states are referred to as the condensed states. Since the molecules are much closer, they interact much more strongly, and the strengths of those interactions dictate the properties of the substance.

7.4-1. Introduction to Solids

The energy of interaction between particles in the solid state is much greater than thermal energy, so the particles in a solid sit in fixed positions relative to one another that maximize their interactions. Figure 7.14 shows the structure of NaCl, which has ions in each position. Each ion interacts with six ions of opposite charge, and the arrangement that is shown is the one that maximizes those interactions. Each particle has a fixed position, so solids have their own shape and volume, which are independent of the container. The kinetic (thermal) energy of each particle is in the form of oscillations about these positions. Thus, solids are characterized by an ordered array of particles, and that order allows us to understand their structures and properties. The structure/property relationships of solids are discussed in much more detail in the next chapter.
salt crystal structure
Figure 7.14

7.5 Liquids

Introduction

The intermolecular interactions in the liquid state are comparable to thermal energy, so the molecules remain relatively close to one another while undergoing random, but restricted, motion. This combination of hindered motion without fixed positions makes understanding liquids much more difficult than gases or solids. Liquids adopt the shape of the bottom of the container but maintain their own volume, which depends only slightly on the temperature. In this section, we introduce a few properties of liquids and relate them to the strengths of the intermolecular interactions.

Objectives

7.5-1. Viscosity

Viscosity, which is the resistance of liquids to flow, depends upon the energy of interaction relative to thermal energy. In order for a liquid to flow, molecules must slide past one another, breaking and forming intermolecular interactions. If thermal energy is relatively low compared to the energy of interaction, flow is retarded and the liquid is viscous because the molecules cannot readily break the interactions. Thus, viscosity decreases as the liquid is heated. Viscosity also depends upon the shape of the molecule: viscosity increases as the ease with which molecules become entangled increases.

7.5-2. Surface Tension

Molecules in a liquid minimize their energy by minimizing their surface area, and the surface tension is the energy required to increase the surface area of a liquid by a fixed amount.
Intermolecular interactions lower the energy of a molecule, so the more interactions that a molecule experiences, the lower is its energy. Molecules on the surface of a liquid cannot interact with as many other molecules as those within the bulk, which causes the liquid to "bead."
There is no net force acting on a molecule in the bulk.
Figure 7.15a: Surface Tension
Molecules in bulk of a liquid experience forces in all directions, which cancel one another. Thus, there is no net force acting on the molecule.
There is a net, inward force acting on a molecule at the surface.
Figure 7.15b: Surface Tension
There are no molecules above one at the surface so there is no upward force to cancel the downward force. Thus, there is a net inward force exerted on molecules at the surface that causes liquid to minimize its surface area.
Alternatively, molecules on the surface are at a higher potential energy (fewer interactions to lower it). Consequently, liquids minimize the number of high-energy molecules on the surface by minimizing their surface, which is why water "beads" into distorted spheres when it is placed on a surface with which it does not interact. Energy is required to increase the surface area of a liquid, and the surface tension is the energy that is required to increase the surface area of a liquid by a fixed amount. Surface tension increases with increasing intermolecular forces. Surface tension is the reason that some insects can walk on water.

7.5-3. Meniscus

A meniscus is concave when the adhesive forces between a liquid and its container are strong, but it is convex when they are weak.
Forces between like molecules are called cohesive forces, while those between unlike molecules are said to be adhesive. Viscosity and surface tension both result from cohesive forces. However, when a liquid is placed in a container, there may also be adhesive forces between the liquid and the walls of the container, and the balance between the cohesive forces that tend to "bead" the liquid and the cohesive forces that tend to "wet" the walls of the container dictates the curved shape of the top of the liquid, which is known as the meniscus.
A concave meniscus results when adhesive forces are strong.
Figure 7.16a: Meniscus
Glass contains many Si–O bonds, and water can hydrogen bond to the oxygens on the surface of the wall. The strong adhesive forces between water and glass tend to draw the liquid up the walls of the glass and produce the concave shape of the meniscus.
A convex meniscus results when adhesive forces are weak.
Figure 7.16b: Meniscus
Mercury atoms do not interact well with glass, so the adhesive forces are very weak. Consequently, mercury forms a convex meniscus to maximize Hg-Hg interactions and minimize interactions with the walls of the container.

7.5-4. Summary of Properties

It is the interplay between thermal energy and the energy of interaction that dictates the state of a substance.
The strength of a dipole does not change with temperature, so two polar molecules do interact when they collide. However, if they are moving fast enough, their kinetic energy is sufficient to overcome the potential energy of their interaction and they do not "stick" to one another. As they are cooled, however, their kinetic energy is decreased, and, when they are cold enough, they no longer have sufficient kinetic energy to escape their interaction. Thus, it is the interplay between temperature and the strength of the intermolecular interaction that dictates the state of a substance. Table 7.3 summarizes some of the differences between the three states of matter. Note that solids are taken up in some detail in the next chapter.
Gas Liquid Solid
Shape shape and volume of the container shape of the bottom of container, but have own volume own shape and volume
Intermolecular Distance great enough that intermolecular forces can be neglected much smaller in a liquid than in a gas similar to those in a liquid, but the molecules are oriented so as to optimize their interactions
Thermal vs. Interaction Energy thermal >> interaction thermal ~ interaction thermal << interaction
Molecular Motion straight line until a collision random but restricted oscillations about fixed positions
Table 7.3: A Comparison of the Gas, Liquid, and Solid States

7.6 Changes in State

Introduction

We have all seen ice melt and liquid water boil as a result of increasing the temperature, but ice can also be melted by applying pressure, and liquid water can be made to boil by reducing the pressure. Thus, ice melts under a skater's weight, and water that boils in Denver will not boil at the same temperature at sea level. In this section, we discuss the various changes in state at a molecular level.

Prerequisites

Objectives

Introductory Overview

7.6-1. Video on the Molecular View of Changes in State

View a video describing changes in state using kinetic molecular theory.

  • Viewing the Video
  • View the video in this window by selecting the play button.
  • Use the video controls to view the video in full screen.
  • View the video in text format by scrolling down.

Energy Considerations

7.6-2. Heats of Change

The potential energy of a particle varies as solid < liquid << gas.
The potential energies of the particles in each of the three states are different because the strength of the intermolecular forces is different. Thus, particles in the solid are at the lowest (most negative) energy because the strength of the intermolecular forces is the strongest in the solid. There are no intermolecular interactions in an ideal gas because the particles are far apart and do not interact. In the liquid, the particles are close together, but interactions are not optimized as they are in the solid. Consequently, the potential energy of a particle varies as solid < liquid << gas. Note that, due to these relative energies,
Figure 7.17: Relative Energies of the Vapor, Liquid, and Solid States
The solid and liquid states have similar energies, but the energy of the vapor is much higher.

7.6-3. Heats of State Changes Defined

The heats of fusion, vaporization, and sublimation are always positive.
The energy required to melt a solid at its melting point is called its heat of fusion, ΔHfus. The heat of fusion is simply the difference in the potential energies of the liquid and solid states. Heats of fusion are always positive because the liquid is always at higher energy (less negative) than the solid.
energy diagram: solid to liquid
Figure 7.18a: Heat of Fusion
The energy required to vaporize a liquid at a given temperature is called the liquid's heat of vaporization, ΔHvap, at that temperature. The heat of vaporization is the potential energy difference between the liquid and gas. Since the potential energy of the gas is much greater than that of the liquid, a substance's heat of vaporization is always positive and greater than its heat of fusion.
energy diagram: liquid to vapor
Figure 7.18b: Heat of Vaporization
The energy required to convert the solid directly to the vapor at a given temperature is called the heat of sublimation, ΔHsub. It is the sum of the heats of fusion and vaporization.
energy diagram: solid to vapor
Figure 7.18c: Heat of Sublimation

Solid-Liquid Equilibria

7.6-4. Solid-Liquid Equilibria

The melting point is the temperature at which the solid and liquid are in equilibrium.
When the temperature is such that the kinetic energy of the molecules is low compared to the potential energy of interaction, the molecules arrange themselves in a rigid array that optimizes the interactions. The rigid array is the solid state. Thermal energy in the solid is in the form of oscillations of the molecules about their fixed positions in the crystal. Warming a solid causes the molecules to oscillate faster and with a greater amplitude. When the temperature reaches the point where the intermolecular interactions are no longer sufficient to keep the particles in position, the crystal melts. The melting point is the temperature at which the solid and liquid are in dynamic equilibrium, i.e., it is the temperature at which the rates of melting and freezing are the same. The dynamic equilibrium is expressed as follows.
solid (melting/freezing) liquid
Figure 7.19
A high melting point means that the thermal energy required to overcome the intermolecular interactions is high, so substances with high melting points have strong intermolecular interactions.

7.6-5. Graph of Solid-Liquid Equilibrium

The more dense phase can be determined from the slope of the P vs. T line for the solid-liquid equilibrium.
A plot of the pressure versus melting point is nearly a straight line with a steep slope. The slope is steep because large changes in pressure are required to produce small changes in melting point. The slope of the line depends upon whether the solid or liquid is the more dense state because the application of pressure to a solid-liquid equilibrium shifts the equilibrium to the more dense state. Consider the two possibilities shown in Figures 7.20a and 7.20b.
pressure versus temperature
Figure 7.20a: Melting Point as a Function of Pressure
The slope of the line is negative, so increasing the pressure (indicated by the arrow) on the solid-liquid equilibrium converts the solid into the liquid (melts the solid). In this case, the liquid is the more dense state, which is the situation for water.
pressure versus temperature
Figure 7.20b: Melting Point as a Function of Pressure
The slope of the line is positive, so increasing the pressure on the solid-liquid equilibrium converts liquid into solid (freezes the liquid). In this case, the solid is more dense than the liquid. This is the more common situation.

Liquid-Vapor Equilibria

7.6-6. Liquid-Vapor Equilibria

Vapor pressure increases with temperature and decreases with the strength of the intermolecular force.
The same intermolecular forces that are responsible for the solid state are at play in the liquid state as well, but the kinetic energy of the particles is sufficient to keep the molecules from being locked into position. The molecules in the liquid can slip past one another. As a liquid is heated, its temperature rises as the molecules gain kinetic energy, but temperature is a measure of the average kinetic energy of the molecules. Thus, some molecules have much more kinetic energy than others. Indeed, some of the molecules have enough kinetic energy to break their intermolecular interactions completely and enter the gas or vapor state in a process known as evaporation, or vaporization. If the container is open, the liquid evaporates, but if it is closed, the concentration of molecules in the vapor reaches a point where some of the vapor undergoes condensation. When the rate of vaporization equals the rate of condensation, the system reaches equilibrium. The pressure of vapor (gas) in equilibrium with the liquid at any temperature is known as the vapor pressure of the liquid at that temperature. Vapor pressures increase with temperature because the fraction of molecules with sufficient kinetic energy to break away increases as the temperature rises. Vapor pressures also depend upon intermolecular interactions—strong intermolecular interactions result in low vapor pressures. The liquid and vapor reach a dynamic equilibrium when molecules evaporate and condense at the same rate, so the process is written as shown in Figure 7.21.
liquid (vaporization/condensation) vapor
Figure 7.21

7.6-7. Vapor Pressure of Water, Dew Point, and Relative Humidity

Table 7.4 gives the vapor pressure of water at various temperatures. Note that the vapor pressure increases with temperature, as it must for all substances. Weather reports normally indicate the partial pressure of water in the atmosphere in terms of the relative humidity or dew point. The relative humidity is 100% times the ratio of the observed partial pressure of water in the atmosphere to the vapor pressure of water at that temperature. The dew point is the temperature at which the atmospheric water would begin to condense, i.e., the temperature at which the vapor pressure of H2O equals its partial pressure in the atmosphere.
Example:
For example, consider a day on which the temperature is 25 °C and the partial pressure of H2O is 12.8 torr. We note from Table 7.4 that the vapor pressure of water is 12.8 torr at 15 °C, so 15 °C is the dew point. The vapor pressure of water at 25 °C is 23.8 torr, so:
relative humidity =
12.8
23.8
(100%) = 53.8%
T
(°C)
P
(torr)
T
(°C)
P
(torr)
0 4.6 50 92.5
5 6.5 55 118.0
10 9.2 60 149.4
15 12.8 65 187.5
20 17.5 70 233.7
25 23.8 75 289.1
30 31.8 80 355.1
35 41.2 85 433.6
40 55.3 90 525.8
45 71.9 95 633.9
Table 7.4: Vapor Pressure of Water at Various Temperatures

7.6-8. Graph of Liquid-Vapor Equilibrium

The vapor pressure of a compound at a given temperature depends upon the strengths of its intermolecular interactions. Compounds with stronger intermolecular interactions have lower vapor pressures than those that interact only weakly because stronger interactions favor the more condensed state. Figure 7.22 shows the vapor pressure of CCl4 and H2O as a function of temperature. From the graph, we can see that the vapor pressure of CCl4 is 760 torr at 77 °C, while H2O must be heated to 100 °C to obtain the same vapor pressure. We also note that the vapor pressure of H2O is less than that of CCl4 at every temperature, so the intermolecular interactions are stronger in H2O. Although the dispersion forces are greater in CCl4, water also has dipolar and hydrogen bonding forces, which results in its relatively low vapor pressure.
vapor pressure versus temperature
Figure 7.22: Vapor Pressure of CCl4 and H2O as a Function of Temperature
Both substances are liquids at temperatures and pressures in the green region and both are gases at conditions in the orange region. CCl4 is a gas while H2O is a liquid in the blue region.

7.6-9. Boiling Points

Molecules with sufficient kinetic energy escape into the gas phase if they are on the surface of the liquid, but what about those that are not on the surface? The highly energetic molecules in the bulk of the liquid also attempt to escape into the gas by aggregating into bubbles, but the bubbles cannot withstand pressures greater than the vapor pressure of the gas at that temperature, so the bubbles cannot form if the external pressure exceeds the vapor pressure. However, the bubbles are sustained at the temperature where the vapor pressure of the liquid equals the external pressure, so we define the boiling point as that temperature at which the vapor pressure equals the external pressure. A liquid can be made to boil at other temperatures by reducing the external pressure to the vapor pressure of the liquid at that temperature, so Table 7.4 and Figure 7.22 can also be viewed as the boiling points as a function of pressure. Thus, the boiling point of water at 4.6 torr is 0 °C, and at 23.8 torr, it is 25 °C. Typically, substances are boiled in a container open to the atmosphere, so the external pressure is 1 atm, and it is this boiling point that is assumed when a boiling point is given without reference to a pressure. We define the normal boiling point as the boiling point at 1 atm pressure. Whenever the boiling point of a substance is given without reference to an external pressure, it is assumed to be the normal boiling point. The normal boiling point of water is
100 °C.

7.6-10. Normal Boiling Points and Intermolecular Interactions

The higher the boiling point of a substance, the stronger are its intermolecular interactions.
Normal boiling points are good indicators of relative strengths of intermolecular forces because molecules with stronger intermolecular interactions have higher normal boiling points. For example, dispersion forces are the only intermolecular forces acting in the halogens and noble gases, and the fact that their normal boiling points increase with molar mass (Figure 7.23) is strong support for our conclusion that dispersion forces increase with molar mass. Fluorine is a gas at room temperature (boiling point = 85 K), while the much larger iodine molecule is a solid at room temperature (boiling point = 457 K). Helium has the weakest intermolecular interaction of any substance and the lowest boiling point (4 K). It is more difficult to quantify dipolar interactions because they never occur in the absence of dispersion forces. However, the dispersion forces in CO and N2 should be very similar because they are isoelectronic (they have the same number of valence electrons) and have the same molar mass, but CO is polar while N2 is not. The normal boiling points are 77 K for N2 and 81 K for CO, so the impact of the dipole is not very great in this case. The difference between F2 and HCl, which have very similar molar masses, is more dramatic. The boiling point of polar HCl is over 100 K higher than that of nonpolar F2 even though the molar mass of HCl is slightly less.
line graph boiling points versus molar mass
Figure 7.23: Boiling Points of the Halogens and the Noble Gases
Hydrogen bonding is a relatively strong interaction, and molecules that interact through hydrogen bonds have unusually high boiling points. The case of the boiling points of the hydrides of Groups 4, 5, 6, and 7 is considered in Figures 7.24a, 7.24b, 7.24c, and 7.24d.
line graph boiling points versus molecular mass
Figure 7.24a: Boiling Points of the Group 4 Hydrides as a Function of Molar Mass
Group 4A: None of the hydrides of Group 4A exhibit hydrogen bonding, so their boiling points decrease as their molar masses decrease.
line graph boiling points versus molecular mass
Figure 7.24b: Boiling Points of the Group 5 Hydrides as a Function of Molar Mass
Group 5A: Dispersion forces are the main differences between the intermolecular forces of SbH3, AsH3, and PH3, so their boiling points decrease with decreasing molar mass. However, NH3 exhibits hydrogen bonding, so its high boiling point is much higher than would be predicted based on the boiling points of the other members of the group.
line graph boiling points versus molecular mass
Figure 7.24c: Boiling Points of the Group 6 Hydrides as a Function of Molar Mass
Group 6A: Water can hydrogen bond through two hydrogen atoms. Consequently, its boiling point is much higher than that of H2S, which has stronger dispersion forces. Based solely on molar mass, water would be predicted to boil around 200 K rather than 373 K. Thus, water would be a gas rather than a liquid at normal conditions on Earth if it were not for the hydrogen bond.
line graph boiling points versus molecular mass
Figure 7.24d: Boiling Points of the Group 7 Hydrides as a Function of Molar Mass
Group 7A: Note that HF exhibits hydrogen bonding, but, based on the above graph, HCl does not. Chlorine is electronegative enough, but it is too large.

7.6-11. Predicting Relative Boiling Points Exercise

Exercise 7.11:

Indicate the molecule with the higher boiling point in each pair.
  • N2
  • CO Both molecules have the same molar mass, so the dispersion forces are similar. However, N2 is not polar, while CO is. Therefore, CO is predicted to have a higher boiling point because the dipole-dipole interactions are in addition to the dispersion forces.
  • CH2Cl2 Both molecules have the same molar mass, so the dispersion forces are similar. However, CF4 is not polar, so it has only dispersion forces, while CH2Cl2 is polar and displays both dispersion and dipole-dipole forces. Consequently, CH2Cl2 is predicted to have a higher boiling point because it has the stronger intermolecular interactions.
  • CF4
  • CS2 The molecules are both linear and are therefore not polar, so neither displays dipole-dipole interactions. The molar mass of CS2 is greater than that of CO2, so CS2 is predicted to have a higher boiling point due to stronger dispersion forces.
  • CO2

7.6-12. Critical Point

Liquid-vapor equilibria, like the one shown for CCl4 and H2O in Figure 7.22, are established in closed containers (the liquid would evaporate before establishing equilibrium in an open container). However, as the liquid-vapor equilibrium is heated, the density of the liquid decreases, while the density of the gas increases. At a temperature known as the critical temperature, Tc, the densities are the same, there is no longer a boundary between the two phases, and the liquid-vapor line ends. The pressure at this point is called the critical pressure, Pc. The point established by the critical temperature and pressure is called the critical point. Thus, the critical temperature is the highest temperature at which the liquid phase can exist, and the critical pressure is the pressure required to liquefy a gas at the critical temperature. The critical temperature is also a function of intermolecular forces and increases in the same manner as the boiling point. Above the critical point, the substance is neither a gas nor a liquid and has properties of both phases. Under these conditions, the substance is a supercritical fluid.
the critical temperature and critical pressure define the critical point
Figure 7.25

Solid-Vapor Equilibria

7.6-13. Solid-Vapor Equilibria

Under certain conditions, the vapor and solid states are in equilibrium. Sublimation is the process in which a solid is converted directly to a gas. The reverse process is called deposition. Ice sublimes slowly, which is why snow can disappear without melting when the temperature does not rise above 0 °C. A dynamic equilibrium is established when the rate of sublimation equals the rate of deposition.
solid (sublimation/deposition) vapor
Figure 7.26
Dry ice, which is solid CO2, does not exist as a liquid at pressures less than 5.1 atm, so it does not melt at 1 atm. Consequently, dry ice sublimes rapidly at room conditions. Figure 7.27 shows the temperatures and pressures at which the solid equilibrium arrow vapor equilibrium is established. Increasing the temperature increases the pressure of vapor above a solid, i.e., it causes the solid to sublime. Increasing the pressure of the gas at a point on the line causes more solid to deposit.
pressure versus temperature
Figure 7.27: Sublimation Pressure as a Function of Temperature

Phase Diagrams

7.6-14. Phase Diagram of Water

A phase diagram shows the stable phases of a substance as a function of temperature and pressure. For example, consider the phase diagram of water shown in Figure 7.28. The lines separating the states represent the equilibria discussed in the previous sections.
phase diagram
Figure 7.28: Phase Diagram of Water
The above phase diagram shows how the state of matter of water varies with conditions of temperature and pressure. Water is a solid in the yellow region, a liquid in the green region, and a gas or vapor in the orange region. The temperatures and pressures are not to scale.

7.6-15. Phase Diagram of Carbon Dioxide

Consider the phase diagram of CO2 shown in Figure 7.29. The slope of the solid equilibrium arrow liquid line indicates that the solid is more dense than the liquid. Finally, carbon dioxide does not have a normal boiling point because the liquid does not exist at pressures below 5.1 atm (the triple point pressure).
phase diagram of CO2
Figure 7.29: Phase Diagram of Carbon Dioxide

7.6-16. Phase Diagram Exercise

Exercise 7.12:

In the following exercises, give all temperatures to the nearest 25 °C and all pressures to the nearest
0.1 atm.
phase diagram
What is the normal boiling point of X in °C?
175___ The normal boiling point is the temperature at which the vapor pressure is 1 atm, which is
175 °C in the above graph.
°C
What is the melting point of X in °C?
-25___ The melting point is the temperature at which solid and liquid are in equilibrium at 1 atm pressure, which is –25 °C in the above graph. °C
What is the temperature in °C at the triple point?
0___ The triple point is the temperature and pressure where the solid, liquid, and vapor are all in equilibrium. The temperature where this occurs in the above phase diagram is 0 °C. °C
What is the pressure in atm at the triple point?
0.5___ The triple point is the temperature and pressure where the solid, liquid, and vapor are all in equilibrium. The pressure where this occurs in the above phase diagram is 0.5 atm. atm
    What state of matter is X in at point A?
  • solid
  • liquid The substance is in the liquid region at point A.
  • gas
    What is the most dense state of X?
  • solid
  • liquid Increasing the pressure on the solid-liquid line moves the substance into the liquid region, so the liquid is the most dense state.
  • gas
What is the critical temperature of X in °C?
250___ The critical temperature is the temperature at the critical point, which is the end of the liquid-vapor line. In this example, it occurs at 250 °C. °C
What is the critical pressure of X in atm?
1.4___ The critical pressure is the pressure at the critical point, which is the end of the liquid-vapor line. In this example, it occurs at 1.2 atm. atm
Indicate the phase change that is caused by each of the following.
    increasing T at point B
  • vaporization Increasing the temperature at point B moves the substance into the vapor region. Therefore, the substance vaporizes.
  • condensation Increasing the temperature moves the substance from the liquid-vapor line into the vapor region.
  • sublimation Increasing the temperature moves the substance from the liquid-vapor line into the vapor region.
  • deposition Increasing the temperature moves the substance from the liquid-vapor line into the vapor region.
  • melting Increasing the temperature moves the substance from the liquid-vapor line into the vapor region.
  • freezing Increasing the temperature moves the substance from the liquid-vapor line into the vapor region.
    increasing T at point C
  • vaporization Increasing the temperature moves the substance from the solid-vapor line into the vapor region.
  • condensation Increasing the temperature moves the substance from the solid-vapor line into the vapor region.
  • sublimation Increasing the temperature moves the substance from the solid-vapor line into the vapor region, so the substance sublimes.
  • deposition Increasing the temperature moves the substance from the solid-vapor line into the vapor region.
  • melting Increasing the temperature moves the substance from the solid-vapor line into the vapor region.
  • freezing Increasing the temperature moves the substance from the solid-vapor line into the vapor region.
    increasing P at point D
  • vaporization Increasing the pressure moves the substance from the solid-liquid equilibrium into the liquid region.
  • condensation Increasing the pressure moves the substance from the solid-liquid equilibrium into the liquid region.
  • sublimation Increasing the pressure moves the substance from the solid-liquid equilibrium into the liquid region.
  • deposition Increasing the pressure moves the substance from the solid-liquid equilibrium into the liquid region.
  • melting Increasing the pressure moves the substance from the solid-liquid equilibrium into the liquid region, so the substance melts.
  • freezing Increasing the pressure moves the substance from the solid-liquid equilibrium into the liquid region.

7.7 Exercises and Solutions

Select the links to view either the end-of-chapter exercises or the solutions to the odd exercises.