1275. Colligative Properties of Electrolytes -
TABLE 11.3

| Formula unit | Classification | Dissolution products | $\boldsymbol{i}$ (predicted) | $\boldsymbol{i}$ (measured) |
| :---: | :---: | :---: | :---: | :---: |
| $\mathrm{MgCl}_{2}$ | Strong electrolyte | $\mathrm{Mg}^{2+}, 2 \mathrm{Cl}^{-}$ | 3 | 2.7 |
| $\mathrm{FeCl}_{3}$ | Strong electrolyte | $\mathrm{Fe}^{3+}, 3 \mathrm{Cl}^{-}$ | 4 | 3.4 |

TABLE 11.3

In 1923, the chemists Peter Debye and Erich Hückel proposed a theory to explain the apparent incomplete ionization of strong electrolytes. They suggested that although interionic attraction in an aqueous solution is very greatly reduced by solvation of the ions and the insulating action of the polar solvent, it is not completely nullified. The residual attractions prevent the ions from behaving as totally independent particles (Figure 11.28). In some cases, a positive and negative ion may actually touch, giving a solvated unit called an ion pair. Thus, the activity, or the effective concentration, of any particular kind of ion is less than that indicated by the actual concentration. Ions become more and more widely separated the more dilute the solution, and the residual interionic attractions become less and less. Thus, in extremely dilute solutions, the effective concentrations of the ions (their activities) are essentially equal to the actual concentrations. Note that the van't Hoff factors for the electrolytes in Table 11.3 are for 0.05 m solutions, at which concentration the value of $i$ for NaCl is 1.9 , as opposed to an ideal value of 2 .

FIGURE 11.28 Dissociation of ionic compounds in water is not always complete due to the formation of ion pairs.


1277. The Freezing Point of a Solution of an Electrolyte -
The concentration of ions in seawater is approximately the same as that in a solution containing 4.2 g of NaCl dissolved in 125 g of water. Use this information and a predicted value for the van't Hoff factor (Table 11.3) to
determine the freezing temperature the solution (assume ideal solution behavior).


1278. Solution -
Solve this problem using the following series of steps.

Step 1. Convert from grams to moles of NaCl using the molar mass of NaCl in the unit conversion factor. Result: 0.072 mol NaCl
Step 2. Determine the number of moles of ions present in the solution using the number of moles of ions in 1 mole of NaCl as the conversion factor ( 2 mol ions $/ 1 \mathrm{~mol} \mathrm{NaCl}$ ).
Result: 0.14 mol ions
Step 3. Determine the molality of the ions in the solution from the number of moles of ions and the mass of solvent, in kilograms.
Result: 1.2 m
Step 4. Use the direct proportionality between the change in freezing point and molal concentration to determine how much the freezing point changes.
Result: $2.1^{\circ} \mathrm{C}$
Step 5. Determine the new freezing point from the freezing point of the pure solvent and the change.
Result: $-2.1^{\circ} \mathrm{C}$
Check each result as a self-assessment, taking care to avoid rounding errors by retaining guard digits in each step's result for computing the next step's result.


1279. Check Your Learning -
Assuming complete dissociation and ideal solution behavior, calculate the freezing point of a solution of 0.724 g of $\mathrm{CaCl}_{2}$ in 175 g of water.


1280. Answer: -
$-0.208{ }^{\circ} \mathrm{C}$


1281. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Describe the composition and properties of colloidal dispersions
- List and explain several technological applications of colloids

As a child, you may have made suspensions such as mixtures of mud and water, flour and water, or a suspension of solid pigments in water, known as tempera paint. These suspensions are heterogeneous mixtures composed of relatively large particles that are visible (or that can be seen with a magnifying glass). They are cloudy, and the suspended particles settle out after mixing. On the other hand, a solution is a homogeneous mixture in which no settling occurs and in which the dissolved species are molecules or ions. Solutions exhibit completely different behavior from suspensions. A solution may be colored, but it is transparent, the molecules or ions are invisible, and they do not settle out on standing. Another class of mixtures called colloids (or colloidal dispersions) exhibit properties intermediate between those of suspensions and solutions (Figure 11.29). The particles in a colloid are larger than most simple molecules;
however, colloidal particles are small enough that they do not settle out upon standing.

FIGURE 11.29 (a) A solution is a homogeneous mixture that appears clear, such as the saltwater in this aquarium. (b) In a colloid, such as milk, the particles are much larger but remain dispersed and do not settle. (c) A suspension, such as mud, is a heterogeneous mixture of suspended particles that appears cloudy and in which the particles can settle. (credit a photo: modification of work by Adam Wimsatt; credit b photo: modification of work by Melissa Wiese; credit c photo: modification of work by Peter Burgess)

The particles in a colloid are large enough to scatter light, a phenomenon called the Tyndall effect. This can make colloidal mixtures appear cloudy or opaque, such as the searchlight beams shown in Figure 11.30. Clouds are colloidal mixtures. They are composed of water droplets that are much larger than molecules, but that are small enough that they do not settle out.

FIGURE 11.30 The paths of searchlight beams are made visible when light is scattered by colloidal-size particles in the air (fog, smoke, etc.). (credit: "Bahman"/Wikimedia Commons)


1281. LEARNING OBJECTIVES -
FIGURE 11.30 The paths of searchlight beams are made visible when light is scattered by colloidal-size particles in the air (fog, smoke, etc.). (credit: "Bahman"/Wikimedia Commons)

The term "colloid"-from the Greek words kolla, meaning "glue," and eidos, meaning "like"-was first used in 1861 by Thomas Graham to classify mixtures such as starch in water and gelatin. Many colloidal particles are aggregates of hundreds or thousands of molecules, but others (such as proteins and polymer molecules) consist of a single extremely large molecule. The protein and synthetic polymer molecules that form colloids may have molecular masses ranging from a few thousand to many million atomic mass units.

Analogous to the identification of solution components as "solute" and "solvent," the components of a colloid are likewise classified according to their relative amounts. The particulate component typically present in a relatively minor amount is called the dispersed phase and the substance or solution throughout which the particulate is dispersed is called the dispersion medium. Colloids may involve virtually any combination of physical states (gas in liquid, liquid in solid, solid in gas, etc.), as illustrated by the examples of colloidal systems given in Table 11.4.

Examples of Colloidal Systems

| Dispersed Phase | Dispersion Medium | Common Examples | Name |
| :--- | :--- | :--- | :--- |
| solid | gas | smoke, dust | - |
| solid | liquid | starch in water, some inks, paints, milk of magnesia | sol |
| solid | solid | some colored gems, some alloys | - |
| liquid | liquid | clouds, fogs, mists, sprays | aerosol |
| liquid | solid | milk, mayonnaise, butter | emulsion |
| liquid | jellies, gels, pearl, opal $\left(\mathrm{H}_{2} \mathrm{O}\right.$ in $\left.\mathrm{SiO}_{2}\right)$ | gel |  |

TABLE 11.4

| Dispersed Phase | Dispersion Medium | Common Examples | Name |
| :--- | :--- | :--- | :--- |
| gas | liquid | foams, whipped cream, beaten egg whites | foam |
| gas | solid | pumice, floating soaps | - |


1281. LEARNING OBJECTIVES -
TABLE 11.4


1282. Preparation of Colloidal Systems -
Colloids are prepared by producing particles of colloidal dimensions and distributing these particles throughout a dispersion medium. Particles of colloidal size are formed by two methods:

1. Dispersion methods: breaking down larger particles. For example, paint pigments are produced by dispersing large particles by grinding in special mills.
2. Condensation methods: growth from smaller units, such as molecules or ions. For example, clouds form when water molecules condense and form very small droplets.

A few solid substances, when brought into contact with water, disperse spontaneously and form colloidal systems. Gelatin, glue, starch, and dehydrated milk powder behave in this manner. The particles are already of colloidal size; the water simply disperses them. Powdered milk particles of colloidal size are produced by dehydrating milk spray. Some atomizers produce colloidal dispersions of a liquid in air.

An emulsion may be prepared by shaking together or blending two immiscible liquids. This breaks one liquid into droplets of colloidal size, which then disperse throughout the other liquid. Oil spills in the ocean may be difficult to clean up, partly because wave action can cause the oil and water to form an emulsion. In many emulsions, however, the dispersed phase tends to coalesce, form large drops, and separate. Therefore, emulsions are usually stabilized by an emulsifying agent, a substance that inhibits the coalescence of the dispersed liquid. For example, a little soap will stabilize an emulsion of kerosene in water. Milk is an emulsion of butterfat in water, with the protein casein serving as the emulsifying agent. Mayonnaise is an emulsion of oil in vinegar, with egg yolk components as the emulsifying agents.

Condensation methods form colloidal particles by aggregation of molecules or ions. If the particles grow beyond the colloidal size range, drops or precipitates form, and no colloidal system results. Clouds form when water molecules aggregate and form colloid-sized particles. If these water particles coalesce to form adequately large water drops of liquid water or crystals of solid water, they settle from the sky as rain, sleet, or snow. Many condensation methods involve chemical reactions. A red colloidal suspension of iron(III) hydroxide may be prepared by mixing a concentrated solution of iron(III) chloride with hot water:


1282. Preparation of Colloidal Systems -
$$
\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{Cl}^{-}(\mathrm{aq})+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)+3 \mathrm{Cl}^{-}(a q) .
$$

A colloidal gold sol results from the reduction of a very dilute solution of gold(III) chloride by a reducing agent such as formaldehyde, tin(II) chloride, or iron(II) sulfate:

$$
\mathrm{Au}^{3+}+3 \mathrm{e}^{-} \longrightarrow \mathrm{Au}
$$

Some gold sols prepared in 1857 are still intact (the particles have not coalesced and settled), illustrating the long-term stability of many colloids.


1283. Soaps and Detergents -
Pioneers made soap by boiling fats with a strongly basic solution made by leaching potassium carbonate, $\mathrm{K}_{2} \mathrm{CO}_{3}$, from wood ashes with hot water. Animal fats contain polyesters of fatty acids (long-chain carboxylic acids). When animal fats are treated with a base like potassium carbonate or sodium hydroxide, glycerol and salts of fatty acids such as palmitic, oleic, and stearic acid are formed. The salts of fatty acids are called soaps. The sodium salt of stearic acid, sodium stearate, has the formula $\mathrm{C}_{17} \mathrm{H}_{35} \mathrm{CO}_{2} \mathrm{Na}$ and contains an uncharged nonpolar hydrocarbon chain, the $\mathrm{C}_{17} \mathrm{H}_{35}$ - unit, and an ionic carboxylate group, the $-\mathrm{CO}_{2}{ }^{-}$unit (Figure 11.31).

sodium stearate (soap)
FIGURE 11.31 Soaps contain a nonpolar hydrocarbon end (blue) and an ionic end (red). The ionic end is a carboxylate group. The length of the hydrocarbon end can vary from soap to soap.

Detergents (soap substitutes) also contain nonpolar hydrocarbon chains, such as $\mathrm{C}_{12} \mathrm{H}_{25}$-, and an ionic group, such as a sulfate- $\mathrm{OSO}_{3}{ }^{-}$, or a sulfonate $-\mathrm{SO}_{3}{ }^{-}$(Figure 11.32). Soaps form insoluble calcium and magnesium compounds in hard water; detergents form water-soluble products-a definite advantage for detergents.

FIGURE 11.32 Detergents contain a nonpolar hydrocarbon end (blue) and an ionic end (red). The ionic end can be either a sulfate or a sulfonate. The length of the hydrocarbon end can vary from detergent to detergent.


1283. Soaps and Detergents -
The cleaning action of soaps and detergents can be explained in terms of the structures of the molecules involved. The hydrocarbon (nonpolar) end of a soap or detergent molecule dissolves in, or is attracted to, nonpolar substances such as oil, grease, or dirt particles. The ionic end is attracted by water (polar), illustrated in Figure 11.33. As a result, the soap or detergent molecules become oriented at the interface between the dirt particles and the water so they act as a kind of bridge between two different kinds of matter, nonpolar and polar. Molecules such as this are termed amphiphilic since they have both a hydrophobic ("water-fearing") part and a hydrophilic ("water-loving") part. As a consequence, dirt particles become suspended as colloidal particles and are readily washed away.

FIGURE 11.33 This diagrammatic cross section of an emulsified drop of oil in water shows how soap or detergent acts as an emulsifier.


1285. Deepwater Horizon Oil Spill -
The blowout of the Deepwater Horizon oil drilling rig on April 20, 2010, in the Gulf of Mexico near Mississippi began the largest marine oil spill in the history of the petroleum industry. In the 87 days following the blowout, an estimated 4.9 million barrels ( 210 million gallons) of oil flowed from the ruptured well 5000 feet below the water's surface. The well was finally declared sealed on September 19, 2010.

Crude oil is immiscible with and less dense than water, so the spilled oil rose to the surface of the water.

Floating booms, skimmer ships, and controlled burns were used to remove oil from the water's surface in an attempt to protect beaches and wetlands along the Gulf coast. In addition to removal of the oil, attempts were also made to lessen its environmental impact by rendering it "soluble" (in the loose sense of the term) and thus allowing it to be diluted to hopefully less harmful levels by the vast volume of ocean water. This approach used 1.84 million gallons of the oil dispersant Corexit 9527, most of which was injected underwater at the site of the leak, with small amounts being sprayed on top of the spill. Corexit 9527 contains 2-butoxyethanol $\left(\mathrm{C}_{6} \mathrm{H}_{14} \mathrm{O}_{2}\right)$, an amphiphilic molecule whose polar and nonpolar ends are useful for emulsifying oil into small droplets, increasing the surface area of the oil and making it more available to marine bacteria for digestion (Figure 11.34). While this approach avoids many of the immediate hazards that bulk oil poses to marine and coastal ecosystems, it introduces the possibility of long-term effects resulting from the introduction of the complex and potential toxic components of petroleum into the ocean's food chain. A number of organizations are involved in monitoring the extended impact of this oil spill, including the National Oceanic and Atmospheric Administration (visit this website (http://openstax.org/l/16gulfspill) for additional details).


1285. Deepwater Horizon Oil Spill -
FIGURE 11.34 (a) This NASA satellite image shows the oil slick from the Deepwater Horizon spill. (b) A US Air Force plane sprays Corexit, a dispersant. (c) The molecular structure of 2-butoxyethanol is shown. (credit a: modification of work by "NASA, FT2, demis.nl"/Wikimedia Commons; credit b: modification of work by "NASA/ MODIS Rapid Response Team"/Wikimedia Commons)


1286. Electrical Properties of Colloidal Particles -
Dispersed colloidal particles are often electrically charged. A colloidal particle of iron(III) hydroxide, for example, does not contain enough hydroxide ions to compensate exactly for the positive charges on the iron(III) ions. Thus, each individual colloidal particle bears a positive charge, and the colloidal dispersion consists of charged colloidal particles and some free hydroxide ions, which keep the dispersion electrically neutral. Most metal hydroxide colloids have positive charges, whereas most metals and metal sulfides form negatively charged dispersions. All colloidal particles in any one system have charges of the same sign. This helps keep them dispersed because particles containing like charges repel each other.

The charged nature of some colloidal particles may be exploited to remove them from a variety of mixtures. For example, the particles comprising smoke are often colloidally dispersed and electrically charged.
Frederick Cottrell, an American chemist, developed a process to remove these particles. The charged particles are attracted to highly charged electrodes, where they are neutralized and deposited as dust (Figure 11.36). This is one of the important methods used to clean up the smoke from a variety of industrial processes. The process is also important in the recovery of valuable products from the smoke and flue dust of smelters, furnaces, and kilns. There are also similar electrostatic air filters designed for home use to improve indoor air quality.


1288. Frederick Gardner Cottrell -
(a)

(b)

FIGURE 11.35 (a) Frederick Cottrell developed (b) the electrostatic precipitator, a device designed to curb air pollution by removing colloidal particles from air. (credit b: modification of work by "SpLot"/Wikimedia Commons)

Born in Oakland, CA, in 1877, Frederick Cottrell devoured textbooks as if they were novels and graduated from high school at the age of 16. He then entered the University of California (UC), Berkeley, completing a Bachelor's degree in three years. He saved money from his \$1200 annual salary as a chemistry teacher at Oakland High School to fund his studies in chemistry in Berlin with Nobel prize winner Jacobus Henricus van't Hoff, and in Leipzig with Wilhelm Ostwald, another Nobel awardee. After earning his PhD in physical chemistry, he returned to the United States to teach at UC Berkeley. He also consulted for the DuPont Company, where he developed the electrostatic precipitator, a device designed to curb air pollution by removing colloidal particles from air. Cottrell used the proceeds from his invention to fund a nonprofit research corporation to finance scientific research.

FIGURE 11.36 In a Cottrell precipitator, positively and negatively charged particles are attracted to highly charged electrodes, where they are neutralized and deposited as dust.


1289. Gels -
Gelatin desserts, such as Jell-O, are a type of colloid (Figure 11.37). Gelatin sets on cooling because the hot aqueous mixture of gelatin coagulates as it cools, yielding an extremely viscous body known as a gel. A gel is a colloidal dispersion of a liquid phase throughout a solid phase. It appears that the fibers of the dispersing medium form a complex three-dimensional network, the interstices being filled with the liquid medium or a dilute solution of the dispersing medium.

FIGURE 11.37 Gelatin desserts are colloids in which an aqueous solution of sweeteners and flavors is dispersed throughout a medium of solid proteins. (credit photo: modification of work by Steven Depolo)

Pectin, a carbohydrate from fruit juices, is a gel-forming substance important in jelly making. Silica gel, a colloidal dispersion of hydrated silicon dioxide, is formed when dilute hydrochloric acid is added to a dilute solution of sodium silicate. Canned Heat is a flammable gel made by mixing alcohol and a saturated aqueous solution of calcium acetate.


1290. Key Terms -
alloy solid mixture of a metallic element and one or more additional elements
amphiphilic molecules possessing both hydrophobic (nonpolar) and a hydrophilic (polar) parts
boiling point elevation elevation of the boiling point of a liquid by addition of a solute
boiling point elevation constant the proportionality constant in the equation relating boiling point elevation to solute molality; also known as the ebullioscopic constant
colligative property property of a solution that depends only on the concentration of a solute species
colloid (also, colloidal dispersion) mixture in which relatively large solid or liquid particles are dispersed uniformly throughout a gas, liquid, or solid
crenation process whereby biological cells become shriveled due to loss of water by osmosis
dispersed phase substance present as relatively large solid or liquid particles in a colloid
dispersion medium solid, liquid, or gas in which colloidal particles are dispersed
dissociation physical process accompanying the dissolution of an ionic compound in which the compound's constituent ions are solvated and dispersed throughout the solution
electrolyte substance that produces ions when dissolved in water
emulsifying agent amphiphilic substance used to stabilize the particles of some emulsions
emulsion colloid formed from immiscible liquids
freezing point depression lowering of the freezing point of a liquid by addition of a solute
freezing point depression constant (also, cryoscopic constant) proportionality constant in the equation relating freezing point depression to solute molality
gel colloidal dispersion of a liquid in a solid
hemolysis rupture of red blood cells due to the accumulation of excess water by osmosis
Henry's law the proportional relationship between the concentration of dissolved gas in a solution and the partial pressure of the gas in contact with the solution
hypertonic of greater osmotic pressure
hypotonic of less osmotic pressure
ideal solution solution that forms with no accompanying energy change
immiscible of negligible mutual solubility; typically refers to liquid substances
ion pair solvated anion/cation pair held together by moderate electrostatic attraction
ion-dipole attraction electrostatic attraction between an ion and a polar molecule
isotonic of equal osmotic pressure
miscible mutually soluble in all proportions; typically refers to liquid substances


1290. Key Terms -
ion-dipole attraction electrostatic attraction between an ion and a polar molecule
isotonic of equal osmotic pressure
miscible mutually soluble in all proportions; typically refers to liquid substances
molality ( $\boldsymbol{m}$ ) a concentration unit defined as the ratio of the numbers of moles of solute to the mass of the solvent in kilograms
nonelectrolyte substance that does not produce ions when dissolved in water
osmosis diffusion of solvent molecules through a semipermeable membrane
osmotic pressure (II) opposing pressure required to prevent bulk transfer of solvent molecules through a semipermeable membrane
partially miscible of moderate mutual solubility; typically refers to liquid substances
Raoult's law the relationship between a solution's vapor pressure and the vapor pressures and concentrations of its components
saturated of concentration equal to solubility; containing the maximum concentration of solute possible for a given temperature and pressure
semipermeable membrane a membrane that selectively permits passage of certain ions or molecules
solubility extent to which a solute may be dissolved in water, or any solvent
solvation exothermic process in which intermolecular attractive forces between the solute and solvent in a solution are established
spontaneous process physical or chemical change that occurs without the addition of energy from an external source
strong electrolyte substance that dissociates or ionizes completely when dissolved in water
supersaturated of concentration that exceeds solubility; a nonequilibrium state
suspension heterogeneous mixture in which relatively large component particles are temporarily dispersed but settle out over time
Tyndall effect scattering of visible light by a colloidal dispersion
unsaturated of concentration less than solubility
van't Hoff factor (i) the ratio of the number of moles of particles in a solution to the number of moles of formula units dissolved in the solution
weak electrolyte substance that ionizes only partially when dissolved in water


1290. Key Terms -
Key Equations
$C_{\mathrm{g}}=k P_{\mathrm{g}}$
$\left(P_{\mathrm{A}}=X_{\mathrm{A}} P_{\mathrm{A}}^{*}\right)$
$P_{\text {solution }}=\sum_{i} P_{i}=\sum_{i} X_{i} P_{i}^{*}$
$P_{\text {solution }}=X_{\text {solvent }} P_{\text {solvent }}^{*}$
$\Delta T_{\mathrm{b}}=K_{\mathrm{b}} m$
$\Delta T_{\mathrm{f}}=K_{\mathrm{f}} m$
$\Pi=M R T$


1291. Summary - 1291.1. The Dissolution Process
A solution forms when two or more substances combine physically to yield a mixture that is homogeneous at the molecular level. The solvent is the most concentrated component and determines the physical state of the solution. The solutes are the other components typically present at concentrations less than that of the solvent. Solutions may form endothermically or exothermically, depending upon the relative magnitudes of solute and solvent intermolecular attractive forces. Ideal solutions form with no appreciable change in energy.


1291. Summary - 1291.2. Electrolytes
Substances that dissolve in water to yield ions are called electrolytes. Electrolytes may be covalent compounds that chemically react with water to produce ions (for example, acids and bases), or they may be ionic compounds that dissociate to yield their constituent cations and anions, when dissolved. Dissolution of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules. Soluble ionic substances and strong acids ionize completely and are strong electrolytes, while weak acids and bases ionize to only a small extent and are weak electrolytes. Nonelectrolytes are substances that do not produce ions when dissolved in water.


1291. Summary - 1291.3. Solubility
The extent to which one substance will dissolve in another is determined by several factors, including the types and relative strengths of intermolecular attractive forces that may exist between the substances' atoms, ions, or molecules. This tendency to dissolve is quantified as a substance's solubility, its maximum concentration in a solution at equilibrium under specified conditions. A
saturated solution contains solute at a concentration equal to its solubility. A supersaturated solution is one in which a solute's concentration exceeds its solubility-a nonequilibrium (unstable) condition that will result in solute precipitation when the solution is appropriately perturbed. Miscible liquids are soluble in all proportions, and immiscible liquids exhibit very low mutual solubility. Solubilities for gaseous solutes decrease with increasing temperature, while those for most, but not all, solid solutes increase with temperature. The concentration of a gaseous solute in a solution is proportional to the partial pressure of the gas to which the solution is exposed, a relation known as Henry's law.


1291. Summary - 1291.4. Colligative Properties
Properties of a solution that depend only on the concentration of solute particles are called colligative properties. They include changes in the vapor pressure, boiling point, and freezing point of the solvent in the solution. The magnitudes of these properties depend only on the total concentration of solute particles in solution, not on the type of particles. The total concentration of solute particles in a solution also determines its osmotic pressure. This is the pressure that must be applied to the solution to prevent diffusion of molecules of pure solvent through a semipermeable membrane into the solution. Ionic compounds may not completely dissociate in solution due to activity effects, in which case observed colligative effects may be less than predicted.


1291. Summary - 1291.5. Colloids
Colloids are mixtures in which one or more substances are dispersed as relatively large solid particles or liquid droplets throughout a solid, liquid, or gaseous medium. The particles of a colloid remain dispersed and do not settle due to gravity,
and they are often electrically charged. Colloids are widespread in nature and are involved in many
technological applications.


1292. Exercises - 1292.1. The Dissolution Process
1. How do solutions differ from compounds? From other mixtures?
2. Which of the principal characteristics of solutions are evident in the solutions of $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ shown in Figure 11.2?
3. When $\mathrm{KNO}_{3}$ is dissolved in water, the resulting solution is significantly colder than the water was originally.
(a) Is the dissolution of $\mathrm{KNO}_{3}$ an endothermic or an exothermic process?
(b) What conclusions can you draw about the intermolecular attractions involved in the process?
(c) Is the resulting solution an ideal solution?
4. Give an example of each of the following types of solutions:
(a) a gas in a liquid
(b) a gas in a gas
(c) a solid in a solid
5. Indicate the most important types of intermolecular attractions in each of the following solutions:
(a) The solution in Figure 11.2.
(b) $\mathrm{NO}(I)$ in $\mathrm{CO}(I)$
(c) $\mathrm{Cl}_{2}(\mathrm{~g})$ in $\mathrm{Br}_{2}(\mathrm{I})$
(d) $\mathrm{HCl}(g)$ in benzene $\mathrm{C}_{6} \mathrm{H}_{6}(\mathrm{I})$
(e) Methanol $\mathrm{CH}_{3} \mathrm{OH}(\mathrm{I})$ in $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$
6. Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as heptane ( $\mathrm{C}_{7} \mathrm{H}_{16}$, nonpolar solvent):
(a) vegetable oil (nonpolar)
(b) isopropyl alcohol (polar)
(c) potassium bromide (ionic)
7. Heat is released when some solutions form; heat is absorbed when other solutions form. Provide a molecular explanation for the difference between these two types of spontaneous processes.


1292. Exercises - 1292.1. The Dissolution Process
(c) potassium bromide (ionic)
7. Heat is released when some solutions form; heat is absorbed when other solutions form. Provide a molecular explanation for the difference between these two types of spontaneous processes.
8. Solutions of hydrogen in palladium may be formed by exposing Pd metal to $\mathrm{H}_{2}$ gas. The concentration of hydrogen in the palladium depends on the pressure of $\mathrm{H}_{2}$ gas applied, but in a more complex fashion than can be described by Henry's law. Under certain conditions, 0.94 g of hydrogen gas is dissolved in 215 g of palladium metal (solution density $=10.8 \mathrm{~g} \mathrm{~cm}^{3}$ ).
(a) Determine the molarity of this solution.
(b) Determine the molality of this solution.
(c) Determine the percent by mass of hydrogen atoms in this solution.


1292. Exercises - 1292.2. Electrolytes
9. Explain why the ions $\mathrm{Na}^{+}$and $\mathrm{Cl}^{-}$are strongly solvated in water but not in hexane, a solvent composed of nonpolar molecules.
10. Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive.
11. Consider the solutions presented:
(a) Which of the following sketches best represents the ions in a solution of $\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}(\mathrm{aq})$ ?

(x)

(y)


1292. Exercises - 1292.2. Electrolytes
(x)

(y)

(z)
(b) Write a balanced chemical equation showing the products of the dissolution of $\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{3}$.
12. Compare the processes that occur when methanol $\left(\mathrm{CH}_{3} \mathrm{OH}\right)$, hydrogen chloride $(\mathrm{HCl})$, and sodium hydroxide $(\mathrm{NaOH})$ dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution.
13. What is the expected electrical conductivity of the following solutions?
(a) $\mathrm{NaOH}(a q)$
(b) $\mathrm{HCl}(a q)$
(c) $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(a q)$ (glucose)
(d) $\mathrm{NH}_{3}(\mathrm{aq})$
14. Why are most solid ionic compounds electrically nonconductive, whereas aqueous solutions of ionic compounds are good conductors? Would you expect a liquid (molten) ionic compound to be electrically conductive or nonconductive? Explain.
15. Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions:
(a) the solutions in Figure 11.7
(b) methanol, $\mathrm{CH}_{3} \mathrm{OH}$, dissolved in ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$
(c) methane, $\mathrm{CH}_{4}$, dissolved in benzene, $\mathrm{C}_{6} \mathrm{H}_{6}$
(d) the polar halocarbon $\mathrm{CF}_{2} \mathrm{Cl}_{2}$ dissolved in the polar halocarbon $\mathrm{CF}_{2} \mathrm{ClCFCl}_{2}$
(e) $\mathrm{O}_{2}(I)$ in $\mathrm{N}_{2}(I)$


1292. Exercises - 1292.3. Solubility
16. Suppose you are presented with a clear solution of sodium thiosulfate, $\mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3}$. How could you determine whether the solution is unsaturated, saturated, or supersaturated?
17. Supersaturated solutions of most solids in water are prepared by cooling saturated solutions. Supersaturated solutions of most gases in water are prepared by heating saturated solutions. Explain the reasons for the difference in the two procedures.
18. Suggest an explanation for the observations that ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$, is completely miscible with water and that ethanethiol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{SH}$, is soluble only to the extent of 1.5 g per 100 mL of water.
19. Calculate the percent by mass of KBr in a saturated solution of KBr in water at $10^{\circ} \mathrm{C}$. See Figure 11.16 for useful data, and report the computed percentage to one significant digit.
20. Which of the following gases is expected to be most soluble in water? Explain your reasoning.
(a) $\mathrm{CH}_{4}$
(b) $\mathrm{CCl}_{4}$
(c) $\mathrm{CHCl}_{3}$
21. At $0^{\circ} \mathrm{C}$ and 1.00 atm , as much as 0.70 g of $\mathrm{O}_{2}$ can dissolve in 1 L of water. At $0^{\circ} \mathrm{C}$ and 4.00 atm , how many grams of $\mathrm{O}_{2}$ dissolve in 1 L of water?
22. Refer to Figure 11.10 .
(a) How did the concentration of dissolved $\mathrm{CO}_{2}$ in the beverage change when the bottle was opened?
(b) What caused this change?
(c) Is the beverage unsaturated, saturated, or supersaturated with $\mathrm{CO}_{2}$ ?


1292. Exercises - 1292.3. Solubility
(b) What caused this change?
(c) Is the beverage unsaturated, saturated, or supersaturated with $\mathrm{CO}_{2}$ ?
23. The Henry's law constant for $\mathrm{CO}_{2}$ is $3.4 \times 10^{-2} \mathrm{M} / \mathrm{atm}$ at $25^{\circ} \mathrm{C}$. Assuming ideal solution behavior, what pressure of carbon dioxide is needed to maintain a $\mathrm{CO}_{2}$ concentration of 0.10 M in a can of lemon-lime soda?
24. The Henry's law constant for $\mathrm{O}_{2}$ is $1.3 \times 10^{-3} \mathrm{M} / \mathrm{atm}$ at $25^{\circ} \mathrm{C}$. Assuming ideal solution behavior, what mass of oxygen would be dissolved in a $40-\mathrm{L}$ aquarium at $25^{\circ} \mathrm{C}$, assuming an atmospheric pressure of 1.00 atm , and that the partial pressure of $\mathrm{O}_{2}$ is 0.21 atm ?
25. Assuming ideal solution behavior, how many liters of HCl gas, measured at $30.0^{\circ} \mathrm{C}$ and 745 torr, are required to prepare 1.25 L of a $3.20-M$ solution of hydrochloric acid?


1292. Exercises - 1292.4. Colligative Properties
26. Which is/are part of the macroscopic domain of solutions and which is/are part of the microscopic domain: boiling point elevation, Henry's law, hydrogen bond, ion-dipole attraction, molarity, nonelectrolyte, nonstoichiometric compound, osmosis, solvated ion?
27. What is the microscopic explanation for the macroscopic behavior illustrated in Figure 11.14?
28. Sketch a qualitative graph of the pressure versus time for water vapor above a sample of pure water and a sugar solution, as the liquids evaporate to half their original volume.
29. A solution of potassium nitrate, an electrolyte, and a solution of glycerin $\left(\mathrm{C}_{3} \mathrm{H}_{5}(\mathrm{OH})_{3}\right)$, a nonelectrolyte, both boil at $100.3^{\circ} \mathrm{C}$. What other physical properties of the two solutions are identical?
30. What are the mole fractions of $\mathrm{H}_{3} \mathrm{PO}_{4}$ and water in a solution of 14.5 g of $\mathrm{H}_{3} \mathrm{PO}_{4}$ in 125 g of water?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
31. What are the mole fractions of $\mathrm{HNO}_{3}$ and water in a concentrated solution of nitric acid $\left(68.0 \% \mathrm{HNO}_{3}\right.$ by mass)?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
32. Calculate the mole fraction of each solute and solvent:
(a) 583 g of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in 1.50 kg of water-the acid solution used in an automobile battery
(b) 0.86 g of NaCl in $1.00 \times 10^{2} \mathrm{~g}$ of water-a solution of sodium chloride for intravenous injection


1292. Exercises - 1292.4. Colligative Properties
(b) 0.86 g of NaCl in $1.00 \times 10^{2} \mathrm{~g}$ of water-a solution of sodium chloride for intravenous injection
(c) 46.85 g of codeine, $\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}$, in 125.5 g of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$
(d) 25 g of $\mathrm{I}_{2}$ in 125 g of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$
33. Calculate the mole fraction of each solute and solvent:
(a) 0.710 kg of sodium carbonate (washing soda), $\mathrm{Na}_{2} \mathrm{CO}_{3}$, in 10.0 kg of water-a saturated solution at $0{ }^{\circ} \mathrm{C}$
(b) 125 g of $\mathrm{NH}_{4} \mathrm{NO}_{3}$ in 275 g of water-a mixture used to make an instant ice pack
(c) 25 g of $\mathrm{Cl}_{2}$ in 125 g of dichloromethane, $\mathrm{CH}_{2} \mathrm{Cl}_{2}$
(d) 0.372 g of tetrahydropyridine, $\mathrm{C}_{5} \mathrm{H}_{9} \mathrm{~N}$, in 125 g of chloroform, $\mathrm{CHCl}_{3}$
34. Calculate the mole fractions of methanol, $\mathrm{CH}_{3} \mathrm{OH}$; ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$; and water in a solution that is $40 \%$ methanol, $40 \%$ ethanol, and $20 \%$ water by mass. (Assume the data are good to two significant figures.)
35. What is the difference between a $1 M$ solution and a 1 m solution?


1292. Exercises - 1292.4. Colligative Properties
35. What is the difference between a $1 M$ solution and a 1 m solution?
36. What is the molality of phosphoric acid, $\mathrm{H}_{3} \mathrm{PO}_{4}$, in a solution of 14.5 g of $\mathrm{H}_{3} \mathrm{PO}_{4}$ in 125 g of water?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
37. What is the molality of nitric acid in a concentrated solution of nitric acid ( $68.0 \% \mathrm{HNO}_{3}$ by mass)?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
38. Calculate the molality of each of the following solutions:
(a) 583 g of $\mathrm{H}_{2} \mathrm{SO}_{4}$ in 1.50 kg of water-the acid solution used in an automobile battery
(b) 0.86 g of NaCl in $1.00 \times 10^{2} \mathrm{~g}$ of water-a solution of sodium chloride for intravenous injection
(c) 46.85 g of codeine, $\mathrm{C}_{18} \mathrm{H}_{21} \mathrm{NO}_{3}$, in 125.5 g of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$
(d) 25 g of $\mathrm{I}_{2}$ in 125 g of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$
39. Calculate the molality of each of the following solutions:
(a) 0.710 kg of sodium carbonate (washing soda), $\mathrm{Na}_{2} \mathrm{CO}_{3}$, in 10.0 kg of water-a saturated solution at $0^{\circ} \mathrm{C}$
(b) 125 g of $\mathrm{NH}_{4} \mathrm{NO}_{3}$ in 275 g of water-a mixture used to make an instant ice pack
(c) 25 g of $\mathrm{Cl}_{2}$ in 125 g of dichloromethane, $\mathrm{CH}_{2} \mathrm{Cl}_{2}$


1292. Exercises - 1292.4. Colligative Properties
(c) 25 g of $\mathrm{Cl}_{2}$ in 125 g of dichloromethane, $\mathrm{CH}_{2} \mathrm{Cl}_{2}$
(d) 0.372 g of tetrahydropyridine, $\mathrm{C}_{5} \mathrm{H}_{9} \mathrm{~N}$, in 125 g of chloroform, $\mathrm{CHCl}_{3}$
40. The concentration of glucose, $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$, in normal spinal fluid is $\frac{75 \mathrm{mg}}{100 \mathrm{~g}}$. What is the molality of the solution?
41. A $13.0 \%$ solution of $\mathrm{K}_{2} \mathrm{CO}_{3}$ by mass has a density of $1.09 \mathrm{~g} / \mathrm{cm}^{3}$. Calculate the molality of the solution.
42. Why does 1 mol of sodium chloride depress the freezing point of 1 kg of water almost twice as much as 1 mol of glycerin?
43. Assuming ideal solution behavior, what is the boiling point of a solution of 115.0 g of nonvolatile sucrose, $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$, in 350.0 g of water?
(a) Outline the steps necessary to answer the question
(b) Answer the question
44. Assuming ideal solution behavior, what is the boiling point of a solution of 9.04 g of $\mathrm{I}_{2}$ in 75.5 g of benzene, assuming the $\mathrm{I}_{2}$ is nonvolatile?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
45. Assuming ideal solution behavior, what is the freezing temperature of a solution of 115.0 g of sucrose, $\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}$, in 350.0 g of water?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.


1292. Exercises - 1292.4. Colligative Properties
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
46. Assuming ideal solution behavior, what is the freezing point of a solution of 9.04 g of $\mathrm{I}_{2}$ in 75.5 g of benzene?
(a) Outline the steps necessary to answer the following question.
(b) Answer the question.
47. Assuming ideal solution behavior, what is the osmotic pressure of an aqueous solution of 1.64 g of $\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}$ in water at $25^{\circ} \mathrm{C}$ ? The volume of the solution is 275 mL .
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
48. Assuming ideal solution behavior, what is osmotic pressure of a solution of bovine insulin (molar mass, $5700 \mathrm{~g} \mathrm{~mol}^{-1}$ ) at $18^{\circ} \mathrm{C}$ if 100.0 mL of the solution contains 0.103 g of the insulin?
(a) Outline the steps necessary to answer the question.
(b) Answer the question.
49. Assuming ideal solution behavior, what is the molar mass of a solution of 5.00 g of a compound in 25.00 g of carbon tetrachloride (bp $76.8^{\circ} \mathrm{C} ; K_{\mathrm{b}}=5.02^{\circ} \mathrm{C} / \mathrm{m}$ ) that boils at $81.5^{\circ} \mathrm{C}$ at 1 atm ?
(a) Outline the steps necessary to answer the question.
(b) Solve the problem.
50. A sample of an organic compound (a nonelectrolyte) weighing 1.35 g lowered the freezing point of 10.0 g of benzene by $3.66^{\circ} \mathrm{C}$. Assuming ideal solution behavior, calculate the molar mass of the compound.
51. A 1.0 m solution of HCl in benzene has a freezing point of $0.4^{\circ} \mathrm{C}$. Is HCl an electrolyte in benzene? Explain.


1292. Exercises - 1292.4. Colligative Properties
51. A 1.0 m solution of HCl in benzene has a freezing point of $0.4^{\circ} \mathrm{C}$. Is HCl an electrolyte in benzene? Explain.
52. A solution contains 5.00 g of urea, $\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2}$, a nonvolatile compound, dissolved in 0.100 kg of water. If the vapor pressure of pure water at $25^{\circ} \mathrm{C}$ is 23.7 torr, what is the vapor pressure of the solution (assuming ideal solution behavior)?
53. A $12.0-\mathrm{g}$ sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at $-1.94{ }^{\circ} \mathrm{C}$. Assuming ideal solution behavior, calculate the molar mass of the substance.
54. Arrange the following solutions in order by their decreasing freezing points: $0.1 \mathrm{~m} \mathrm{Na}_{3} \mathrm{PO}_{4}, 0.1 \mathrm{~m}_{2} \mathrm{H}_{5} \mathrm{OH}$, $0.01 \mathrm{~m} \mathrm{CO}_{2}, 0.15 \mathrm{~m} \mathrm{NaCl}$, and $0.2 \mathrm{~m} \mathrm{CaCl}_{2}$.
55. Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of $\mathrm{NaCl}, 0.020 \mathrm{~mol}$ of $\mathrm{Na}_{2} \mathrm{SO}_{4}$, and 0.030 mol of $\mathrm{MgCl}_{2}$, assuming complete dissociation of these electrolytes and ideal solution behavior.
56. How could you prepare a 3.08 m aqueous solution of glycerin, $\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}$ ? Assuming ideal solution behavior, what is the freezing point of this solution?


1292. Exercises - 1292.4. Colligative Properties
57. A sample of sulfur weighing 0.210 g was dissolved in 17.8 g of carbon disulfide, $\mathrm{CS}_{2}\left(K_{\mathrm{b}}=2.34^{\circ} \mathrm{C} / \mathrm{m}\right)$. If the boiling point elevation was $0.107^{\circ} \mathrm{C}$, what is the formula of a sulfur molecule in carbon disulfide (assuming ideal solution behavior)?
58. In a significant experiment performed many years ago, 5.6977 g of cadmium iodide in 44.69 g of water raised the boiling point $0.181{ }^{\circ} \mathrm{C}$. What does this suggest about the nature of a solution of $\mathrm{CdI}_{2}$ ?
59. Lysozyme is an enzyme that cleaves cell walls. A $0.100-\mathrm{L}$ sample of a solution of lysozyme that contains 0.0750 g of the enzyme exhibits an osmotic pressure of $1.32 \times 10^{-3}$ atm at $25^{\circ} \mathrm{C}$. Assuming ideal solution behavior, what is the molar mass of lysozyme?
60. The osmotic pressure of a solution containing 7.0 g of insulin per liter is 23 torr at $25^{\circ} \mathrm{C}$. Assuming ideal solution behavior, what is the molar mass of insulin?
61. The osmotic pressure of human blood is 7.6 atm at $37^{\circ} \mathrm{C}$. What mass of glucose, $\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}$, is required to make 1.00 L of aqueous solution for intravenous feeding if the solution must have the same osmotic pressure as blood at body temperature, $37^{\circ} \mathrm{C}$ (assuming ideal solution behavior)?


1292. Exercises - 1292.4. Colligative Properties
62. Assuming ideal solution behavior, what is the freezing point of a solution of dibromobenzene, $\mathrm{C}_{6} \mathrm{H}_{4} \mathrm{Br}_{2}$, in 0.250 kg of benzene, if the solution boils at $83.5^{\circ} \mathrm{C}$ ?
63. Assuming ideal solution behavior, what is the boiling point of a solution of NaCl in water if the solution freezes at $-0.93^{\circ} \mathrm{C}$ ?
64. The sugar fructose contains $40.0 \% \mathrm{C}, 6.7 \% \mathrm{H}$, and $53.3 \% \mathrm{O}$ by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of $78.59^{\circ} \mathrm{C}$. The boiling point of ethanol is $78.35^{\circ} \mathrm{C}$, and $K_{\mathrm{b}}$ for ethanol is $1.20^{\circ} \mathrm{C} / \mathrm{m}$. Assuming ideal solution behavior, what is the molecular formula of fructose?
65. The vapor pressure of methanol, $\mathrm{CH}_{3} \mathrm{OH}$, is 94 torr at $20^{\circ} \mathrm{C}$. The vapor pressure of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}$, is 44 torr at the same temperature.
(a) Calculate the mole fraction of methanol and of ethanol in a solution of 50.0 g of methanol and 50.0 g of ethanol.
(b) Ethanol and methanol form a solution that behaves like an ideal solution. Calculate the vapor pressure of methanol and of ethanol above the solution at $20^{\circ} \mathrm{C}$.
(c) Calculate the mole fraction of methanol and of ethanol in the vapor above the solution.
66. The triple point of air-free water is defined as 273.16 K . Why is it important that the water be free of air?


1292. Exercises - 1292.4. Colligative Properties
66. The triple point of air-free water is defined as 273.16 K . Why is it important that the water be free of air?
67. Meat can be classified as fresh (not frozen) even though it is stored at $-1^{\circ} \mathrm{C}$. Why wouldn’t meat freeze at this temperature?
68. An organic compound has a composition of $93.46 \% \mathrm{C}$ and $6.54 \% \mathrm{H}$ by mass. A solution of 0.090 g of this compound in 1.10 g of camphor melts at $158.4^{\circ} \mathrm{C}$. The melting point of pure camphor is $178.4^{\circ} \mathrm{C} . K_{\mathrm{f}}$ for camphor is $37.7^{\circ} \mathrm{C} / \mathrm{m}$. Assuming ideal solution behavior, what is the molecular formula of the solute? Show your calculations.
69. A sample of $\mathrm{HgCl}_{2}$ weighing 9.41 g is dissolved in 32.75 g of ethanol, $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\left(\mathrm{K}_{\mathrm{b}}=1.20^{\circ} \mathrm{C} / \mathrm{m}\right)$. The boiling point elevation of the solution is $1.27^{\circ} \mathrm{C}$. Is $\mathrm{HgCl}_{2}$ an electrolyte in ethanol? Show your calculations.
70. A salt is known to be an alkali metal fluoride. A quick approximate determination of freezing point indicates that 4 g of the salt dissolved in 100 g of water produces a solution that freezes at about $-1.4^{\circ} \mathrm{C}$. Assuming ideal solution behavior, what is the formula of the salt? Show your calculations.


1292. Exercises - 1292.5. Colloids
71. Identify the dispersed phase and the dispersion medium in each of the following colloidal systems: starch dispersion, smoke, fog, pearl, whipped cream, floating soap, jelly, milk, and ruby.
72. Distinguish between dispersion methods and condensation methods for preparing colloidal systems.
73. How do colloids differ from solutions with regard to dispersed particle size and homogeneity?
74. Explain the cleansing action of soap.
75. How can it be demonstrated that colloidal particles are electrically charged?


1293. CHAPTER 12 <br> Thermodynamics -
Figure 12.1 Geysers are a dramatic display of thermodynamic principles in nature. Water deep within the underground channels of the geyser is under high pressure and heated to high temperature by magma. When a pocket of water near the surface reaches boiling point and is expelled, the resulting drop in pressure causes larger volumes of water to flash boil, forcefully ejecting steam and water in an impressive eruption. (credit: modification of work by Yellowstone National Park)


1294. CHAPTER OUTLINE - 1294.1. Spontaneity
12.2 Entropy
12.3 The Second and Third Laws of Thermodynamics
12.4 Free Energy

INTRODUCTION Among the many capabilities of chemistry is its ability to predict if a process will occur under specified conditions. Thermodynamics, the study of relationships between the energy and work associated with chemical and physical processes, provides this predictive ability. Previous chapters in this text have described various applications of thermochemistry, an important aspect of thermodynamics concerned with the heat flow accompanying chemical reactions and phase transitions. This chapter will introduce additional thermodynamic concepts, including those that enable the prediction of any chemical or physical changes under a given set of conditions.


1295. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Distinguish between spontaneous and nonspontaneous processes
- Describe the dispersal of matter and energy that accompanies certain spontaneous processes

Processes have a natural tendency to occur in one direction under a given set of conditions. Water will naturally flow downhill, but uphill flow requires outside intervention such as the use of a pump. Iron exposed to the earth's atmosphere will corrode, but rust is not converted to iron without intentional chemical treatment. A spontaneous process is one that occurs naturally under certain conditions. A nonspontaneous process, on the other hand, will not take place unless it is "driven" by the continual input of energy from an external source. A process that is spontaneous in one direction under a particular set of conditions is nonspontaneous in the reverse direction. At room temperature and typical atmospheric pressure, for example, ice will spontaneously melt, but water will not spontaneously freeze.

The spontaneity of a process is not correlated to the speed of the process. A spontaneous change may be so rapid that it is essentially instantaneous or so slow that it cannot be observed over any practical period of time. To illustrate this concept, consider the decay of radioactive isotopes, a topic more thoroughly treated in the chapter on nuclear chemistry. Radioactive decay is by definition a spontaneous process in which the nuclei of unstable isotopes emit radiation as they are converted to more stable nuclei. All the decay processes occur spontaneously, but the rates at which different isotopes decay vary widely. Technetium-99m is a popular radioisotope for medical imaging studies that undergoes relatively rapid decay and exhibits a half-life of about six hours. Uranium-238 is the most abundant isotope of uranium, and its decay occurs much more slowly, exhibiting a half-life of more than four billion years (Figure 12.2).

FIGURE 12.2 Both U-238 and Tc-99m undergo spontaneous radioactive decay, but at drastically different rates. Over the course of one week, essentially all of a Tc-99m sample and none of a U-238 sample will have decayed.

As another example, consider the conversion of diamond into graphite (Figure 12.3).

$$
\mathrm{C}(s, \text { diamond }) \longrightarrow \mathrm{C}(s, \text { graphite })
$$


1295. LEARNING OBJECTIVES -
As another example, consider the conversion of diamond into graphite (Figure 12.3).

$$
\mathrm{C}(s, \text { diamond }) \longrightarrow \mathrm{C}(s, \text { graphite })
$$

The phase diagram for carbon indicates that graphite is the stable form of this element under ambient atmospheric pressure, while diamond is the stable allotrope at very high pressures, such as those present during its geologic formation. Thermodynamic calculations of the sort described in the last section of this chapter indicate that the conversion of diamond to graphite at ambient pressure occurs spontaneously, yet diamonds are observed to exist, and persist, under these conditions. Though the process is spontaneous under typical ambient conditions, its rate is extremely slow; so, for all practical purposes diamonds are indeed "forever." Situations such as these emphasize the important distinction between the thermodynamic and the kinetic aspects of a process. In this particular case, diamonds are said to be thermodynamically unstable but kinetically stable under ambient conditions.

FIGURE 12.3 The conversion of carbon from the diamond allotrope to the graphite allotrope is spontaneous at ambient pressure, but its rate is immeasurably slow at low to moderate temperatures. This process is known as graphitization, and its rate can be increased to easily measurable values at temperatures in the 1000-2000 K range. (credit "diamond" photo: modification of work by "Fancy Diamonds"/Flickr; credit "graphite" photo: modification of work by images-of-elements.com/carbon.php)


1296. Dispersal of Matter and Energy -
Extending the discussion of thermodynamic concepts toward the objective of predicting spontaneity, consider now an isolated system consisting of two flasks connected with a closed valve. Initially there is an ideal gas in one flask and the other flask is empty $(P=0)$. (Figure 12.4). When the valve is opened, the gas spontaneously expands to fill both flasks equally. Recalling the definition of pressure-volume work from the chapter on thermochemistry, note that no work has been done because the pressure in a vacuum is zero.

$$
w=-P \Delta V=0 \quad(P=0 \text { in a vacuum })
$$

Note as well that since the system is isolated, no heat has been exchanged with the surroundings ( $q=0$ ). The first law of thermodynamics confirms that there has been no change in the system's internal energy as a result of this process.

$$
\Delta U=q+w=0+0=0
$$

The spontaneity of this process is therefore not a consequence of any change in energy that accompanies the process. Instead, the driving force appears to be related to the greater, more uniform dispersal of matter that results when the gas is allowed to expand. Initially, the system was comprised of one flask containing matter and another flask containing nothing. After the spontaneous expansion took place, the matter was distributed both more widely (occupying twice its original volume) and more uniformly (present in equal amounts in each flask).

FIGURE 12.4 An isolated system consists of an ideal gas in one flask that is connected by a closed valve to a
second flask containing a vacuum. Once the valve is opened, the gas spontaneously becomes evenly distributed between the flasks.

Now consider two objects at different temperatures: object X at temperature $T_{\mathrm{X}}$ and object Y at temperature $T_{\mathrm{Y}}$, with $T_{\mathrm{X}}>T_{\mathrm{Y}}$ (Figure 12.5). When these objects come into contact, heat spontaneously flows from the hotter object $(\mathrm{X})$ to the colder one $(\mathrm{Y})$. This corresponds to a loss of thermal energy by X and a gain of thermal energy by Y.


1296. Dispersal of Matter and Energy -
$$
q_{\mathrm{X}}<0 \quad \text { and } \quad q_{\mathrm{Y}}=-q_{\mathrm{X}}>0
$$

From the perspective of this two-object system, there was no net gain or loss of thermal energy, rather the available thermal energy was redistributed among the two objects. This spontaneous process resulted in a more uniform dispersal of energy.

FIGURE 12.5 When two objects at different temperatures come in contact, heat spontaneously flows from the hotter to the colder object.

As illustrated by the two processes described, an important factor in determining the spontaneity of a process is the extent to which it changes the dispersal or distribution of matter and/or energy. In each case, a spontaneous process took place that resulted in a more uniform distribution of matter or energy.


1298. Redistribution of Matter during a Spontaneous Process -
Describe how matter is redistributed when the following spontaneous processes take place:
(a) A solid sublimes.
(b) A gas condenses.
(c) A drop of food coloring added to a glass of water forms a solution with uniform color.


1299. Solution -
FIGURE 12.6 (credit a: modification of work by Jenny Downing; credit b: modification of work by "Fuzzy Gerdes"/Flickr; credit c: modification of work by Paul A. Flowers)
(a) Sublimation is the conversion of a solid (relatively high density) to a gas (much lesser density). This process yields a much greater dispersal of matter, since the molecules will occupy a much greater volume after the solid-to-gas transition.
(b) Condensation is the conversion of a gas (relatively low density) to a liquid (much greater density). This process yields a much lesser dispersal of matter, since the molecules will occupy a much lesser volume after
the gas-to-liquid transition.
(c) The process in question is diffusion. This process yields a more uniform dispersal of matter, since the initial state of the system involves two regions of different dye concentrations (high in the drop of dye, zero in the water), and the final state of the system contains a single dye concentration throughout.


1300. Check Your Learning -
Describe how energy is redistributed when a spoon at room temperature is placed in a cup of hot coffee.


1301. Answer: -
Heat will spontaneously flow from the hotter object (coffee) to the colder object (spoon), resulting in a more uniform distribution of thermal energy as the spoon warms and the coffee cools.


1302. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Define entropy
- Explain the relationship between entropy and the number of microstates
- Predict the sign of the entropy change for chemical and physical processes

In 1824, at the age of 28, Nicolas Léonard Sadi Carnot (Figure 12.7) published the results of an extensive study regarding the efficiency of steam heat engines. A later review of Carnot's findings by Rudolf Clausius introduced a new thermodynamic property that relates the spontaneous heat flow accompanying a process to the temperature at which the process takes place. This new property was expressed as the ratio of the reversible heat ( $q_{\mathrm{rev}}$ ) and the kelvin temperature ( $T$ ). In thermodynamics, a reversible process is one that takes place at such a slow rate that it is always at equilibrium and its direction can be changed (it can be "reversed") by an infinitesimally small change in some condition. Note that the idea of a reversible process is a formalism required to support the development of various thermodynamic concepts; no real processes are truly reversible, rather they are classified as irreversible.

(a)

(b)

FIGURE 12.7 (a) Nicholas Léonard Sadi Carnot's research into steam-powered machinery and (b) Rudolf Clausius's later study of those findings led to groundbreaking discoveries about spontaneous heat flow processes.

Similar to other thermodynamic properties, this new quantity is a state function, so its change depends only upon the initial and final states of a system. In 1865, Clausius named this property entropy (S) and defined its change for any process as the following:

$$
\Delta S=\frac{q_{\mathrm{rev}}}{T}
$$

The entropy change for a real, irreversible process is then equal to that for the theoretical reversible process that involves the same initial and final states.


1303. Entropy and Microstates -
Following the work of Carnot and Clausius, Ludwig Boltzmann developed a molecular-scale statistical model that related the entropy of a system to the number of microstates ( $W$ ) possible for the system. A microstate is a specific configuration of all the locations and energies of the atoms or molecules that make up a system. The relation between a system's entropy and the number of possible microstates is

$$
S=k \ln W
$$

where $k$ is the Boltzmann constant, $1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K}$.
As for other state functions, the change in entropy for a process is the difference between its final ( $S_{\mathrm{f}}$ ) and initial $\left(S_{\mathrm{i}}\right)$ values:

$$
\Delta S=S_{\mathrm{f}}-S_{\mathrm{i}}=k \ln W_{\mathrm{f}}-k \ln W_{\mathrm{i}}=k \ln \frac{W_{\mathrm{f}}}{W_{\mathrm{i}}}
$$

For processes involving an increase in the number of microstates, $W_{\mathrm{f}}>W_{\mathrm{i}}$, the entropy of the system increases and $\Delta S>0$. Conversely, processes that reduce the number of microstates, $W_{\mathrm{f}}<W_{\mathrm{i}}$, yield a decrease in system entropy, $\Delta S<0$. This molecular-scale interpretation of entropy provides a link to the probability that a process will occur as illustrated in the next paragraphs.

Consider the general case of a system comprised of $N$ particles distributed among $n$ boxes. The number of microstates possible for such a system is $n^{N}$. For example, distributing four particles among two boxes will result in $2^{4}=16$ different microstates as illustrated in Figure 12.8. Microstates with equivalent particle arrangements (not considering individual particle identities) are grouped together and are called distributions. The probability that a system will exist with its components in a given distribution is proportional to the number of microstates within the distribution. Since entropy increases logarithmically with the number of microstates, the most probable distribution is therefore the one of greatest entropy.
(a)

(b)

(c)

(d)

(e)


1303. Entropy and Microstates -
(b)

(c)

(d)

(e)

FIGURE 12.8 The sixteen microstates associated with placing four particles in two boxes are shown. The microstates are collected into five distributions-(a), (b), (c), (d), and (e)-based on the numbers of particles in each box.

For this system, the most probable configuration is one of the six microstates associated with distribution (c) where the particles are evenly distributed between the boxes, that is, a configuration of two particles in each
box. The probability of finding the system in this configuration is $\frac{6}{16}$ or $\frac{3}{8}$. The least probable configuration of the system is one in which all four particles are in one box, corresponding to distributions (a) and (e), each with a probability of $\frac{1}{16}$. The probability of finding all particles in only one box (either the left box or right box) is then $\left(\frac{1}{16}+\frac{1}{16}\right)=\frac{2}{16}$ or $\frac{1}{8}$.
As you add more particles to the system, the number of possible microstates increases exponentially $\left(2^{N}\right)$. A macroscopic (laboratory-sized) system would typically consist of moles of particles ( $N \sim 10^{23}$ ), and the corresponding number of microstates would be staggeringly huge. Regardless of the number of particles in the system, however, the distributions in which roughly equal numbers of particles are found in each box are always the most probable configurations.

This matter dispersal model of entropy is often described qualitatively in terms of the disorder of the system. By this description, microstates in which all the particles are in a single box are the most ordered, thus possessing the least entropy. Microstates in which the particles are more evenly distributed among the boxes are more disordered, possessing greater entropy.


1303. Entropy and Microstates -
The previous description of an ideal gas expanding into a vacuum (Figure 12.4) is a macroscopic example of this particle-in-a-box model. For this system, the most probable distribution is confirmed to be the one in which the matter is most uniformly dispersed or distributed between the two flasks. Initially, the gas molecules are confined to just one of the two flasks. Opening the valve between the flasks increases the volume available to the gas molecules and, correspondingly, the number of microstates possible for the system. Since $W_{\mathrm{f}}>W_{\mathrm{i}}$, the expansion process involves an increase in entropy $(\Delta S>0)$ and is spontaneous.

A similar approach may be used to describe the spontaneous flow of heat. Consider a system consisting of two objects, each containing two particles, and two units of thermal energy (represented as "*") in Figure 12.9. The hot object is comprised of particles $\mathbf{A}$ and $\mathbf{B}$ and initially contains both energy units. The cold object is comprised of particles $\mathbf{C}$ and $\mathbf{D}$, which initially has no energy units. Distribution (a) shows the three microstates possible for the initial state of the system, with both units of energy contained within the hot object. If one of the two energy units is transferred, the result is distribution (b) consisting of four microstates. If both energy units are transferred, the result is distribution (c) consisting of three microstates. Thus, we may describe this system by a total of ten microstates. The probability that the heat does not flow when the two objects are brought into contact, that is, that the system remains in distribution (a), is $\frac{3}{10}$. More likely is the flow of heat to yield one of the other two distribution, the combined probability being $\frac{7}{10}$. The most likely result is the flow of heat to yield the uniform dispersal of energy represented by distribution (b), the probability of this configuration being $\frac{4}{10}$. This supports the common observation that placing hot and cold objects in contact results in spontaneous heat flow that ultimately equalizes the objects' temperatures. And, again, this spontaneous process is also characterized by an increase in system entropy.


1303. Entropy and Microstates -
FIGURE 12.9 This shows a microstate model describing the flow of heat from a hot object to a cold object. (a) Before the heat flow occurs, the object comprised of particles $\mathbf{A}$ and $\mathbf{B}$ contains both units of energy and as represented by a distribution of three microstates. (b) If the heat flow results in an even dispersal of energy (one energy unit transferred), a distribution of four microstates results. (c) If both energy units are transferred, the resulting distribution has three microstates.


1305. Determination of $\Delta S$ -
Calculate the change in entropy for the process depicted below.


1306. Solution -
The initial number of microstates is one, the final six:

$$
\Delta S=k \ln \frac{W_{\mathrm{c}}}{W_{\mathrm{a}}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \times \ln \frac{6}{1}=2.47 \times 10^{-23} \mathrm{~J} / \mathrm{K}
$$

The sign of this result is consistent with expectation; since there are more microstates possible for the final state than for the initial state, the change in entropy should be positive.


1307. Check Your Learning -
Consider the system shown in Figure 12.9. What is the change in entropy for the process where all the energy is transferred from the hot object ( $\mathbf{A B}$ ) to the cold object ( $\mathbf{C D}$ )?


1308. Answer: -
$0 \mathrm{~J} / \mathrm{K}$


1309. Predicting the Sign of $\Delta S$ -
The relationships between entropy, microstates, and matter/energy dispersal described previously allow us to make generalizations regarding the relative entropies of substances and to predict the sign of entropy changes for chemical and physical processes. Consider the phase changes illustrated in Figure 12.10. In the solid phase, the atoms or molecules are restricted to nearly fixed positions with respect to each other and are capable of only modest oscillations about these positions. With essentially fixed locations for the system's component particles, the number of microstates is relatively small. In the liquid phase, the atoms or molecules are free to move over and around each other, though they remain in relatively close proximity to one another. This increased freedom of motion results in a greater variation in possible particle locations, so the number of microstates is correspondingly greater than for the solid. As a result, $S_{\text {liquid }}>S_{\text {solid }}$ and the process of converting a substance from solid to liquid (melting) is characterized by an increase in entropy, $\Delta S>0$. By the same logic, the reciprocal process (freezing) exhibits a decrease in entropy, $\Delta S<0$.

FIGURE 12.10 The entropy of a substance increases ( $\Delta S>0$ ) as it transforms from a relatively ordered solid, to a less-ordered liquid, and then to a still less-ordered gas. The entropy decreases $(\Delta S<0)$ as the substance transforms from a gas to a liquid and then to a solid.

Now consider the gaseous phase, in which a given number of atoms or molecules occupy a much greater volume than in the liquid phase. Each atom or molecule can be found in many more locations, corresponding to a much greater number of microstates. Consequently, for any substance, $S_{\text {gas }}>S_{\text {liquid }}>S_{\text {solid }}$, and the processes of vaporization and sublimation likewise involve increases in entropy, $\Delta S>0$. Likewise, the reciprocal phase transitions, condensation and deposition, involve decreases in entropy, $\Delta S<0$.


1309. Predicting the Sign of $\Delta S$ -
According to kinetic-molecular theory, the temperature of a substance is proportional to the average kinetic energy of its particles. Raising the temperature of a substance will result in more extensive vibrations of the particles in solids and more rapid translations of the particles in liquids and gases. At higher temperatures, the distribution of kinetic energies among the atoms or molecules of the substance is also broader (more dispersed) than at lower temperatures. Thus, the entropy for any substance increases with temperature (Figure 12.11).

FIGURE 12.11 Entropy increases as the temperature of a substance is raised, which corresponds to the greater spread of kinetic energies. When a substance undergoes a phase transition, its entropy changes significantly.


1310. LINK TO LEARNING -
Try this simulator (http://openstax.org/l/16freemotion) with interactive visualization of the dependence of
particle location and freedom of motion on physical state and temperature.

The entropy of a substance is influenced by the structure of the particles (atoms or molecules) that comprise the substance. With regard to atomic substances, heavier atoms possess greater entropy at a given temperature than lighter atoms, which is a consequence of the relation between a particle's mass and the spacing of quantized translational energy levels (a topic beyond the scope of this text). For molecules, greater numbers of atoms increase the number of ways in which the molecules can vibrate and thus the number of possible microstates and the entropy of the system.

Finally, variations in the types of particles affects the entropy of a system. Compared to a pure substance, in which all particles are identical, the entropy of a mixture of two or more different particle types is greater. This is because of the additional orientations and interactions that are possible in a system comprised of nonidentical components. For example, when a solid dissolves in a liquid, the particles of the solid experience both a greater freedom of motion and additional interactions with the solvent particles. This corresponds to a more uniform dispersal of matter and energy and a greater number of microstates. The process of dissolution therefore involves an increase in entropy, $\Delta S>0$.

Considering the various factors that affect entropy allows us to make informed predictions of the sign of $\Delta S$ for various chemical and physical processes as illustrated in Example 12.3.


1312. Predicting the Sign of $\Delta S$ -
Predict the sign of the entropy change for the following processes. Indicate the reason for each of your predictions.
(a) One mole liquid water at room temperature $\longrightarrow$ one mole liquid water at $50^{\circ} \mathrm{C}$
(b) $\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s)$
(c) $\mathrm{C}_{6} \mathrm{H}_{6}(l)+\frac{15}{2} \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(l)$
$(d) \mathrm{NH}_{3}(s) \longrightarrow \mathrm{NH}_{3}(l)$


1313. Solution -
(a) positive, temperature increases
(b) negative, reduction in the number of ions (particles) in solution, decreased dispersal of matter
(c) negative, net decrease in the amount of gaseous species
(d) positive, phase transition from solid to liquid, net increase in dispersal of matter


1314. Check Your Learning -
Predict the sign of the entropy change for the following processes. Give a reason for your prediction.
(a) $\mathrm{NaNO}_{3}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{NO}_{3}{ }^{-}(a q)$
(b) the freezing of liquid water
(c) $\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g)$
(d) $\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaO}(s)+\mathrm{CO}_{2}(g)$


1315. Answer: -
(a) Positive; The solid dissolves to give an increase of mobile ions in solution. (b) Negative; The liquid becomes a more ordered solid. (c) Positive; The relatively ordered solid becomes a gas. (d) Positive; There is a net increase in the amount of gaseous species.


1315. Answer: - 1315.1. The Second and Third Laws of Thermodynamics
LEARNING OBJECTIVES
By the end of this section, you will be able to:

- State and explain the second and third laws of thermodynamics
- Calculate entropy changes for phase transitions and chemical reactions under standard conditions


1316. The Second Law of Thermodynamics -
In the quest to identify a property that may reliably predict the spontaneity of a process, a promising candidate has been identified: entropy. Processes that involve an increase in entropy of the system ( $\Delta S>0$ ) are very often spontaneous; however, examples to the contrary are plentiful. By expanding consideration of entropy changes to include the surroundings, we may reach a significant conclusion regarding the relation between this property and spontaneity. In thermodynamic models, the system and surroundings comprise everything, that is, the universe, and so the following is true:

$$
\Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}
$$

To illustrate this relation, consider again the process of heat flow between two objects, one identified as the system and the other as the surroundings. There are three possibilities for such a process:

1. The objects are at different temperatures, and heat flows from the hotter to the cooler object. This is always observed to occur spontaneously. Designating the hotter object as the system and invoking the definition of entropy yields the following:

$$
\Delta S_{\mathrm{sys}}=\frac{-q_{\mathrm{rev}}}{T_{\mathrm{sys}}} \quad \text { and } \quad \Delta S_{\mathrm{surr}}=\frac{q_{\mathrm{rev}}}{T_{\mathrm{surr}}}
$$

The magnitudes of $-q_{\mathrm{rev}}$ and $q_{\mathrm{rev}}$ are equal, their opposite arithmetic signs denoting loss of heat by the system and gain of heat by the surroundings. Since $T_{\text {sys }}>T_{\text {surr }}$ in this scenario, the entropy decrease of the system will be less than the entropy increase of the surroundings, and so the entropy of the universe will increase:

$$
\begin{aligned}
& \left|\Delta S_{\text {sys }}\right|<\left|\Delta S_{\text {surr }}\right| \\
& \Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surr }}>0
\end{aligned}
$$


1316. The Second Law of Thermodynamics -
2. The objects are at different temperatures, and heat flows from the cooler to the hotter object. This is never observed to occur spontaneously. Again designating the hotter object as the system and invoking the definition of entropy yields the following:

$$
\Delta S_{\mathrm{sys}}=\frac{q_{\mathrm{rev}}}{T_{\mathrm{sys}}} \quad \text { and } \quad \Delta S_{\mathrm{surr}}=\frac{-q_{\mathrm{rev}}}{T_{\mathrm{surr}}}
$$

The arithmetic signs of $q_{\mathrm{rev}}$ denote the gain of heat by the system and the loss of heat by the surroundings. The magnitude of the entropy change for the surroundings will again be greater than that for the system, but in this case, the signs of the heat changes (that is, the direction of the heat flow) will yield a negative value for $\Delta S_{\text {univ. }}$. This process involves a decrease in the entropy of the universe.
3. The objects are at essentially the same temperature, $T_{\text {sys }} \approx T_{\text {surr }}$, and so the magnitudes of the entropy changes are essentially the same for both the system and the surroundings. In this case, the entropy change of the universe is zero, and the system is at equilibrium.

$$
\begin{aligned}
& \left|\Delta S_{\text {sys }}\right| \approx\left|\Delta S_{\text {surr }}\right| \\
& \Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\Delta S_{\text {surf }}=0
\end{aligned}
$$

These results lead to a profound statement regarding the relation between entropy and spontaneity known as the second law of thermodynamics: all spontaneous changes cause an increase in the entropy of the universe. A summary of these three relations is provided in Table 12.1.

The Second Law of Thermodynamics

| $\Delta S_{\text {univ }}>0$ | spontaneous |
| :--- | :--- |
| $\Delta S_{\text {univ }}<0$ | nonspontaneous (spontaneous in opposite direction) |
| $\Delta S_{\text {univ }}=0$ | at equilibrium |


1316. The Second Law of Thermodynamics -
TABLE 12.1

For many realistic applications, the surroundings are vast in comparison to the system. In such cases, the heat gained or lost by the surroundings as a result of some process represents a very small, nearly infinitesimal, fraction of its total thermal energy. For example, combustion of a fuel in air involves transfer of heat from a system (the fuel and oxygen molecules undergoing reaction) to surroundings that are infinitely more massive (the earth's atmosphere). As a result, $q_{\text {surr }}$ is a good approximation of $q_{\text {rev }}$, and the second law may be stated as the following:

$$
\Delta S_{\mathrm{univ}}=\Delta S_{\mathrm{sys}}+\Delta S_{\mathrm{surr}}=\Delta S_{\mathrm{sys}}+\frac{q_{\mathrm{surr}}}{T}
$$

We may use this equation to predict the spontaneity of a process as illustrated in Example 12.4.


1318. Will Ice Spontaneously Melt? -
The entropy change for the process

$$
\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)
$$

is $22.1 \mathrm{~J} / \mathrm{K}$ and requires that the surroundings transfer 6.00 kJ of heat to the system. Is the process spontaneous at $-10.00^{\circ} \mathrm{C}$ ? Is it spontaneous at $+10.00^{\circ} \mathrm{C}$ ?


1319. Solution -
We can assess the spontaneity of the process by calculating the entropy change of the universe. If $\Delta S_{\text {univ }}$ is positive, then the process is spontaneous. At both temperatures, $\Delta S_{\text {sys }}=22.1 \mathrm{~J} / \mathrm{K}$ and $q_{\text {surr }}=-6.00 \mathrm{~kJ}$.

At $-10.00{ }^{\circ} \mathrm{C}(263.15 \mathrm{~K})$, the following is true:

$$
\begin{aligned}
\Delta S_{\text {univ }} & =\Delta S_{\text {sys }}+\Delta S_{\text {surr }}=\Delta S_{\text {sys }}+\frac{q_{\text {surr }}}{T} \\
& =22.1 \mathrm{~J} / \mathrm{K}+\frac{-6.00 \times 10^{3} \mathrm{~J}}{263.15 \mathrm{~K}}=-0.7 \mathrm{~J} / \mathrm{K}
\end{aligned}
$$

$S_{\text {univ }}<0$, so melting is nonspontaneous (not spontaneous) at $-10.0^{\circ} \mathrm{C}$.
At $10.00^{\circ} \mathrm{C}(283.15 \mathrm{~K})$, the following is true:

$$
\begin{gathered}
\Delta S_{\text {univ }}=\Delta S_{\text {sys }}+\frac{q_{\text {surr }}}{T} \\
=22.1 \mathrm{~J} / \mathrm{K}+\frac{-6.00 \times 10^{3} \mathrm{~J}}{283.15 \mathrm{~K}}=+0.9 \mathrm{~J} / \mathrm{K}
\end{gathered}
$$

$S_{\text {univ }}>0$, so melting is spontaneous at $10.00^{\circ} \mathrm{C}$.


1320. Check Your Learning -
Using this information, determine if liquid water will spontaneously freeze at the same temperatures. What can you say about the values of $S_{\text {univ }}$ ?


1321. Answer: -
Entropy is a state function, so $\Delta S_{\text {freezing }}=-\Delta S_{\text {melting }}=-22.1 \mathrm{~J} / \mathrm{K}$ and $q_{\text {surr }}=+6.00 \mathrm{~kJ}$. At $-10.00^{\circ} \mathrm{C}$ spontaneous, $+0.7 \mathrm{~J} / \mathrm{K}$; at $+10.00^{\circ} \mathrm{C}$ nonspontaneous, $-0.9 \mathrm{~J} / \mathrm{K}$.


1322. The Third Law of Thermodynamics -
The previous section described the various contributions of matter and energy dispersal that contribute to the entropy of a system. With these contributions in mind, consider the entropy of a pure, perfectly crystalline solid possessing no kinetic energy (that is, at a temperature of absolute zero, 0 K ). This system may be described by a single microstate, as its purity, perfect crystallinity and complete lack of motion means there is but one possible location for each identical atom or molecule comprising the crystal ( $W=1$ ). According to the Boltzmann equation, the entropy of this system is zero.

$$
S=k \ln W=k \ln (1)=0
$$

This limiting condition for a system's entropy represents the third law of thermodynamics: the entropy of a pure, perfect crystalline substance at 0 K is zero.

Careful calorimetric measurements can be made to determine the temperature dependence of a substance's entropy and to derive absolute entropy values under specific conditions. Standard entropies ( $\boldsymbol{S}^{\circ}$ ) are for one mole of substance under standard conditions (a pressure of 1 bar and a temperature of 298.15 K ; see details regarding standard conditions in the thermochemistry chapter of this text). The standard entropy change ( $\boldsymbol{\Delta} \boldsymbol{S}^{\circ}$ ) for a reaction may be computed using standard entropies as shown below:

$$
\Delta S^{\circ}=\sum v S^{\circ}(\text { products })-\sum \nu S^{\circ}(\text { reactants })
$$

where $\nu$ represents stoichiometric coefficients in the balanced equation representing the process. For example, $\Delta S^{\circ}$ for the following reaction at room temperature

$$
m \mathrm{~A}+n \mathrm{~B} \longrightarrow x \mathrm{C}+y \mathrm{D}
$$

is computed as:

$$
=\left[x S^{\circ}(\mathrm{C})+y S^{\circ}(\mathrm{D})\right]-\left[m S^{\circ}(\mathrm{A})+n S^{\circ}(\mathrm{B})\right]
$$


1322. The Third Law of Thermodynamics -
A partial listing of standard entropies is provided in Table 12.2, and additional values are provided in Appendix G. The example exercises that follow demonstrate the use of $S^{\circ}$ values in calculating standard entropy changes for physical and chemical processes.

| Substance | $S^{\circ}\left(\mathbf{J ~ m o l}^{\mathbf{- 1}} \mathbf{K}^{\mathbf{- 1}}\right)$ |  |
| :--- | :--- | :---: |
| carbon |  |  |
| $\mathrm{C}(s$, graphite $)$ | 5.740 |  |
| $\mathrm{C}(s$, diamond $)$ | 2.38 |  |
| $\mathrm{CO}(\mathrm{g})$ | 197.7 |  |
| $\mathrm{CO}_{2}(g)$ | 213.8 |  |
| $\mathrm{CH}_{4}(\mathrm{~g})$ | 186.3 |  |
| $\mathrm{C}_{2} \mathrm{H}_{4}(\mathrm{~g})$ | 219.5 |  |
| $\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{~g})$ | 229.5 |  |

| $\mathrm{CH}_{3} \mathrm{OH}(\mathrm{I})$ | 126.8 |
| :--- | :---: |
| $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(\mathrm{l})$ | 160.7 |
| hydrogen |  |
| $\mathrm{H}_{2}(g)$ | 130.57 |
| $\mathrm{H}(g)$ | 114.6 |
| $\mathrm{H}_{2} \mathrm{O}(g)$ | 188.71 |
| $\mathrm{H}_{2} \mathrm{O}(\mathrm{l})$ | 69.91 |
| $\mathrm{HCI}(g)$ | 186.8 |
| $\mathrm{H}_{2} \mathrm{~S}(g)$ | 205.7 |
| oxygen | 205.03 |
| $\mathrm{O}_{2}(g)$ |  |


1322. The Third Law of Thermodynamics -
TABLE 12.2 Standard entropies for selected substances measured at 1 atm and 298.15 K. (Values are approximately equal to those measured at 1 bar, the currently accepted standard state pressure.)


1324. Determination of $\boldsymbol{\Delta} \boldsymbol{S}^{\circ}$ -
Calculate the standard entropy change for the following process:

$$
\mathrm{H}_{2} \mathrm{O}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)
$$


1325. Solution -
Calculate the entropy change using standard entropies as shown above:

$$
\Delta S^{\circ}=(1 \mathrm{~mol})\left(70.0 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)-(1 \mathrm{~mol})\left(188.8 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right)=-118.8 \mathrm{~J} / \mathrm{K}
$$

The value for $\Delta S^{\circ}$ is negative, as expected for this phase transition (condensation), which the previous section discussed.


1326. Check Your Learning -
Calculate the standard entropy change for the following process:

$$
\mathrm{H}_{2}(g)+\mathrm{C}_{2} \mathrm{H}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)
$$


1327. Answer: -
$-120.6 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$


1329. Determination of $\boldsymbol{\Delta} \boldsymbol{S}^{\circ}$ -
Calculate the standard entropy change for the combustion of methanol, $\mathrm{CH}_{3} \mathrm{OH}$ :

$$
2 \mathrm{CH}_{3} \mathrm{OH}(l)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+4 \mathrm{H}_{2} \mathrm{O}(l)
$$


1330. Solution -
Calculate the entropy change using standard entropies as shown above:

$$
\begin{gathered}
\Delta S^{\circ}=\sum v S^{\circ}(\text { products })-\sum \nu S^{\circ}(\text { reactants }) \\
{\left[2 \mathrm{~mol} \times S^{\circ}\left(\mathrm{CO}_{2}(g)\right)+4 \mathrm{~mol} \times S^{\circ}\left(\mathrm{H}_{2} \mathrm{O}(l)\right)\right]-\left[2 \mathrm{~mol} \times S^{\circ}\left(\mathrm{CH}_{3} \mathrm{OH}(l)\right)+3 \mathrm{~mol} \times S^{\circ}\left(\mathrm{O}_{2}(g)\right)\right]} \\
=\{[2(213.8)+4 \times 70.0]-[2(126.8)+3(205.03)]\}=-161.1 \mathrm{~J} / \mathrm{K}
\end{gathered}
$$


1331. Check Your Learning -
Calculate the standard entropy change for the following reaction:

$$
\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{~s}) \longrightarrow \mathrm{CaO}(s)+\mathrm{H}_{2} \mathrm{O}(l)
$$


1332. Answer: -
$24.7 \mathrm{~J} / \mathrm{K}$


1333. LEARNING OBJECTIVES -
By the end of this section, you will be able to:

- Define Gibbs free energy, and describe its relation to spontaneity
- Calculate free energy change for a process using free energies of formation for its reactants and products
- Calculate free energy change for a process using enthalpies of formation and the entropies for its reactants and products

One of the challenges of using the second law of thermodynamics to determine if a process is spontaneous is that it requires measurements of the entropy change for the system and the entropy change for the surroundings. An alternative approach involving a new thermodynamic property defined in terms of system properties only was introduced in the late nineteenth century by American mathematician Josiah Willard Gibbs. This new property is called the Gibbs free energy ( $\boldsymbol{G}$ ) (or simply the free energy), and it is defined in terms of a system's enthalpy and entropy as the following:

$$
G=H-T S
$$

Free energy is a state function, and at constant temperature and pressure, the free energy change ( $\boldsymbol{\Delta} \boldsymbol{G}$ ) may be expressed as the following:

$$
\Delta G=\Delta H-T \Delta S
$$

(For simplicity's sake, the subscript "sys" will be omitted henceforth.)
The relationship between this system property and the spontaneity of a process may be understood by recalling the previously derived second law expression:

$$
\Delta S_{\mathrm{univ}}=\Delta S+\frac{q_{\mathrm{surr}}}{T}
$$

The first law requires that $q_{\text {surr }}=-q_{\text {sys }}$, and at constant pressure $q_{\text {sys }}=\Delta H$, so this expression may be rewritten as:

$$
\Delta S_{\mathrm{univ}}=\Delta S-\frac{\Delta H}{T}
$$

Multiplying both sides of this equation by $-T$, and rearranging yields the following:

$$
-T \Delta S_{\mathrm{univ}}=\Delta H-T \Delta S
$$

Comparing this equation to the previous one for free energy change shows the following relation:

$$
\Delta G=-T \Delta S_{\text {univ }}
$$


1333. LEARNING OBJECTIVES -
Comparing this equation to the previous one for free energy change shows the following relation:

$$
\Delta G=-T \Delta S_{\text {univ }}
$$

The free energy change is therefore a reliable indicator of the spontaneity of a process, being directly related to the previously identified spontaneity indicator, $\Delta S_{\text {univ }}$. Table 12.3 summarizes the relation between the spontaneity of a process and the arithmetic signs of these indicators.

Relation between Process Spontaneity and Signs of Thermodynamic Properties

| $\Delta S_{\text {univ }}>0$ | $\Delta G<0$ | spontaneous |
| :---: | :---: | :---: |
| $\Delta S_{\text {univ }}<0$ | $\Delta G>0$ | nonspontaneous |
| $\Delta S_{\text {univ }}=0$ | $\Delta G=0$ | at equilibrium |

TABLE 12.3


1334. What's "Free" about $\Delta G$ ? -
In addition to indicating spontaneity, the free energy change also provides information regarding the amount of useful work ( $w$ ) that may be accomplished by a spontaneous process. Although a rigorous treatment of this subject is beyond the scope of an introductory chemistry text, a brief discussion is helpful for gaining a better perspective on this important thermodynamic property.

For this purpose, consider a spontaneous, exothermic process that involves a decrease in entropy. The free energy, as defined by

$$
\Delta G=\Delta H-T \Delta S
$$

may be interpreted as representing the difference between the energy produced by the process, $\Delta H$, and the energy lost to the surroundings, $T \Delta S$. The difference between the energy produced and the energy lost is the energy available (or "free") to do useful work by the process, $\Delta G$. If the process somehow could be made to take place under conditions of thermodynamic reversibility, the amount of work that could be done would be maximal:

$$
\Delta G=w_{\max }
$$

However, as noted previously in this chapter, such conditions are not realistic. In addition, the technologies used to extract work from a spontaneous process (e.g., automobile engine, steam turbine) are never 100\% efficient, and so the work done by these processes is always less than the theoretical maximum. Similar reasoning may be applied to a nonspontaneous process, for which the free energy change represents the minimum amount of work that must be done on the system to carry out the process.


1335. Calculating Free Energy Change -
Free energy is a state function, so its value depends only on the conditions of the initial and final states of the system. A convenient and common approach to the calculation of free energy changes for physical and chemical reactions is by use of widely available compilations of standard state thermodynamic data. One method involves the use of standard enthalpies and entropies to compute standard free energy changes, $\boldsymbol{\Delta} \boldsymbol{G}^{\circ}$, according to the following relation:

$$
\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}
$$


1337. Using Standard Enthalpy and Entropy Changes to Calculate $\boldsymbol{\Delta} \boldsymbol{G}^{\circ}$ -
Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the vaporization of water at room temperature ( 298 K ). What does the computed value for $\Delta G^{\circ}$ say about the spontaneity of this process?


1338. Solution -
The process of interest is the following:

$$
\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g)
$$

The standard change in free energy may be calculated using the following equation:

$$
\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}
$$

From Appendix G:

| Substance | $\Delta H_{\mathrm{f}}^{\circ}(\mathrm{kJ} / \mathrm{mol})$ |  |
| :---: | :---: | :---: |
| $\mathrm{H}_{2} \mathrm{O}(\mathrm{I})$ | -286.83 | 70.0 |
| $\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$ | -241.82 | 188.8 |

Using the appendix data to calculate the standard enthalpy and entropy changes yields:


1338. Solution -
Using the appendix data to calculate the standard enthalpy and entropy changes yields:

$$
\begin{gathered}
\Delta H^{\circ}=\Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}(g)\right)-\Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}(l)\right) \\
=[-241.82 \mathrm{~kJ} / \mathrm{mol}-(-286.83)] \mathrm{kJ} / \mathrm{mol}=45.01 \mathrm{~kJ} \\
\Delta S^{\circ}=1 \mathrm{~mol} \times S^{\circ}\left(\mathrm{H}_{2} \mathrm{O}(g)\right)-1 \mathrm{~mol} \times S^{\circ}\left(\mathrm{H}_{2} \mathrm{O}(l)\right) \\
=(1 \mathrm{~mol}) 188.8 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{~K}-(1 \mathrm{~mol}) 70.0 \mathrm{~J} / \mathrm{mol} \mathrm{~K}=118.8 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{~K} \\
\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}
\end{gathered}
$$

Substitution into the standard free energy equation yields:

$$
\begin{gathered}
\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ} \\
=45.01 \mathrm{~kJ}-(298 \mathrm{~K} \times 118.8 \mathrm{~J} / \mathrm{K}) \times \frac{1 \mathrm{~kJ}}{1000 \mathrm{~J}} \\
45.01 \mathrm{~kJ}-35.4 \mathrm{~kJ}=9.6 \mathrm{~kJ}
\end{gathered}
$$


1338. Solution -
At $298 \mathrm{~K}\left(25^{\circ} \mathrm{C}\right) \Delta G^{\circ}>0$, so boiling is nonspontaneous (not spontaneous).


1339. Check Your Learning -
Use standard enthalpy and entropy data from Appendix G to calculate the standard free energy change for the reaction shown here ( 298 K ). What does the computed value for $\Delta G^{\circ}$ say about the spontaneity of this process?

$$
\mathrm{C}_{2} \mathrm{H}_{6}(g) \longrightarrow \mathrm{H}_{2}(g)+\mathrm{C}_{2} \mathrm{H}_{4}(g)
$$


1340. Answer: -
$\Delta G^{\circ}=102.0 \mathrm{~kJ} / \mathrm{mol}$; the reaction is nonspontaneous (not spontaneous) at $25^{\circ} \mathrm{C}$.

The standard free energy change for a reaction may also be calculated from standard free energy of formation $\Delta \mathbf{G}^{\mathbf{o}} \mathbf{f}$ values of the reactants and products involved in the reaction. The standard free energy of formation is the free energy change that accompanies the formation of one mole of a substance from its elements in their standard states. Similar to the standard enthalpy of formation, $\Delta G_{\mathrm{f}}^{\circ}$ is by definition zero for elemental substances in their standard states. The approach used to calculate $\Delta G^{\circ}$ for a reaction from $\Delta G_{\mathrm{f}}^{\circ}$ values is the same as that demonstrated previously for enthalpy and entropy changes. For the reaction

$$
m \mathrm{~A}+n \mathrm{~B} \longrightarrow x \mathrm{C}+y \mathrm{D}
$$

the standard free energy change at room temperature may be calculated as

$$
\begin{gathered}
\Delta G^{\circ}=\sum \nu \Delta G^{\circ} \text { (products) }-\sum \nu \Delta G^{\circ}(\text { reactants }) \\
=\left[x \Delta G_{\mathrm{f}}^{\circ}(\mathrm{C})+y \Delta G_{\mathrm{f}}^{\circ}(\mathrm{D})\right]-\left[m \Delta G_{\mathrm{f}}^{\circ}(\mathrm{A})+n \Delta G_{\mathrm{f}}^{\circ}(\mathrm{B})\right] .
\end{gathered}
$$


1342. Using Standard Free Energies of Formation to Calculate $\Delta \boldsymbol{G}^{\circ}$ -
Consider the decomposition of yellow mercury(II) oxide.

$$
\mathrm{HgO}(s, \text { yellow }) \longrightarrow \mathrm{Hg}(l)+\frac{1}{2} \mathrm{O}_{2}(g)
$$

Calculate the standard free energy change at room temperature, $\Delta G^{\circ}$, using (a) standard free energies of formation and (b) standard enthalpies of formation and standard entropies. Do the results indicate the reaction to be spontaneous or nonspontaneous under standard conditions?


1343. Solution -
The required data are available in Appendix G and are shown here.

| Compound | $\Delta G_{\mathrm{f}}^{\circ}(\mathrm{kJ} / \mathrm{mol})$ | $\Delta H_{\mathrm{f}}^{\circ}(\mathrm{kJ} / \mathrm{mol})$ | $S^{\circ}(\mathrm{J} / \mathrm{K} \cdot \mathrm{mol})$ |
| :---: | :---: | :---: | :---: |
| $\mathrm{HgO}(s$, yellow $)$ | -58.43 | -90.46 | 71.13 |
| $\mathrm{Hg}(\mathrm{l})$ | 0 | 0 | 75.9 |
| $\mathrm{O}_{2}(g)$ | 0 | 0 | 205.2 |

(a) Using free energies of formation:

$$
\begin{gathered}
\Delta G^{\circ}=\sum \nu G_{\mathrm{f}}^{\circ}(\text { products })-\sum \nu \Delta G_{\mathrm{f}}^{\circ}(\text { reactants }) \\
=\left[1 \Delta G_{\mathrm{f}}^{\circ} \operatorname{Hg}(l)+\frac{1}{2} \Delta G_{\mathrm{f}}^{\circ} \mathrm{O}_{2}(g)\right]-1 \Delta G_{\mathrm{f}}^{\circ} \mathrm{HgO}(s, \text { yellow }) \\
=\left[1 \mathrm{~mol}(0 \mathrm{~kJ} / \mathrm{mol})+\frac{1}{2} \operatorname{mol}(0 \mathrm{~kJ} / \mathrm{mol})\right]-1 \mathrm{~mol}(-58.43 \mathrm{~kJ} / \mathrm{mol})=58.43 \mathrm{~kJ} / \mathrm{mol}
\end{gathered}
$$

(b) Using enthalpies and entropies of formation:


1343. Solution -
(b) Using enthalpies and entropies of formation:

$$
\begin{gathered}
\Delta H^{\circ}=\sum \nu \Delta H_{\mathrm{f}}^{\circ}(\text { products })-\sum \nu \Delta H_{\mathrm{f}}^{\circ}(\text { reactants }) \\
=\left[1 \Delta H_{\mathrm{f}}^{\circ} \mathrm{Hg}(l)+\frac{1}{2} \Delta H_{\mathrm{f}}^{\circ} \mathrm{O}_{2}(g)\right]-1 \Delta H_{\mathrm{f}}^{\circ} \mathrm{HgO}(s, \text { yellow }) \\
=\left[1 \mathrm{~mol}(0 \mathrm{~kJ} / \mathrm{mol})+\frac{1}{2} \operatorname{mol}(0 \mathrm{~kJ} / \mathrm{mol})\right]-1 \mathrm{~mol}(-90.46 \mathrm{~kJ} / \mathrm{mol})=90.46 \mathrm{~kJ} / \mathrm{mol} \\
\Delta S^{\circ}=\sum \nu \Delta S^{\circ}(\text { products })-\sum \nu \Delta S^{\circ}(\text { reactants }) \\
=\left[1 \Delta S^{\circ} \mathrm{Hg}(l)+\frac{1}{2} \Delta S^{\circ} \mathrm{O}_{2}(g)\right]-1 \Delta S^{\circ} \mathrm{HgO}(s, \text { yellow })
\end{gathered}
$$