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Measuring Temperature with a Volume Change
Temperature is sometimes measured with a gas thermometer by observing the change in the volume of the gas as the temperature changes at constant pressure. The hydrogen in a particular hydrogen gas thermometer has a volume of 150.0 cm3 when immersed in a mixture of ice and water (0.00 °C). When immersed in boiling liquid ammonia, the volume of the hydrogen, at the same pressure, is 131.7 cm3. Find the temperature of boiling ammonia on the kelvin and Celsius scales.
Solution
A volume change caused by a temperature change at constant pressure means we should use Charles’s law. Taking V1 and T1 as the initial values, T2 as the temperature at which the volume is unknown and V2 as the unknown volume, and converting °C into K we have:
Rearrangement gives
Subtracting 273.15 from 239.8 K, we find that the temperature of the boiling ammonia on the Celsius scale is –33.4 °C.
Check Your Learning
What is the volume of a sample of ethane at 467 K and 1.1 atm if it occupies 405 mL at 298 K and 1.1 atm?
Answer:
635 mL
Volume and Pressure: Boyle’s Law
If we partially fill an airtight syringe with air, the syringe contains a specific amount of air at constant temperature, say 25 °C. If we slowly push in the plunger while keeping temperature constant, the gas in the syringe is compressed into a smaller volume and its pressure increases; if we pull out the plunger, the volume increases and the pressure decreases. This example of the effect of volume on the pressure of a given amount of a confined gas is true in general. Decreasing the volume of a contained gas will increase its pressure, and increasing its volume will decrease its pressure. In fact, if the volume increases by a certain factor, the pressure decreases by the same factor, and vice versa. Volume-pressure data for an air sample at room temperature are graphed in .
FIGURE 9.13 When a gas occupies a smaller volume, it exerts a higher pressure; when it occupies a larger volume, it exerts a lower pressure (assuming the amount of gas and the temperature do not change). Since P and V are inversely proportional, a graph of vs. V is linear.
Unlike the P-T and V-T relationships, pressure and volume are not directly proportional to each other. Instead, P and V exhibit inverse proportionality: Increasing the pressure results in a decrease of the volume of the gas. Mathematically this can be written:
with k being a constant. Graphically, this relationship is shown by the straight line that results when plotting the inverse of the pressure versus the volume (V), or the inverse of volume versus the pressure (P). Graphs with curved lines are difficult to read accurately at low or high values of the variables, and they are more difficult to use in fitting theoretical equations and parameters to experimental data. For those reasons,
scientists often try to find a way to “linearize” their data. If we plot P versus V, we obtain a hyperbola (see
).
FIGURE 9.14 The relationship between pressure and volume is inversely proportional. (a) The graph of P vs. V is a hyperbola, whereas (b) the graph of vs. V is linear.
The relationship between the volume and pressure of a given amount of gas at constant temperature was first published by the English natural philosopher Robert Boyle over 300 years ago. It is summarized in the statement now known as Boyle’s law: The volume of a given amount of gas held at constant temperature is inversely proportional to the pressure under which it is measured.
Volume of a Gas Sample
The sample of gas in has a volume of 15.0 mL at a pressure of 13.0 psi. Determine the pressure of the gas at a volume of 7.5 mL, using:
the P-V graph in
the vs. V graph in
the Boyle’s law equation
Comment on the likely accuracy of each method.
Solution
Estimating from the P-V graph gives a value for P somewhere around 27 psi.
Estimating from the versus V graph give a value of about 26 psi.
From Boyle’s law, we know that the product of pressure and volume (PV) for a given sample of gas at a constant temperature is always equal to the same value. Therefore we have P1V1 = k and P2V2 = k which means that P1V1 = P2V2.
Using P1 and V1 as the known values 13.0 psi and 15.0 mL, P2 as the pressure at which the volume is unknown, and V2 as the unknown volume, we have: |
It was more difficult to estimate well from the P-V graph, so (a) is likely more inaccurate than (b) or (c). The calculation will be as accurate as the equation and measurements allow.
Check Your Learning
The sample of gas in has a volume of 30.0 mL at a pressure of 6.5 psi. Determine the volume of the gas at a pressure of 11.0 psi, using:
the P-V graph in
the vs. V graph in
the Boyle’s law equation
Comment on the likely accuracy of each method.
Answer:
(a) about 17–18 mL; (b) ~18 mL; (c) 17.7 mL; it was more difficult to estimate well from the P-V graph, so (a) is likely more inaccurate than (b); the calculation will be as accurate as the equation and measurements allow
Moles of Gas and Volume: Avogadro’s Law
The Italian scientist Amedeo Avogadro advanced a hypothesis in 1811 to account for the behavior of gases, stating that equal volumes of all gases, measured under the same conditions of temperature and pressure, contain the same number of molecules. Over time, this relationship was supported by many experimental observations as expressed by Avogadro’s law: For a confined gas, the volume (V) and number of moles (n) are directly proportional if the pressure and temperature both remain constant.
In equation form, this is written as:
Mathematical relationships can also be determined for the other variable pairs, such as P versus n, and n
versus T.
LINK TO LEARNING
Visit this to investigate the relationships between pressure, volume, temperature, and amount of gas. Use the simulation to examine the effect of changing one parameter on another while holding the other parameters constant (as described in the preceding sections on the various gas laws).
The Ideal Gas Law
To this point, four separate laws have been discussed that relate pressure, volume, temperature, and the number of moles of the gas:
Boyle’s law: PV = constant at constant T and n
Amontons’s law: = constant at constant V and n
Charles’s law: = constant at constant P and n
Avogadro’s law: = constant at constant P and T
Combining these four laws yields the ideal gas law, a relation between the pressure, volume, temperature, and number of moles of a gas:
where P is the pressure of a gas, V is its volume, n is the number of moles of the gas, T is its temperature on the kelvin scale, and R is a constant called the ideal gas constant or the universal gas constant. The units used to express pressure, volume, and temperature will determine the proper form of the gas constant as required by dimensional analysis, the most commonly encountered values being 0.08206 L atm mol–1 K–1 and 8.314 kPa L mol–1 K–1.
Gases whose properties of P, V, and T are accurately described by the ideal gas law (or the other gas laws) are said to exhibit ideal behavior or to approximate the traits of an ideal gas. An ideal gas is a hypothetical construct that may be used along with kinetic molecular theory to effectively explain the gas laws as will be described in a later module of this chapter. Although all the calculations presented in this module assume ideal behavior, this assumption is only reasonable for gases under conditions of relatively low pressure and high temperature. In the final module of this chapter, a modified gas law will be introduced that accounts for the non-ideal behavior observed for many gases at relatively high pressures and low temperatures.
The ideal gas equation contains five terms, the gas constant R and the variable properties P, V, n, and T. Specifying any four of these terms will permit use of the ideal gas law to calculate the fifth term as demonstrated in the following example exercises.
Using the Ideal Gas Law
Methane, CH4, is being considered for use as an alternative automotive fuel to replace gasoline. One gallon of gasoline could be replaced by 655 g of CH4. What is the volume of this much methane at 25 °C and 745 torr?
Solution
We must rearrange PV = nRT to solve for V:
If we choose to use R = 0.08206 L atm mol–1 K–1, then the amount must be in moles, temperature must be in kelvin, and pressure must be in atm.
Converting into the “right” units:
It would require 1020 L (269 gal) of gaseous methane at about 1 atm of pressure to replace 1 gal of gasoline. It requires a large container to hold enough methane at 1 atm to replace several gallons of gasoline.
Check Your Learning
Calculate the pressure in bar of 2520 moles of hydrogen gas stored at 27 °C in the 180-L storage tank of a modern hydrogen-powered car.
Answer:
350 bar
If the number of moles of an ideal gas are kept constant under two different sets of conditions, a useful mathematical relationship called the combined gas law is obtained: using units of atm, L, and
K. Both sets of conditions are equal to the product of n R (where n = the number of moles of the gas and R is the ideal gas law constant).
Using the Combined Gas Law
When filled with air, a typical scuba tank with a volume of 13.2 L has a pressure of 153 atm (). If the water temperature is 27 °C, how many liters of air will such a tank provide to a diver’s lungs at a depth of approximately 70 feet in the ocean where the pressure is 3.13 atm?
FIGURE 9.16 Scuba divers use compressed air to breathe while underwater. (credit: modification of work by Mark Goodchild)
Letting 1 represent the air in the scuba tank and 2 represent the air in the lungs, and noting that body temperature (the temperature the air will be in the lungs) is 37 °C, we have: |
(Note: Be advised that this particular example is one in which the assumption of ideal gas behavior is not very reasonable, since it involves gases at relatively high pressures and low temperatures. Despite this limitation, the calculated volume can be viewed as a good “ballpark” estimate.)
Check Your Learning
A sample of ammonia is found to occupy 0.250 L under laboratory conditions of 27 °C and 0.850 atm. Find the volume of this sample at 0 °C and 1.00 atm.
Answer:
0.193 L
FIGURE 9.17 Scuba divers, whether at the Great Barrier Reef or in the Caribbean, must be aware of buoyancy, pressure equalization, and the amount of time they spend underwater, to avoid the risks associated with pressurized gases in the body. (credit: Kyle Taylor)
Pressure increases with ocean depth, and the pressure changes most rapidly as divers reach the surface. The pressure a diver experiences is the sum of all pressures above the diver (from the water and the air). Most pressure measurements are given in units of atmospheres, expressed as “atmospheres absolute” or ATA in the diving community: Every 33 feet of salt water represents 1 ATA of pressure in addition to 1 ATA of pressure from the atmosphere at sea level. As a diver descends, the increase in pressure causes the body’s air pockets in the ears and lungs to compress; on the ascent, the decrease in pressure causes these air pockets to expand, potentially rupturing eardrums or bursting the lungs. Divers must therefore undergo equalization by adding air to body airspaces on the descent by breathing normally and adding air to the mask by breathing out of the nose or adding air to the ears and sinuses by equalization techniques; the corollary is also true on ascent, divers must release air from the body to maintain equalization.
Buoyancy, or the ability to control whether a diver sinks or floats, is controlled by the buoyancy compensator (BCD). If a diver is ascending, the air in their BCD expands because of lower pressure according to Boyle’s law (decreasing the pressure of gases increases the volume). The expanding air increases the buoyancy of the diver, and they begin to ascend. The diver must vent air from the BCD or risk an uncontrolled ascent that could rupture the lungs. In descending, the increased pressure causes the air in the BCD to compress and the diver sinks much more quickly; the diver must add air to the BCD or risk an uncontrolled descent, facing much higher pressures near the ocean floor. The pressure also impacts how long a diver can stay underwater before ascending. The deeper a diver dives, the more compressed the air that is breathed because of increased pressure: If a diver dives 33 feet, the pressure is 2 ATA and the air would be compressed to one-half of its original volume. The diver uses up available air twice as fast as at the surface.
Standard Conditions of Temperature and Pressure
We have seen that the volume of a given quantity of gas and the number of molecules (moles) in a given volume of gas vary with changes in pressure and temperature. Chemists sometimes make comparisons against a standard temperature and pressure (STP) for reporting properties of gases: 273.15 K and 1 atm (101.325 kPa). At STP, one mole of an ideal gas has a volume of about 22.4 L—this is referred to as the standard molar volume (). |
FIGURE 9.18 Regardless of its chemical identity, one mole of gas behaving ideally occupies a volume of ~22.4 L at STP.
Stoichiometry of Gaseous Substances, Mixtures, and Reactions
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Use the ideal gas law to compute gas densities and molar masses
Perform stoichiometric calculations involving gaseous substances
State Dalton’s law of partial pressures and use it in calculations involving gaseous mixtures
The study of the chemical behavior of gases was part of the basis of perhaps the most fundamental chemical revolution in history. French nobleman Antoine Lavoisier, widely regarded as the “father of modern chemistry,” changed chemistry from a qualitative to a quantitative science through his work with gases. He discovered the law of conservation of matter, discovered the role of oxygen in combustion reactions, determined the composition of air, explained respiration in terms of chemical reactions, and more. He was a casualty of the French Revolution, guillotined in 1794. Of his death, mathematician and astronomer Joseph- Louis Lagrange said, “It took the mob only a moment to remove his head; a century will not suffice to reproduce it.” Much of the knowledge we do have about Lavoisier's contributions is due to his wife, Marie- Anne Paulze Lavoisier, who worked with him in his lab. A trained artist fluent in several languages, she created detailed illustrations of the equipment in his lab, and translated texts from foreign scientists to complement his knowledge. After his execution, she was instrumental in publishing Lavoisier's major treatise, which unified many concepts of chemistry and laid the groundwork for significant further study.
As described in an earlier chapter of this text, we can turn to chemical stoichiometry for answers to many of the questions that ask “How much?” The essential property involved in such use of stoichiometry is the amount of substance, typically measured in moles (n). For gases, molar amount can be derived from convenient experimental measurements of pressure, temperature, and volume. Therefore, these measurements are useful in assessing the stoichiometry of pure gases, gas mixtures, and chemical reactions involving gases. This section will not introduce any new material or ideas, but will provide examples of applications and ways to integrate concepts already discussed. |
Gas Density and Molar Mass
The ideal gas law described previously in this chapter relates the properties of pressure P, volume V, temperature T, and molar amount n. This law is universal, relating these properties in identical fashion regardless of the chemical identity of the gas:
The density d of a gas, on the other hand, is determined by its identity. As described in another chapter of this text, the density of a substance is a characteristic property that may be used to identify the substance. |
Measuring Gas Density
What is the density of molecular nitrogen gas at STP?
Solution
The molar mass of molecular nitrogen, N2, is 28.01 g/mol. Substituting this value along with standard temperature and pressure into the gas density equation yields |
When the identity of a gas is unknown, measurements of the mass, pressure, volume, and temperature of a sample can be used to calculate the molar mass of the gas (a useful property for identification purposes).
Combining the ideal gas equation |
Determining the Molecular Formula of a Gas from its Molar Mass and Empirical Formula Cyclopropane, a gas once used with oxygen as a general anesthetic, is composed of 85.7% carbon and 14.3% hydrogen by mass. Find the empirical formula. If 1.56 g of cyclopropane occupies a volume of 1.00 L at 0.984 atm and 50 °C, what is the molecular formula for cyclopropane?
Solution
First determine the empirical formula of the gas. Assume 100 g and convert the percentage of each element into grams. Determine the number of moles of carbon and hydrogen in the 100-g sample of cyclopropane. Divide by the smallest number of moles to relate the number of moles of carbon to the number of moles of hydrogen. In the last step, realize that the smallest whole number ratio is the empirical formula: |
Check Your Learning
Acetylene, a fuel used welding torches, is composed of 92.3% C and 7.7% H by mass. Find the empirical formula. If 1.10 g of acetylene occupies of volume of 1.00 L at 1.15 atm and 59.5 °C, what is the molecular formula for acetylene?
Answer:
Empirical formula, CH; Molecular formula, C2H2
Determining the Molar Mass of a Volatile Liquid
The approximate molar mass of a volatile liquid can be determined by:
Heating a sample of the liquid in a flask with a tiny hole at the top, which converts the liquid into gas that
may escape through the hole
Removing the flask from heat at the instant when the last bit of liquid becomes gas, at which time the flask will be filled with only gaseous sample at ambient pressure
Sealing the flask and permitting the gaseous sample to condense to liquid, and then weighing the flask to determine the sample’s mass (see )
FIGURE 9.19 When the volatile liquid in the flask is heated past its boiling point, it becomes gas and drives air out of the flask. At the flask is filled with volatile liquid gas at the same pressure as the atmosphere. If the flask is then cooled to room temperature, the gas condenses and the mass of the gas that filled the flask, and is now liquid, can be measured. (credit: modification of work by Mark Ott)
Using this procedure, a sample of chloroform gas weighing 0.494 g is collected in a flask with a volume of 129 cm3 at 99.6 °C when the atmospheric pressure is 742.1 mm Hg. What is the approximate molar mass of chloroform?
Solution
Since ℳ and substituting and rearranging gives ℳ
then |
Check Your Learning
A sample of phosphorus that weighs 3.243 10−2 g exerts a pressure of 31.89 kPa in a 56.0-mL bulb at 550 °C. What are the molar mass and molecular formula of phosphorus vapor?
Answer:
124 g/mol P4
The Pressure of a Mixture of Gases: Dalton’s Law
Unless they chemically react with each other, the individual gases in a mixture of gases do not affect each other’s pressure. Each individual gas in a mixture exerts the same pressure that it would exert if it were present alone in the container (). The pressure exerted by each individual gas in a mixture is called its partial pressure. This observation is summarized by Dalton’s law of partial pressures: The total pressure of a mixture of ideal gases is equal to the sum of the partial pressures of the component gases: |
FIGURE 9.20 If equal-volume cylinders containing gasses at pressures of 300 kPa, 450 kPa, and 600 kPa are all combined in the same-size cylinder, the total pressure of the gas mixture is 1350 kPa.
The partial pressure of gas A is related to the total pressure of the gas mixture via its mole fraction (X), a unit of concentration defined as the number of moles of a component of a solution divided by the total number of moles of all components: |
The Pressure of a Mixture of Gases
A 10.0-L vessel contains 2.50 10−3 mol of H2, 1.00 10−3 mol of He, and 3.00 10−4 mol of Ne at 35 °C.
What are the partial pressures of each of the gases?
What is the total pressure in atmospheres?
Solution
The gases behave independently, so the partial pressure of each gas can be determined from the ideal gas equation, using : |
The Pressure of a Mixture of Gases
A gas mixture used for anesthesia contains 2.83 mol oxygen, O2, and 8.41 mol nitrous oxide, N2O. The total pressure of the mixture is 192 kPa.
What are the mole fractions of O2 and N2O?
What are the partial pressures of O2 and N2O?
Solution
The mole fraction is given by and the partial pressure is PA = XA PTotal. For O2, |
Collection of Gases over Water
A simple way to collect gases that do not react with water is to capture them in a bottle that has been filled with water and inverted into a dish filled with water. The pressure of the gas inside the bottle can be made equal to the air pressure outside by raising or lowering the bottle. When the water level is the same both inside and outside the bottle (), the pressure of the gas is equal to the atmospheric pressure, which can be measured with a barometer.
FIGURE 9.21 When a reaction produces a gas that is collected above water, the trapped gas is a mixture of the gas produced by the reaction and water vapor. If the collection flask is appropriately positioned to equalize the water levels both within and outside the flask, the pressure of the trapped gas mixture will equal the atmospheric pressure outside the flask (see the earlier discussion of manometers).
However, there is another factor we must consider when we measure the pressure of the gas by this method. Water evaporates and there is always gaseous water (water vapor) above a sample of liquid water. As a gas is collected over water, it becomes saturated with water vapor and the total pressure of the mixture equals the partial pressure of the gas plus the partial pressure of the water vapor. The pressure of the pure gas is therefore equal to the total pressure minus the pressure of the water vapor—this is referred to as the “dry” gas pressure, that is, the pressure of the gas only, without water vapor. The vapor pressure of water, which is the pressure exerted by water vapor in equilibrium with liquid water in a closed container, depends on the temperature (); more detailed information on the temperature dependence of water vapor can be found in , and vapor pressure will be discussed in more detail in the next chapter on liquids. |
Pressure of a Gas Collected Over Water
If 0.200 L of argon is collected over water at a temperature of 26 °C and a pressure of 750 torr in a system like that shown in , what is the partial pressure of argon?
Solution
According to Dalton’s law, the total pressure in the bottle (750 torr) is the sum of the partial pressure of argon and the partial pressure of gaseous water: |
Check Your Learning
A sample of oxygen collected over water at a temperature of 29.0 °C and a pressure of 764 torr has a volume of
0.560 L. What volume would the dry oxygen from this sample have under the same conditions of temperature |
Chemical Stoichiometry and Gases
Chemical stoichiometry describes the quantitative relationships between reactants and products in chemical reactions.
We have previously measured quantities of reactants and products using masses for solids and volumes in conjunction with the molarity for solutions; now we can also use gas volumes to indicate quantities. If we know the volume, pressure, and temperature of a gas, we can use the ideal gas equation to calculate how many moles of the gas are present. If we know how many moles of a gas are involved, we can calculate the volume of a gas at any temperature and pressure.
Avogadro’s Law Revisited
Sometimes we can take advantage of a simplifying feature of the stoichiometry of gases that solids and solutions do not exhibit: All gases that show ideal behavior contain the same number of molecules in the same volume (at the same temperature and pressure). Thus, the ratios of volumes of gases involved in a chemical reaction are given by the coefficients in the equation for the reaction, provided that the gas volumes are measured at the same temperature and pressure.
We can extend Avogadro’s law (that the volume of a gas is directly proportional to the number of moles of the gas) to chemical reactions with gases: Gases combine, or react, in definite and simple proportions by volume, provided that all gas volumes are measured at the same temperature and pressure. For example, since nitrogen and hydrogen gases react to produce ammonia gas according to a given volume of nitrogen gas reacts with three times that volume of hydrogen gas to produce two times that volume of ammonia gas, if pressure and temperature remain constant.
The explanation for this is illustrated in . According to Avogadro’s law, equal volumes of gaseous N2, H2, and NH3, at the same temperature and pressure, contain the same number of molecules. Because one molecule of N2 reacts with three molecules of H2 to produce two molecules of NH3, the volume of H2 required is three times the volume of N2, and the volume of NH3 produced is two times the volume of N2. |
Reaction of Gases
Propane, C3H8(g), is used in gas grills to provide the heat for cooking. What volume of O2(g) measured at 25 °C and 760 torr is required to react with 2.7 L of propane measured under the same conditions of temperature and pressure? Assume that the propane undergoes complete combustion.
Solution
The ratio of the volumes of C3H8 and O2 will be equal to the ratio of their coefficients in the balanced equation for the reaction: |
A volume of 13.5 L of O2 will be required to react with 2.7 L of C3H8.
Check Your Learning
An acetylene tank for an oxyacetylene welding torch provides 9340 L of acetylene gas, C2H2, at 0 °C and 1 atm. How many tanks of oxygen, each providing 7.00 103 L of O2 at 0 °C and 1 atm, will be required to burn the acetylene? |
Volumes of Reacting Gases
Ammonia is an important fertilizer and industrial chemical. Suppose that a volume of 683 billion cubic feet of gaseous ammonia, measured at 25 °C and 1 atm, was manufactured. What volume of H2(g), measured under the same conditions, was required to prepare this amount of ammonia by reaction with N2?
Solution
Because equal volumes of H2 and NH3 contain equal numbers of molecules and each three molecules of H2 that react produce two molecules of NH3, the ratio of the volumes of H2 and NH3 will be equal to 3:2. Two volumes of NH3, in this case in units of billion ft3, will be formed from three volumes of H2:
The manufacture of 683 billion ft3 of NH3 required 1020 billion ft3 of H2. (At 25 °C and 1 atm, this is the volume of a cube with an edge length of approximately 1.9 miles.)
Check Your Learning
What volume of O2(g) measured at 25 °C and 760 torr is required to react with 17.0 L of ethylene, C2H4(g), measured under the same conditions of temperature and pressure? The products are CO2 and water vapor. |
Convert the provided temperature and pressure values to appropriate units (K and atm, respectively), and then use the molar amount of hydrogen gas and the ideal gas equation to calculate the volume of gas:
Check Your Learning
Sulfur dioxide is an intermediate in the preparation of sulfuric acid. What volume of SO2 at 343 °C and 1.21 atm is produced by burning l.00 kg of sulfur in excess oxygen?
Answer:
1.30 103 L
Greenhouse Gases and Climate Change
The thin skin of our atmosphere keeps the earth from being an ice planet and makes it habitable. In fact, this is due to less than 0.5% of the air molecules. Of the energy from the sun that reaches the earth, almost is reflected back into space, with the rest absorbed by the atmosphere and the surface of the earth. Some of the energy that the earth absorbs is re-emitted as infrared (IR) radiation, a portion of which passes back out through the atmosphere into space. Most if this IR radiation, however, is absorbed by certain atmospheric gases, effectively trapping heat within the atmosphere in a phenomenon known as the greenhouse effect. This effect maintains global temperatures within the range needed to sustain life on earth. Without our atmosphere, the earth's average temperature would be lower by more than 30 °C (nearly 60 °F). The major greenhouse gases (GHGs) are water vapor, carbon dioxide, methane, and ozone. Since the Industrial Revolution, human activity has been increasing the concentrations of GHGs, which have changed the energy balance and are significantly altering the earth’s climate ().
FIGURE 9.24 Greenhouse gases trap enough of the sun’s energy to make the planet habitable—this is known as the greenhouse effect. Human activities are increasing greenhouse gas levels, warming the planet and causing more extreme weather events.
There is strong evidence from multiple sources that higher atmospheric levels of CO2 are caused by human activity, with fossil fuel burning accounting for about of the recent increase in CO2. Reliable data from ice cores reveals that CO2 concentration in the atmosphere is at the highest level in the past 800,000 years; other evidence indicates that it may be at its highest level in 20 million years. In recent years, the CO2 concentration has increased preindustrial levels of ~280 ppm to more than 400 ppm today (). |
Effusion and Diffusion of Gases
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Define and explain effusion and diffusion
State Graham’s law and use it to compute relevant gas properties
If you have ever been in a room when a piping hot pizza was delivered, you have been made aware of the fact that gaseous molecules can quickly spread throughout a room, as evidenced by the pleasant aroma that soon reaches your nose. Although gaseous molecules travel at tremendous speeds (hundreds of meters per second), they collide with other gaseous molecules and travel in many different directions before reaching the desired target. At room temperature, a gaseous molecule will experience billions of collisions per second. The mean free path is the average distance a molecule travels between collisions. The mean free path increases with decreasing pressure; in general, the mean free path for a gaseous molecule will be hundreds of times the diameter of the molecule
In general, we know that when a sample of gas is introduced to one part of a closed container, its molecules very quickly disperse throughout the container; this process by which molecules disperse in space in response to differences in concentration is called diffusion (shown in ). The gaseous atoms or molecules are, of course, unaware of any concentration gradient, they simply move randomly—regions of higher concentration have more particles than regions of lower concentrations, and so a net movement of species from high to low concentration areas takes place. In a closed environment, diffusion will ultimately result in equal concentrations of gas throughout, as depicted in . The gaseous atoms and molecules continue to move, but since their concentrations are the same in both bulbs, the rates of transfer between the bulbs are equal (no net transfer of molecules occurs). |
The lighter gas, H2, passes through the opening faster than O2, so just after the stopcock is opened, more H2 molecules move to the O2 side than O2 molecules move to the H2 side. (c) After a short time, both the slower- moving O2 molecules and the faster-moving H2 molecules have distributed themselves evenly on both sides of the vessel.
We are often interested in the rate of diffusion, the amount of gas passing through some area per unit time:
The diffusion rate depends on several factors: the concentration gradient (the increase or decrease in concentration from one point to another); the amount of surface area available for diffusion; and the distance the gas particles must travel. Note also that the time required for diffusion to occur is inversely proportional to the rate of diffusion, as shown in the rate of diffusion equation.
A process involving movement of gaseous species similar to diffusion is effusion, the escape of gas molecules through a tiny hole such as a pinhole in a balloon into a vacuum (). Although diffusion and effusion rates both depend on the molar mass of the gas involved, their rates are not equal; however, the ratios of their rates are the same.
FIGURE 9.28 Diffusion involves the unrestricted dispersal of molecules throughout space due to their random motion. When this process is restricted to passage of molecules through very small openings in a physical barrier, the process is called effusion.
If a mixture of gases is placed in a container with porous walls, the gases effuse through the small openings in the walls. The lighter gases pass through the small openings more rapidly (at a higher rate) than the heavier ones (). In 1832, Thomas Graham studied the rates of effusion of different gases and formulated Graham’s law of effusion: The rate of effusion of a gas is inversely proportional to the square root of the mass of its particles:
This means that if two gases A and B are at the same temperature and pressure, the ratio of their effusion rates is inversely proportional to the ratio of the square roots of the masses of their particles:
FIGURE 9.29 The left photograph shows two balloons inflated with different gases, helium (orange) and argon (blue).The right-side photograph shows the balloons approximately 12 hours after being filled, at which time the helium balloon has become noticeably more deflated than the argon balloon, due to the greater effusion rate of the lighter helium gas. (credit: modification of work by Paul Flowers) |
Hydrogen effuses four times as rapidly as oxygen.
Check Your Learning
At a particular pressure and temperature, nitrogen gas effuses at the rate of 79 mL/s. Under the same conditions, at what rate will sulfur dioxide effuse?
Answer:
52 mL/s
Effusion Time Calculations
It takes 243 s for 4.46 10−5 mol Xe to effuse through a tiny hole. Under the same conditions, how long will it take 4.46 10−5 mol Ne to effuse?
Solution
It is important to resist the temptation to use the times directly, and to remember how rate relates to time as well as how it relates to mass. Recall the definition of rate of effusion: |
Note that this answer is reasonable: Since Ne is lighter than Xe, the effusion rate for Ne will be larger than that for Xe, which means the time of effusion for Ne will be smaller than that for Xe.
Check Your Learning
A party balloon filled with helium deflates to of its original volume in 8.0 hours. How long will it take an identical balloon filled with the same number of moles of air (ℳ = 28.2 g/mol) to deflate to of its original volume?
Answer:
32 h
Determining Molar Mass Using Graham’s Law
An unknown gas effuses 1.66 times more rapidly than CO2. What is the molar mass of the unknown gas? Can you make a reasonable guess as to its identity?
Solution
From Graham’s law, we have: |
The gas could well be CH4, the only gas with this molar mass.
Check Your Learning
Hydrogen gas effuses through a porous container 8.97-times faster than an unknown gas. Estimate the molar mass of the unknown gas.
Answer:
163 g/mol
Use of Diffusion for Nuclear Energy Applications: Uranium Enrichment
Gaseous diffusion has been used to produce enriched uranium for use in nuclear power plants and weapons. Naturally occurring uranium contains only 0.72% of 235U, the kind of uranium that is “fissile,” that is, capable of sustaining a nuclear fission chain reaction. Nuclear reactors require fuel that is 2–5% 235U, and nuclear bombs need even higher concentrations. One way to enrich uranium to the desired levels is to take advantage of Graham’s law. In a gaseous diffusion enrichment plant, uranium hexafluoride (UF6, the only uranium compound that is volatile enough to work) is slowly pumped through large cylindrical vessels called diffusers, which contain porous barriers with microscopic openings. The process is one of diffusion because the other side of the barrier is not evacuated. The 235UF6 molecules have a higher average speed and diffuse through the barrier a little faster than the heavier 238UF6 molecules. The gas that has passed through the barrier is slightly enriched in 235UF6 and the residual gas is slightly depleted. The small difference in molecular weights between 235UF6 and 238UF6 only about 0.4% enrichment, is achieved in one diffuser (). But by connecting many diffusers in a sequence of stages (called a cascade), the desired level of enrichment can be attained.
FIGURE 9.30 In a diffuser, gaseous UF6 is pumped through a porous barrier, which partially separates 235UF6 from
238UF6 The UF6 must pass through many large diffuser units to achieve sufficient enrichment in 235U.
The large scale separation of gaseous 235UF6 from 238UF6 was first done during the World War II, at the atomic energy installation in Oak Ridge, Tennessee, as part of the Manhattan Project (the development of the first atomic bomb). Although the theory is simple, this required surmounting many daunting technical challenges to make it work in practice. The barrier must have tiny, uniform holes (about 10–6 cm in diameter) and be porous enough to produce high flow rates. All materials (the barrier, tubing, surface coatings, lubricants, and gaskets) need to be able to contain, but not react with, the highly reactive and corrosive UF6.
Because gaseous diffusion plants require very large amounts of energy (to compress the gas to the high
pressures required and drive it through the diffuser cascade, to remove the heat produced during compression, and so on), it is now being replaced by gas centrifuge technology, which requires far less energy. A current hot political issue is how to deny this technology to Iran, to prevent it from producing enough enriched uranium for them to use to make nuclear weapons.
The Kinetic-Molecular Theory
LEARNING OBJECTIVES
State the postulates of the kinetic-molecular theory
Use this theory’s postulates to explain the gas laws
The gas laws that we have seen to this point, as well as the ideal gas equation, are empirical, that is, they have been derived from experimental observations. The mathematical forms of these laws closely describe the macroscopic behavior of most gases at pressures less than about 1 or 2 atm. Although the gas laws describe relationships that have been verified by many experiments, they do not tell us why gases follow these relationships.
The kinetic molecular theory (KMT) is a simple microscopic model that effectively explains the gas laws described in previous modules of this chapter. This theory is based on the following five postulates described here. (Note: The term “molecule” will be used to refer to the individual chemical species that compose the gas, although some gases are composed of atomic species, for example, the noble gases.)
Gases are composed of molecules that are in continuous motion, travelling in straight lines and changing direction only when they collide with other molecules or with the walls of a container.
The molecules composing the gas are negligibly small compared to the distances between them.
The pressure exerted by a gas in a container results from collisions between the gas molecules and the container walls.
Gas molecules exert no attractive or repulsive forces on each other or the container walls; therefore, their collisions are elastic (do not involve a loss of energy).
The average kinetic energy of the gas molecules is proportional to the kelvin temperature of the gas.
The test of the KMT and its postulates is its ability to explain and describe the behavior of a gas. The various gas laws can be derived from the assumptions of the KMT, which have led chemists to believe that the assumptions of the theory accurately represent the properties of gas molecules. We will first look at the individual gas laws (Boyle’s, Charles’s, Amontons’s, Avogadro’s, and Dalton’s laws) conceptually to see how the KMT explains them. Then, we will more carefully consider the relationships between molecular masses, speeds, and kinetic energies with temperature, and explain Graham’s law.
The Kinetic-Molecular Theory Explains the Behavior of Gases, Part I
Recalling that gas pressure is exerted by rapidly moving gas molecules and depends directly on the number of molecules hitting a unit area of the wall per unit of time, we see that the KMT conceptually explains the behavior of a gas as follows:
Amontons’s law. If the temperature is increased, the average speed and kinetic energy of the gas molecules increase. If the volume is held constant, the increased speed of the gas molecules results in more frequent and more forceful collisions with the walls of the container, therefore increasing the pressure ().
Charles’s law. If the temperature of a gas is increased, a constant pressure may be maintained only if the
volume occupied by the gas increases. This will result in greater average distances traveled by the molecules to reach the container walls, as well as increased wall surface area. These conditions will decrease the both the frequency of molecule-wall collisions and the number of collisions per unit area, the combined effects of which balance the effect of increased collision forces due to the greater kinetic energy at the higher temperature.
Boyle’s law. If the gas volume volume of a given amount of gas at a given temperature is decreased (that is,
if the gas is compressed), the molecules will be exposed to a decreased container wall area. Collisions with
the container wall will therefore occur more frequently and the pressure exerted by the gas will increase ().
Avogadro’s law. At constant pressure and temperature, the frequency and force of molecule-wall collisions
are constant. Under such conditions, increasing the number of gaseous molecules will require a proportional increase in the container volume in order to yield a decrease in the number of collisions per unit area to compensate for the increased frequency of collisions ().
Dalton’s Law. Because of the large distances between them, the molecules of one gas in a mixture bombard
the container walls with the same frequency whether other gases are present or not, and the total pressure of a gas mixture equals the sum of the (partial) pressures of the individual gases.
FIGURE 9.31 (a) When gas temperature increases, gas pressure increases due to increased force and frequency of molecular collisions. (b) When volume decreases, gas pressure increases due to increased frequency of molecular collisions. (c) When the amount of gas increases at a constant pressure, volume increases to yield a constant number of collisions per unit wall area per unit time.
molecular speeds and Kinetic Energy
The previous discussion showed that the KMT qualitatively explains the behaviors described by the various gas laws. The postulates of this theory may be applied in a more quantitative fashion to derive these individual laws. To do this, we must first look at speeds and kinetic energies of gas molecules, and the temperature of a gas sample.
In a gas sample, individual molecules have widely varying speeds; however, because of the vast number of molecules and collisions involved, the molecular speed distribution and average speed are constant. This molecular speed distribution is known as a Maxwell-Boltzmann distribution, and it depicts the relative numbers of molecules in a bulk sample of gas that possesses a given speed ().
FIGURE 9.32 The molecular speed distribution for oxygen gas at 300 K is shown here. Very few molecules move at either very low or very high speeds. The number of molecules with intermediate speeds increases rapidly up to a maximum, which is the most probable speed, then drops off rapidly. Note that the most probable speed, νp, is a little less than 400 m/s, while the root mean square speed, urms, is closer to 500 m/s.
The kinetic energy (KE) of a particle of mass (m) and speed (u) is given by:
Expressing mass in kilograms and speed in meters per second will yield energy values in units of joules (J = kg m2 s–2). To deal with a large number of gas molecules, we use averages for both speed and kinetic energy. In the KMT, the root mean square speed of a particle, urms, is defined as the square root of the average of the squares of the speeds with n = the number of particles: |
where R is the gas constant and T is the kelvin temperature. When used in this equation, the appropriate form of the gas constant is 8.314 J/mol⋅K (8.314 kg m2s–2mol–1K–1). These two separate equations for KEavg may be combined and rearranged to yield a relation between molecular speed and temperature: |
If the temperature of a gas increases, its KEavg increases, more molecules have higher speeds and fewer molecules have lower speeds, and the distribution shifts toward higher speeds overall, that is, to the right. If temperature decreases, KEavg decreases, more molecules have lower speeds and fewer molecules have higher speeds, and the distribution shifts toward lower speeds overall, that is, to the left. This behavior is illustrated for nitrogen gas in .
FIGURE 9.33 The molecular speed distribution for nitrogen gas (N2) shifts to the right and flattens as the temperature increases; it shifts to the left and heightens as the temperature decreases.
At a given temperature, all gases have the same KEavg for their molecules. Gases composed of lighter molecules have more high-speed particles and a higher urms, with a speed distribution that peaks at relatively higher speeds. Gases consisting of heavier molecules have more low-speed particles, a lower urms, and a speed distribution that peaks at relatively lower speeds. This trend is demonstrated by the data for a series of noble gases shown in .
FIGURE 9.34 molecular speed is directly related to molecular mass. At a given temperature, lighter molecules move faster on average than heavier molecules.
LINK TO LEARNING
The may be used to examine the effect of temperature on molecular speeds. Examine the simulator’s “energy histograms” (molecular speed distributions) and “species information” (which gives average speed values) for molecules of different masses at various temperatures.
The Kinetic-Molecular Theory Explains the Behavior of Gases, Part II
According to Graham’s law, the molecules of a gas are in rapid motion and the molecules themselves are small. The average distance between the molecules of a gas is large compared to the size of the molecules. As a consequence, gas molecules can move past each other easily and diffuse at relatively fast rates.
The rate of effusion of a gas depends directly on the (average) speed of its molecules: |
The ratio of the rates of effusion is thus derived to be inversely proportional to the ratio of the square roots of their masses. This is the same relation observed experimentally and expressed as Graham’s law.
Non-Ideal Gas Behavior
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Describe the physical factors that lead to deviations from ideal gas behavior
Explain how these factors are represented in the van der Waals equation
Define compressibility (Z) and describe how its variation with pressure reflects non-ideal behavior
Quantify non-ideal behavior by comparing computations of gas properties using the ideal gas law and the van der Waals equation
Thus far, the ideal gas law, PV = nRT, has been applied to a variety of different types of problems, ranging from
reaction stoichiometry and empirical and molecular formula problems to determining the density and molar mass of a gas. As mentioned in the previous modules of this chapter, however, the behavior of a gas is often non-ideal, meaning that the observed relationships between its pressure, volume, and temperature are not accurately described by the gas laws. In this section, the reasons for these deviations from ideal gas behavior are considered.
One way in which the accuracy of PV = nRT can be judged is by comparing the actual volume of 1 mole of gas (its molar volume, Vm) to the molar volume of an ideal gas at the same temperature and pressure. This ratio is called the compressibility factor (Z) with:
Ideal gas behavior is therefore indicated when this ratio is equal to 1, and any deviation from 1 is an indication of non-ideal behavior. shows plots of Z over a large pressure range for several common gases.
FIGURE 9.35 A graph of the compressibility factor (Z) vs. pressure shows that gases can exhibit significant deviations from the behavior predicted by the ideal gas law.
As is apparent from , the ideal gas law does not describe gas behavior well at relatively high pressures. To determine why this is, consider the differences between real gas properties and what is expected of a hypothetical ideal gas.
Particles of a hypothetical ideal gas have no significant volume and do not attract or repel each other. In general, real gases approximate this behavior at relatively low pressures and high temperatures. However, at high pressures, the molecules of a gas are crowded closer together, and the amount of empty space between the molecules is reduced. At these higher pressures, the volume of the gas molecules themselves becomes appreciable relative to the total volume occupied by the gas. The gas therefore becomes less compressible at these high pressures, and although its volume continues to decrease with increasing pressure, this decrease is not proportional as predicted by Boyle’s law.
At relatively low pressures, gas molecules have practically no attraction for one another because they are (on average) so far apart, and they behave almost like particles of an ideal gas. At higher pressures, however, the force of attraction is also no longer insignificant. This force pulls the molecules a little closer together, slightly decreasing the pressure (if the volume is constant) or decreasing the volume (at constant pressure) ( ). This change is more pronounced at low temperatures because the molecules have lower KE relative to the attractive forces, and so they are less effective in overcoming these attractions after colliding with one another.
FIGURE 9.36 (a) Attractions between gas molecules serve to decrease the gas volume at constant pressure compared to an ideal gas whose molecules experience no attractive forces. (b) These attractive forces will decrease the force of collisions between the molecules and container walls, therefore reducing the pressure exerted at constant volume compared to an ideal gas.
There are several different equations that better approximate gas behavior than does the ideal gas law. The first, and simplest, of these was developed by the Dutch scientist Johannes van der Waals in 1879. The van der Waals equation improves upon the ideal gas law by adding two terms: one to account for the volume of the gas molecules and another for the attractive forces between them.
The constant a corresponds to the strength of the attraction between molecules of a particular gas, and the constant b corresponds to the size of the molecules of a particular gas. The “correction” to the pressure term in
the ideal gas law is and the “correction” to the volume is nb. Note that when V is relatively large and n is
relatively small, both of these correction terms become negligible, and the van der Waals equation reduces to the ideal gas law, PV = nRT. Such a condition corresponds to a gas in which a relatively low number of molecules is occupying a relatively large volume, that is, a gas at a relatively low pressure. Experimental values for the van der Waals constants of some common gases are given in . |
At low pressures, the correction for intermolecular attraction, a, is more important than the one for molecular volume, b. At high pressures and small volumes, the correction for the volume of the molecules becomes important because the molecules themselves are incompressible and constitute an appreciable fraction of the total volume. At some intermediate pressure, the two corrections have opposing influences and the gas appears to follow the relationship given by PV = nRT over a small range of pressures. This behavior is reflected by the “dips” in several of the compressibility curves shown in . The attractive force between molecules initially makes the gas more compressible than an ideal gas, as pressure is raised (Z decreases with increasing P). At very high pressures, the gas becomes less compressible (Z increases with P), as the gas molecules begin to occupy an increasingly significant fraction of the total gas volume.
Strictly speaking, the ideal gas equation functions well when intermolecular attractions between gas molecules are negligible and the gas molecules themselves do not occupy an appreciable part of the whole volume. These criteria are satisfied under conditions of low pressure and high temperature. Under such conditions, the gas is said to behave ideally, and deviations from the gas laws are small enough that they may be disregarded—this is, however, very often not the case.
Comparison of Ideal Gas Law and van der Waals Equation
A 4.25-L flask contains 3.46 mol CO2 at 229 °C. Calculate the pressure of this sample of CO2:
from the ideal gas law
from the van der Waals equation
Explain the reason(s) for the difference.
Solution
From the ideal gas law: |
This finally yields P = 32.4 atm.
This is not very different from the value from the ideal gas law because the pressure is not very high and the temperature is not very low. The value is somewhat different because CO2 molecules do have some volume and attractions between molecules, and the ideal gas law assumes they do not have volume or attractions.
Check your Learning
A 560-mL flask contains 21.3 g N2 at 145 °C. Calculate the pressure of N2:
from the ideal gas law
from the van der Waals equation
Explain the reason(s) for the difference.
Answer:
(a) 46.562 atm; (b) 46.594 atm; (c) The van der Waals equation takes into account the volume of the gas molecules themselves as well as intermolecular attractions.
Key Terms
absolute zero temperature at which the volume of a gas would be zero according to Charles’s law.
Amontons’s law (also, Gay-Lussac’s law) pressure of a given number of moles of gas is directly proportional to its kelvin temperature when the volume is held constant
atmosphere (atm) unit of pressure; 1 atm =
101,325 Pa
Avogadro’s law volume of a gas at constant temperature and pressure is proportional to the number of gas molecules
bar (bar or b) unit of pressure; 1 bar = 100,000 Pa
barometer device used to measure atmospheric pressure
Boyle’s law volume of a given number of moles of gas held at constant temperature is inversely proportional to the pressure under which it is measured
Charles’s law volume of a given number of moles
of gas is directly proportional to its kelvin temperature when the pressure is held constant
compressibility factor (Z) ratio of the
experimentally measured molar volume for a gas to its molar volume as computed from the ideal gas equation
Dalton’s law of partial pressures total pressure of
a mixture of ideal gases is equal to the sum of the partial pressures of the component gases
diffusion movement of an atom or molecule from a
region of relatively high concentration to one of relatively low concentration (discussed in this chapter with regard to gaseous species, but applicable to species in any phase)
effusion transfer of gaseous atoms or molecules
from a container to a vacuum through very small openings
Graham’s law of effusion rates of diffusion and
effusion of gases are inversely proportional to the square roots of their molecular masses
hydrostatic pressure pressure exerted by a fluid
due to gravity
ideal gas hypothetical gas whose physical properties are perfectly described by the gas laws |
ideal gas constant (R) constant derived from the ideal gas equation R = 0.08206 L atm mol–1 K–1 or
8.314 L kPa mol–1 K–1
ideal gas law relation between the pressure, volume, amount, and temperature of a gas under conditions derived by combination of the simple gas laws
kinetic molecular theory theory based on simple
principles and assumptions that effectively explains ideal gas behavior
manometer device used to measure the pressure
of a gas trapped in a container
mean free path average distance a molecule travels between collisions
mole fraction (X) concentration unit defined as the ratio of the molar amount of a mixture component to the total number of moles of all mixture components
partial pressure pressure exerted by an individual
gas in a mixture
pascal (Pa) SI unit of pressure; 1 Pa = 1 N/m2
pounds per square inch (psi) unit of pressure common in the US
pressure force exerted per unit area
rate of diffusion amount of gas diffusing through a given area over a given time
root mean square speed (urms) measure of average speed for a group of particles calculated as the square root of the average squared speed
standard conditions of temperature and pressure
(STP) 273.15 K (0 °C) and 1 atm (101.325 kPa)
standard molar volume volume of 1 mole of gas at STP, approximately 22.4 L for gases behaving ideally
torr unit of pressure;
van der Waals equation modified version of the ideal gas equation containing additional terms to account for non-ideal gas behavior
vapor pressure of water pressure exerted by
water vapor in equilibrium with liquid water in a closed container at a specific temperature |
Gases exert pressure, which is force per unit area. The pressure of a gas may be expressed in the SI unit of pascal or kilopascal, as well as in many other units including torr, atmosphere, and bar.
Atmospheric pressure is measured using a barometer; other gas pressures can be measured using one of several types of manometers.
The behavior of gases can be described by several laws based on experimental observations of their properties. The pressure of a given amount of gas is directly proportional to its absolute temperature, provided that the volume does not change (Amontons’s law). The volume of a given gas sample is directly proportional to its absolute temperature at constant pressure (Charles’s law). The volume of a given amount of gas is inversely proportional to its pressure when temperature is held constant (Boyle’s law). Under the same conditions of temperature and pressure, equal volumes of all gases contain the same number of molecules (Avogadro’s law).
The equations describing these laws are special cases of the ideal gas law, PV = nRT, where P is the pressure of the gas, V is its volume, n is the number of moles of the gas, T is its kelvin temperature, and R is the ideal (universal) gas constant. |
quantities to properties of interest for gaseous substances and mixtures. Appropriate rearrangement of the ideal gas equation may be made to permit the calculation of gas densities and molar masses. Dalton’s law of partial pressures may be used to relate measured gas pressures for gaseous mixtures to their compositions. Avogadro’s law may be used in stoichiometric computations for chemical reactions involving gaseous reactants or products.
Gaseous atoms and molecules move freely and randomly through space. Diffusion is the process whereby gaseous atoms and molecules are transferred from regions of relatively high concentration to regions of relatively low concentration. Effusion is a similar process in which gaseous species pass from a container to a vacuum through very small orifices. The rates of effusion of gases are inversely proportional to the square roots of their densities or to the square roots of their atoms/molecules’ masses (Graham’s law).
The kinetic molecular theory is a simple but very effective model that effectively explains ideal gas behavior. The theory assumes that gases consist of widely separated molecules of negligible volume that are in constant motion, colliding elastically with one another and the walls of their container with average speeds determined by their absolute temperatures. The individual molecules of a gas exhibit a range of speeds, the distribution of these speeds being dependent on the temperature of the |
Gas molecules possess a finite volume and experience forces of attraction for one another. Consequently, gas behavior is not necessarily described well by the ideal gas law. Under conditions of low pressure and high temperature, these factors are negligible, the ideal gas equation is an accurate |
description of gas behavior, and the gas is said to exhibit ideal behavior. However, at lower temperatures and higher pressures, corrections for molecular volume and molecular attractions are required to account for finite molecular size and attractive forces. The van der Waals equation is a modified version of the ideal gas law that can be used to account for the non-ideal behavior of gases under these conditions.
Why are sharp knives more effective than dull knives? (Hint: Think about the definition of pressure.)
Why do some small bridges have weight limits that depend on how many wheels or axles the crossing vehicle has?
Why should you roll or belly-crawl rather than walk across a thinly-frozen pond?
A typical barometric pressure in Redding, California, is about 750 mm Hg. Calculate this pressure in atm and kPa.
A typical barometric pressure in Denver, Colorado, is 615 mm Hg. What is this pressure in atmospheres and kilopascals?
A typical barometric pressure in Kansas City is 740 torr. What is this pressure in atmospheres, in millimeters of mercury, and in kilopascals?
Canadian tire pressure gauges are marked in units of kilopascals. What reading on such a gauge corresponds to 32 psi?
During the Viking landings on Mars, the atmospheric pressure was determined to be on the average about
6.50 millibars (1 bar = 0.987 atm). What is that pressure in torr and kPa?
The pressure of the atmosphere on the surface of the planet Venus is about 88.8 atm. Compare that pressure in psi to the normal pressure on earth at sea level in psi.
A medical laboratory catalog describes the pressure in a cylinder of a gas as 14.82 MPa. What is the pressure of this gas in atmospheres and torr?
Consider this scenario and answer the following questions: On a mid-August day in the northeastern United States, the following information appeared in the local newspaper: atmospheric pressure at sea level 29.97 in. Hg, 1013.9 mbar.
What was the pressure in kPa?
The pressure near the seacoast in the northeastern United States is usually reported near 30.0 in. Hg. During a hurricane, the pressure may fall to near 28.0 in. Hg. Calculate the drop in pressure in torr.
Why is it necessary to use a nonvolatile liquid in a barometer or manometer? |
The pressure of a sample of gas is measured with an open-end manometer, partially shown to the right. The liquid in the manometer is mercury. Assuming atmospheric pressure is 29.92 in. Hg, determine the pressure of the gas in:
torr
Pa
bar |
Sometimes leaving a bicycle in the sun on a hot day will cause a blowout. Why?
Explain how the volume of the bubbles exhausted by a scuba diver () change as they rise to the surface, assuming that they remain intact.
One way to state Boyle’s law is “All other things being equal, the pressure of a gas is inversely proportional to its volume.” (a) What is the meaning of the term “inversely proportional?” (b) What are the “other things” that must be equal?
An alternate way to state Avogadro’s law is “All other things being equal, the number of molecules in a gas
is directly proportional to the volume of the gas.” (a) What is the meaning of the term “directly proportional?” (b) What are the “other things” that must be equal?
How would the graph in change if the number of moles of gas in the sample used to determine
the curve were doubled?
How would the graph in change if the number of moles of gas in the sample used to determine the curve were doubled?
In addition to the data found in , what other information do we need to find the mass of the sample of air used to determine the graph?
Determine the volume of 1 mol of CH4 gas at 150 K and 1 atm, using .
Determine the pressure of the gas in the syringe shown in when its volume is 12.5 mL, using:
the appropriate graph
Boyle’s law
A spray can is used until it is empty except for the propellant gas, which has a pressure of 1344 torr at 23
°C. If the can is thrown into a fire (T = 475 °C), what will be the pressure in the hot can?
What is the temperature of an 11.2-L sample of carbon monoxide, CO, at 744 torr if it occupies 13.3 L at 55
°C and 744 torr?
A 2.50-L volume of hydrogen measured at –196 °C is warmed to 100 °C. Calculate the volume of the gas at the higher temperature, assuming no change in pressure.
A balloon inflated with three breaths of air has a volume of 1.7 L. At the same temperature and pressure, what is the volume of the balloon if five more same-sized breaths are added to the balloon? |
The volume of an automobile air bag was 66.8 L when inflated at 25 °C with 77.8 g of nitrogen gas. What was the pressure in the bag in kPa?
How many moles of gaseous boron trifluoride, BF3, are contained in a 4.3410-L bulb at 788.0 K if the pressure is 1.220 atm? How many grams of BF3?
Iodine, I2, is a solid at room temperature but sublimes (converts from a solid into a gas) when warmed. What is the temperature in a 73.3-mL bulb that contains 0.292 g of I2 vapor at a pressure of 0.462 atm?
How many grams of gas are present in each of the following cases?
0.100 L of CO2 at 307 torr and 26 °C
8.75 L of C2H4, at 378.3 kPa and 483 K
221 mL of Ar at 0.23 torr and –54 °C
A high altitude balloon is filled with 1.41 104 L of hydrogen at a temperature of 21 °C and a pressure of 745 torr. What is the volume of the balloon at a height of 20 km, where the temperature is –48 °C and the pressure is 63.1 torr?
A cylinder of medical oxygen has a volume of 35.4 L, and contains O2 at a pressure of 151 atm and a
temperature of 25 °C. What volume of O2 does this correspond to at normal body conditions, that is, 1 atm and 37 °C?
A large scuba tank () with a volume of 18 L is rated for a pressure of 220 bar. The tank is filled
at 20 °C and contains enough air to supply 1860 L of air to a diver at a pressure of 2.37 atm (a depth of 45 feet). Was the tank filled to capacity at 20 °C?
A 20.0-L cylinder containing 11.34 kg of butane, C4H10, was opened to the atmosphere. Calculate the mass
of the gas remaining in the cylinder if it were opened and the gas escaped until the pressure in the cylinder was equal to the atmospheric pressure, 0.983 atm, and a temperature of 27 °C.
While resting, the average 70-kg human male consumes 14 L of pure O2 per hour at 25 °C and 100 kPa.
How many moles of O2 are consumed by a 70 kg man while resting for 1.0 h?
For a given amount of gas showing ideal behavior, draw labeled graphs of:
the variation of P with V
the variation of V with T
the variation of P with T
the variation of with V
A liter of methane gas, CH4, at STP contains more atoms of hydrogen than does a liter of pure hydrogen gas, H2, at STP. Using Avogadro’s law as a starting point, explain why.
The effect of chlorofluorocarbons (such as CCl2F2) on the depletion of the ozone layer is well known. The use of substitutes, such as CH3CH2F(g), for the chlorofluorocarbons, has largely corrected the problem. Calculate the volume occupied by 10.0 g of each of these compounds at STP:
CCl2F2(g)
CH3CH2F(g)
As 1 g of the radioactive element radium decays over 1 year, it produces 1.16 1018 alpha particles (helium nuclei). Each alpha particle becomes an atom of helium gas. What is the pressure in pascal of the helium gas produced if it occupies a volume of 125 mL at a temperature of 25 °C?
A balloon with a volume of 100.21 L at 21 °C and 0.981 atm is released and just barely clears the top of
Mount Crumpit in British Columbia. If the final volume of the balloon is 144.53 L at a temperature of 5.24
°C, what is the pressure experienced by the balloon as it clears Mount Crumpet?
If the temperature of a fixed amount of a gas is doubled at constant volume, what happens to the pressure?
If the volume of a fixed amount of a gas is tripled at constant temperature, what happens to the pressure?
What is the density of laughing gas, dinitrogen monoxide, N2O, at a temperature of 325 K and a pressure of
113.0 kPa?
Calculate the density of Freon 12, CF2Cl2, at 30.0 °C and 0.954 atm.
Which is denser at the same temperature and pressure, dry air or air saturated with water vapor? Explain.
A cylinder of O2(g) used in breathing by patients with emphysema has a volume of 3.00 L at a pressure of
atm. If the temperature of the cylinder is 28.0 °C, what mass of oxygen is in the cylinder?
What is the molar mass of a gas if 0.0494 g of the gas occupies a volume of 0.100 L at a temperature 26 °C and a pressure of 307 torr?
What is the molar mass of a gas if 0.281 g of the gas occupies a volume of 125 mL at a temperature 126 °C and a pressure of 777 torr?
How could you show experimentally that the molecular formula of propene is C3H6, not CH2?
The density of a certain gaseous fluoride of phosphorus is 3.93 g/L at STP. Calculate the molar mass of this fluoride and determine its molecular formula.
Consider this question: What is the molecular formula of a compound that contains 39% C, 45% N, and 16% H if 0.157 g of the compound occupies l25 mL with a pressure of 99.5 kPa at 22 °C?
Outline the steps necessary to answer the question.
Answer the question.
A 36.0–L cylinder of a gas used for calibration of blood gas analyzers in medical laboratories contains 350 g CO2, 805 g O2, and 4,880 g N2. At 25 degrees C, what is the pressure in the cylinder in atmospheres?
A cylinder of a gas mixture used for calibration of blood gas analyzers in medical laboratories contains 5.0% CO2, 12.0% O2, and the remainder N2 at a total pressure of 146 atm. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
A sample of gas isolated from unrefined petroleum contains 90.0% CH4, 8.9% C2H6, and 1.1% C3H8 at a
total pressure of 307.2 kPa. What is the partial pressure of each component of this gas? (The percentages given indicate the percent of the total pressure that is due to each component.)
A mixture of 0.200 g of H2, 1.00 g of N2, and 0.820 g of Ar is stored in a closed container at STP. Find the
volume of the container, assuming that the gases exhibit ideal behavior.
Most mixtures of hydrogen gas with oxygen gas are explosive. However, a mixture that contains less than
3.0 % O2 is not. If enough O2 is added to a cylinder of H2 at 33.2 atm to bring the total pressure to 34.5 atm, is the mixture explosive?
A commercial mercury vapor analyzer can detect, in air, concentrations of gaseous Hg atoms (which are
poisonous) as low as 2 10−6 mg/L of air. At this concentration, what is the partial pressure of gaseous mercury if the atmospheric pressure is 733 torr at 26 °C?
A sample of carbon monoxide was collected over water at a total pressure of 756 torr and a temperature of
18 °C. What is the pressure of the carbon monoxide? (See for the vapor pressure of water.)
In an experiment in a general chemistry laboratory, a student collected a sample of a gas over water. The volume of the gas was 265 mL at a pressure of 753 torr and a temperature of 27 °C. The mass of the gas was 0.472 g. What was the molar mass of the gas?
Joseph Priestley first prepared pure oxygen by heating mercuric oxide, HgO:
Outline the steps necessary to answer the following question: What volume of O2 at 23 °C and 0.975 atm is produced by the decomposition of 5.36 g of HgO?
Answer the question.
Cavendish prepared hydrogen in 1766 by the novel method of passing steam through a red-hot gun barrel:
Outline the steps necessary to answer the following question: What volume of H2 at a pressure of 745 torr and a temperature of 20 °C can be prepared from the reaction of 15.O g of H2O?
Answer the question.
The chlorofluorocarbon CCl2F2 can be recycled into a different compound by reaction with hydrogen to produce CH2F2(g), a compound useful in chemical manufacturing:
Outline the steps necessary to answer the following question: What volume of hydrogen at 225 atm and
35.5 °C would be required to react with 1 ton (1.000 103 kg) of CCl2F2?
Answer the question.
Automobile air bags are inflated with nitrogen gas, which is formed by the decomposition of solid sodium azide (NaN3). The other product is sodium metal. Calculate the volume of nitrogen gas at 27 °C and 756 torr formed by the decomposition of 125 g of sodium azide.
Lime, CaO, is produced by heating calcium carbonate, CaCO3; carbon dioxide is the other product.
Outline the steps necessary to answer the following question: What volume of carbon dioxide at 875 K and 0.966 atm is produced by the decomposition of 1 ton (1.000 103 kg) of calcium carbonate?
Answer the question.
Before small batteries were available, carbide lamps were used for bicycle lights. Acetylene gas, C2H2, and solid calcium hydroxide were formed by the reaction of calcium carbide, CaC2, with water. The ignition of the acetylene gas provided the light. Currently, the same lamps are used by some cavers, and calcium carbide is used to produce acetylene for carbide cannons.
Outline the steps necessary to answer the following question: What volume of C2H2 at 1.005 atm and
12.2 °C is formed by the reaction of 15.48 g of CaC2 with water?
Answer the question.
Calculate the volume of oxygen required to burn 12.00 L of ethane gas, C2H6, to produce carbon dioxide and water, if the volumes of C2H6 and O2 are measured under the same conditions of temperature and pressure.
What volume of O2 at STP is required to oxidize 8.0 L of NO at STP to NO2? What volume of NO2 is
produced at STP?
Consider the following questions:
What is the total volume of the CO2(g) and H2O(g) at 600 °C and 0.888 atm produced by the combustion of 1.00 L of C2H6(g) measured at STP?
What is the partial pressure of H2O in the product gases?
Methanol, CH3OH, is produced industrially by the following reaction:
Assuming that the gases behave as ideal gases, find the ratio of the total volume of the reactants to the final volume.
What volume of oxygen at 423.0 K and a pressure of 127.4 kPa is produced by the decomposition of 129.7
g of BaO2 to BaO and O2?
A 2.50-L sample of a colorless gas at STP decomposed to give 2.50 L of N2 and 1.25 L of O2 at STP. What is the colorless gas?
Ethanol, C2H5OH, is produced industrially from ethylene, C2H4, by the following sequence of reactions:
What volume of ethylene at STP is required to produce 1.000 metric ton (1000 kg) of ethanol if the overall yield of ethanol is 90.1%?
One molecule of hemoglobin will combine with four molecules of oxygen. If 1.0 g of hemoglobin combines
with 1.53 mL of oxygen at body temperature (37 °C) and a pressure of 743 torr, what is the molar mass of hemoglobin?
A sample of a compound of xenon and fluorine was confined in a bulb with a pressure of 18 torr. Hydrogen
was added to the bulb until the pressure was 72 torr. Passage of an electric spark through the mixture produced Xe and HF. After the HF was removed by reaction with solid KOH, the final pressure of xenon and unreacted hydrogen in the bulb was 36 torr. What is the empirical formula of the xenon fluoride in the original sample? (Note: Xenon fluorides contain only one xenon atom per molecule.)
One method of analyzing amino acids is the van Slyke method. The characteristic amino groups (−NH2) in protein material are allowed to react with nitrous acid, HNO2, to form N2 gas. From the volume of the gas, the amount of amino acid can be determined. A 0.0604-g sample of a biological sample containing glycine, CH2(NH2)COOH, was analyzed by the van Slyke method and yielded 3.70 mL of N2 collected over water at a pressure of 735 torr and 29 °C. What was the percentage of glycine in the sample?
A balloon filled with helium gas takes 6 hours to deflate to 50% of its original volume. How long will it take for an identical balloon filled with the same volume of hydrogen gas (instead of helium) to decrease its volume by 50%?
Explain why the numbers of molecules are not identical in the left- and right-hand bulbs shown in the
center illustration of .
Starting with the definition of rate of effusion and Graham’s finding relating rate and molar mass, show how to derive the Graham’s law equation, relating the relative rates of effusion for two gases to their molecular masses.
Heavy water, D2O (molar mass = 20.03 g mol–1), can be separated from ordinary water, H2O (molar mass =
18.01), as a result of the difference in the relative rates of diffusion of the molecules in the gas phase. Calculate the relative rates of diffusion of H2O and D2O.
Which of the following gases diffuse more slowly than oxygen? F2, Ne, N2O, C2H2, NO, Cl2, H2S
During the discussion of gaseous diffusion for enriching uranium, it was claimed that 235UF6 diffuses 0.4% faster than 238UF6. Show the calculation that supports this value. The molar mass of 235UF6 = 235.043930 + 6 18.998403 = 349.034348 g/mol, and the molar mass of 238UF6 = 238.050788 + 6
18.998403 = 352.041206 g/mol.
Calculate the relative rate of diffusion of 1H2 (molar mass 2.0 g/mol) compared with 2H2 (molar mass 4.0 g/mol) and the relative rate of diffusion of O2 (molar mass 32 g/mol) compared with O3 (molar mass 48 g/ mol).
A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide
diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.
When two cotton plugs, one moistened with ammonia and the other with hydrochloric acid, are simultaneously inserted into opposite ends of a glass tube that is 87.0 cm long, a white ring of NH4Cl forms where gaseous NH3 and gaseous HCl first come into contact. At approximately what distance from the ammonia moistened plug does this occur? (Hint: Calculate the rates of diffusion for both NH3 and HCl, and find out how much faster NH3 diffuses than HCl.)
Using the postulates of the kinetic molecular theory, explain why a gas uniformly fills a container of any shape.
Can the speed of a given molecule in a gas double at constant temperature? Explain your answer.
Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:
The pressure of the gas is increased by reducing the volume at constant temperature.
The pressure of the gas is increased by increasing the temperature at constant volume.
The average speed of the molecules is increased by a factor of 2.
The distribution of molecular speeds in a sample of helium is shown in . If the sample is cooled, will the distribution of speeds look more like that of H2 or of H2O? Explain your answer.
What is the ratio of the average kinetic energy of a SO2 molecule to that of an O2 molecule in a mixture of two gases? What is the ratio of the root mean square speeds, urms, of the two gases?
A 1-L sample of CO initially at STP is heated to 546 K, and its volume is increased to 2 L.
What effect do these changes have on the number of collisions of the molecules of the gas per unit area of the container wall?
What is the effect on the average kinetic energy of the molecules?
What is the effect on the root mean square speed of the molecules?
The root mean square speed of H2 molecules at 25 °C is about 1.6 km/s. What is the root mean square speed of a N2 molecule at 25 °C?
Answer the following questions:
Is the pressure of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
Is the density of the gas in the hot-air balloon shown at the opening of this chapter greater than, less than, or equal to that of the atmosphere outside the balloon?
At a pressure of 1 atm and a temperature of 20 °C, dry air has a density of 1.2256 g/L. What is the (average) molar mass of dry air?
The average temperature of the gas in a hot-air balloon is 1.30 102 °F. Calculate its density, assuming
the molar mass equals that of dry air.
The lifting capacity of a hot-air balloon is equal to the difference in the mass of the cool air displaced by the balloon and the mass of the gas in the balloon. What is the difference in the mass of 1.00 L of the cool air in part (c) and the hot air in part (d)?
An average balloon has a diameter of 60 feet and a volume of 1.1 105 ft3. What is the lifting power of
such a balloon? If the weight of the balloon and its rigging is 500 pounds, what is its capacity for carrying passengers and cargo?
A balloon carries 40.0 gallons of liquid propane (density 0.5005 g/L). What volume of CO2 and H2O gas is produced by the combustion of this propane?
A balloon flight can last about 90 minutes. If all of the fuel is burned during this time, what is the approximate rate of heat loss (in kJ/min) from the hot air in the bag during the flight?
Show that the ratio of the rate of diffusion of Gas 1 to the rate of diffusion of Gas 2, is the same at 0 °C and 100 °C. |
Under which of the following sets of conditions does a real gas behave most like an ideal gas, and for which conditions is a real gas expected to deviate from ideal behavior? Explain.
high pressure, small volume
high temperature, low pressure
low temperature, high pressure
Describe the factors responsible for the deviation of the behavior of real gases from that of an ideal gas.
For which of the following gases should the correction for the molecular volume be largest: CO, CO2, H2, He, NH3, SF6?
A 0.245-L flask contains 0.467 mol CO2 at 159 °C. Calculate the pressure:
using the ideal gas law
using the van der Waals equation
Explain the reason for the difference.
Identify which correction (that for P or V) is dominant and why.
Answer the following questions:
If XX behaved as an ideal gas, what would its graph of Z vs. P look like?
For most of this chapter, we performed calculations treating gases as ideal. Was this justified?
What is the effect of the volume of gas molecules on Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.
What is the effect of intermolecular attractions on the value of Z? Under what conditions is this effect small? When is it large? Explain using an appropriate diagram.
In general, under what temperature conditions would you expect Z to have the largest deviations from the Z for an ideal gas? |
INTRODUCTION Leprosy has been a devastating disease throughout much of human history. Aside from the symptoms and complications of the illness, its social stigma led sufferers to be cast out of communities and isolated in colonies; in some regions this practice lasted well into the twentieth century. At that time, the best potential treatment for leprosy was oil from the chaulmoogra tree, but the oil was extremely thick, causing blisters and making usage painful and ineffective. Healthcare professionals seeking a better application contacted Alice Ball, a young chemist at the University of Hawaii, who had focused her masters thesis on a similar plant. Ball initiated a sequence of procedures (repeated acidification and purification to change the characteristics of the oil and isolate the active substances (esters, discussed later in this text). The "Ball Method" as it later came to be called, became the standard treatment for leprosy for decades. In the liquid and solid states, atomic and molecular interactions are of considerable strength and play an important role in determining a number of physical properties of the substance. For example, the thickness, or viscosity, of the chaulmoogra oil was due to its intermolecular forces. In this chapter, the nature of these interactions and their effects on various physical properties of liquid and solid phases will be examined.
Intermolecular Forces
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Describe the types of intermolecular forces possible between atoms or molecules in condensed phases (dispersion forces, dipole-dipole attractions, and hydrogen bonding)
Identify the types of intermolecular forces experienced by specific molecules based on their structures
Explain the relation between the intermolecular forces present within a substance and the temperatures associated with changes in its physical state
As was the case for gaseous substances, the kinetic molecular theory may be used to explain the behavior of solids and liquids. In the following description, the term particle will be used to refer to an atom, molecule, or ion. Note that we will use the popular phrase “intermolecular attraction” to refer to attractive forces between the particles of a substance, regardless of whether these particles are molecules, atoms, or ions.
Consider these two aspects of the molecular-level environments in solid, liquid, and gaseous matter:
Particles in a solid are tightly packed together and often arranged in a regular pattern; in a liquid, they are close together with no regular arrangement; in a gas, they are far apart with no regular arrangement.
Particles in a solid vibrate about fixed positions and do not generally move in relation to one another; in a liquid, they move past each other but remain in essentially constant contact; in a gas, they move independently of one another except when they collide.
The differences in the properties of a solid, liquid, or gas reflect the strengths of the attractive forces between the atoms, molecules, or ions that make up each phase. The phase in which a substance exists depends on the relative extents of its intermolecular forces (IMFs) and the kinetic energies (KE) of its molecules. IMFs are the various forces of attraction that may exist between the atoms and molecules of a substance due to electrostatic phenomena, as will be detailed in this module. These forces serve to hold particles close together, whereas the particles’ KE provides the energy required to overcome the attractive forces and thus increase the distance between particles. illustrates how changes in physical state may be induced by changing the temperature, hence, the average KE, of a given substance.
FIGURE 10.2 Transitions between solid, liquid, and gaseous states of a substance occur when conditions of temperature or pressure favor the associated changes in intermolecular forces. (Note: The space between particles in the gas phase is much greater than shown.)
As an example of the processes depicted in this figure, consider a sample of water. When gaseous water is cooled sufficiently, the attractions between H2O molecules will be capable of holding them together when they come into contact with each other; the gas condenses, forming liquid H2O. For example, liquid water forms on the outside of a cold glass as the water vapor in the air is cooled by the cold glass, as seen in .
FIGURE 10.3 Condensation forms when water vapor in the air is cooled enough to form liquid water, such as (a) on the outside of a cold beverage glass or (b) in the form of fog. (credit a: modification of work by Jenny Downing; credit b: modification of work by Cory Zanker)
We can also liquefy many gases by compressing them, if the temperature is not too high. The increased pressure brings the molecules of a gas closer together, such that the attractions between the molecules become strong relative to their KE. Consequently, they form liquids. Butane, C4H10, is the fuel used in disposable lighters and is a gas at standard temperature and pressure. Inside the lighter’s fuel compartment, the butane is compressed to a pressure that results in its condensation to the liquid state, as shown in .
FIGURE 10.4 Gaseous butane is compressed within the storage compartment of a disposable lighter, resulting in its condensation to the liquid state. (credit: modification of work by “Sam-Cat”/Flickr)
Finally, if the temperature of a liquid becomes sufficiently low, or the pressure on the liquid becomes sufficiently high, the molecules of the liquid no longer have enough KE to overcome the IMF between them, and a solid forms. A more thorough discussion of these and other changes of state, or phase transitions, is provided in a later module of this chapter.
LINK TO LEARNING
Access this on states of matter, phase transitions, and intermolecular forces. This simulation is useful for visualizing concepts introduced throughout this chapter.
Forces between Molecules
Under appropriate conditions, the attractions between all gas molecules will cause them to form liquids or solids. This is due to intermolecular forces, not intramolecular forces. Intramolecular forces are those within the molecule that keep the molecule together, for example, the bonds between the atoms. Intermolecular forces are the attractions between molecules, which determine many of the physical properties of a substance. illustrates these different molecular forces. The strengths of these attractive forces vary widely, though usually the IMFs between small molecules are weak compared to the intramolecular forces that bond atoms together within a molecule. For example, to overcome the IMFs in one mole of liquid HCl and convert it into gaseous HCl requires only about 17 kilojoules. However, to break the covalent bonds between the hydrogen and chlorine atoms in one mole of HCl requires about 25 times more energy—430 kilojoules.
FIGURE 10.5 Intramolecular forces keep a molecule intact. Intermolecular forces hold multiple molecules together and determine many of a substance’s properties.
All of the attractive forces between neutral atoms and molecules are known as van der Waals forces, although they are usually referred to more informally as intermolecular attraction. We will consider the various types of IMFs in the next three sections of this module.
Dispersion Forces
One of the three van der Waals forces is present in all condensed phases, regardless of the nature of the atoms or molecules composing the substance. This attractive force is called the London dispersion force in honor of German-born American physicist Fritz London who, in 1928, first explained it. This force is often referred to as simply the dispersion force. Because the electrons of an atom or molecule are in constant motion (or, alternatively, the electron’s location is subject to quantum-mechanical variability), at any moment in time, an atom or molecule can develop a temporary, instantaneous dipole if its electrons are distributed asymmetrically. The presence of this dipole can, in turn, distort the electrons of a neighboring atom or molecule, producing an induced dipole. These two rapidly fluctuating, temporary dipoles thus result in a relatively weak electrostatic attraction between the species—a so-called dispersion force like that illustrated in .
FIGURE 10.6 Dispersion forces result from the formation of temporary dipoles, as illustrated here for two nonpolar diatomic molecules.
Dispersion forces that develop between atoms in different molecules can attract the two molecules to each other. The forces are relatively weak, however, and become significant only when the molecules are very close.
Larger and heavier atoms and molecules exhibit stronger dispersion forces than do smaller and lighter atoms and molecules. F2 and Cl2 are gases at room temperature (reflecting weaker attractive forces); Br2 is a liquid, and I2 is a solid (reflecting stronger attractive forces). Trends in observed melting and boiling points for the halogens clearly demonstrate this effect, as seen in . |
The increase in melting and boiling points with increasing atomic/molecular size may be rationalized by considering how the strength of dispersion forces is affected by the electronic structure of the atoms or molecules in the substance. In a larger atom, the valence electrons are, on average, farther from the nuclei than in a smaller atom. Thus, they are less tightly held and can more easily form the temporary dipoles that produce the attraction. The measure of how easy or difficult it is for another electrostatic charge (for example, a nearby ion or polar molecule) to distort a molecule’s charge distribution (its electron cloud) is known as polarizability. A molecule that has a charge cloud that is easily distorted is said to be very polarizable and will have large dispersion forces; one with a charge cloud that is difficult to distort is not very polarizable and will have small dispersion forces.
London Forces and Their Effects
Order the following compounds of a group 14 element and hydrogen from lowest to highest boiling point: CH4, SiH4, GeH4, and SnH4. Explain your reasoning.
Solution
Applying the skills acquired in the chapter on chemical bonding and molecular geometry, all of these compounds are predicted to be nonpolar, so they may experience only dispersion forces: the smaller the molecule, the less polarizable and the weaker the dispersion forces; the larger the molecule, the larger the dispersion forces. The molar masses of CH4, SiH4, GeH4, and SnH4 are approximately 16 g/mol, 32 g/mol, 77 g/ mol, and 123 g/mol, respectively. Therefore, CH4 is expected to have the lowest boiling point and SnH4 the highest boiling point. The ordering from lowest to highest boiling point is expected to be CH4 < SiH4 < GeH4 < SnH4.
A graph of the actual boiling points of these compounds versus the period of the group 14 element shows this prediction to be correct:
Check Your Learning
Order the following hydrocarbons from lowest to highest boiling point: C2H6, C3H8, and C4H10.
Answer:
C2H6 < C3H8 < C4H10. All of these compounds are nonpolar and only have London dispersion forces: the larger the molecule, the larger the dispersion forces and the higher the boiling point. The ordering from lowest to highest boiling point is therefore C2H6 < C3H8 < C4H10.
The shapes of molecules also affect the magnitudes of the dispersion forces between them. For example, boiling points for the isomers n-pentane, isopentane, and neopentane (shown in ) are 36 °C, 27 °C, and 9.5 °C, respectively. Even though these compounds are composed of molecules with the same chemical formula, C5H12, the difference in boiling points suggests that dispersion forces in the liquid phase are different, being greatest for n-pentane and least for neopentane. The elongated shape of n-pentane provides a greater surface area available for contact between molecules, resulting in correspondingly stronger dispersion forces. The more compact shape of isopentane offers a smaller surface area available for intermolecular contact and, therefore, weaker dispersion forces. Neopentane molecules are the most compact of the three, offering the least available surface area for intermolecular contact and, hence, the weakest dispersion forces. This behavior is analogous to the connections that may be formed between strips of VELCRO brand fasteners: the greater the area of the strip’s contact, the stronger the connection. |
Chemistry in Everyday Life
Geckos and Intermolecular Forces
Geckos have an amazing ability to adhere to most surfaces. They can quickly run up smooth walls and across ceilings that have no toe-holds, and they do this without having suction cups or a sticky substance on their toes. And while a gecko can lift its feet easily as it walks along a surface, if you attempt to pick it up, it sticks to the surface. How are geckos (as well as spiders and some other insects) able to do this? Although this phenomenon has been investigated for hundreds of years, scientists only recently uncovered the details of the process that allows geckos’ feet to behave this way.
Geckos’ toes are covered with hundreds of thousands of tiny hairs known as setae, with each seta, in turn, branching into hundreds of tiny, flat, triangular tips called spatulae. The huge numbers of spatulae on its setae provide a gecko, shown in , with a large total surface area for sticking to a surface. In 2000, Kellar Autumn, who leads a multi-institutional gecko research team, found that geckos adhered equally well to both polar silicon dioxide and nonpolar gallium arsenide. This proved that geckos stick to surfaces because of dispersion forces—weak intermolecular attractions arising from temporary, synchronized charge distributions between adjacent molecules. Although dispersion forces are very weak, the total attraction over millions of spatulae is large enough to support many times the gecko’s weight.
In 2014, two scientists developed a model to explain how geckos can rapidly transition from “sticky” to “non-sticky.” Alex Greaney and Congcong Hu at Oregon State University described how geckos can achieve this by changing the angle between their spatulae and the surface. Geckos’ feet, which are normally nonsticky, become sticky when a small shear force is applied. By curling and uncurling their toes, geckos can alternate between sticking and unsticking from a surface, and thus easily move across it. Later research led by Alyssa Stark at University of Akron showed that geckos can maintain their hold on hydrophobic surfaces (similar to the leaves in their habitats) equally well whether the surfaces were wet or dry. Stark's experiment used a ribbon to gently pull the geckos until they slipped, so that the researchers could determine the geckos' ability to hold various surfaces under wet and dry conditions. Further investigations may eventually lead to the development of better adhesives and other applications. |
Dipole-Dipole Attractions
Recall from the chapter on chemical bonding and molecular geometry that polar molecules have a partial positive charge on one side and a partial negative charge on the other side of the molecule—a separation of charge called a dipole. Consider a polar molecule such as hydrogen chloride, HCl. In the HCl molecule, the more electronegative Cl atom bears the partial negative charge, whereas the less electronegative H atom bears the partial positive charge. An attractive force between HCl molecules results from the attraction between the positive end of one HCl molecule and the negative end of another. This attractive force is called a dipole-dipole attraction—the electrostatic force between the partially positive end of one polar molecule and the partially negative end of another, as illustrated in .
FIGURE 10.9 This image shows two arrangements of polar molecules, such as HCl, that allow an attraction between the partial negative end of one molecule and the partial positive end of another.
The effect of a dipole-dipole attraction is apparent when we compare the properties of HCl molecules to nonpolar F2 molecules. Both HCl and F2 consist of the same number of atoms and have approximately the same molecular mass. At a temperature of 150 K, molecules of both substances would have the same average KE. However, the dipole-dipole attractions between HCl molecules are sufficient to cause them to “stick together” to form a liquid, whereas the relatively weaker dispersion forces between nonpolar F2 molecules are not, and so this substance is gaseous at this temperature. The higher normal boiling point of HCl (188 K) compared to F2 (85 K) is a reflection of the greater strength of dipole-dipole attractions between HCl molecules, compared to the attractions between nonpolar F2 molecules. We will often use values such as boiling or freezing points, or enthalpies of vaporization or fusion, as indicators of the relative strengths of IMFs of attraction present within different substances.
Dipole-Dipole Forces and Their Effects
Predict which will have the higher boiling point: N2 or CO. Explain your reasoning.
Solution
CO and N2 are both diatomic molecules with masses of about 28 amu, so they experience similar London dispersion forces. Because CO is a polar molecule, it experiences dipole-dipole attractions. Because N2 is nonpolar, its molecules cannot exhibit dipole-dipole attractions. The dipole-dipole attractions between CO molecules are comparably stronger than the dispersion forces between nonpolar N2 molecules, so CO is expected to have the higher boiling point.
Check Your Learning
Predict which will have the higher boiling point: ICl or Br2. Explain your reasoning.
Answer:
ICl. ICl and Br2 have similar masses (~160 amu) and therefore experience similar London dispersion forces. ICl is polar and thus also exhibits dipole-dipole attractions; Br2 is nonpolar and does not. The relatively stronger dipole-dipole attractions require more energy to overcome, so ICl will have the higher boiling point.
Hydrogen Bonding
Nitrosyl fluoride (ONF, molecular mass 49 amu) is a gas at room temperature. Water (H2O, molecular mass 18 amu) is a liquid, even though it has a lower molecular mass. We clearly cannot attribute this difference between the two compounds to dispersion forces. Both molecules have about the same shape and ONF is the heavier and larger molecule. It is, therefore, expected to experience more significant dispersion forces.
Additionally, we cannot attribute this difference in boiling points to differences in the dipole moments of the molecules. Both molecules are polar and exhibit comparable dipole moments. The large difference between the boiling points is due to a particularly strong dipole-dipole attraction that may occur when a molecule contains a hydrogen atom bonded to a fluorine, oxygen, or nitrogen atom (the three most electronegative elements). The very large difference in electronegativity between the H atom (2.1) and the atom to which it is bonded (4.0 for an F atom, 3.5 for an O atom, or 3.0 for a N atom), combined with the very small size of a H atom and the relatively small sizes of F, O, or N atoms, leads to highly concentrated partial charges with these atoms. Molecules with F-H, O-H, or N-H moieties are very strongly attracted to similar moieties in nearby molecules, a particularly strong type of dipole-dipole attraction called hydrogen bonding. Examples of hydrogen bonds include HF⋯HF, H2O⋯HOH, and H3N⋯HNH2, in which the hydrogen bonds are denoted by dots. illustrates hydrogen bonding between water molecules. |
Despite use of the word “bond,” keep in mind that hydrogen bonds are intermolecular attractive forces, not intramolecular attractive forces (covalent bonds). Hydrogen bonds are much weaker than covalent bonds, only about 5 to 10% as strong, but are generally much stronger than other dipole-dipole attractions and dispersion forces.
Hydrogen bonds have a pronounced effect on the properties of condensed phases (liquids and solids). For example, consider the trends in boiling points for the binary hydrides of group 15 (NH3, PH3, AsH3, and SbH3), group 16 hydrides (H2O, H2S, H2Se, and H2Te), and group 17 hydrides (HF, HCl, HBr, and HI). The boiling points of the heaviest three hydrides for each group are plotted in . As we progress down any of these groups, the polarities of the molecules decrease slightly, whereas the sizes of the molecules increase substantially. The effect of increasingly stronger dispersion forces dominates that of increasingly weaker dipole-dipole attractions, and the boiling points are observed to increase steadily.
FIGURE 10.11 For the group 15, 16, and 17 hydrides, the boiling points for each class of compounds increase with increasing molecular mass for elements in periods 3, 4, and 5.
If we use this trend to predict the boiling points for the lightest hydride for each group, we would expect NH3 to boil at about −120 °C, H2O to boil at about −80 °C, and HF to boil at about −110 °C. However, when we measure the boiling points for these compounds, we find that they are dramatically higher than the trends would predict, as shown in . The stark contrast between our naïve predictions and reality provides compelling evidence for the strength of hydrogen bonding. |
Effect of Hydrogen Bonding on Boiling Points
Consider the compounds dimethylether (CH3OCH3), ethanol (CH3CH2OH), and propane (CH3CH2CH3). Their boiling points, not necessarily in order, are −42.1 °C, −24.8 °C, and 78.4 °C. Match each compound with its boiling point. Explain your reasoning.
Solution
The VSEPR-predicted shapes of CH3OCH3, CH3CH2OH, and CH3CH2CH3 are similar, as are their molar masses (46 g/mol, 46 g/mol, and 44 g/mol, respectively), so they will exhibit similar dispersion forces. Since CH3CH2CH3 is nonpolar, it may exhibit only dispersion forces. Because CH3OCH3 is polar, it will also experience dipole-dipole attractions. Finally, CH3CH2OH has an −OH group, and so it will experience the uniquely strong dipole-dipole attraction known as hydrogen bonding. So the ordering in terms of strength of IMFs, and thus boiling points, is CH3CH2CH3 < CH3OCH3 < CH3CH2OH. The boiling point of propane is −42.1
°C, the boiling point of dimethylether is −24.8 °C, and the boiling point of ethanol is 78.5 °C.
Check Your Learning
Ethane (CH3CH3) has a melting point of −183 °C and a boiling point of −89 °C. Predict the melting and boiling points for methylamine (CH3NH2). Explain your reasoning.
Answer:
The melting point and boiling point for methylamine are predicted to be significantly greater than those of ethane. CH3CH3 and CH3NH2 are similar in size and mass, but methylamine possesses an −NH group and therefore may exhibit hydrogen bonding. This greatly increases its IMFs, and therefore its melting and boiling points. It is difficult to predict values, but the known values are a melting point of −93 °C and a boiling point of
−6 °C.
Hydrogen Bonding and DNA
Deoxyribonucleic acid (DNA) is found in every living organism and contains the genetic information that determines the organism’s characteristics, provides the blueprint for making the proteins necessary for life, and serves as a template to pass this information on to the organism’s offspring. A DNA molecule consists of two (anti-)parallel chains of repeating nucleotides, which form its well-known double helical structure, as shown in .
FIGURE 10.13 Two separate DNA molecules form a double-stranded helix in which the molecules are held together via hydrogen bonding. (credit: modification of work by Jerome Walker, Dennis Myts)
Each nucleotide contains a (deoxyribose) sugar bound to a phosphate group on one side, and one of four nitrogenous bases on the other. Two of the bases, cytosine (C) and thymine (T), are single-ringed structures known as pyrimidines. The other two, adenine (A) and guanine (G), are double-ringed structures called purines. These bases form complementary base pairs consisting of one purine and one pyrimidine, with adenine pairing with thymine, and cytosine with guanine. Each base pair is held together by hydrogen bonding. A and T share two hydrogen bonds, C and G share three, and both pairings have a similar shape and structure .
FIGURE 10.14 The geometries of the base molecules result in maximum hydrogen bonding between adenine and thymine (AT) and between guanine and cytosine (GC), so-called “complementary base pairs.”
The cumulative effect of millions of hydrogen bonds effectively holds the two strands of DNA together. Importantly, the two strands of DNA can relatively easily “unzip” down the middle since hydrogen bonds are relatively weak compared to the covalent bonds that hold the atoms of the individual DNA molecules together. This allows both strands to function as a template for replication.
Properties of Liquids
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Distinguish between adhesive and cohesive forces
Define viscosity, surface tension, and capillary rise
Describe the roles of intermolecular attractive forces in each of these properties/phenomena
When you pour a glass of water, or fill a car with gasoline, you observe that water and gasoline flow freely. But when you pour syrup on pancakes or add oil to a car engine, you note that syrup and motor oil do not flow as readily. The viscosity of a liquid is a measure of its resistance to flow. Water, gasoline, and other liquids that flow freely have a low viscosity. Honey, syrup, motor oil, and other liquids that do not flow freely, like those shown in , have higher viscosities. We can measure viscosity by measuring the rate at which a metal ball falls through a liquid (the ball falls more slowly through a more viscous liquid) or by measuring the rate at which a liquid flows through a narrow tube (more viscous liquids flow more slowly).
FIGURE 10.15 (a) Honey and (b) motor oil are examples of liquids with high viscosities; they flow slowly. (credit a: modification of work by Scott Bauer; credit b: modification of work by David Nagy)
The IMFs between the molecules of a liquid, the size and shape of the molecules, and the temperature determine how easily a liquid flows. As shows, the more structurally complex are the molecules in a liquid and the stronger the IMFs between them, the more difficult it is for them to move past each other and the greater is the viscosity of the liquid. As the temperature increases, the molecules move more rapidly and their kinetic energies are better able to overcome the forces that hold them together; thus, the viscosity of the liquid decreases. |
The various IMFs between identical molecules of a substance are examples of cohesive forces. The molecules within a liquid are surrounded by other molecules and are attracted equally in all directions by the cohesive forces within the liquid. However, the molecules on the surface of a liquid are attracted only by about one-half as many molecules. Because of the unbalanced molecular attractions on the surface molecules, liquids contract to form a shape that minimizes the number of molecules on the surface—that is, the shape with the minimum surface area. A small drop of liquid tends to assume a spherical shape, as shown in , because in a sphere, the ratio of surface area to volume is at a minimum. Larger drops are more greatly affected by gravity, air resistance, surface interactions, and so on, and as a result, are less spherical.
FIGURE 10.16 Attractive forces result in a spherical water drop that minimizes surface area; cohesive forces hold the sphere together; adhesive forces keep the drop attached to the web. (credit photo: modification of work by “OliBac”/Flickr)
Surface tension is defined as the energy required to increase the surface area of a liquid, or the force required to increase the length of a liquid surface by a given amount. This property results from the cohesive forces between molecules at the surface of a liquid, and it causes the surface of a liquid to behave like a stretched rubber membrane. Surface tensions of several liquids are presented in . Among common liquids, water exhibits a distinctly high surface tension due to strong hydrogen bonding between its molecules. As a result of this high surface tension, the surface of water represents a relatively “tough skin” that can withstand considerable force without breaking. A steel needle carefully placed on water will float. Some insects, like the one shown in , even though they are denser than water, move on its surface because they are supported by the surface tension. |
FIGURE 10.17 Surface tension (right) prevents this insect, a “water strider,” from sinking into the water.
Surface tension is affected by a variety of variables, including the introduction of additional substances on the surface. In the late 1800s, Agnes Pockels, who was initially blocked from pursuing a scientific career but
studied on her own, began investigating the impact and characteristics of soapy and greasy films in water. Using homemade materials, she developed an instrument known as a trough for measuring surface contaminants and their effects. With the support of renowned scientist Lord Rayleigh, her 1891 paper showed that surface contamination significantly reduces surface tension, and also that changing the characteristics of the surface (compressing or expanding it) also affects surface tension. Decades later, Irving Langmuir and Katharine Blodgett built on Pockels' work in their own trough and important advances in surface chemistry. Langmuir pioneered methods for producing single-molecule layers of film; Blodgett applied these to the development of non-reflective glass (critical for film-making and other applications), and also studied methods related to cleaning surfaces, which are important in semiconductor fabrication.
The IMFs of attraction between two different molecules are called adhesive forces. Consider what happens when water comes into contact with some surface. If the adhesive forces between water molecules and the molecules of the surface are weak compared to the cohesive forces between the water molecules, the water does not “wet” the surface. For example, water does not wet waxed surfaces or many plastics such as polyethylene. Water forms drops on these surfaces because the cohesive forces within the drops are greater than the adhesive forces between the water and the plastic. Water spreads out on glass because the adhesive force between water and glass is greater than the cohesive forces within the water. When water is confined in a glass tube, its meniscus (surface) has a concave shape because the water wets the glass and creeps up the side of the tube. On the other hand, the cohesive forces between mercury atoms are much greater than the adhesive forces between mercury and glass. Mercury therefore does not wet glass, and it forms a convex meniscus when confined in a tube because the cohesive forces within the mercury tend to draw it into a drop ().
FIGURE 10.18 Differences in the relative strengths of cohesive and adhesive forces result in different meniscus shapes for mercury (left) and water (right) in glass tubes. (credit: Mark Ott)
If you place one end of a paper towel in spilled wine, as shown in , the liquid wicks up the paper towel. A similar process occurs in a cloth towel when you use it to dry off after a shower. These are examples of capillary action—when a liquid flows within a porous material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules. The adhesive forces between the liquid and the porous material, combined with the cohesive forces within the liquid, may be strong enough to move the liquid upward against gravity.
FIGURE 10.19 Wine wicks up a paper towel (left) because of the strong attractions of water (and ethanol) molecules to the −OH groups on the towel’s cellulose fibers and the strong attractions of water molecules to other water (and ethanol) molecules (right). (credit photo: modification of work by Mark Blaser)
Towels soak up liquids like water because the fibers of a towel are made of molecules that are attracted to water molecules. Most cloth towels are made of cotton, and paper towels are generally made from paper pulp. Both consist of long molecules of cellulose that contain many −OH groups. Water molecules are attracted to these −OH groups and form hydrogen bonds with them, which draws the H2O molecules up the cellulose molecules. The water molecules are also attracted to each other, so large amounts of water are drawn up the cellulose fibers.
Capillary action can also occur when one end of a small diameter tube is immersed in a liquid, as illustrated in . If the liquid molecules are strongly attracted to the tube molecules, the liquid creeps up the inside of the tube until the weight of the liquid and the adhesive forces are in balance. The smaller the diameter of the tube is, the higher the liquid climbs. It is partly by capillary action occurring in plant cells called xylem that water and dissolved nutrients are brought from the soil up through the roots and into a plant. Capillary action is the basis for thin layer chromatography, a laboratory technique commonly used to separate small quantities of mixtures. You depend on a constant supply of tears to keep your eyes lubricated and on capillary action to pump tear fluid away.
FIGURE 10.20 Depending upon the relative strengths of adhesive and cohesive forces, a liquid may rise (such as water) or fall (such as mercury) in a glass capillary tube. The extent of the rise (or fall) is directly proportional to the surface tension of the liquid and inversely proportional to the density of the liquid and the radius of the tube.
The height to which a liquid will rise in a capillary tube is determined by several factors as shown in the |
In this equation, h is the height of the liquid inside the capillary tube relative to the surface of the liquid outside the tube, T is the surface tension of the liquid, θ is the contact angle between the liquid and the tube, r is the radius of the tube, ρ is the density of the liquid, and g is the acceleration due to gravity, 9.8 m/s2. When the tube is made of a material to which the liquid molecules are strongly attracted, they will spread out completely on the surface, which corresponds to a contact angle of 0°. This is the situation for water rising in a glass tube.
Capillary Rise
At 25 °C, how high will water rise in a glass capillary tube with an inner diameter of 0.25 mm? For water, T = 71.99 mN/m and ρ = 1.0 g/cm3.
Solution
The liquid will rise to a height h given by:
The Newton is defined as a kg m/s2, and so the provided surface tension is equivalent to 0.07199 kg/s2. The provided density must be converted into units that will cancel appropriately: ρ = 1000 kg/m3. The diameter of the tube in meters is 0.00025 m, so the radius is 0.000125 m. For a glass tube immersed in water, the contact angle is θ = 0°, so cos θ = 1. Finally, acceleration due to gravity on the earth is g = 9.8 m/s2. Substituting these values into the equation, and cancelling units, we have: |
Phase Transitions
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Define phase transitions and phase transition temperatures
Explain the relation between phase transition temperatures and intermolecular attractive forces
Describe the processes represented by typical heating and cooling curves, and compute heat flows and enthalpy changes accompanying these processes
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 module, 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 , and using a manometer to measure the increase in pressure that is due to the vapor in equilibrium with the condensed phase. |
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. |
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 −OH 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
At 20 °C, the vapor pressures of several alcohols are given in this table. Explain these vapor pressures in terms of types and extents of IMFs for these alcohols:
Answer:
All these compounds exhibit hydrogen bonding; these strong IMFs are difficult for the molecules to overcome, so the vapor pressures are relatively low. As the size of molecule increases from methanol to butanol, dispersion forces increase, which means that the vapor pressures decrease as observed:
Pmethanol > Pethanol > Ppropanol > Pbutanol.
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 . 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.
FIGURE 10.23 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). 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.
FIGURE 10.24 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.)
A Boiling Point at Reduced Pressure
A typical atmospheric pressure in Leadville, Colorado (elevation 10,200 feet) is 68 kPa. Use the graph in
to determine the boiling point of water at this elevation.
Solution
The graph of the vapor pressure of water versus temperature in 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 boiling point of ethyl ether was measured to be 10 °C at a base camp on the slopes of Mount Everest. Use to determine the approximate atmospheric pressure at the camp.
Answer:
Approximately 40 kPa (0.4 atm) |
where ΔHvap is the enthalpy of vaporization for the liquid, R is the gas constant, and A is a constant whose value depends on the chemical identity of the substance. Temperature T must be in Kelvin in this equation. This equation is often rearranged into logarithmic form to yield the linear equation:
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 P2, the corresponding linear equations are: |
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.
Solution
The enthalpy of vaporization, ΔHvap, can be determined by using the Clausius-Clapeyron equation: |
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
At 20.0 °C, the vapor pressure of ethanol is 5.95 kPa, and at 63.5 °C, its vapor pressure is 53.3 kPa. Use this information to estimate the enthalpy of vaporization for ethanol.
Answer:
41,360 J/mol or 41.4 kJ/mol
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?
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:
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, ΔHvap = 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: |
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, ΔHvap. For example, the vaporization of water at standard temperature is represented by: |
Using Enthalpy of Vaporization
One way our body is cooled is by evaporation of the water in sweat (). 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); ΔHvap = 43.46 kJ/mol at 37 °C.
FIGURE 10.25 Evaporation of sweat helps cool the body. (credit: “Kullez”/Flickr)
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:
Thus, 3600 kJ of heat are removed by the evaporation of 1.5 L of water.
Check Your Learning
How much heat is required to evaporate 100.0 g of liquid ammonia, NH3, at its boiling point if its enthalpy of vaporization is 4.8 kJ/mol?
Answer:
28 kJ
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 |
FIGURE 10.26 (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 processes 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, ΔHfus of the substance. The enthalpy of fusion of ice is 6.0 kJ/mol at 0 °C. Fusion (melting) is an endothermic process: |
Sublimation and Deposition
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 CO2) 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 (). 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.
FIGURE 10.27 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)
Like vaporization, the process of sublimation requires an input of energy to overcome intermolecular attractions. The enthalpy of sublimation, ΔHsub, 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: |
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 . For example:
FIGURE 10.28 For a given substance, the sum of its enthalpy of fusion and enthalpy of vaporization is approximately equal to its enthalpy of sublimation.
Heating and Cooling Curves
In the chapter on thermochemistry, the relation between the amount of heat absorbed or released by a substance, q, and its accompanying temperature change, ΔT, was introduced:
where m is the mass of the substance and c is its specific heat. The relation applies to matter being heated or cooled, but not undergoing a change in state. When a substance being heated or cooled reaches a temperature corresponding to one of its phase transitions, further gain or loss of heat is a result of diminishing or enhancing intermolecular attractions, instead of increasing or decreasing molecular kinetic energies. While a substance is undergoing a change in state, its temperature remains constant. shows a typical heating curve.
Consider the example of heating a pot of water to boiling. A stove burner will supply heat at a roughly constant rate; initially, this heat serves to increase the water’s temperature. When the water reaches its boiling point, the temperature remains constant despite the continued input of heat from the stove burner. This same temperature is maintained by the water as long as it is boiling. If the burner setting is increased to provide heat at a greater rate, the water temperature does not rise, but instead the boiling becomes more vigorous (rapid).
This behavior is observed for other phase transitions as well: For example, temperature remains constant while the change of state is in progress.
FIGURE 10.29 A typical heating curve for a substance depicts changes in temperature that result as the substance absorbs increasing amounts of heat. Plateaus in the curve (regions of constant temperature) are exhibited when the substance undergoes phase transitions.
Total Heat Needed to Change Temperature and Phase for a Substance
How much heat is required to convert 135 g of ice at −15 °C into water vapor at 120 °C?
Solution
The transition described involves the following steps:
Heat ice from −15 °C to 0 °C
Melt ice
Heat water from 0 °C to 100 °C
Boil water
Heat steam from 100 °C to 120 °C
The heat needed to change the temperature of a given substance (with no change in phase) is: q = m c ΔT (see previous chapter on thermochemistry). The heat needed to induce a given change in phase is given by q = n ΔH.
Using these equations with the appropriate values for specific heat of ice, water, and steam, and enthalpies of fusion and vaporization, we have: |
Phase Diagrams
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Explain the construction and use of a typical phase diagram
Use phase diagrams to identify stable phases at given temperatures and pressures, and to describe phase transitions resulting from changes in these properties
Describe the supercritical fluid phase of matter
In the previous module, the variation of a liquid’s equilibrium vapor pressure with temperature was described. Considering the definition of boiling point, plots of vapor pressure versus temperature represent how the boiling point of the liquid varies with pressure. Also described was the use of heating and cooling curves to determine a substance’s melting (or freezing) point. Making such measurements over a wide range of pressures yields data that may be presented graphically as a phase diagram. A phase diagram combines plots of pressure versus temperature for the liquid-gas, solid-liquid, and solid-gas phase-transition equilibria of a substance. These diagrams indicate the physical states that exist under specific conditions of pressure and temperature, and also provide the pressure dependence of the phase-transition temperatures (melting points, sublimation points, boiling points). A typical phase diagram for a pure substance is shown in .
FIGURE 10.30 The physical state of a substance and its phase-transition temperatures are represented graphically in a phase diagram.
To illustrate the utility of these plots, consider the phase diagram for water shown in .
FIGURE 10.31 The pressure and temperature axes on this phase diagram of water are not drawn to constant scale in order to illustrate several important properties.
We can use the phase diagram to identify the physical state of a sample of water under specified conditions of pressure and temperature. For example, a pressure of 50 kPa and a temperature of −10 °C correspond to the region of the diagram labeled “ice.” Under these conditions, water exists only as a solid (ice). A pressure of 50 kPa and a temperature of 50 °C correspond to the “water” region—here, water exists only as a liquid. At 25 kPa and 200 °C, water exists only in the gaseous state. Note that on the H2O phase diagram, the pressure and temperature axes are not drawn to a constant scale in order to permit the illustration of several important features as described here.
The curve BC in is the plot of vapor pressure versus temperature as described in the previous module of this chapter. This “liquid-vapor” curve separates the liquid and gaseous regions of the phase diagram and provides the boiling point for water at any pressure. For example, at 1 atm, the boiling point is 100 °C. Notice that the liquid-vapor curve terminates at a temperature of 374 °C and a pressure of 218 atm,
indicating that water cannot exist as a liquid above this temperature, regardless of the pressure. The physical properties of water under these conditions are intermediate between those of its liquid and gaseous phases. This unique state of matter is called a supercritical fluid, a topic that will be described in the next section of this module.
The solid-vapor curve, labeled AB in , indicates the temperatures and pressures at which ice and water vapor are in equilibrium. These temperature-pressure data pairs correspond to the sublimation, or deposition, points for water. If we could zoom in on the solid-gas line in , we would see that ice has a vapor pressure of about 0.20 kPa at −10 °C. Thus, if we place a frozen sample in a vacuum with a pressure less than 0.20 kPa, ice will sublime. This is the basis for the “freeze-drying” process often used to preserve foods, such as the ice cream shown in .
FIGURE 10.32 Freeze-dried foods, like this ice cream, are dehydrated by sublimation at pressures below the triple point for water. (credit: ʺlwaoʺ/Flickr)
The solid-liquid curve labeled BD shows the temperatures and pressures at which ice and liquid water are in equilibrium, representing the melting/freezing points for water. Note that this curve exhibits a slight negative slope (greatly exaggerated for clarity), indicating that the melting point for water decreases slightly as pressure increases. Water is an unusual substance in this regard, as most substances exhibit an increase in melting point with increasing pressure. This behavior is partly responsible for the movement of glaciers, like the one shown in . The bottom of a glacier experiences an immense pressure due to its weight that can melt some of the ice, forming a layer of liquid water on which the glacier may more easily slide.
FIGURE 10.33 The immense pressures beneath glaciers result in partial melting to produce a layer of water that provides lubrication to assist glacial movement. This satellite photograph shows the advancing edge of the Perito Moreno glacier in Argentina. (credit: NASA)
The point of intersection of all three curves is labeled B in . At the pressure and temperature represented by this point, all three phases of water coexist in equilibrium. This temperature-pressure data |
Determining the State of Water
Using the phase diagram for water given in , determine the state of water at the following temperatures and pressures:
−10 °C and 50 kPa
25 °C and 90 kPa
50 °C and 40 kPa
80 °C and 5 kPa
−10 °C and 0.3 kPa
50 °C and 0.3 kPa
Solution
Using the phase diagram for water, we can determine that the state of water at each temperature and pressure given are as follows: (a) solid; (b) liquid; (c) liquid; (d) gas; (e) solid; (f ) gas.
Check Your Learning
What phase changes can water undergo as the temperature changes if the pressure is held at 0.3 kPa? If the pressure is held at 50 kPa?
Answer:
At 0.3 kPa: at −58 °C. At 50 kPa: at 0 °C, l ⟶ g at 78 °C
Consider the phase diagram for carbon dioxide shown in as another example. The solid-liquid curve exhibits a positive slope, indicating that the melting point for CO2 increases with pressure as it does for most substances (water being a notable exception as described previously). Notice that the triple point is well above 1 atm, indicating that carbon dioxide cannot exist as a liquid under ambient pressure conditions.
Instead, cooling gaseous carbon dioxide at 1 atm results in its deposition into the solid state. Likewise, solid carbon dioxide does not melt at 1 atm pressure but instead sublimes to yield gaseous CO2. Finally, notice that the critical point for carbon dioxide is observed at a relatively modest temperature and pressure in comparison to water. |
Determining the State of Carbon Dioxide
Using the phase diagram for carbon dioxide shown in , determine the state of CO2 at the following temperatures and pressures:
−30 °C and 2000 kPa
−90 °C and 1000 kPa
−60 °C and 100 kPa
−40 °C and 1500 kPa
0 °C and 100 kPa
20 °C and 100 kPa
Solution
Using the phase diagram for carbon dioxide provided, we can determine that the state of CO2 at each temperature and pressure given are as follows: (a) liquid; (b) solid; (c) gas; (d) liquid; (e) gas; (f ) gas.
Check Your Learning
Identify the phase changes that carbon dioxide will undergo as its temperature is increased from −100 °C while holding its pressure constant at 1500 kPa. At 50 kPa. At what approximate temperatures do these phase changes occur?
Answer:
at 1500 kPa: at −55 °C, at −10 °C;
at 50 kPa: at −60 °C
Supercritical Fluids
If we place a sample of water in a sealed container at 25 °C, remove the air, and let the vaporization- condensation equilibrium establish itself, we are left with a mixture of liquid water and water vapor at a pressure of 0.03 atm. A distinct boundary between the more dense liquid and the less dense gas is clearly
observed. As we increase the temperature, the pressure of the water vapor increases, as described by the liquid-gas curve in the phase diagram for water (), and a two-phase equilibrium of liquid and gaseous phases remains. At a temperature of 374 °C, the vapor pressure has risen to 218 atm, and any further increase in temperature results in the disappearance of the boundary between liquid and vapor phases. All of the water in the container is now present in a single phase whose physical properties are intermediate between those of the gaseous and liquid states. This phase of matter is called a supercritical fluid, and the temperature and pressure above which this phase exists is the critical point (). Above its critical temperature, a gas cannot be liquefied no matter how much pressure is applied. The pressure required to liquefy a gas at its critical temperature is called the critical pressure. The critical temperatures and critical pressures of some common substances are given in the following table. |
FIGURE 10.35 (a) A sealed container of liquid carbon dioxide slightly below its critical point is heated, resulting in
(b) the formation of the supercritical fluid phase. Cooling the supercritical fluid lowers its temperature and pressure below the critical point, resulting in the reestablishment of separate liquid and gaseous phases (c and d). Colored floats illustrate differences in density between the liquid, gaseous, and supercritical fluid states. (credit: modification of work by “mrmrobin”/YouTube)
LINK TO LEARNING
Observe the for carbon dioxide.
Like a gas, a supercritical fluid will expand and fill a container, but its density is much greater than typical gas densities, typically being close to those for liquids. Similar to liquids, these fluids are capable of dissolving nonvolatile solutes. They exhibit essentially no surface tension and very low viscosities, however, so they can more effectively penetrate very small openings in a solid mixture and remove soluble components. These properties make supercritical fluids extremely useful solvents for a wide range of applications. For example, supercritical carbon dioxide has become a very popular solvent in the food industry, being used to decaffeinate coffee, remove fats from potato chips, and extract flavor and fragrance compounds from citrus |
The Critical Temperature of Carbon Dioxide
If we shake a carbon dioxide fire extinguisher on a cool day (18 °C), we can hear liquid CO2 sloshing around inside the cylinder. However, the same cylinder appears to contain no liquid on a hot summer day (35 °C). Explain these observations.
Solution
On the cool day, the temperature of the CO2 is below the critical temperature of CO2, 304 K or 31 °C, so liquid CO2 is present in the cylinder. On the hot day, the temperature of the CO2 is greater than its critical temperature of 31 °C. Above this temperature no amount of pressure can liquefy CO2 so no liquid CO2 exists in the fire extinguisher.
Check Your Learning
Ammonia can be liquefied by compression at room temperature; oxygen cannot be liquefied under these conditions. Why do the two gases exhibit different behavior?
Answer:
The critical temperature of ammonia is 405.5 K, which is higher than room temperature. The critical temperature of oxygen is below room temperature; thus oxygen cannot be liquefied at room temperature.
Chemistry in Everyday Life
Decaffeinating Coffee Using Supercritical CO2
Coffee is the world’s second most widely traded commodity, following only petroleum. Across the globe, people love coffee’s aroma and taste. Many of us also depend on one component of coffee—caffeine—to help us get going in the morning or stay alert in the afternoon. But late in the day, coffee’s stimulant effect can keep you from sleeping, so you may choose to drink decaffeinated coffee in the evening.
Since the early 1900s, many methods have been used to decaffeinate coffee. All have advantages and disadvantages, and all depend on the physical and chemical properties of caffeine. Because caffeine is a somewhat polar molecule, it dissolves well in water, a polar liquid. However, since many of the other 400-plus compounds that contribute to coffee’s taste and aroma also dissolve in H2O, hot water decaffeination processes can also remove some of these compounds, adversely affecting the smell and
taste of the decaffeinated coffee. Dichloromethane (CH2Cl2) and ethyl acetate (CH3CO2C2H5) have similar polarity to caffeine, and are therefore very effective solvents for caffeine extraction, but both also remove some flavor and aroma components, and their use requires long extraction and cleanup times. Because both of these solvents are toxic, health concerns have been raised regarding the effect of residual solvent remaining in the decaffeinated coffee.
Supercritical fluid extraction using carbon dioxide is now being widely used as a more effective and environmentally friendly decaffeination method (). At temperatures above 304.2 K and pressures above 7376 kPa, CO2 is a supercritical fluid, with properties of both gas and liquid. Like a gas, it penetrates deep into the coffee beans; like a liquid, it effectively dissolves certain substances. Supercritical carbon dioxide extraction of steamed coffee beans removes 97−99% of the caffeine, leaving coffee’s flavor and aroma compounds intact. Because CO2 is a gas under standard conditions, its removal from the extracted coffee beans is easily accomplished, as is the recovery of the caffeine from the extract. The caffeine recovered from coffee beans via this process is a valuable product that can be used subsequently as an additive to other foods or drugs.
The Solid State of Matter
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Define and describe the bonding and properties of ionic, molecular, metallic, and covalent network crystalline solids
Describe the main types of crystalline solids: ionic solids, metallic solids, covalent network solids, and molecular solids
Explain the ways in which crystal defects can occur in a solid
When most liquids are cooled, they eventually freeze and form crystalline solids, solids in which the atoms, ions, or molecules are arranged in a definite repeating pattern. It is also possible for a liquid to freeze before its molecules become arranged in an orderly pattern. The resulting materials are called amorphous solids or noncrystalline solids (or, sometimes, glasses). The particles of such solids lack an ordered internal structure and are randomly arranged (). |
randomly (amorphous).
Metals and ionic compounds typically form ordered, crystalline solids. Substances that consist of large molecules, or a mixture of molecules whose movements are more restricted, often form amorphous solids. For examples, candle waxes are amorphous solids composed of large hydrocarbon molecules. Some substances, such as silicon dioxide (shown in ), can form either crystalline or amorphous solids, depending on the conditions under which it is produced. Also, amorphous solids may undergo a transition to the crystalline state under appropriate conditions.
FIGURE 10.38 (a) Silicon dioxide, SiO2, is abundant in nature as one of several crystalline forms of the mineral quartz. (b) Rapid cooling of molten SiO2 yields an amorphous solid known as “fused silica”.
Crystalline solids are generally classified according to the nature of the forces that hold its particles together. These forces are primarily responsible for the physical properties exhibited by the bulk solids. The following sections provide descriptions of the major types of crystalline solids: ionic, metallic, covalent network, and molecular.
Ionic Solids
Ionic solids, such as sodium chloride and nickel oxide, are composed of positive and negative ions that are held together by electrostatic attractions, which can be quite strong (). Many ionic crystals also have high melting points. This is due to the very strong attractions between the ions—in ionic compounds, the attractions between full charges are (much) larger than those between the partial charges in polar molecular compounds. This will be looked at in more detail in a later discussion of lattice energies. Although they are hard, they also tend to be brittle, and they shatter rather than bend. Ionic solids do not conduct electricity; however, they do conduct when molten or dissolved because their ions are free to move. Many simple compounds formed by the reaction of a metallic element with a nonmetallic element are ionic. |
Metallic solids such as crystals of copper, aluminum, and iron are formed by metal atoms . The structure of metallic crystals is often described as a uniform distribution of atomic nuclei within a “sea” of delocalized electrons. The atoms within such a metallic solid are held together by a unique force known as metallic bonding that gives rise to many useful and varied bulk properties. All exhibit high thermal and electrical conductivity, metallic luster, and malleability. Many are very hard and quite strong. Because of their
malleability (the ability to deform under pressure or hammering), they do not shatter and, therefore, make useful construction materials. The melting points of the metals vary widely. Mercury is a liquid at room temperature, and the alkali metals melt below 200 °C. Several post-transition metals also have low melting points, whereas the transition metals melt at temperatures above 1000 °C. These differences reflect differences in strengths of metallic bonding among the metals. |
Covalent network solids include crystals of diamond, silicon, some other nonmetals, and some covalent compounds such as silicon dioxide (sand) and silicon carbide (carborundum, the abrasive on sandpaper). Many minerals have networks of covalent bonds. The atoms in these solids are held together by a network of covalent bonds, as shown in . To break or to melt a covalent network solid, covalent bonds must be broken. Because covalent bonds are relatively strong, covalent network solids are typically characterized by hardness, strength, and high melting points. For example, diamond is one of the hardest substances known and melts above 3500 °C.
FIGURE 10.41 A covalent crystal contains a three-dimensional network of covalent bonds, as illustrated by the structures of diamond, silicon dioxide, silicon carbide, and graphite. Graphite is an exceptional example, composed of planar sheets of covalent crystals that are held together in layers by noncovalent forces. Unlike typical covalent solids, graphite is very soft and electrically conductive.
Molecular Solid
Molecular solids, such as ice, sucrose (table sugar), and iodine, as shown in , are composed of neutral molecules. The strengths of the attractive forces between the units present in different crystals vary widely, as indicated by the melting points of the crystals. Small symmetrical molecules (nonpolar molecules), such as H2, N2, O2, and F2, have weak attractive forces and form molecular solids with very low melting points (below −200 °C). Substances consisting of larger, nonpolar molecules have larger attractive forces and melt at higher temperatures. Molecular solids composed of molecules with permanent dipole moments (polar |
FIGURE 10.42 Carbon dioxide (CO2) consists of small, nonpolar molecules and forms a molecular solid with a melting point of −78 °C. Iodine (I2) consists of larger, nonpolar molecules and forms a molecular solid that melts at 114 °C.
Properties of Solids
A crystalline solid, like those listed in , has a precise melting temperature because each atom or molecule of the same type is held in place with the same forces or energy. Thus, the attractions between the units that make up the crystal all have the same strength and all require the same amount of energy to be broken. The gradual softening of an amorphous material differs dramatically from the distinct melting of a crystalline solid. This results from the structural nonequivalence of the molecules in the amorphous solid. Some forces are weaker than others, and when an amorphous material is heated, the weakest intermolecular attractions break first. As the temperature is increased further, the stronger attractions are broken. Thus amorphous materials soften over a range of temperatures. |
Graphene: Material of the Future
Carbon is an essential element in our world. The unique properties of carbon atoms allow the existence of carbon-based life forms such as ourselves. Carbon forms a huge variety of substances that we use on a daily basis, including those shown in . You may be familiar with diamond and graphite, the two most common allotropes of carbon. (Allotropes are different structural forms of the same element.) Diamond is one of the hardest-known substances, whereas graphite is soft enough to be used as pencil lead. These very different properties stem from the different arrangements of the carbon atoms in the different allotropes.
FIGURE 10.43 Diamond is extremely hard because of the strong bonding between carbon atoms in all directions. Graphite (in pencil lead) rubs off onto paper due to the weak attractions between the carbon layers. An image of a graphite surface shows the distance between the centers of adjacent carbon atoms. (credit left photo: modification of work by Steve Jurvetson; credit middle photo: modification of work by United States Geological Survey)
You may be less familiar with a recently discovered form of carbon: graphene. Graphene was first isolated in 2004 by using tape to peel off thinner and thinner layers from graphite. It is essentially a single sheet (one atom thick) of graphite. Graphene, illustrated in , is not only strong and lightweight, but it is also an excellent conductor of electricity and heat. These properties may prove very useful in a wide range of applications, such as vastly improved computer chips and circuits, better batteries and solar cells, and stronger and lighter structural materials. The 2010 Nobel Prize in Physics was awarded to Andre Geim and Konstantin Novoselov for their pioneering work with graphene. |
Crystal Defects
In a crystalline solid, the atoms, ions, or molecules are arranged in a definite repeating pattern, but occasional defects may occur in the pattern. Several types of defects are known, as illustrated in . Vacancies are defects that occur when positions that should contain atoms or ions are vacant. Less commonly, some atoms or ions in a crystal may occupy positions, called interstitial sites, located between the regular positions for atoms. Other distortions are found in impure crystals, as, for example, when the cations, anions, or molecules of the impurity are too large to fit into the regular positions without distorting the structure. Trace amounts of impurities are sometimes added to a crystal (a process known as doping) in order to create defects in the structure that yield desirable changes in its properties. For example, silicon crystals are doped with varying amounts of different elements to yield suitable electrical properties for their use in the manufacture of semiconductors and computer chips.
FIGURE 10.45 Types of crystal defects include vacancies, interstitial atoms, and substitutions impurities.
Lattice Structures in Crystalline Solids
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Describe the arrangement of atoms and ions in crystalline structures
Compute ionic radii using unit cell dimensions
Explain the use of X-ray diffraction measurements in determining crystalline structures
Over 90% of naturally occurring and man-made solids are crystalline. Most solids form with a regular arrangement of their particles because the overall attractive interactions between particles are maximized, and the total intermolecular energy is minimized, when the particles pack in the most efficient manner. The regular arrangement at an atomic level is often reflected at a macroscopic level. In this module, we will explore some of the details about the structures of metallic and ionic crystalline solids, and learn how these structures are determined experimentally.
The Structures of Metals
We will begin our discussion of crystalline solids by considering elemental metals, which are relatively simple because each contains only one type of atom. A pure metal is a crystalline solid with metal atoms packed closely together in a repeating pattern. Some of the properties of metals in general, such as their malleability and ductility, are largely due to having identical atoms arranged in a regular pattern. The different properties of one metal compared to another partially depend on the sizes of their atoms and the specifics of their spatial arrangements. We will explore the similarities and differences of four of the most common metal crystal geometries in the sections that follow.
Unit Cells of Metals
The structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell. The unit cell consists of lattice points that represent the locations of atoms or ions. The entire structure then consists of this unit cell repeating in three dimensions, as illustrated in .
FIGURE 10.46 A unit cell shows the locations of lattice points repeating in all directions.
Let us begin our investigation of crystal lattice structure and unit cells with the most straightforward structure and the most basic unit cell. To visualize this, imagine taking a large number of identical spheres, such as tennis balls, and arranging them uniformly in a container. The simplest way to do this would be to make layers in which the spheres in one layer are directly above those in the layer below, as illustrated in . This arrangement is called simple cubic structure, and the unit cell is called the simple cubic unit cell or primitive cubic unit cell.
FIGURE 10.47 When metal atoms are arranged with spheres in one layer directly above or below spheres in another layer, the lattice structure is called simple cubic. Note that the spheres are in contact.
In a simple cubic structure, the spheres are not packed as closely as they could be, and they only “fill” about 52% of the volume of the container. This is a relatively inefficient arrangement, and only one metal (polonium, Po) crystallizes in a simple cubic structure. As shown in , a solid with this type of arrangement consists of planes (or layers) in which each atom contacts only the four nearest neighbors in its layer; one atom directly above it in the layer above; and one atom directly below it in the layer below. The number of other particles that each particle in a crystalline solid contacts is known as its coordination number. For a polonium atom in a simple cubic array, the coordination number is, therefore, six.
FIGURE 10.48 An atom in a simple cubic lattice structure contacts six other atoms, so it has a coordination number of six.
In a simple cubic lattice, the unit cell that repeats in all directions is a cube defined by the centers of eight atoms, as shown in . Atoms at adjacent corners of this unit cell contact each other, so the edge length of this cell is equal to two atomic radii, or one atomic diameter. A cubic unit cell contains only the parts of these atoms that are within it. Since an atom at a corner of a simple cubic unit cell is contained by a total of eight unit cells, only one-eighth of that atom is within a specific unit cell. And since each simple cubic unit cell has one atom at each of its eight “corners,” there is atom within one simple cubic unit cell. |
Calculation of Atomic Radius and Density for Metals, Part 1
The edge length of the unit cell of alpha polonium is 336 pm.
(a) Determine the radius of a polonium atom.
(b) Determine the density of alpha polonium.
Solution
Alpha polonium crystallizes in a simple cubic unit cell:
(a) Two adjacent Po atoms contact each other, so the edge length of this cell is equal to two Po atomic radii: l = 2r. Therefore, the radius of Po is
(b) Density is given by The density of polonium can be found by determining the density of its unit cell (the mass contained within a unit cell divided by the volume of the unit cell). Since a Po unit cell contains one-eighth of a Po atom at each of its eight corners, a unit cell contains one Po atom.
The mass of a Po unit cell can be found by: |
Check Your Learning
The edge length of the unit cell for nickel is 0.3524 nm. The density of Ni is 8.90 g/cm3. Does nickel crystallize in a simple cubic structure? Explain.
Answer:
No. If Ni was simple cubic, its density would be given by: |
Most metal crystals are one of the four major types of unit cells. For now, we will focus on the three cubic unit cells: simple cubic (which we have already seen), body-centered cubic unit cell, and face-centered cubic unit cell—all of which are illustrated in . (Note that there are actually seven different lattice systems, some of which have more than one type of lattice, for a total of 14 different types of unit cells. We leave the more complicated geometries for later in this module.)
FIGURE 10.50 Cubic unit cells of metals show (in the upper figures) the locations of lattice points and (in the lower figures) metal atoms located in the unit cell.
Some metals crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and an atom in the center, as shown in . This is called a body-centered cubic (BCC) solid. Atoms in the corners of a BCC unit cell do not contact each other but contact the atom in the center. A BCC unit cell contains two atoms: one-eighth of an atom at each of the eight corners atom from the corners) plus one atom from the center. Any atom in this structure touches four atoms in the layer above it and four atoms in the layer below it. Thus, an atom in a BCC structure has a coordination number of eight.
FIGURE 10.51 In a body-centered cubic structure, atoms in a specific layer do not touch each other. Each atom touches four atoms in the layer above it and four atoms in the layer below it.
Atoms in BCC arrangements are much more efficiently packed than in a simple cubic structure, occupying about 68% of the total volume. Isomorphous metals with a BCC structure include K, Ba, Cr, Mo, W, and Fe at room temperature. (Elements or compounds that crystallize with the same structure are said to be isomorphous.)
Many other metals, such as aluminum, copper, and lead, crystallize in an arrangement that has a cubic unit cell with atoms at all of the corners and at the centers of each face, as illustrated in . This arrangement is called a face-centered cubic (FCC) solid. A FCC unit cell contains four atoms: one-eighth of an atom at each of the eight corners atom from the corners) and one-half of an atom on each of the six faces atoms from the faces). The atoms at the corners touch the atoms in the centers of the adjacent faces along the face diagonals of the cube. Because the atoms are on identical lattice points, they have identical environments.
FIGURE 10.52 A face-centered cubic solid has atoms at the corners and, as the name implies, at the centers of the faces of its unit cells.
Atoms in an FCC arrangement are packed as closely together as possible, with atoms occupying 74% of the volume. This structure is also called cubic closest packing (CCP). In CCP, there are three repeating layers of hexagonally arranged atoms. Each atom contacts six atoms in its own layer, three in the layer above, and three in the layer below. In this arrangement, each atom touches 12 near neighbors, and therefore has a coordination number of 12. The fact that FCC and CCP arrangements are equivalent may not be immediately obvious, but why they are actually the same structure is illustrated in .
FIGURE 10.53 A CCP arrangement consists of three repeating layers (ABCABC…) of hexagonally arranged atoms. Atoms in a CCP structure have a coordination number of 12 because they contact six atoms in their layer, plus three atoms in the layer above and three atoms in the layer below. By rotating our perspective, we can see that a CCP structure has a unit cell with a face containing an atom from layer A at one corner, atoms from layer B across a diagonal (at two corners and in the middle of the face), and an atom from layer C at the remaining corner. This is the same as a face-centered cubic arrangement.
Because closer packing maximizes the overall attractions between atoms and minimizes the total intermolecular energy, the atoms in most metals pack in this manner. We find two types of closest packing in simple metallic crystalline structures: CCP, which we have already encountered, and hexagonal closest packing (HCP) shown in . Both consist of repeating layers of hexagonally arranged atoms. In both types, a second layer (B) is placed on the first layer (A) so that each atom in the second layer is in contact with three atoms in the first layer. The third layer is positioned in one of two ways. In HCP, atoms in the third layer are directly above atoms in the first layer (i.e., the third layer is also type A), and the stacking consists of alternating type A and type B close-packed layers (i.e., ABABAB⋯). In CCP, atoms in the third layer are not above atoms in either of the first two layers (i.e., the third layer is type C), and the stacking consists of alternating type A, type B, and type C close-packed layers (i.e., ABCABCABC⋯). About two–thirds of all metals crystallize in closest-packed arrays with coordination numbers of 12. Metals that crystallize in an HCP structure include Cd, Co, Li, Mg, Na, and Zn, and metals that crystallize in a CCP structure include Ag, Al, Ca, Cu, Ni, Pb, and Pt.
FIGURE 10.54 In both types of closest packing, atoms are packed as compactly as possible. Hexagonal closest packing consists of two alternating layers (ABABAB…). Cubic closest packing consists of three alternating layers (ABCABCABC…).
Calculation of Atomic Radius and Density for Metals, Part 2
Calcium crystallizes in a face-centered cubic structure. The edge length of its unit cell is 558.8 pm.
(a) What is the atomic radius of Ca in this structure?
(b) Calculate the density of Ca.
Solution
(a) In an FCC structure, Ca atoms contact each other across the diagonal of the face, so the length of the diagonal is equal to four Ca atomic radii (d = 4r). Two adjacent edges and the diagonal of the face form a right triangle, with the length of each side equal to 558.8 pm and the length of the hypotenuse equal to four Ca atomic radii:
Solving this gives
(b) Density is given by The density of calcium can be found by determining the density of its unit cell: for example, the mass contained within a unit cell divided by the volume of the unit cell. A face- centered Ca unit cell has one-eighth of an atom at each of the eight corners atom) and one-half of an atom on each of the six faces atoms), for a total of four atoms in the unit cell.
The mass of the unit cell can be found by: |
(Note that the edge length was converted from pm to cm to get the usual volume units for density.) Then, the density of
Check Your Learning
Silver crystallizes in an FCC structure. The edge length of its unit cell is 409 pm.
(a) What is the atomic radius of Ag in this structure?
(b) Calculate the density of Ag.
Answer:
(a) 144 pm; (b) 10.5 g/cm3
In general, a unit cell is defined by the lengths of three axes (a, b, and c) and the angles (α, β, and γ) between them, as illustrated in . The axes are defined as being the lengths between points in the space lattice. Consequently, unit cell axes join points with identical environments.
FIGURE 10.55 A unit cell is defined by the lengths of its three axes (a, b, and c) and the angles (α, β, and γ) between the axes.
There are seven different lattice systems, some of which have more than one type of lattice, for a total of fourteen different unit cells, which have the shapes shown in .
FIGURE 10.56 There are seven different lattice systems and 14 different unit cells.
The Structures of Ionic Crystals
Ionic crystals consist of two or more different kinds of ions that usually have different sizes. The packing of these ions into a crystal structure is more complex than the packing of metal atoms that are the same size.
Most monatomic ions behave as charged spheres, and their attraction for ions of opposite charge is the same in every direction. Consequently, stable structures for ionic compounds result (1) when ions of one charge are
surrounded by as many ions as possible of the opposite charge and (2) when the cations and anions are in contact with each other. Structures are determined by two principal factors: the relative sizes of the ions and the ratio of the numbers of positive and negative ions in the compound.
In simple ionic structures, we usually find the anions, which are normally larger than the cations, arranged in a closest-packed array. (As seen previously, additional electrons attracted to the same nucleus make anions larger and fewer electrons attracted to the same nucleus make cations smaller when compared to the atoms from which they are formed.) The smaller cations commonly occupy one of two types of holes (or interstices) remaining between the anions. The smaller of the holes is found between three anions in one plane and one anion in an adjacent plane. The four anions surrounding this hole are arranged at the corners of a tetrahedron, so the hole is called a tetrahedral hole. The larger type of hole is found at the center of six anions (three in one layer and three in an adjacent layer) located at the corners of an octahedron; this is called an octahedral hole. illustrates both of these types of holes.
FIGURE 10.57 Cations may occupy two types of holes between anions: octahedral holes or tetrahedral holes.
Depending on the relative sizes of the cations and anions, the cations of an ionic compound may occupy tetrahedral or octahedral holes, as illustrated in . Relatively small cations occupy tetrahedral holes, and larger cations occupy octahedral holes. If the cations are too large to fit into the octahedral holes, the anions may adopt a more open structure, such as a simple cubic array. The larger cations can then occupy the larger cubic holes made possible by the more open spacing.
FIGURE 10.58 A cation’s size and the shape of the hole occupied by the compound are directly related.
There are two tetrahedral holes for each anion in either an HCP or CCP array of anions. A compound that crystallizes in a closest-packed array of anions with cations in the tetrahedral holes can have a maximum cation:anion ratio of 2:1; all of the tetrahedral holes are filled at this ratio. Examples include Li2O, Na2O, Li2S,
and Na2S. Compounds with a ratio of less than 2:1 may also crystallize in a closest-packed array of anions with cations in the tetrahedral holes, if the ionic sizes fit. In these compounds, however, some of the tetrahedral holes remain vacant.
Occupancy of Tetrahedral Holes
Zinc sulfide is an important industrial source of zinc and is also used as a white pigment in paint. Zinc sulfide crystallizes with zinc ions occupying one-half of the tetrahedral holes in a closest-packed array of sulfide ions. What is the formula of zinc sulfide?
Solution
Because there are two tetrahedral holes per anion (sulfide ion) and one-half of these holes are occupied by zinc ions, there must be or 1, zinc ion per sulfide ion. Thus, the formula is ZnS.
Check Your Learning
Lithium selenide can be described as a closest-packed array of selenide ions with lithium ions in all of the tetrahedral holes. What it the formula of lithium selenide?
Answer:
Li2Se
The ratio of octahedral holes to anions in either an HCP or CCP structure is 1:1. Thus, compounds with cations in octahedral holes in a closest-packed array of anions can have a maximum cation:anion ratio of 1:1. In NiO, MnS, NaCl, and KH, for example, all of the octahedral holes are filled. Ratios of less than 1:1 are observed when some of the octahedral holes remain empty.
Stoichiometry of Ionic Compounds
Sapphire is aluminum oxide. Aluminum oxide crystallizes with aluminum ions in two-thirds of the octahedral holes in a closest-packed array of oxide ions. What is the formula of aluminum oxide?
Solution
Because there is one octahedral hole per anion (oxide ion) and only two-thirds of these holes are occupied, the ratio of aluminum to oxygen must be :1, which would give The simplest whole number ratio is 2:3, so the formula is Al2O3.
Check Your Learning
The white pigment titanium oxide crystallizes with titanium ions in one-half of the octahedral holes in a closest-packed array of oxide ions. What is the formula of titanium oxide?
Answer:
TiO2
In a simple cubic array of anions, there is one cubic hole that can be occupied by a cation for each anion in the array. In CsCl, and in other compounds with the same structure, all of the cubic holes are occupied. Half of the cubic holes are occupied in SrH2, UO2, SrCl2, and CaF2.
Different types of ionic compounds often crystallize in the same structure when the relative sizes of their ions and their stoichiometries (the two principal features that determine structure) are similar.
Unit Cells of Ionic Compounds
Many ionic compounds crystallize with cubic unit cells, and we will use these compounds to describe the general features of ionic structures.
When an ionic compound is composed of cations and anions of similar size in a 1:1 ratio, it typically forms a simple cubic structure. Cesium chloride, CsCl, (illustrated in ) is an example of this, with Cs+ and Cl− having radii of 174 pm and 181 pm, respectively. We can think of this as chloride ions forming a simple cubic unit cell, with a cesium ion in the center; or as cesium ions forming a unit cell with a chloride ion in the center; or as simple cubic unit cells formed by Cs+ ions overlapping unit cells formed by Cl− ions. Cesium ions and chloride ions touch along the body diagonals of the unit cells. One cesium ion and one chloride ion are present per unit cell, giving the l:l stoichiometry required by the formula for cesium chloride. Note that there is no lattice point in the center of the cell, and CsCl is not a BCC structure because a cesium ion is not identical to a chloride ion.
FIGURE 10.59 Ionic compounds with similar-sized cations and anions, such as CsCl, usually form a simple cubic structure. They can be described by unit cells with either cations at the corners or anions at the corners.
We have said that the location of lattice points is arbitrary. This is illustrated by an alternate description of the CsCl structure in which the lattice points are located in the centers of the cesium ions. In this description, the cesium ions are located on the lattice points at the corners of the cell, and the chloride ion is located at the center of the cell. The two unit cells are different, but they describe identical structures.
When an ionic compound is composed of a 1:1 ratio of cations and anions that differ significantly in size, it typically crystallizes with an FCC unit cell, like that shown in . Sodium chloride, NaCl, is an example of this, with Na+ and Cl− having radii of 102 pm and 181 pm, respectively. We can think of this as chloride ions forming an FCC cell, with sodium ions located in the octahedral holes in the middle of the cell edges and in the center of the cell. The sodium and chloride ions touch each other along the cell edges. The unit cell contains four sodium ions and four chloride ions, giving the 1:1 stoichiometry required by the formula, NaCl.
FIGURE 10.60 Ionic compounds with anions that are much larger than cations, such as NaCl, usually form an FCC structure. They can be described by FCC unit cells with cations in the octahedral holes.
The cubic form of zinc sulfide, zinc blende, also crystallizes in an FCC unit cell, as illustrated in . This structure contains sulfide ions on the lattice points of an FCC lattice. (The arrangement of sulfide ions is identical to the arrangement of chloride ions in sodium chloride.) The radius of a zinc ion is only about 40% of the radius of a sulfide ion, so these small Zn2+ ions are located in alternating tetrahedral holes, that is, in one half of the tetrahedral holes. There are four zinc ions and four sulfide ions in the unit cell, giving the empirical formula ZnS.
FIGURE 10.61 ZnS, zinc sulfide (or zinc blende) forms an FCC unit cell with sulfide ions at the lattice points and much smaller zinc ions occupying half of the tetrahedral holes in the structure.
A calcium fluoride unit cell, like that shown in , is also an FCC unit cell, but in this case, the cations are located on the lattice points; equivalent calcium ions are located on the lattice points of an FCC lattice. All of the tetrahedral sites in the FCC array of calcium ions are occupied by fluoride ions. There are four calcium ions and eight fluoride ions in a unit cell, giving a calcium:fluorine ratio of 1:2, as required by the chemical formula, CaF2. Close examination of will reveal a simple cubic array of fluoride ions with calcium ions in one half of the cubic holes. The structure cannot be described in terms of a space lattice of points on the fluoride ions because the fluoride ions do not all have identical environments. The orientation of the four calcium ions about the fluoride ions differs.
FIGURE 10.62 Calcium fluoride, CaF2, forms an FCC unit cell with calcium ions (green) at the lattice points and fluoride ions (red) occupying all of the tetrahedral sites between them.
Calculation of Ionic Radii
If we know the edge length of a unit cell of an ionic compound and the position of the ions in the cell, we can calculate ionic radii for the ions in the compound if we make assumptions about individual ionic shapes and contacts.
Calculation of Ionic Radii
The edge length of the unit cell of LiCl (NaCl-like structure, FCC) is 0.514 nm or 5.14 Å. Assuming that the lithium ion is small enough so that the chloride ions are in contact, as in , calculate the ionic radius for the chloride ion.
Note: The length unit angstrom, Å, is often used to represent atomic-scale dimensions and is equivalent to 10−10 m.
Solution
On the face of a LiCl unit cell, chloride ions contact each other across the diagonal of the face:
Drawing a right triangle on the face of the unit cell, we see that the length of the diagonal is equal to four chloride radii (one radius from each corner chloride and one diameter—which equals two radii—from the chloride ion in the center of the face), so d = 4r. From the Pythagorean theorem, we have: |
Check Your Learning
The edge length of the unit cell of KCl (NaCl-like structure, FCC) is 6.28 Å. Assuming anion-cation contact along the cell edge, calculate the radius of the potassium ion. The radius of the chloride ion is 1.82 Å.
Answer:
The radius of the potassium ion is 1.33 Å.
It is important to realize that values for ionic radii calculated from the edge lengths of unit cells depend on numerous assumptions, such as a perfect spherical shape for ions, which are approximations at best. Hence, such calculated values are themselves approximate and comparisons cannot be pushed too far. Nevertheless, this method has proved useful for calculating ionic radii from experimental measurements such as X-ray crystallographic determinations.
X-Ray Crystallography
The size of the unit cell and the arrangement of atoms in a crystal may be determined from measurements of the diffraction of X-rays by the crystal, termed X-ray crystallography. Diffraction is the change in the direction of travel experienced by an electromagnetic wave when it encounters a physical barrier whose dimensions are comparable to those of the wavelength of the light. X-rays are electromagnetic radiation with wavelengths about as long as the distance between neighboring atoms in crystals (on the order of a few Å).
When a beam of monochromatic X-rays strikes a crystal, its rays are scattered in all directions by the atoms within the crystal. When scattered waves traveling in the same direction encounter one another, they undergo interference, a process by which the waves combine to yield either an increase or a decrease in amplitude (intensity) depending upon the extent to which the combining waves’ maxima are separated (see ).
FIGURE 10.63 Light waves occupying the same space experience interference, combining to yield waves of greater
(a) or lesser (b) intensity, depending upon the separation of their maxima and minima.
When X-rays of a certain wavelength, λ, are scattered by atoms in adjacent crystal planes separated by a distance, d, they may undergo constructive interference when the difference between the distances traveled by the two waves prior to their combination is an integer factor, n, of the wavelength. This condition is satisfied when the angle of the diffracted beam, θ, is related to the wavelength and interatomic distance by the equation:
This relation is known as the Bragg equation in honor of W. H. Bragg, the English physicist who first explained this phenomenon. illustrates two examples of diffracted waves from the same two crystal planes. The figure on the left depicts waves diffracted at the Bragg angle, resulting in constructive interference, while that on the right shows diffraction and a different angle that does not satisfy the Bragg condition, resulting in destructive interference.
FIGURE 10.64 The diffraction of X-rays scattered by the atoms within a crystal permits the determination of the distance between the atoms. The top image depicts constructive interference between two scattered waves and a resultant diffracted wave of high intensity. The bottom image depicts destructive interference and a low intensity diffracted wave.
LINK TO LEARNING
Visit this for more details on the Bragg equation and a simulator that allows you to explore the effect of each variable on the intensity of the diffracted wave.
An X-ray diffractometer, such as the one illustrated in , may be used to measure the angles at which X-rays are diffracted when interacting with a crystal as described earlier. From such measurements, the Bragg equation may be used to compute distances between atoms as demonstrated in the following example exercise. |
Using the Bragg Equation
In a diffractometer, X-rays with a wavelength of 0.1315 nm were used to produce a diffraction pattern for copper. The first order diffraction (n = 1) occurred at an angle θ = 25.25°. Determine the spacing between the diffracting planes in copper.
Solution
The distance between the planes is found by solving the Bragg equation, nλ = 2d sin θ, for d. This gives:
Check Your Learning
A crystal with spacing between planes equal to 0.394 nm diffracts X-rays with a wavelength of 0.147 nm. What is the angle for the first order diffraction?
Answer:
10.8°
Key Terms
adhesive force force of attraction between molecules of different chemical identities
amorphous solid (also, noncrystalline solid) solid in which the particles lack an ordered internal structure
body-centered cubic (BCC) solid crystalline
structure that has a cubic unit cell with lattice points at the corners and in the center of the cell
body-centered cubic unit cell simplest repeating
unit of a body-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of the cube
boiling point temperature at which the vapor
pressure of a liquid equals the pressure of the gas above it
Bragg equation equation that relates the angles at
which X-rays are diffracted by the atoms within a crystal
capillary action flow of liquid within a porous
material due to the attraction of the liquid molecules to the surface of the material and to other liquid molecules
Clausius-Clapeyron equation mathematical
relationship between the temperature, vapor pressure, and enthalpy of vaporization for a substance
cohesive force force of attraction between
identical molecules
condensation change from a gaseous to a liquid state
coordination number number of atoms closest to any given atom in a crystal or to the central metal atom in a complex
covalent network solid solid whose particles are
held together by covalent bonds
critical point temperature and pressure above which a gas cannot be condensed into a liquid
crystalline solid solid in which the particles are arranged in a definite repeating pattern
cubic closest packing (CCP) crystalline structure in which planes of closely packed atoms or ions are stacked as a series of three alternating layers of different relative orientations (ABC)
deposition change from a gaseous state directly to
a solid state
diffraction redirection of electromagnetic radiation that occurs when it encounters a physical barrier of appropriate dimensions
dipole-dipole attraction intermolecular attraction
between two permanent dipoles
dispersion force (also, London dispersion force) attraction between two rapidly fluctuating,
temporary dipoles; significant only when particles are very close together
dynamic equilibrium state of a system in which
reciprocal processes are occurring at equal rates
face-centered cubic (FCC) solid crystalline structure consisting of a cubic unit cell with lattice points on the corners and in the center of each face
face-centered cubic unit cell simplest repeating
unit of a face-centered cubic crystal; it is a cube containing lattice points at each corner and in the center of each face
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
hexagonal closest packing (HCP) crystalline structure in which close packed layers of atoms or ions are stacked as a series of two alternating layers of different relative orientations (AB)
hole (also, interstice) space between atoms within
a crystal
hydrogen bonding occurs when exceptionally strong dipoles attract; bonding that exists when hydrogen is bonded to one of the three most electronegative elements: F, O, or N
induced dipole temporary dipole formed when the
electrons of an atom or molecule are distorted by the instantaneous dipole of a neighboring atom or molecule
instantaneous dipole temporary dipole that
occurs for a brief moment in time when the electrons of an atom or molecule are distributed asymmetrically
intermolecular force noncovalent attractive force
between atoms, molecules, and/or ions
interstitial sites spaces between the regular particle positions in any array of atoms or ions
ionic solid solid composed of positive and negative ions held together by strong electrostatic attractions
isomorphous possessing the same crystalline
structure
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
metallic solid solid composed of metal atoms
molecular solid solid composed of neutral molecules held together by intermolecular forces of attraction
normal boiling point temperature at which a
liquid’s vapor pressure equals 1 atm (760 torr)
octahedral hole open space in a crystal at the center of six particles located at the corners of an octahedron
phase diagram pressure-temperature graph
summarizing conditions under which the phases of a substance can exist
polarizability measure of the ability of a charge to
distort a molecule’s charge distribution (electron cloud)
simple cubic structure crystalline structure with a
cubic unit cell with lattice points only at the corners
simple cubic unit cell (also, primitive cubic unit
cell) unit cell in the simple cubic structure
space lattice all points within a crystal that have identical environments
sublimation change from solid state directly to gaseous state
supercritical fluid substance at a temperature and pressure higher than its critical point; exhibits properties intermediate between those of gaseous and liquid states
surface tension energy required to increase the
area, or length, of a liquid surface by a given amount |
tetrahedral hole tetrahedral space formed by four atoms or ions in a crystal
triple point temperature and pressure at which the vapor, liquid, and solid phases of a substance are in equilibrium
unit cell smallest portion of a space lattice that is
repeated in three dimensions to form the entire lattice
vacancy defect that occurs when a position that
should contain an atom or ion is vacant
van der Waals force attractive or repulsive force between molecules, including dipole-dipole, dipole-induced dipole, and London dispersion forces; does not include forces due to covalent or ionic bonding, or the attraction between ions and molecules
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
viscosity measure of a liquid’s resistance to flow
X-ray crystallography experimental technique for determining distances between atoms in a crystal by measuring the angles at which X-rays are diffracted when passing through the crystal |
The physical properties of condensed matter (liquids and solids) can be explained in terms of the kinetic molecular theory. In a liquid, intermolecular attractive forces hold the molecules in contact, although they still have sufficient KE to move past each other.
Intermolecular attractive forces, collectively referred to as van der Waals forces, are responsible for the behavior of liquids and solids and are electrostatic in nature. Dipole-dipole attractions result from the electrostatic attraction of the partial negative end of one polar molecule for the partial positive end of another. The temporary dipole that results from the
motion of the electrons in an atom can induce a dipole in an adjacent atom and give rise to the London dispersion force. London forces increase with increasing molecular size. Hydrogen bonds are a special type of dipole-dipole attraction that results when hydrogen is bonded to one of the three most electronegative elements: F, O, or N.
The intermolecular forces between molecules in the liquid state vary depending upon their chemical identities and result in corresponding variations in various physical properties. Cohesive forces between like molecules are responsible for a liquid’s viscosity (resistance to flow) and surface tension
(elasticity of a liquid surface). Adhesive forces between the molecules of a liquid and different molecules composing a surface in contact with the liquid are responsible for phenomena such as surface wetting and capillary rise.
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.
The temperature and pressure conditions at which a substance exists in solid, liquid, and gaseous states are summarized in a phase diagram for that substance. Phase diagrams are combined plots of three pressure-temperature equilibrium curves: solid-liquid, liquid-gas, and solid-gas. These curves represent the relationships between phase- transition temperatures and pressures. The point of intersection of all three curves represents the substance’s triple point—the temperature and pressure at which all three phases are in equilibrium. At pressures below the triple point, a substance cannot exist in the liquid state, regardless of its temperature. The terminus of the liquid-gas curve represents the substance’s critical point, the pressure and temperature above which a liquid phase cannot exist. |
of particles in a very organized structure; others form amorphous (noncrystalline) solids with an internal structure that is not ordered. The main types of crystalline solids are ionic solids, metallic solids, covalent network solids, and molecular solids. The properties of the different kinds of crystalline solids are due to the types of particles of which they consist, the arrangements of the particles, and the strengths of the attractions between them. Because their particles experience identical attractions, crystalline solids have distinct melting temperatures; the particles in amorphous solids experience a range of interactions, so they soften gradually and melt over a range of temperatures. Some crystalline solids have defects in the definite repeating pattern of their particles. These defects (which include vacancies, atoms or ions not in the regular positions, and impurities) change physical properties such as electrical conductivity, which is exploited in the silicon crystals used to manufacture computer chips.
The structures of crystalline metals and simple ionic compounds can be described in terms of packing of spheres. Metal atoms can pack in hexagonal closest- packed structures, cubic closest-packed structures, body-centered structures, and simple cubic structures. The anions in simple ionic structures commonly adopt one of these structures, and the cations occupy the spaces remaining between the anions. Small cations usually occupy tetrahedral holes in a closest-packed array of anions. Larger cations usually occupy octahedral holes. Still larger cations can occupy cubic holes in a simple cubic array of anions. The structure of a solid can be described by indicating the size and shape of a unit cell and the contents of the cell. The type of structure and dimensions of the unit cell can be determined by X-ray diffraction measurements.
In terms of their bulk properties, how do liquids and solids differ? How are they similar?
In terms of the kinetic molecular theory, in what ways are liquids similar to solids? In what ways are liquids different from solids?
In terms of the kinetic molecular theory, in what ways are liquids similar to gases? In what ways are liquids different from gases?
Explain why liquids assume the shape of any container into which they are poured, whereas solids are rigid and retain their shape.
What is the evidence that all neutral atoms and molecules exert attractive forces on each other?
Open the to answer the following questions:
Select the Solid, Liquid, Gas tab. Explore by selecting different substances, heating and cooling the systems, and changing the state. What similarities do you notice between the four substances for each phase (solid, liquid, gas)? What differences do you notice?
For each substance, select each of the states and record the given temperatures. How do the given temperatures for each state correlate with the strengths of their intermolecular attractions? Explain.
Select the Interaction Potential tab, and use the default neon atoms. Move the Ne atom on the right and observe how the potential energy changes. Select the Total Force button, and move the Ne atom as before. When is the total force on each atom attractive and large enough to matter? Then select the Component Forces button, and move the Ne atom. When do the attractive (van der Waals) and repulsive (electron overlap) forces balance? How does this relate to the potential energy versus the distance between atoms graph? Explain.
Define the following and give an example of each:
dispersion force
dipole-dipole attraction
hydrogen bond
The types of intermolecular forces in a substance are identical whether it is a solid, a liquid, or a gas. Why then does a substance change phase from a gas to a liquid or to a solid?
Why do the boiling points of the noble gases increase in the order He < Ne < Ar < Kr < Xe?
Neon and HF have approximately the same molecular masses.
Explain why the boiling points of Neon and HF differ.
Compare the change in the boiling points of Ne, Ar, Kr, and Xe with the change of the boiling points of HF, HCl, HBr, and HI, and explain the difference between the changes with increasing atomic or molecular mass.
Arrange each of the following sets of compounds in order of increasing boiling point temperature:
HCl, H2O, SiH4
F2, Cl2, Br2
CH4, C2H6, C3H8
O2, NO, N2
The molecular mass of butanol, C4H9OH, is 74.14; that of ethylene glycol, CH2(OH)CH2OH, is 62.08, yet their boiling points are 117.2 °C and 174 °C, respectively. Explain the reason for the difference.
On the basis of intermolecular attractions, explain the differences in the boiling points of n–butane (−1 °C) and chloroethane (12 °C), which have similar molar masses.
On the basis of dipole moments and/or hydrogen bonding, explain in a qualitative way the differences in the boiling points of acetone (56.2 °C) and 1-propanol (97.4 °C), which have similar molar masses.
The melting point of H2O(s) is 0 °C. Would you expect the melting point of H2S(s) to be −85 °C, 0 °C, or 185
°C? Explain your answer.
Silane (SiH4), phosphine (PH3), and hydrogen sulfide (H2S) melt at −185 °C, −133 °C, and −85 °C, respectively. What does this suggest about the polar character and intermolecular attractions of the three compounds?
Explain why a hydrogen bond between two water molecules is weaker than a hydrogen bond between two
hydrogen fluoride molecules. |
Draw a dimer of acetic acid, showing how two CH3COOH molecules are held together, and stating the type of IMF that is responsible.
Proteins are chains of amino acids that can form in a variety of arrangements, one of which is a helix.
What kind of IMF is responsible for holding the protein strand in this shape? On the protein image, show the locations of the IMFs that hold the protein together:
The density of liquid NH3 is 0.64 g/mL; the density of gaseous NH3 at STP is 0.0007 g/mL. Explain the difference between the densities of these two phases.
Identify the intermolecular forces present in the following solids:
CH3CH2OH
CH3CH2CH3
CH3CH2Cl
The test tubes shown here contain equal amounts of the specified motor oils. Identical metal spheres were dropped at the same time into each of the tubes, and a brief moment later, the spheres had fallen to the heights indicated in the illustration.
Rank the motor oils in order of increasing viscosity, and explain your reasoning: |
Explain their differences in viscosity in terms of the size and shape of their molecules and their IMFs.
Explain their differences in surface tension in terms of the size and shape of their molecules and their IMFs:
You may have heard someone use the figure of speech “slower than molasses in winter” to describe a
process that occurs slowly. Explain why this is an apt idiom, using concepts of molecular size and shape, molecular interactions, and the effect of changing temperature.
It is often recommended that you let your car engine run idle to warm up before driving, especially on cold
winter days. While the benefit of prolonged idling is dubious, it is certainly true that a warm engine is more fuel efficient than a cold one. Explain the reason for this. |
As temperature increases, what happens to the surface tension of water? Explain why this occurs, in terms of molecular interactions and the effect of changing temperature.
As temperature increases, what happens to the viscosity of water? Explain why this occurs, in terms of molecular interactions and the effect of changing temperature.
At 25 °C, how high will water rise in a glass capillary tube with an inner diameter of 0.63 mm? Refer to
for the required information.
Water rises in a glass capillary tube to a height of 17 cm. What is the diameter of the capillary tube?
Heat is added to boiling water. Explain why the temperature of the boiling water does not change. What does change?
Heat is added to ice at 0 °C. Explain why the temperature of the ice does not change. What does change?
What feature characterizes the dynamic equilibrium between a liquid and its vapor in a closed container?
Identify two common observations indicating some liquids have sufficient vapor pressures to noticeably evaporate?
Identify two common observations indicating some solids, such as dry ice and mothballs, have vapor pressures sufficient to sublime?
What is the relationship between the intermolecular forces in a liquid and its vapor pressure?
What is the relationship between the intermolecular forces in a solid and its melting temperature?
Why does spilled gasoline evaporate more rapidly on a hot day than on a cold day?
Carbon tetrachloride, CCl4, 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 CCl4 is 54.0 kPa, and its enthalpy of vaporization is 33.05 kJ/mol. Use this information to estimate the normal boiling point for CCl4.
When is the boiling point of a liquid equal to its normal boiling point?
How does the boiling of a liquid differ from its evaporation?
Use the information in to estimate the boiling point of water in Denver when the atmospheric pressure is 83.3 kPa.
A syringe at a temperature of 20 °C is filled with liquid ether in such a way that there is no space for any vapor. If the temperature is kept constant and the plunger is withdrawn to create a volume that can be occupied by vapor, what would be the approximate pressure of the vapor produced?
Explain the following observations:
It takes longer to cook an egg in Ft. Davis, Texas (altitude, 5000 feet above sea level) than it does in Boston (at sea level).
Perspiring is a mechanism for cooling the body.
The enthalpy of vaporization of water is larger than its enthalpy of fusion. Explain why.
Explain why the molar enthalpies of vaporization of the following substances increase in the order CH4 < C2H6 < C3H8, even though the type of IMF (dispersion) is the same.
Explain why the enthalpies of vaporization of the following substances increase in the order CH4 < NH3 < H2O, even though all three substances have approximately the same molar mass.
The enthalpy of vaporization of CO2(l) is 9.8 kJ/mol. Would you expect the enthalpy of vaporization of CS2(l) to be 28 kJ/mol, 9.8 kJ/mol, or −8.4 kJ/mol? Discuss the plausibility of each of these answers.
The hydrogen fluoride molecule, HF, is more polar than a water molecule, H2O (for example, has a greater dipole moment), yet the molar enthalpy of vaporization for liquid hydrogen fluoride is lesser than that for water. Explain.
Ethyl chloride (boiling point, 13 °C) is used as a local anesthetic. When the liquid is sprayed on the skin, it
cools the skin enough to freeze and numb it. Explain the cooling effect of liquid ethyl chloride.
Which contains the compounds listed correctly in order of increasing boiling points?
N2 < CS2 < H2O < KCl
H2O < N2 < CS2 < KCl
N2 < KCl < CS2 < H2O
CS2 < N2 < KCl < H2O
KCl < H2O < CS2 < N2
How much heat is required to convert 422 g of liquid H2O at 23.5 °C into steam at 150 °C?
Evaporation of sweat requires energy and thus take excess heat away from the body. Some of the water that you drink may eventually be converted into sweat and evaporate. If you drink a 20-ounce bottle of water that had been in the refrigerator at 3.8 °C, how much heat is needed to convert all of that water into sweat and then to vapor? (Note: Your body temperature is 36.6 °C. For the purpose of solving this problem, assume that the thermal properties of sweat are the same as for water.)
Titanium tetrachloride, TiCl4, has a melting point of −23.2 °C and has a ΔH fusion = 9.37 kJ/mol.
How much energy is required to melt 263.1 g TiCl4?
For TiCl4, which will likely have the larger magnitude: ΔH fusion or ΔH vaporization? Explain your reasoning.
From the phase diagram for water (), determine the state of water at:
35 °C and 85 kPa
−15 °C and 40 kPa
−15 °C and 0.1 kPa
75 °C and 3 kPa
40 °C and 0.1 kPa
60 °C and 50 kPa
What phase changes will take place when water is subjected to varying pressure at a constant temperature of 0.005 °C? At 40 °C? At −40 °C?
Pressure cookers allow food to cook faster because the higher pressure inside the pressure cooker increases the boiling temperature of water. A particular pressure cooker has a safety valve that is set to vent steam if the pressure exceeds 3.4 atm. What is the approximate maximum temperature that can be reached inside this pressure cooker? Explain your reasoning.
From the phase diagram for carbon dioxide in , determine the state of CO2 at:
20 °C and 1000 kPa
10 °C and 2000 kPa
10 °C and 100 kPa
−40 °C and 500 kPa
−80 °C and 1500 kPa
−80 °C and 10 kPa
Determine the phase changes that carbon dioxide undergoes as pressure is increased at a constant temperature of (a) −50 °C and (b) 50 °C. If the temperature is held at −40 °C? At 20 °C? (See the phase diagram in .)
Consider a cylinder containing a mixture of liquid carbon dioxide in equilibrium with gaseous carbon
dioxide at an initial pressure of 65 atm and a temperature of 20 °C. Sketch a plot depicting the change in the cylinder pressure with time as gaseous carbon dioxide is released at constant temperature.
Dry ice, CO2(s), does not melt at atmospheric pressure. It sublimes at a temperature of −78 °C. What is the lowest pressure at which CO2(s) will melt to give CO2(l)? At approximately what temperature will this occur? (See for the phase diagram.)
If a severe storm results in the loss of electricity, it may be necessary to use a clothesline to dry laundry. In many parts of the country in the dead of winter, the clothes will quickly freeze when they are hung on the line. If it does not snow, will they dry anyway? Explain your answer.
Is it possible to liquefy nitrogen at room temperature (about 25 °C)? Is it possible to liquefy sulfur dioxide
at room temperature? Explain your answers.
Elemental carbon has one gas phase, one liquid phase, and two different solid phases, as shown in the phase diagram:
On the phase diagram, label the gas and liquid regions.
Graphite is the most stable phase of carbon at normal conditions. On the phase diagram, label the graphite phase.
If graphite at normal conditions is heated to 2500 K while the pressure is increased to 1010 Pa, it is
converted into diamond. Label the diamond phase.
Circle each triple point on the phase diagram.
In what phase does carbon exist at 5000 K and 108 Pa?
If the temperature of a sample of carbon increases from 3000 K to 5000 K at a constant pressure of 106 Pa, which phase transition occurs, if any?
What types of liquids typically form amorphous solids?
At very low temperatures oxygen, O2, freezes and forms a crystalline solid. Which best describes these crystals?
ionic
covalent network
metallic
amorphous
molecular crystals
As it cools, olive oil slowly solidifies and forms a solid over a range of temperatures. Which best describes the solid?
ionic
covalent network
metallic
amorphous
molecular crystals
Explain why ice, which is a crystalline solid, has a melting temperature of 0 °C, whereas butter, which is an amorphous solid, softens over a range of temperatures.
Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances:
SiO2
KCl
Cu
CO2
C (diamond)
BaSO4
NH3
NH4F
C2H5OH
Identify the type of crystalline solid (metallic, network covalent, ionic, or molecular) formed by each of the following substances:
CaCl2
SiC
N2
Fe
C (graphite)
CH3CH2CH2CH3
HCl
NH4NO3
K3PO4
Classify each substance in the table as either a metallic, ionic, molecular, or covalent network solid: |
Identify the following substances as ionic, metallic, covalent network, or molecular solids:
Substance A is malleable, ductile, conducts electricity well, and has a melting point of 1135 °C. Substance B is brittle, does not conduct electricity as a solid but does when molten, and has a melting point of 2072
°C. Substance C is very hard, does not conduct electricity, and has a melting point of 3440 °C. Substance D is soft, does not conduct electricity, and has a melting point of 185 °C.
Substance A is shiny, conducts electricity well, and melts at 975 °C. Substance A is likely a(n):
ionic solid
metallic solid
molecular solid
covalent network solid
Substance B is hard, does not conduct electricity, and melts at 1200 °C. Substance B is likely a(n):
ionic solid
metallic solid
molecular solid
covalent network solid
Describe the crystal structure of iron, which crystallizes with two equivalent metal atoms in a cubic unit cell.
Describe the crystal structure of Pt, which crystallizes with four equivalent metal atoms in a cubic unit cell.
What is the coordination number of a chromium atom in the body-centered cubic structure of chromium?
What is the coordination number of an aluminum atom in the face-centered cubic structure of aluminum?
Cobalt metal crystallizes in a hexagonal closest packed structure. What is the coordination number of a cobalt atom?
Nickel metal crystallizes in a cubic closest packed structure. What is the coordination number of a nickel atom?
Tungsten crystallizes in a body-centered cubic unit cell with an edge length of 3.165 Å.
What is the atomic radius of tungsten in this structure?
Calculate the density of tungsten.
Platinum (atomic radius = 1.38 Å) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of platinum.
Barium crystallizes in a body-centered cubic unit cell with an edge length of 5.025 Å
What is the atomic radius of barium in this structure?
Calculate the density of barium.
Aluminum (atomic radius = 1.43 Å) crystallizes in a cubic closely packed structure. Calculate the edge length of the face-centered cubic unit cell and the density of aluminum.
The density of aluminum is 2.7 g/cm3; that of silicon is 2.3 g/cm3. Explain why Si has the lower density
even though it has heavier atoms.
The free space in a metal may be found by subtracting the volume of the atoms in a unit cell from the volume of the cell. Calculate the percentage of free space in each of the three cubic lattices if all atoms in each are of equal size and touch their nearest neighbors. Which of these structures represents the most efficient packing? That is, which packs with the least amount of unused space?
Cadmium sulfide, sometimes used as a yellow pigment by artists, crystallizes with cadmium, occupying
one-half of the tetrahedral holes in a closest packed array of sulfide ions. What is the formula of cadmium sulfide? Explain your answer.
A compound of cadmium, tin, and phosphorus is used in the fabrication of some semiconductors. It
crystallizes with cadmium occupying one-fourth of the tetrahedral holes and tin occupying one-fourth of the tetrahedral holes in a closest packed array of phosphide ions. What is the formula of the compound? Explain your answer.
What is the formula of the magnetic oxide of cobalt, used in recording tapes, that crystallizes with cobalt
atoms occupying one-eighth of the tetrahedral holes and one-half of the octahedral holes in a closely packed array of oxide ions?
A compound containing zinc, aluminum, and sulfur crystallizes with a closest-packed array of sulfide
ions. Zinc ions are found in one-eighth of the tetrahedral holes and aluminum ions in one-half of the octahedral holes. What is the empirical formula of the compound?
A compound of thallium and iodine crystallizes in a simple cubic array of iodide ions with thallium ions in
all of the cubic holes. What is the formula of this iodide? Explain your answer.
Which of the following elements reacts with sulfur to form a solid in which the sulfur atoms form a closest- packed array with all of the octahedral holes occupied: Li, Na, Be, Ca, or Al?
What is the percent by mass of titanium in rutile, a mineral that contains titanium and oxygen, if structure can be described as a closest packed array of oxide ions with titanium ions in one-half of the octahedral holes? What is the oxidation number of titanium?
Explain why the chemically similar alkali metal chlorides NaCl and CsCl have different structures,
whereas the chemically different NaCl and MnS have the same structure.
As minerals were formed from the molten magma, different ions occupied the same cites in the crystals. Lithium often occurs along with magnesium in minerals despite the difference in the charge on their ions. Suggest an explanation.
Rubidium iodide crystallizes with a cubic unit cell that contains iodide ions at the corners and a rubidium
ion in the center. What is the formula of the compound?
One of the various manganese oxides crystallizes with a cubic unit cell that contains manganese ions at the corners and in the center. Oxide ions are located at the center of each edge of the unit cell. What is the formula of the compound?
NaH crystallizes with the same crystal structure as NaCl. The edge length of the cubic unit cell of NaH is
4.880 Å.
Calculate the ionic radius of H−. (The ionic radius of Li+ is 0.0.95 Å.)
Calculate the density of NaH.
Thallium(I) iodide crystallizes with the same structure as CsCl. The edge length of the unit cell of TlI is
4.20 Å. Calculate the ionic radius of TI+. (The ionic radius of I− is 2.16 Å.)
A cubic unit cell contains manganese ions at the corners and fluoride ions at the center of each edge.
What is the empirical formula of this compound? Explain your answer.
What is the coordination number of the Mn3+ ion?
Calculate the edge length of the unit cell if the radius of a Mn3+ ion is 0.65 A.
Calculate the density of the compound.
What is the spacing between crystal planes that diffract X-rays with a wavelength of 1.541 nm at an angle
θ of 15.55° (first order reflection)?
A diffractometer using X-rays with a wavelength of 0.2287 nm produced first order diffraction peak for a crystal angle θ = 16.21°. Determine the spacing between the diffracting planes in this crystal.
A metal with spacing between planes equal to 0.4164 nm diffracts X-rays with a wavelength of 0.2879 nm. What is the diffraction angle for the first order diffraction peak?
Gold crystallizes in a face-centered cubic unit cell. The second-order reflection (n = 2) of X-rays for the planes that make up the tops and bottoms of the unit cells is at θ = 22.20°. The wavelength of the X-rays is 1.54 Å. What is the density of metallic gold?
When an electron in an excited molybdenum atom falls from the L to the K shell, an X-ray is emitted.
These X-rays are diffracted at an angle of 7.75° by planes with a separation of 2.64 Å. What is the difference in energy between the K shell and the L shell in molybdenum assuming a first order diffraction? |
INTRODUCTION Coral reefs are home to about 25% of all marine species. They are being threatened by climate change, oceanic acidification, and water pollution, all of which change the composition of the solution known as seawater. Dissolved oxygen in seawater is critical for sea creatures, but as the oceans warm, oxygen becomes less soluble. As the concentration of carbon dioxide in the atmosphere increases, the concentration of carbon dioxide in the oceans increases, contributing to oceanic acidification. Coral reefs are particularly sensitive to the acidification of the ocean, since the exoskeletons of the coral polyps are soluble in acidic solutions. Humans contribute to the changing of seawater composition by allowing agricultural runoff and other forms of pollution to affect our oceans.
Solutions are crucial to the processes that sustain life and to many other processes involving chemical reactions. This chapter considers the nature of solutions and examines factors that determine whether a solution will form and what properties it may have. The properties of colloids—mixtures containing dispersed particles larger than the molecules and ions of typical solutions—are also discussed.
The Dissolution Process
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Describe the basic properties of solutions and how they form
Predict whether a given mixture will yield a solution based on molecular properties of its components
Explain why some solutions either produce or absorb heat when they form
An earlier chapter of this text introduced solutions, defined as homogeneous mixtures of two or more substances. Often, one component of a solution is present at a significantly greater concentration, in which case it is called the solvent. The other components of the solution present in relatively lesser concentrations are called solutes. Sugar is a covalent solid composed of sucrose molecules, C12H22O11. When this compound dissolves in water, its molecules become uniformly distributed among the molecules of water:
The subscript “aq” in the equation signifies that the sucrose molecules are solutes and are therefore individually dispersed throughout the aqueous solution (water is the solvent). Although sucrose molecules are heavier than water molecules, they remain dispersed throughout the solution; gravity does not cause them to “settle out” over time.
Potassium dichromate, K2Cr2O7, is an ionic compound composed of colorless potassium ions, K+, and orange dichromate ions, When a small amount of solid potassium dichromate is added to water, the compound dissolves and dissociates to yield potassium ions and dichromate ions uniformly distributed throughout the mixture (), as indicated in this equation: |
FIGURE 11.2 When potassium dichromate (K2Cr2O7) is mixed with water, it forms a homogeneous orange solution. (credit: modification of work by Mark Ott)
LINK TO LEARNING
Visit this to view simulations of the dissolution of common covalent and ionic substances (sugar and salt) in water.
Water is used so often as a solvent that the word solution has come to imply an aqueous solution to many people. However, almost any gas, liquid, or solid can act as a solvent. Many alloys are solid solutions of one metal dissolved in another; for example, US five-cent coins contain nickel dissolved in copper. Air is a gaseous solution, a homogeneous mixture of nitrogen, oxygen, and several other gases. Oxygen (a gas), alcohol (a liquid), and sugar (a solid) all dissolve in water (a liquid) to form liquid solutions. gives examples of several different solutions and the phases of the solutes and solvents. |
Solutions exhibit these defining traits:
They are homogeneous; after a solution is mixed, it has the same composition at all points throughout (its composition is uniform).
The physical state of a solution—solid, liquid, or gas—is typically the same as that of the solvent, as demonstrated by the examples in .
The components of a solution are dispersed on a molecular scale; they consist of a mixture of separated solute particles (molecules, atoms, and/or ions) each closely surrounded by solvent species.
The dissolved solute in a solution will not settle out or separate from the solvent.
The composition of a solution, or the concentrations of its components, can be varied continuously (within limits determined by the solubility of the components, discussed in detail later in this chapter).
The Formation of Solutions
The formation of a solution is an example of a spontaneous process, a process that occurs under specified conditions without the requirement of energy from some external source. Sometimes a mixture is stirred to speed up the dissolution process, but this is not necessary; a homogeneous solution will form eventually. The topic of spontaneity is critically important to the study of chemical thermodynamics and is treated more thoroughly in a later chapter of this text. For purposes of this chapter’s discussion, it will suffice to consider two criteria that favor, but do not guarantee, the spontaneous formation of a solution:
a decrease in the internal energy of the system (an exothermic change, as discussed in the previous chapter on thermochemistry)
an increased dispersal of matter in the system (which indicates an increase in the entropy of the system,
as you will learn about in the later chapter on thermodynamics)
In the process of dissolution, an internal energy change often, but not always, occurs as heat is absorbed or evolved. An increase in matter dispersal always results when a solution forms from the uniform distribution of solute molecules throughout a solvent.
When the strengths of the intermolecular forces of attraction between solute and solvent species in a solution are no different than those present in the separated components, the solution is formed with no accompanying energy change. Such a solution is called an ideal solution. A mixture of ideal gases (or gases such as helium and argon, which closely approach ideal behavior) is an example of an ideal solution, since the entities comprising these gases experience no significant intermolecular attractions. |
When containers of helium and argon are connected, the gases spontaneously mix due to diffusion and form a solution (). The formation of this solution clearly involves an increase in matter dispersal, since the helium and argon atoms occupy a volume twice as large as that which each occupied before mixing.
FIGURE 11.3 Samples of helium and argon spontaneously mix to give a solution.
Ideal solutions may also form when structurally similar liquids are mixed. For example, mixtures of the alcohols methanol (CH3OH) and ethanol (C2H5OH) form ideal solutions, as do mixtures of the hydrocarbons pentane, C5H12, and hexane, C6H14. Placing methanol and ethanol, or pentane and hexane, in the bulbs shown in will result in the same diffusion and subsequent mixing of these liquids as is observed for the He and Ar gases (although at a much slower rate), yielding solutions with no significant change in energy.
Unlike a mixture of gases, however, the components of these liquid-liquid solutions do, indeed, experience intermolecular attractive forces. But since the molecules of the two substances being mixed are structurally very similar, the intermolecular attractive forces between like and unlike molecules are essentially the same, and the dissolution process, therefore, does not entail any appreciable increase or decrease in energy. These examples illustrate how increased matter dispersal alone can provide the driving force required to cause the spontaneous formation of a solution. In some cases, however, the relative magnitudes of intermolecular forces of attraction between solute and solvent species may prevent dissolution.
Three types of intermolecular attractive forces are relevant to the dissolution process: solute-solute, solvent- solvent, and solute-solvent. As illustrated in , the formation of a solution may be viewed as a stepwise process in which energy is consumed to overcome solute-solute and solvent-solvent attractions (endothermic processes) and released when solute-solvent attractions are established (an exothermic process referred to as solvation). The relative magnitudes of the energy changes associated with these stepwise processes determine whether the dissolution process overall will release or absorb energy. In some cases, solutions do not form because the energy required to separate solute and solvent species is so much greater than the energy released by solvation.
FIGURE 11.4 This schematic representation of dissolution shows a stepwise process involving the endothermic separation of solute and solvent species (Steps 1 and 2) and exothermic solvation (Step 3).
Consider the example of an ionic compound dissolving in water. Formation of the solution requires the electrostatic forces between the cations and anions of the compound (solute–solute) be overcome completely as attractive forces are established between these ions and water molecules (solute–solvent). Hydrogen bonding between a relatively small fraction of the water molecules must also be overcome to accommodate any dissolved solute. If the solute’s electrostatic forces are significantly greater than the solvation forces, the dissolution process is significantly endothermic and the compound may not dissolve to an appreciable extent. Calcium carbonate, the major component of coral reefs, is one example of such an “insoluble” ionic compound (see ). On the other hand, if the solvation forces are much stronger than the compound’s electrostatic forces, the dissolution is significantly exothermic and the compound may be highly soluble. A common example of this type of ionic compound is sodium chloride, commonly known as table salt.
As noted at the beginning of this module, spontaneous solution formation is favored, but not guaranteed, by exothermic dissolution processes. While many soluble compounds do, indeed, dissolve with the release of heat, some dissolve endothermically. Ammonium nitrate (NH4NO3) is one such example and is used to make instant cold packs, like the one pictured in , which are used for treating injuries. A thin-walled plastic bag of water is sealed inside a larger bag with solid NH4NO3. When the smaller bag is broken, a solution of NH4NO3 forms, absorbing heat from the surroundings (the injured area to which the pack is applied) and providing a cold compress that decreases swelling. Endothermic dissolutions such as this one require a greater energy input to separate the solute species than is recovered when the solutes are solvated, but they are spontaneous nonetheless due to the increase in disorder that accompanies formation of the solution.
FIGURE 11.5 An instant cold pack gets cold when certain salts, such as ammonium nitrate, dissolve in water—an endothermic process.
LINK TO LEARNING
Watch this brief illustrating endothermic and exothermic dissolution processes.
Electrolytes
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Define and give examples of electrolytes
Distinguish between the physical and chemical changes that accompany dissolution of ionic and covalent electrolytes
Relate electrolyte strength to solute-solvent attractive forces
When some substances are dissolved in water, they undergo either a physical or a chemical change that yields ions in solution. These substances constitute an important class of compounds called electrolytes. Substances that do not yield ions when dissolved are called nonelectrolytes. If the physical or chemical process that generates the ions is essentially 100% efficient (all of the dissolved compound yields ions), then the substance is known as a strong electrolyte. If only a relatively small fraction of the dissolved substance undergoes the ion-producing process, it is called a weak electrolyte.
Substances may be identified as strong, weak, or nonelectrolytes by measuring the electrical conductance of an aqueous solution containing the substance. To conduct electricity, a substance must contain freely mobile, charged species. Most familiar is the conduction of electricity through metallic wires, in which case the mobile, charged entities are electrons. Solutions may also conduct electricity if they contain dissolved ions, with conductivity increasing as ion concentration increases. Applying a voltage to electrodes immersed in a solution permits assessment of the relative concentration of dissolved ions, either quantitatively, by measuring the electrical current flow, or qualitatively, by observing the brightness of a light bulb included in the circuit |
FIGURE 11.6 Solutions of nonelectrolytes such as ethanol do not contain dissolved ions and cannot conduct electricity. Solutions of electrolytes contain ions that permit the passage of electricity. The conductivity of an electrolyte solution is related to the strength of the electrolyte.
Ionic Electrolytes
Water and other polar molecules are attracted to ions, as shown in . The electrostatic attraction between an ion and a molecule with a dipole is called an ion-dipole attraction. These attractions play an important role in the dissolution of ionic compounds in water.
FIGURE 11.7 As potassium chloride (KCl) dissolves in water, the ions are hydrated. The polar water molecules are attracted by the charges on the K+ and Cl− ions. Water molecules in front of and behind the ions are not shown.
When ionic compounds dissolve in water, the ions in the solid separate and disperse uniformly throughout the solution because water molecules surround and solvate the ions, reducing the strong electrostatic forces between them. This process represents a physical change known as dissociation. Under most conditions, ionic compounds will dissociate nearly completely when dissolved, and so they are classified as strong electrolytes. Even sparingly, soluble ionic compounds are strong electrolytes, since the small amount that does dissolve will dissociate completely.
Consider what happens at the microscopic level when solid KCl is added to water. Ion-dipole forces attract the positive (hydrogen) end of the polar water molecules to the negative chloride ions at the surface of the solid, and they attract the negative (oxygen) ends to the positive potassium ions. The water molecules surround individual K+ and Cl− ions, reducing the strong interionic forces that bind the ions together and letting them move off into solution as solvated ions, as shows. Overcoming the electrostatic attraction permits the independent motion of each hydrated ion in a dilute solution as the ions transition from fixed positions in the undissolved compound to widely dispersed, solvated ions in solution.
Covalent Electrolytes
Pure water is an extremely poor conductor of electricity because it is only very slightly ionized—only about two out of every 1 billion molecules ionize at 25 °C. Water ionizes when one molecule of water gives up a proton (H+ ion) to another molecule of water, yielding hydronium and hydroxide ions.
In some cases, solutions prepared from covalent compounds conduct electricity because the solute molecules react chemically with the solvent to produce ions. For example, pure hydrogen chloride is a gas consisting of covalent HCl molecules. This gas contains no ions. However, an aqueous solution of HCl is a very good conductor, indicating that an appreciable concentration of ions exists within the solution.
Because HCl is an acid, its molecules react with water, transferring H+ ions to form hydronium ions (H3O+) and |
This reaction is essentially 100% complete for HCl (i.e., it is a strong acid and, consequently, a strong electrolyte). Likewise, weak acids and bases that only react partially generate relatively low concentrations of ions when dissolved in water and are classified as weak electrolytes. The reader may wish to review the discussion of strong and weak acids provided in the earlier chapter of this text on reaction classes and stoichiometry.
Solubility
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Describe the effects of temperature and pressure on solubility
State Henry’s law and use it in calculations involving the solubility of a gas in a liquid
Explain the degrees of solubility possible for liquid-liquid solutions
Imagine adding a small amount of sugar to a glass of water, stirring until all the sugar has dissolved, and then adding a bit more. You can repeat this process until the sugar concentration of the solution reaches its natural limit, a limit determined primarily by the relative strengths of the solute-solute, solute-solvent, and solvent- solvent attractive forces discussed in the previous two modules of this chapter. You can be certain that you have reached this limit because, no matter how long you stir the solution, undissolved sugar remains. The concentration of sugar in the solution at this point is known as its solubility.
The solubility of a solute in a particular solvent is the maximum concentration that may be achieved under given conditions when the dissolution process is at equilibrium.
When a solute’s concentration is equal to its solubility, the solution is said to be saturated with that solute. If the solute’s concentration is less than its solubility, the solution is said to be unsaturated. A solution that contains a relatively low concentration of solute is called dilute, and one with a relatively high concentration is called concentrated.
LINK TO LEARNING
Use this to prepare various saturated solutions.
Solutions may be prepared in which a solute concentration exceeds its solubility. Such solutions are said to be supersaturated, and they are interesting examples of nonequilibrium states (a detailed treatment of this important concept is provided in the text chapters on equilibrium). For example, the carbonated beverage in an open container that has not yet “gone flat” is supersaturated with carbon dioxide gas; given time, the CO2 concentration will decrease until it reaches its solubility.
LINK TO LEARNING
Watch this showing the precipitation of sodium acetate from a supersaturated solution.
Solutions of Gases in Liquids
As for any solution, the solubility of a gas in a liquid is affected by the intermolecular attractive forces between solute and solvent species. Unlike solid and liquid solutes, however, there is no solute-solute intermolecular attraction to overcome when a gaseous solute dissolves in a liquid solvent (see ) since the atoms or molecules comprising a gas are far separated and experience negligible interactions. Consequently, solute- solvent interactions are the sole energetic factor affecting solubility. For example, the water solubility of
oxygen is approximately three times greater than that of helium (there are greater dispersion forces between water and the larger oxygen molecules) but 100 times less than the solubility of chloromethane, CHCl3 (polar chloromethane molecules experience dipole–dipole attraction to polar water molecules). Likewise note the solubility of oxygen in hexane, C6H14, is approximately 20 times greater than it is in water because greater dispersion forces exist between oxygen and the larger hexane molecules.
Temperature is another factor affecting solubility, with gas solubility typically decreasing as temperature increases (). This inverse relation between temperature and dissolved gas concentration is responsible for one of the major impacts of thermal pollution in natural waters.
FIGURE 11.8 The solubilities of these gases in water decrease as the temperature increases. All solubilities were measured with a constant pressure of 101.3 kPa (1 atm) of gas above the solutions.
When the temperature of a river, lake, or stream is raised, the solubility of oxygen in the water is decreased. Decreased levels of dissolved oxygen may have serious consequences for the health of the water’s ecosystems and, in severe cases, can result in large-scale fish kills ().
FIGURE 11.9 (a) The small bubbles of air in this glass of chilled water formed when the water warmed to room temperature and the solubility of its dissolved air decreased. (b) The decreased solubility of oxygen in natural waters subjected to thermal pollution can result in large-scale fish kills. (credit a: modification of work by Liz West; credit b: modification of work by U.S. Fish and Wildlife Service)
The solubility of a gaseous solute is also affected by the partial pressure of solute in the gas to which the
solution is exposed. Gas solubility increases as the pressure of the gas increases. Carbonated beverages provide a nice illustration of this relationship. The carbonation process involves exposing the beverage to a relatively high pressure of carbon dioxide gas and then sealing the beverage container, thus saturating the beverage with CO2 at this pressure. When the beverage container is opened, a familiar hiss is heard as the carbon dioxide gas pressure is released, and some of the dissolved carbon dioxide is typically seen leaving solution in the form of small bubbles (). At this point, the beverage is supersaturated with carbon dioxide and, with time, the dissolved carbon dioxide concentration will decrease to its equilibrium value and the beverage will become “flat.”
FIGURE 11.10 Opening the bottle of carbonated beverage reduces the pressure of the gaseous carbon dioxide above the beverage. The solubility of CO2 is thus lowered, and some dissolved carbon dioxide may be seen leaving the solution as small gas bubbles. (credit: modification of work by Derrick Coetzee)
For many gaseous solutes, the relation between solubility, Cg, and partial pressure, Pg, is a proportional one:
where k is a proportionality constant that depends on the identities of the gaseous solute and solvent, and on the solution temperature. This is a mathematical statement of Henry’s law: The quantity of an ideal gas that dissolves in a definite volume of liquid is directly proportional to the pressure of the gas.
Application of Henry’s Law
At 20 °C, the concentration of dissolved oxygen in water exposed to gaseous oxygen at a partial pressure of
101.3 kPa is 1.38 10−3 mol L−1. Use Henry’s law to determine the solubility of oxygen when its partial pressure is 20.7 kPa, the approximate pressure of oxygen in earth’s atmosphere.
Solution
According to Henry’s law, for an ideal solution the solubility, Cg, of a gas (1.38 10−3 mol L−1, in this case) is directly proportional to the pressure, Pg, of the undissolved gas above the solution (101.3 kPa in this case). Because both Cg and Pg are known, this relation can be rearragned and used to solve for k. |
Note that various units may be used to express the quantities involved in these sorts of computations. Any combination of units that yield to the constraints of dimensional analysis are acceptable.
Check Your Learning
Exposing a 100.0 mL sample of water at 0 °C to an atmosphere containing a gaseous solute at 152 torr resulted in the dissolution of 1.45 10−3 g of the solute. Use Henry’s law to determine the solubility of this gaseous solute when its pressure is 760 torr.
Answer:
7.25 10−3 in 100.0 mL or 0.0725 g/L
Thermal Pollution and Oxygen Solubility
A certain species of freshwater trout requires a dissolved oxygen concentration of 7.5 mg/L. Could these fish thrive in a thermally polluted mountain stream (water temperature is 30.0 °C, partial pressure of atmospheric oxygen is 0.17 atm)? Use the data in to estimate a value for the Henry's law constant at this temperature.
Solution
First, estimate the Henry’s law constant for oxygen in water at the specified temperature of 30.0 °C ( indicates the solubility at this temperature is approximately ~1.2 mol/L). |
This concentration is lesser than the required minimum value of 7.5 mg/L, and so these trout would likely not thrive in the polluted stream.
Check Your Learning
What dissolved oxygen concentration is expected for the stream above when it returns to a normal summer time temperature of 15 °C?
Answer:
8.2 mg/L
Chemistry in Everyday Life
Decompression Sickness or “The Bends”
Decompression sickness (DCS), or “the bends,” is an effect of the increased pressure of the air inhaled by scuba divers when swimming underwater at considerable depths. In addition to the pressure exerted by the atmosphere, divers are subjected to additional pressure due to the water above them, experiencing an increase of approximately 1 atm for each 10 m of depth. Therefore, the air inhaled by a diver while submerged contains gases at the corresponding higher ambient pressure, and the concentrations of the gases dissolved in the diver’s blood are proportionally higher per Henry’s law.
As the diver ascends to the surface of the water, the ambient pressure decreases and the dissolved gases becomes less soluble. If the ascent is too rapid, the gases escaping from the diver’s blood may form bubbles that can cause a variety of symptoms ranging from rashes and joint pain to paralysis and death. To avoid DCS, divers must ascend from depths at relatively slow speeds (10 or 20 m/min) or otherwise make several decompression stops, pausing for several minutes at given depths during the ascent. When these preventive measures are unsuccessful, divers with DCS are often provided hyperbaric oxygen therapy in pressurized vessels called decompression (or recompression) chambers (). Researchers are also investigating related body reactions and defenses in order to develop better testing and treatment for decompression sicknetss. For example, Ingrid Eftedal, a barophysiologist specializing in bodily reactions to diving, has shown that white blood cells undergo chemical and genetic changes as a result of the condition; these can potentially be used to create biomarker tests and other methods to manage decompression sickness. |
Deviations from Henry’s law are observed when a chemical reaction takes place between the gaseous solute and the solvent. Thus, for example, the solubility of ammonia in water increases more rapidly with increasing pressure than predicted by the law because ammonia, being a base, reacts to some extent with water to form ammonium ions and hydroxide ions.
Gases can form supersaturated solutions. If a solution of a gas in a liquid is prepared either at low temperature or under pressure (or both), then as the solution warms or as the gas pressure is reduced, the solution may become supersaturated. In 1986, more than 1700 people in Cameroon were killed when a cloud of gas, almost certainly carbon dioxide, bubbled from Lake Nyos (), a deep lake in a volcanic crater. The water at the bottom of Lake Nyos is saturated with carbon dioxide by volcanic activity beneath the lake. It is believed
that the lake underwent a turnover due to gradual heating from below the lake, and the warmer, less-dense water saturated with carbon dioxide reached the surface. Consequently, tremendous quantities of dissolved CO2 were released, and the colorless gas, which is denser than air, flowed down the valley below the lake and suffocated humans and animals living in the valley.
FIGURE 11.12 (a) It is believed that the 1986 disaster that killed more than 1700 people near Lake Nyos in Cameroon resulted when a large volume of carbon dioxide gas was released from the lake. (b) A CO2 vent has since been installed to help outgas the lake in a slow, controlled fashion and prevent a similar catastrophe from happening in the future. (credit a: modification of work by Jack Lockwood; credit b: modification of work by Bill Evans)
Solutions of Liquids in Liquids
Some liquids may be mixed in any proportions to yield solutions; in other words, they have infinite mutual solubility and are said to be miscible. Ethanol, sulfuric acid, and ethylene glycol (popular for use as antifreeze, pictured in ) are examples of liquids that are completely miscible with water. Two-cycle motor oil is miscible with gasoline, mixtures of which are used as lubricating fuels for various types of outdoor power equipment (chainsaws, leaf blowers, and so on).
FIGURE 11.13 Water and antifreeze are miscible; mixtures of the two are homogeneous in all proportions. (credit: “dno1967”/Wikimedia commons)
Miscible liquids are typically those with very similar polarities. Consider, for example, liquids that are polar or capable of hydrogen bonding. For such liquids, the dipole-dipole attractions (or hydrogen bonding) of the solute molecules with the solvent molecules are at least as strong as those between molecules in the pure solute or in the pure solvent. Hence, the two kinds of molecules mix easily. Likewise, nonpolar liquids are miscible with each other because there is no appreciable difference in the strengths of solute-solute, solvent- solvent, and solute-solvent intermolecular attractions. The solubility of polar molecules in polar solvents and of nonpolar molecules in nonpolar solvents is, again, an illustration of the chemical axiom “like dissolves like.”
Two liquids that do not mix to an appreciable extent are called immiscible. Separate layers are formed when immiscible liquids are poured into the same container. Gasoline, oil (), benzene, carbon tetrachloride, some paints, and many other nonpolar liquids are immiscible with water. Relatively weak attractive forces between the polar water molecules and the nonpolar liquid molecules are not adequate to overcome much stronger hydrogen bonding between water molecules. The distinction between immiscibility and miscibility is really one of extent, so that miscible liquids are of infinite mutual solubility, while liquids said to be immiscible are of very low (though not zero) mutual solubility.
FIGURE 11.14 Water and oil are immiscible. Mixtures of these two substances will form two separate layers with the less dense oil floating on top of the water. (credit: “Yortw”/Flickr)
Two liquids, such as bromine and water, that are of moderate mutual solubility are said to be partially miscible. Two partially miscible liquids usually form two layers when mixed. In the case of the bromine and water mixture, the upper layer is water, saturated with bromine, and the lower layer is bromine saturated with water. Since bromine is nonpolar, and, thus, not very soluble in water, the water layer is only slightly discolored by the bright orange bromine dissolved in it. Since the solubility of water in bromine is very low, there is no noticeable effect on the dark color of the bromine layer ().
FIGURE 11.15 Bromine (the deep orange liquid on the left) and water (the clear liquid in the middle) are partially miscible. The top layer in the mixture on the right is a saturated solution of bromine in water; the bottom layer is a saturated solution of water in bromine. (credit: Paul Flowers)
Solutions of Solids in Liquids
The dependence of solubility on temperature for a number of solids in water is shown by the solubility curves in . Reviewing these data indicates a general trend of increasing solubility with temperature, although there are exceptions, as illustrated by the ionic compound cerium sulfate. |
The temperature dependence of solubility can be exploited to prepare supersaturated solutions of certain compounds. A solution may be saturated with the compound at an elevated temperature (where the solute is more soluble) and subsequently cooled to a lower temperature without precipitating the solute. The resultant solution contains solute at a concentration greater than its equilibrium solubility at the lower temperature (i.e., it is supersaturated) and is relatively stable. Precipitation of the excess solute can be initiated by adding a seed crystal (see the video in the Link to Learning earlier in this module) or by mechanically agitating the solution. Some hand warmers, such as the one pictured in , take advantage of this behavior.
FIGURE 11.17 This hand warmer produces heat when the sodium acetate in a supersaturated solution precipitates. Precipitation of the solute is initiated by a mechanical shockwave generated when the flexible metal disk within the solution is “clicked.” (credit: modification of work by “Velela”/Wikimedia Commons)
LINK TO LEARNING
This shows the crystallization process occurring in a hand warmer.
Colligative Properties
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Express concentrations of solution components using mole fraction and molality
Describe the effect of solute concentration on various solution properties (vapor pressure, boiling point, freezing point, and osmotic pressure)
Perform calculations using the mathematical equations that describe these various colligative effects
Describe the process of distillation and its practical applications
Explain the process of osmosis and describe how it is applied industrially and in nature
The properties of a solution are different from those of either the pure solute(s) or solvent. Many solution properties are dependent upon the chemical identity of the solute. Compared to pure water, a solution of hydrogen chloride is more acidic, a solution of ammonia is more basic, a solution of sodium chloride is more dense, and a solution of sucrose is more viscous. There are a few solution properties, however, that depend only upon the total concentration of solute species, regardless of their identities. These colligative properties include vapor pressure lowering, boiling point elevation, freezing point depression, and osmotic pressure.
This small set of properties is of central importance to many natural phenomena and technological applications, as will be described in this module.
Mole Fraction and Molality
Several units commonly used to express the concentrations of solution components were introduced in an earlier chapter of this text, each providing certain benefits for use in different applications. For example, molarity (M) is a convenient unit for use in stoichiometric calculations, since it is defined in terms of the molar amounts of solute species:
Because solution volumes vary with temperature, molar concentrations will likewise vary. When expressed as molarity, the concentration of a solution with identical numbers of solute and solvent species will be different
at different temperatures, due to the contraction/expansion of the solution. More appropriate for calculations involving many colligative properties are mole-based concentration units whose values are not dependent on temperature. Two such units are mole fraction (introduced in the previous chapter on gases) and molality.
The mole fraction, X, of a component is the ratio of its molar amount to the total number of moles of all solution components:
By this definition, the sum of mole fractions for all solution components (the solvent and all solutes) is equal to one.
Molality is a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms:
Since these units are computed using only masses and molar amounts, they do not vary with temperature and, thus, are better suited for applications requiring temperature-independent concentrations, including several colligative properties, as will be described in this chapter module.
Calculating Mole Fraction and Molality
The antifreeze in most automobile radiators is a mixture of equal volumes of ethylene glycol and water, with minor amounts of other additives that prevent corrosion. What are the (a) mole fraction and (b) molality of ethylene glycol, C2H4(OH)2, in a solution prepared from 2.22 103 g of ethylene glycol and 2.00 103 g of water (approximately 2 L of glycol and 2 L of water)?
Solution
The mole fraction of ethylene glycol may be computed by first deriving molar amounts of both solution components and then substituting these amounts into the definition of mole fraction.
Notice that mole fraction is a dimensionless property, being the ratio of properties with identical units (moles).
Derive moles of solute and mass of solvent (in kg).
First, use the given mass of ethylene glycol and its molar mass to find the moles of solute: |
Converting Mole Fraction and Molal Concentrations
Calculate the mole fraction of solute and solvent in a 3.0 m solution of sodium chloride.
Solution
Converting from one concentration unit to another is accomplished by first comparing the two unit definitions. In this case, both units have the same numerator (moles of solute) but different denominators. The provided molal concentration may be written as: |
Molality and Molarity Conversions
Intravenous infusion of a 0.556 M aqueous solution of glucose (density of 1.04 g/mL) is part of some post- operative recovery therapies. What is the molal concentration of glucose in this solution?
Solution
The provided molal concentration may be explicitly written as: |
Vapor Pressure Lowering
As described in the chapter on liquids and solids, the equilibrium vapor pressure of a liquid is the pressure exerted by its gaseous phase when vaporization and condensation are occurring at equal rates:
Dissolving a nonvolatile substance in a volatile liquid results in a lowering of the liquid’s vapor pressure. This phenomenon can be rationalized by considering the effect of added solute molecules on the liquid's vaporization and condensation processes. To vaporize, solvent molecules must be present at the surface of the solution. The presence of solute decreases the surface area available to solvent molecules and thereby reduces the rate of solvent vaporization. Since the rate of condensation is unaffected by the presence of solute, the net result is that the vaporization-condensation equilibrium is achieved with fewer solvent molecules in the vapor phase (i.e., at a lower vapor pressure) (). While this interpretation is useful, it does not account for several important aspects of the colligative nature of vapor pressure lowering. A more rigorous explanation involves the property of entropy, a topic of discussion in a later text chapter on thermodynamics. For purposes of understanding the lowering of a liquid's vapor pressure, it is adequate to note that the more dispersed nature of matter in a solution, compared to separate solvent and solute phases, serves to effectively stabilize the solvent molecules and hinder their vaporization. A lower vapor pressure results, and a correspondingly |
FIGURE 11.18 The presence of nonvolatile solutes lowers the vapor pressure of a solution by impeding the evaporation of solvent molecules.
The relationship between the vapor pressures of solution components and the concentrations of those components is described by Raoult’s law: The partial pressure exerted by any component of an ideal solution is equal to the vapor pressure of the pure component multiplied by its mole fraction in the solution.
where PA is the partial pressure exerted by component A in the solution, is the vapor pressure of pure A, and XA is the mole fraction of A in the solution.
Recalling that the total pressure of a gaseous mixture is equal to the sum of partial pressures for all its components (Dalton’s law of partial pressures), the total vapor pressure exerted by a solution containing i components is |
ethanol, C2H5OH, at 40 °C. The vapor pressure of pure ethanol is 0.178 atm at 40 °C. Glycerin is essentially nonvolatile at this temperature.
Solution
Since the solvent is the only volatile component of this solution, its vapor pressure may be computed per Raoult’s law as: |
Check Your Learning
A solution contains 5.00 g of urea, CO(NH2)2 (a nonvolatile solute) and 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution assuming ideal behavior?
Answer:
23.4 torr
Distillation of Solutions
Solutions whose components have significantly different vapor pressures may be separated by a selective vaporization process known as distillation. Consider the simple case of a mixture of two volatile liquids, A and B, with A being the more volatile liquid. Raoult’s law can be used to show that the vapor above the solution is enriched in component A, that is, the mole fraction of A in the vapor is greater than the mole fraction of A in the liquid (see end-of-chapter Exercise 65). By appropriately heating the mixture, component A may be vaporized, condensed, and collected—effectively separating it from component B.
Distillation is widely applied in both laboratory and industrial settings, being used to refine petroleum, to isolate fermentation products, and to purify water. A typical apparatus for laboratory-scale distillations is shown in .
FIGURE 11.19 A typical laboratory distillation unit is shown in (a) a photograph and (b) a schematic diagram of the components. (credit a: modification of work by “Rifleman82”/Wikimedia commons; credit b: modification of work by “Slashme”/Wikimedia Commons)
Oil refineries use large-scale fractional distillation to separate the components of crude oil. The crude oil is heated to high temperatures at the base of a tall fractionating column, vaporizing many of the components that rise within the column. As vaporized components reach adequately cool zones during their ascent, they condense and are collected. The collected liquids are simpler mixtures of hydrocarbons and other petroleum compounds that are of appropriate composition for various applications (e.g., diesel fuel, kerosene, gasoline), as depicted in .
FIGURE 11.20 Crude oil is a complex mixture that is separated by large-scale fractional distillation to isolate various simpler mixtures.
Boiling Point Elevation
As described in the chapter on liquids and solids, the boiling point of a liquid is the temperature at which its vapor pressure is equal to ambient atmospheric pressure. Since the vapor pressure of a solution is lowered due to the presence of nonvolatile solutes, it stands to reason that the solution’s boiling point will subsequently be increased. Vapor pressure increases with temperature, and so a solution will require a higher temperature than will pure solvent to achieve any given vapor pressure, including one equivalent to that of the surrounding atmosphere. The increase in boiling point observed when nonvolatile solute is dissolved in a solvent, ΔTb, is called boiling point elevation and is directly proportional to the molal concentration of solute species:
where Kb is the boiling point elevation constant, or the ebullioscopic constant and m is the molal concentration (molality) of all solute species.
Boiling point elevation constants are characteristic properties that depend on the identity of the solvent. Values of Kb for several solvents are listed in . |