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The extent to which the vapor pressure of a solvent is lowered and the boiling point is elevated depends on the total number of solute particles present in a given amount of solvent, not on the mass or size or chemical identities of the particles. A 1 m aqueous solution of sucrose (342 g/mol) and a 1 m aqueous solution of ethylene glycol (62 g/mol) will exhibit the same boiling point because each solution has one mole of solute particles (molecules) per kilogram of solvent.
Calculating the Boiling Point of a Solution
Assuming ideal solution behavior, what is the boiling point of a 0.33 m solution of a nonvolatile solute in benzene?
Solution
Use the equation relating boiling point elevation to solute molality to solve this problem in two steps. |
Step 2. Add the boiling point elevation to the pure solvent’s boiling point.
Check Your Learning
Assuming ideal solution behavior, what is the boiling point of the antifreeze described in ?
Answer:
109.2 °C
The Boiling Point of an Iodine Solution
Find the boiling point of a solution of 92.1 g of iodine, I2, in 800.0 g of chloroform, CHCl3, assuming that the iodine is nonvolatile and that the solution is ideal.
Solution
A four-step approach to solving this problem is outlined below.
Step 1. Convert from grams to moles of I2 using the molar mass of I2 in the unit conversion factor.
Result: 0.363 mol
Step 2. Determine the molality of the solution from the number of moles of solute and the mass of solvent, in kilograms.
Result: 0.454 m
Step 3. Use the direct proportionality between the change in boiling point and molal concentration to determine how much the boiling point changes.
Result: 1.65 °C
Step 4. Determine the new boiling point from the boiling point of the pure solvent and the change.
Result: 62.91 °C
Check each result as a self-assessment.
Check Your Learning
What is the boiling point of a solution of 1.0 g of glycerin, C3H5(OH)3, in 47.8 g of water? Assume an ideal solution.
Answer:
100.12 °C
Freezing Point Depression
Solutions freeze at lower temperatures than pure liquids. This phenomenon is exploited in “de-icing” schemes that use salt (), calcium chloride, or urea to melt ice on roads and sidewalks, and in the use of ethylene glycol as an “antifreeze” in automobile radiators. Seawater freezes at a lower temperature than fresh water, and so the Arctic and Antarctic oceans remain unfrozen even at temperatures below 0 °C (as do the body fluids of fish and other cold-blooded sea animals that live in these oceans).
FIGURE 11.21 Rock salt (NaCl), calcium chloride (CaCl2), or a mixture of the two are used to melt ice. (credit: modification of work by Eddie Welker)
The decrease in freezing point of a dilute solution compared to that of the pure solvent, ΔTf, is called the
freezing point depression and is directly proportional to the molal concentration of the solute
where m is the molal concentration of the solute and Kf is called the freezing point depression constant (or
cryoscopic constant). Just as for boiling point elevation constants, these are characteristic properties whose |
Calculation of the Freezing Point of a Solution
Assuming ideal solution behavior, what is the freezing point of the 0.33 m solution of a nonvolatile nonelectrolyte solute in benzene described in ?
Solution
Use the equation relating freezing point depression to solute molality to solve this problem in two steps. |
Step 2. Subtract the freezing point change observed from the pure solvent’s freezing point.
Check Your Learning
Assuming ideal solution behavior, what is the freezing point of a 1.85 m solution of a nonvolatile nonelectrolyte solute in nitrobenzene?
Answer:
−9.3 °C
Phase Diagram for a Solution
The colligative effects on vapor pressure, boiling point, and freezing point described in the previous section are conveniently summarized by comparing the phase diagrams for a pure liquid and a solution derived from that liquid ().
FIGURE 11.23 Phase diagrams for a pure solvent (solid curves) and a solution formed by dissolving nonvolatile solute in the solvent (dashed curves).
The liquid-vapor curve for the solution is located beneath the corresponding curve for the solvent, depicting the vapor pressure lowering, ΔP, that results from the dissolution of nonvolatile solute. Consequently, at any given pressure, the solution’s boiling point is observed at a higher temperature than that for the pure solvent, reflecting the boiling point elevation, ΔTb, associated with the presence of nonvolatile solute. The solid-liquid curve for the solution is displaced left of that for the pure solvent, representing the freezing point depression, ΔTf, that accompanies solution formation. Finally, notice that the solid-gas curves for the solvent and its solution are identical. This is the case for many solutions comprising liquid solvents and nonvolatile solutes. Just as for vaporization, when a solution of this sort is frozen, it is actually just the solvent molecules that
undergo the liquid-to-solid transition, forming pure solid solvent that excludes solute species. The solid and gaseous phases, therefore, are composed of solvent only, and so transitions between these phases are not subject to colligative effects.
Osmosis and Osmotic Pressure of Solutions
A number of natural and synthetic materials exhibit selective permeation, meaning that only molecules or ions of a certain size, shape, polarity, charge, and so forth, are capable of passing through (permeating) the material. Biological cell membranes provide elegant examples of selective permeation in nature, while dialysis tubing used to remove metabolic wastes from blood is a more simplistic technological example. Regardless of how they may be fabricated, these materials are generally referred to as semipermeable membranes.
Consider the apparatus illustrated in , in which samples of pure solvent and a solution are separated by a membrane that only solvent molecules may permeate. Solvent molecules will diffuse across the membrane in both directions. Since the concentration of solvent is greater in the pure solvent than the solution, these molecules will diffuse from the solvent side of the membrane to the solution side at a faster rate than they will in the reverse direction. The result is a net transfer of solvent molecules from the pure solvent to the solution. Diffusion-driven transfer of solvent molecules through a semipermeable membrane is a process known as osmosis.
FIGURE 11.24 (a) A solution and pure solvent are initially separated by an osmotic membrane. (b) Net transfer of solvent molecules to the solution occurs until its osmotic pressure yields equal rates of transfer in both directions.
When osmosis is carried out in an apparatus like that shown in , the volume of the solution increases as it becomes diluted by accumulation of solvent. This causes the level of the solution to rise, increasing its hydrostatic pressure (due to the weight of the column of solution in the tube) and resulting in a faster transfer of solvent molecules back to the pure solvent side. When the pressure reaches a value that yields a reverse solvent transfer rate equal to the osmosis rate, bulk transfer of solvent ceases. This pressure is called the osmotic pressure (Π) of the solution. The osmotic pressure of a dilute solution is related to its solute molarity, M, and absolute temperature, T, according to the equation |
Calculation of Osmotic Pressure
Assuming ideal solution behavior, what is the osmotic pressure (atm) of a 0.30 M solution of glucose in water that is used for intravenous infusion at body temperature, 37 °C?
Solution
Find the osmotic pressure, Π, using the formula Π = MRT, where T is on the Kelvin scale (310 K) and the value of R is expressed in appropriate units (0.08206 L atm/mol K). |
If a solution is placed in an apparatus like the one shown in , applying pressure greater than the osmotic pressure of the solution reverses the osmosis and pushes solvent molecules from the solution into the pure solvent. This technique of reverse osmosis is used for large-scale desalination of seawater and on smaller scales to produce high-purity tap water for drinking. |
Chemistry in Everyday Life
Reverse Osmosis Water Purification
In the process of osmosis, diffusion serves to move water through a semipermeable membrane from a less concentrated solution to a more concentrated solution. Osmotic pressure is the amount of pressure that must be applied to the more concentrated solution to cause osmosis to stop. If greater pressure is applied, the water will go from the more concentrated solution to a less concentrated (more pure) solution. This is called reverse osmosis. Reverse osmosis (RO) is used to purify water in many applications, from desalination plants in coastal cities, to water-purifying machines in grocery stores (), and smaller reverse-osmosis household units. With a hand-operated pump, small RO units can be used in third-world countries, disaster areas, and in lifeboats. Our military forces have a variety of generator- operated RO units that can be transported in vehicles to remote locations.
FIGURE 11.26 Reverse osmosis systems for purifying drinking water are shown here on (a) small and (b) large scales. (credit a: modification of work by Jerry Kirkhart; credit b: modification of work by Willard J. Lathrop)
Examples of osmosis are evident in many biological systems because cells are surrounded by semipermeable membranes. Carrots and celery that have become limp because they have lost water can be made crisp again by placing them in water. Water moves into the carrot or celery cells by osmosis. A cucumber placed in a concentrated salt solution loses water by osmosis and absorbs some salt to become a pickle. Osmosis can also affect animal cells. Solute concentrations are particularly important when solutions are injected into the body. Solutes in body cell fluids and blood serum give these solutions an osmotic pressure of approximately 7.7 atm. Solutions injected into the body must have the same osmotic pressure as blood serum; that is, they should be isotonic with blood serum. If a less concentrated solution, a hypotonic solution, is injected in sufficient quantity to dilute the blood serum, water from the diluted serum passes into the blood cells by osmosis, causing the cells to expand and rupture. This process is called hemolysis. When a more concentrated solution, a hypertonic solution, is injected, the cells lose water to the more concentrated solution, shrivel, and possibly die in a process called crenation. These effects are illustrated in .
FIGURE 11.27 Red blood cell membranes are water permeable and will (a) swell and possibly rupture in a hypotonic solution; (b) maintain normal volume and shape in an isotonic solution; and (c) shrivel and possibly die in a hypertonic solution. (credit a/b/c: modifications of work by “LadyofHats”/Wikimedia commons)
Determination of Molar Masses
Osmotic pressure and changes in freezing point, boiling point, and vapor pressure are directly proportional to the number of solute species present in a given amount of solution. Consequently, measuring one of these properties for a solution prepared using a known mass of solute permits determination of the solute’s molar mass.
Determination of a Molar Mass from a Freezing Point Depression
A solution of 4.00 g of a nonelectrolyte dissolved in 55.0 g of benzene is found to freeze at 2.32 °C. Assuming ideal solution behavior, what is the molar mass of this compound?
Solution
Solve this problem using the following steps. |
Step 4. Determine the molar mass from the mass of the solute and the number of moles in that mass.
Check Your Learning
A solution of 35.7 g of a nonelectrolyte in 220.0 g of chloroform has a boiling point of 64.5 °C. Assuming ideal solution behavior, what is the molar mass of this compound?
Answer:
1.8 102 g/mol
Determination of a Molar Mass from Osmotic Pressure
A 0.500 L sample of an aqueous solution containing 10.0 g of hemoglobin has an osmotic pressure of 5.9 torr at 22 °C. Assuming ideal solution behavior, what is the molar mass of hemoglobin?
Solution
Here is one set of steps that can be used to solve the problem: |
Check Your Learning
Assuming ideal solution behavior, what is the molar mass of a protein if a solution of 0.02 g of the protein in
25.0 mL of solution has an osmotic pressure of 0.56 torr at 25 °C?
Answer:
3 104 g/mol
Colligative Properties of Electrolytes
As noted previously in this module, the colligative properties of a solution depend only on the number, not on the identity, of solute species dissolved. The concentration terms in the equations for various colligative properties (freezing point depression, boiling point elevation, osmotic pressure) pertain to all solute species present in the solution. For the solutions considered thus far in this chapter, the solutes have been nonelectrolytes that dissolve physically without dissociation or any other accompanying process. Each molecule that dissolves yields one dissolved solute molecule. The dissolution of an electroyte, however, is not this simple, as illustrated by the two common examples below:
Considering the first of these examples, and assuming complete dissociation, a 1.0 m aqueous solution of NaCl contains 2.0 mole of ions (1.0 mol Na+ and 1.0 mol Cl−) per each kilogram of water, and its freezing point depression is expected to be
When this solution is actually prepared and its freezing point depression measured, however, a value of 3.4 °C is obtained. Similar discrepancies are observed for other ionic compounds, and the differences between the measured and expected colligative property values typically become more significant as solute concentrations increase. These observations suggest that the ions of sodium chloride (and other strong electrolytes) are not completely dissociated in solution.
To account for this and avoid the errors accompanying the assumption of total dissociation, an experimentally measured parameter named in honor of Nobel Prize-winning German chemist Jacobus Henricus van’t Hoff is used. The van’t Hoff factor (i) is defined as the ratio of solute particles in solution to the number of formula units dissolved: |
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 ( ). 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 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. |
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 () to |
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 mol 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 °C
Step 5. Determine the new freezing point from the freezing point of the pure solvent and the change.
Result: −2.1 °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.
Check Your Learning
Assuming complete dissociation and ideal solution behavior, calculate the freezing point of a solution of 0.724 g of CaCl2 in 175 g of water.
Answer:
−0.208 °C
Colloids
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 (). The particles in a colloid are larger than most simple molecules; |
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 .
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)
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 . |
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:
Dispersion methods: breaking down larger particles. For example, paint pigments are produced by dispersing large particles by grinding in special mills.
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: |
Some gold sols prepared in 1857 are still intact (the particles have not coalesced and settled), illustrating the long-term stability of many colloids.
Soaps and Detergents
Pioneers made soap by boiling fats with a strongly basic solution made by leaching potassium carbonate, K2CO3, 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 C17H35CO2Na and contains an uncharged nonpolar hydrocarbon chain, the C17H35— unit, and an ionic carboxylate group, the — unit ( ).
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 C12H25—, and an ionic group, such as a sulfate— or a sulfonate— (). 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.
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 . 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. |
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 (C6H14O2), 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 (). 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 for additional details).
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)
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 ().
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. |
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.
Gels
Gelatin desserts, such as Jell-O, are a type of colloid (). 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.
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
molality (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 (Π) 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 |
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.
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.
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.
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.
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, |
How do solutions differ from compounds? From other mixtures?
Which of the principal characteristics of solutions are evident in the solutions of K2Cr2O7 shown in ?
When KNO3 is dissolved in water, the resulting solution is significantly colder than the water was originally.
Is the dissolution of KNO3 an endothermic or an exothermic process?
What conclusions can you draw about the intermolecular attractions involved in the process?
Is the resulting solution an ideal solution?
Give an example of each of the following types of solutions:
a gas in a liquid
a gas in a gas
a solid in a solid
Indicate the most important types of intermolecular attractions in each of the following solutions:
The solution in .
NO(l) in CO(l)
Cl2(g) in Br2(l)
HCl(g) in benzene C6H6(l)
Methanol CH3OH(l) in H2O(l)
Predict whether each of the following substances would be more soluble in water (polar solvent) or in a hydrocarbon such as heptane (C7H16, nonpolar solvent):
vegetable oil (nonpolar)
isopropyl alcohol (polar)
potassium bromide (ionic)
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.
Solutions of hydrogen in palladium may be formed by exposing Pd metal to H2 gas. The concentration of hydrogen in the palladium depends on the pressure of H2 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 g cm3).
Determine the molarity of this solution.
Determine the molality of this solution.
Determine the percent by mass of hydrogen atoms in this solution.
Explain why the ions Na+ and Cl− are strongly solvated in water but not in hexane, a solvent composed of nonpolar molecules.
Explain why solutions of HBr in benzene (a nonpolar solvent) are nonconductive, while solutions in water (a polar solvent) are conductive. |
Write a balanced chemical equation showing the products of the dissolution of Fe(NO3)3.
Compare the processes that occur when methanol (CH3OH), hydrogen chloride (HCl), and sodium hydroxide (NaOH) dissolve in water. Write equations and prepare sketches showing the form in which each of these compounds is present in its respective solution.
What is the expected electrical conductivity of the following solutions?
NaOH(aq)
HCl(aq)
C6H12O6(aq) (glucose)
NH3(aq)
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.
Indicate the most important type of intermolecular attraction responsible for solvation in each of the following solutions:
the solutions in
methanol, CH3OH, dissolved in ethanol, C2H5OH
methane, CH4, dissolved in benzene, C6H6
the polar halocarbon CF2Cl2 dissolved in the polar halocarbon CF2ClCFCl2
O2(l) in N2(l)
Suppose you are presented with a clear solution of sodium thiosulfate, Na2S2O3. How could you determine whether the solution is unsaturated, saturated, or supersaturated?
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.
Suggest an explanation for the observations that ethanol, C2H5OH, is completely miscible with water and
that ethanethiol, C2H5SH, is soluble only to the extent of 1.5 g per 100 mL of water.
Calculate the percent by mass of KBr in a saturated solution of KBr in water at 10 °C. See for useful data, and report the computed percentage to one significant digit.
Which of the following gases is expected to be most soluble in water? Explain your reasoning.
CH4
CCl4
CHCl3
At 0 °C and 1.00 atm, as much as 0.70 g of O2 can dissolve in 1 L of water. At 0 °C and 4.00 atm, how many grams of O2 dissolve in 1 L of water?
Refer to .
How did the concentration of dissolved CO2 in the beverage change when the bottle was opened?
What caused this change?
Is the beverage unsaturated, saturated, or supersaturated with CO2?
The Henry’s law constant for CO2 is 3.4 10−2 M/atm at 25 °C. Assuming ideal solution behavior, what pressure of carbon dioxide is needed to maintain a CO2 concentration of 0.10 M in a can of lemon-lime soda?
The Henry’s law constant for O2 is 1.3 10−3 M/atm at 25 °C. Assuming ideal solution behavior, what mass of oxygen would be dissolved in a 40-L aquarium at 25 °C, assuming an atmospheric pressure of 1.00 atm, and that the partial pressure of O2 is 0.21 atm?
Assuming ideal solution behavior, how many liters of HCl gas, measured at 30.0 °C and 745 torr, are
required to prepare 1.25 L of a 3.20-M solution of hydrochloric acid?
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?
What is the microscopic explanation for the macroscopic behavior illustrated in ?
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.
A solution of potassium nitrate, an electrolyte, and a solution of glycerin (C3H5(OH)3), a nonelectrolyte, both boil at 100.3 °C. What other physical properties of the two solutions are identical?
What are the mole fractions of H3PO4 and water in a solution of 14.5 g of H3PO4 in 125 g of water?
Outline the steps necessary to answer the question.
Answer the question.
What are the mole fractions of HNO3 and water in a concentrated solution of nitric acid (68.0% HNO3 by mass)?
Outline the steps necessary to answer the question.
Answer the question.
Calculate the mole fraction of each solute and solvent:
583 g of H2SO4 in 1.50 kg of water—the acid solution used in an automobile battery
0.86 g of NaCl in 1.00 102 g of water—a solution of sodium chloride for intravenous injection
46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH
25 g of I2 in 125 g of ethanol, C2H5OH
Calculate the mole fraction of each solute and solvent:
0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg of water—a saturated solution at 0 °C
125 g of NH4NO3 in 275 g of water—a mixture used to make an instant ice pack
25 g of Cl2 in 125 g of dichloromethane, CH2Cl2
0.372 g of tetrahydropyridine, C5H9N, in 125 g of chloroform, CHCl3
Calculate the mole fractions of methanol, CH3OH; ethanol, C2H5OH; 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.)
What is the difference between a 1 M solution and a 1 m solution?
What is the molality of phosphoric acid, H3PO4, in a solution of 14.5 g of H3PO4 in 125 g of water?
Outline the steps necessary to answer the question.
Answer the question.
What is the molality of nitric acid in a concentrated solution of nitric acid (68.0% HNO3 by mass)?
Outline the steps necessary to answer the question.
Answer the question.
Calculate the molality of each of the following solutions:
583 g of H2SO4 in 1.50 kg of water—the acid solution used in an automobile battery
0.86 g of NaCl in 1.00 102 g of water—a solution of sodium chloride for intravenous injection
46.85 g of codeine, C18H21NO3, in 125.5 g of ethanol, C2H5OH
25 g of I2 in 125 g of ethanol, C2H5OH
Calculate the molality of each of the following solutions:
0.710 kg of sodium carbonate (washing soda), Na2CO3, in 10.0 kg of water—a saturated solution at 0°C
125 g of NH4NO3 in 275 g of water—a mixture used to make an instant ice pack
25 g of Cl2 in 125 g of dichloromethane, CH2Cl2
0.372 g of tetrahydropyridine, C5H9N, in 125 g of chloroform, CHCl3
The concentration of glucose, C6H12O6, in normal spinal fluid is What is the molality of the
solution?
A 13.0% solution of K2CO3 by mass has a density of 1.09 g/cm3. Calculate the molality of the solution.
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?
Assuming ideal solution behavior, what is the boiling point of a solution of 115.0 g of nonvolatile sucrose, C12H22O11, in 350.0 g of water?
Outline the steps necessary to answer the question
Answer the question
Assuming ideal solution behavior, what is the boiling point of a solution of 9.04 g of I2 in 75.5 g of benzene, assuming the I2 is nonvolatile?
Outline the steps necessary to answer the question.
Answer the question.
Assuming ideal solution behavior, what is the freezing temperature of a solution of 115.0 g of sucrose, C12H22O11, in 350.0 g of water?
Outline the steps necessary to answer the question.
Answer the question.
Assuming ideal solution behavior, what is the freezing point of a solution of 9.04 g of I2 in 75.5 g of benzene?
Outline the steps necessary to answer the following question.
Answer the question.
Assuming ideal solution behavior, what is the osmotic pressure of an aqueous solution of 1.64 g of Ca(NO3)2 in water at 25 °C? The volume of the solution is 275 mL.
Outline the steps necessary to answer the question.
Answer the question.
Assuming ideal solution behavior, what is osmotic pressure of a solution of bovine insulin (molar mass, 5700 g mol−1) at 18 °C if 100.0 mL of the solution contains 0.103 g of the insulin?
Outline the steps necessary to answer the question.
Answer the question.
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 °C; Kb = 5.02 °C/m) that boils at 81.5 °C at 1 atm?
Outline the steps necessary to answer the question.
Solve the problem.
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 °C. Assuming ideal solution behavior, calculate the molar mass of the compound.
A 1.0 m solution of HCl in benzene has a freezing point of 0.4 °C. Is HCl an electrolyte in benzene? Explain.
A solution contains 5.00 g of urea, CO(NH2)2, a nonvolatile compound, dissolved in 0.100 kg of water. If the vapor pressure of pure water at 25 °C is 23.7 torr, what is the vapor pressure of the solution (assuming ideal solution behavior)?
A 12.0-g sample of a nonelectrolyte is dissolved in 80.0 g of water. The solution freezes at −1.94 °C.
Assuming ideal solution behavior, calculate the molar mass of the substance.
Arrange the following solutions in order by their decreasing freezing points: 0.1 m Na3PO4, 0.1 m C2H5OH,
0.01 m CO2, 0.15 m NaCl, and 0.2 m CaCl2.
Calculate the boiling point elevation of 0.100 kg of water containing 0.010 mol of NaCl, 0.020 mol of Na2SO4, and 0.030 mol of MgCl2, assuming complete dissociation of these electrolytes and ideal solution behavior.
How could you prepare a 3.08 m aqueous solution of glycerin, C3H8O3? Assuming ideal solution behavior,
what is the freezing point of this solution?
A sample of sulfur weighing 0.210 g was dissolved in 17.8 g of carbon disulfide, CS2 (Kb = 2.34 °C/m). If the boiling point elevation was 0.107 °C, what is the formula of a sulfur molecule in carbon disulfide (assuming ideal solution behavior)?
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 °C. What does this suggest about the nature of a solution of CdI2?
Lysozyme is an enzyme that cleaves cell walls. A 0.100-L sample of a solution of lysozyme that contains 0.0750 g of the enzyme exhibits an osmotic pressure of 1.32 10−3 atm at 25 °C. Assuming ideal solution behavior, what is the molar mass of lysozyme?
The osmotic pressure of a solution containing 7.0 g of insulin per liter is 23 torr at 25 °C. Assuming ideal solution behavior, what is the molar mass of insulin?
The osmotic pressure of human blood is 7.6 atm at 37 °C. What mass of glucose, C6H12O6, 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 °C (assuming ideal solution behavior)?
Assuming ideal solution behavior, what is the freezing point of a solution of dibromobenzene, C6H4Br2, in
0.250 kg of benzene, if the solution boils at 83.5 °C?
Assuming ideal solution behavior, what is the boiling point of a solution of NaCl in water if the solution freezes at −0.93 °C?
The sugar fructose contains 40.0% C, 6.7% H, and 53.3% O by mass. A solution of 11.7 g of fructose in 325 g of ethanol has a boiling point of 78.59 °C. The boiling point of ethanol is 78.35 °C, and Kb for ethanol is
1.20 °C/m. Assuming ideal solution behavior, what is the molecular formula of fructose?
The vapor pressure of methanol, CH3OH, is 94 torr at 20 °C. The vapor pressure of ethanol, C2H5OH, is 44 torr at the same temperature.
Calculate the mole fraction of methanol and of ethanol in a solution of 50.0 g of methanol and 50.0 g of ethanol.
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 °C.
Calculate the mole fraction of methanol and of ethanol in the vapor above the solution.
The triple point of air-free water is defined as 273.16 K. Why is it important that the water be free of air?
Meat can be classified as fresh (not frozen) even though it is stored at −1 °C. Why wouldn’t meat freeze at this temperature?
An organic compound has a composition of 93.46% C and 6.54% H by mass. A solution of 0.090 g of this compound in 1.10 g of camphor melts at 158.4 °C. The melting point of pure camphor is 178.4 °C. Kf for camphor is 37.7 °C/m. Assuming ideal solution behavior, what is the molecular formula of the solute? Show your calculations.
A sample of HgCl2 weighing 9.41 g is dissolved in 32.75 g of ethanol, C2H5OH (Kb = 1.20 °C/m). The boiling point elevation of the solution is 1.27 °C. Is HgCl2 an electrolyte in ethanol? Show your calculations.
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 °C. Assuming ideal solution behavior, what is the formula of the salt? Show your calculations.
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.
Distinguish between dispersion methods and condensation methods for preparing colloidal systems.
How do colloids differ from solutions with regard to dispersed particle size and homogeneity?
Explain the cleansing action of soap.
How can it be demonstrated that colloidal particles are electrically charged? |
INTRODUCTION The lizard in the photograph is not simply enjoying the sunshine or working on its tan. The heat from the sun’s rays is critical to the lizard’s survival. A warm lizard can move faster than a cold one because the chemical reactions that allow its muscles to move occur more rapidly at higher temperatures. A cold lizard is a slower lizard and an easier meal for predators.
From baking a cake to determining the useful lifespan of a bridge, rates of chemical reactions play important roles in our understanding of processes that involve chemical changes. Two questions are typically posed when planning to carry out a chemical reaction. The first is: “Will the reaction produce the desired products in useful quantities?” The second question is: “How rapidly will the reaction occur?” A third question is often asked when investigating reactions in greater detail: “What specific molecular-level processes take place as the reaction occurs?” Knowing the answer to this question is of practical importance when the yield or rate of a reaction needs to be controlled.
The study of chemical kinetics concerns the second and third questions—that is, the rate at which a reaction yields products and the molecular-scale means by which a reaction occurs. This chapter examines the factors
that influence the rates of chemical reactions, the mechanisms by which reactions proceed, and the quantitative techniques used to describe the rates at which reactions occur.
Chemical Reaction Rates
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Define chemical reaction rate
Derive rate expressions from the balanced equation for a given chemical reaction
Calculate reaction rates from experimental data
A rate is a measure of how some property varies with time. Speed is a familiar rate that expresses the distance traveled by an object in a given amount of time. Wage is a rate that represents the amount of money earned by a person working for a given amount of time. Likewise, the rate of a chemical reaction is a measure of how much reactant is consumed, or how much product is produced, by the reaction in a given amount of time.
The rate of reaction is the change in the amount of a reactant or product per unit time. Reaction rates are therefore determined by measuring the time dependence of some property that can be related to reactant or product amounts. Rates of reactions that consume or produce gaseous substances, for example, are conveniently determined by measuring changes in volume or pressure. For reactions involving one or more colored substances, rates may be monitored via measurements of light absorption. For reactions involving aqueous electrolytes, rates may be measured via changes in a solution’s conductivity.
For reactants and products in solution, their relative amounts (concentrations) are conveniently used for purposes of expressing reaction rates. For example, the concentration of hydrogen peroxide, H2O2, in an aqueous solution changes slowly over time as it decomposes according to the equation: |
This mathematical representation of the change in species concentration over time is the rate expression for the reaction. The brackets indicate molar concentrations, and the symbol delta (Δ) indicates “change in.” Thus,
represents the molar concentration of hydrogen peroxide at some time t1; likewise,
represents the molar concentration of hydrogen peroxide at a later time t2; and Δ[H2O2] represents the change in molar concentration of hydrogen peroxide during the time interval Δt (that is, t2 − t1). Since the reactant concentration decreases as the reaction proceeds, Δ[H2O2] is a negative quantity. Reaction rates are, by convention, positive quantities, and so this negative change in concentration is multiplied by −1. provides an example of data collected during the decomposition of H2O2.
FIGURE 12.2 The rate of decomposition of H2O2 in an aqueous solution decreases as the concentration of H2O2 decreases.
To obtain the tabulated results for this decomposition, the concentration of hydrogen peroxide was measured every 6 hours over the course of a day at a constant temperature of 40 °C. Reaction rates were computed for each time interval by dividing the change in concentration by the corresponding time increment, as shown here for the first 6-hour period: |
This behavior indicates the reaction continually slows with time. Using the concentrations at the beginning and end of a time period over which the reaction rate is changing results in the calculation of an average rate for the reaction over this time interval. At any specific time, the rate at which a reaction is proceeding is known as its instantaneous rate. The instantaneous rate of a reaction at “time zero,” when the reaction commences, is its initial rate. Consider the analogy of a car slowing down as it approaches a stop sign. The vehicle’s initial rate—analogous to the beginning of a chemical reaction—would be the speedometer reading at the moment the driver begins pressing the brakes (t0). A few moments later, the instantaneous rate at a specific moment—call it t1—would be somewhat slower, as indicated by the speedometer reading at that point in time. As time passes, the instantaneous rate will continue to fall until it reaches zero, when the car (or reaction) stops. Unlike instantaneous speed, the car’s average speed is not indicated by the speedometer; but it can be calculated as the ratio of the distance traveled to the time required to bring the vehicle to a complete stop (Δt). Like the decelerating car, the average rate of a chemical reaction will fall somewhere between its initial and final rates.
The instantaneous rate of a reaction may be determined one of two ways. If experimental conditions permit the measurement of concentration changes over very short time intervals, then average rates computed as described earlier provide reasonably good approximations of instantaneous rates. Alternatively, a graphical procedure may be used that, in effect, yields the results that would be obtained if short time interval measurements were possible. In a plot of the concentration of hydrogen peroxide against time, the instantaneous rate of decomposition of H2O2 at any time t is given by the slope of a straight line that is tangent to the curve at that time (). These tangent line slopes may be evaluated using calculus, but the procedure for doing so is beyond the scope of this chapter.
FIGURE 12.3 This graph shows a plot of concentration versus time for a 1.000 M solution of H2O2. The rate at any time is equal to the negative of the slope of a line tangent to the curve at that time. Tangents are shown at t = 0 h (“initial rate”) and at t = 12 h (“instantaneous rate” at 12 h).
Chemistry in Everyday Life
Reaction Rates in Analysis: Test Strips for Urinalysis
Physicians often use disposable test strips to measure the amounts of various substances in a patient’s urine (). These test strips contain various chemical reagents, embedded in small pads at various locations along the strip, which undergo changes in color upon exposure to sufficient concentrations of specific substances. The usage instructions for test strips often stress that proper read time is critical for optimal results. This emphasis on read time suggests that kinetic aspects of the chemical reactions occurring on the test strip are important considerations.
The test for urinary glucose relies on a two-step process represented by the chemical equations shown here:
The first equation depicts the oxidation of glucose in the urine to yield glucolactone and hydrogen peroxide. The hydrogen peroxide produced subsequently oxidizes colorless iodide ion to yield brown iodine, which may be visually detected. Some strips include an additional substance that reacts with iodine to produce a more distinct color change.
The two test reactions shown above are inherently very slow, but their rates are increased by special enzymes embedded in the test strip pad. This is an example of catalysis, a topic discussed later in this chapter. A typical glucose test strip for use with urine requires approximately 30 seconds for completion of the color-forming reactions. Reading the result too soon might lead one to conclude that the glucose concentration of the urine sample is lower than it actually is (a false-negative result). Waiting too long to assess the color change can lead to a false positive due to the slower (not catalyzed) oxidation of iodide ion by other substances found in urine.
Relative Rates of Reaction
The rate of a reaction may be expressed as the change in concentration of any reactant or product. For any given reaction, these rate expressions are all related simply to one another according to the reaction stoichiometry. The rate of the general reaction |
Note that a negative sign has been included as a factor to account for the opposite signs of the two amount changes (the reactant amount is decreasing while the product amount is increasing). For homogeneous reactions, both the reactants and products are present in the same solution and thus occupy the same volume, so the molar amounts may be replaced with molar concentrations: |
and hydrogen at 1100 °C. Slopes of the tangent lines at t = 500 s show that the instantaneous rates derived from all three species involved in the reaction are related by their stoichiometric factors. The rate of hydrogen production, for example, is observed to be three times greater than that for nitrogen production:
FIGURE 12.5 Changes in concentrations of the reactant and products for the reaction The rates of change of the three concentrations are related by the reaction stoichiometry, as shown by the different slopes of the tangents at t = 500 s. |
Write the equations that relate the rates of consumption of the reactants and the rates of formation of the products.
Solution
Considering the stoichiometry of this homogeneous reaction, the rates for the consumption of reactants and formation of products are: |
Factors Affecting Reaction Rates
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Describe the effects of chemical nature, physical state, temperature, concentration, and catalysis on reaction rates
The rates at which reactants are consumed and products are formed during chemical reactions vary greatly. Five factors typically affecting the rates of chemical reactions will be explored in this section: the chemical nature of the reacting substances, the state of subdivision (one large lump versus many small particles) of the reactants, the temperature of the reactants, the concentration of the reactants, and the presence of a catalyst.
The Chemical Nature of the Reacting Substances
The rate of a reaction depends on the nature of the participating substances. Reactions that appear similar may have different rates under the same conditions, depending on the identity of the reactants. For example, when small pieces of the metals iron and sodium are exposed to air, the sodium reacts completely with air
overnight, whereas the iron is barely affected. The active metals calcium and sodium both react with water to form hydrogen gas and a base. Yet calcium reacts at a moderate rate, whereas sodium reacts so rapidly that the reaction is almost explosive.
The Physical States of the Reactants
A chemical reaction between two or more substances requires intimate contact between the reactants. When reactants are in different physical states, or phases (solid, liquid, gaseous, dissolved), the reaction takes place only at the interface between the phases. Consider the heterogeneous reaction between a solid phase and either a liquid or gaseous phase. Compared with the reaction rate for large solid particles, the rate for smaller particles will be greater because the surface area in contact with the other reactant phase is greater. For example, large pieces of iron react more slowly with acids than they do with finely divided iron powder ( ). Large pieces of wood smolder, smaller pieces burn rapidly, and saw dust burns explosively.
FIGURE 12.6 (a) Iron powder reacts rapidly with dilute hydrochloric acid and produces bubbles of hydrogen gas: 2Fe(s) + 6HCl(aq) 2FeCl3(aq) + 3H2(g). (b) An iron nail reacts more slowly because the surface area exposed to the acid is much less.
LINK TO LEARNING
Watch to see the reaction of cesium with water in slow motion and a discussion of how the state of reactants and particle size affect reaction rates.
Temperature of the Reactants
Chemical reactions typically occur faster at higher temperatures. Food can spoil quickly when left on the kitchen counter. However, the lower temperature inside of a refrigerator slows that process so that the same food remains fresh for days. Gas burners, hot plates, and ovens are often used in the laboratory to increase the speed of reactions that proceed slowly at ordinary temperatures. For many chemical processes, reaction rates are approximately doubled when the temperature is raised by 10 °C.
Concentrations of the Reactants
The rates of many reactions depend on the concentrations of the reactants. Rates usually increase when the concentration of one or more of the reactants increases. For example, calcium carbonate (CaCO3) deteriorates as a result of its reaction with the pollutant sulfur dioxide. The rate of this reaction depends on the amount of sulfur dioxide in the air (). An acidic oxide, sulfur dioxide combines with water vapor in the air to produce sulfurous acid in the following reaction: |
In a polluted atmosphere where the concentration of sulfur dioxide is high, calcium carbonate deteriorates more rapidly than in less polluted air. Similarly, phosphorus burns much more rapidly in an atmosphere of pure oxygen than in air, which is only about 20% oxygen.
FIGURE 12.7 Statues made from carbonate compounds such as limestone and marble typically weather slowly over time due to the actions of water, and thermal expansion and contraction. However, pollutants like sulfur dioxide can accelerate weathering. As the concentration of air pollutants increases, deterioration of limestone occurs more rapidly. (credit: James P Fisher III)
LINK TO LEARNING
Phosphorus burns rapidly in air, but it will burn even more rapidly if the concentration of oxygen is higher. Watch this to see an example.
The Presence of a Catalyst
Relatively dilute aqueous solutions of hydrogen peroxide, H2O2, are commonly used as topical antiseptics. Hydrogen peroxide decomposes to yield water and oxygen gas according to the equation:
Under typical conditions, this decomposition occurs very slowly. When dilute H2O2(aq) is poured onto an open wound, however, the reaction occurs rapidly and the solution foams because of the vigorous production of oxygen gas. This dramatic difference is caused by the presence of substances within the wound’s exposed tissues that accelerate the decomposition process. Substances that function to increase the rate of a reaction are called catalysts, a topic treated in greater detail later in this chapter.
LINK TO LEARNING
Chemical reactions occur when molecules collide with each other and undergo a chemical transformation. Before physically performing a reaction in a laboratory, scientists can use molecular modeling simulations to predict how the parameters discussed earlier will influence the rate of a reaction. Use the to explore how temperature, concentration, and the nature of the reactants affect reaction rates.
Rate Laws
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Explain the form and function of a rate law
Use rate laws to calculate reaction rates
Use rate and concentration data to identify reaction orders and derive rate laws
As described in the previous module, the rate of a reaction is often affected by the concentrations of reactants. Rate laws (sometimes called differential rate laws) or rate equations are mathematical expressions that describe the relationship between the rate of a chemical reaction and the concentration of its reactants. As an example, consider the reaction described by the chemical equation |
in which [A] and [B] represent the molar concentrations of reactants, and k is the rate constant, which is specific for a particular reaction at a particular temperature. The exponents m and n are the reaction orders and are typically positive integers, though they can be fractions, negative, or zero. The rate constant k and the reaction orders m and n must be determined experimentally by observing how the rate of a reaction changes as the concentrations of the reactants are changed. The rate constant k is independent of the reactant concentrations, but it does vary with temperature.
The reaction orders in a rate law describe the mathematical dependence of the rate on reactant concentrations. Referring to the generic rate law above, the reaction is m order with respect to A and n order with respect to B. For example, if m = 1 and n = 2, the reaction is first order in A and second order in B. The overall reaction order is simply the sum of orders for each reactant. For the example rate law here, the reaction is third order overall (1 + 2 = 3). A few specific examples are shown below to further illustrate this concept.
The rate law: |
Remember that a number raised to the zero power is equal to 1, thus [CO]0 = 1, which is why the CO concentration term may be omitted from the rate law: the rate of reaction is solely dependent on the concentration of NO2. A later chapter section on reaction mechanisms will explain how a reactant’s concentration can have no effect on a reaction rate despite being involved in the reaction.
Check Your Learning
The rate law for the reaction: |
overall order of the reaction?
Answer:
order in NO = 2; order in H2 = 1; overall order = 3
Check Your Learning
In a transesterification reaction, a triglyceride reacts with an alcohol to form an ester and glycerol. Many students learn about the reaction between methanol (CH3OH) and ethyl acetate (CH3CH2OCOCH3) as a sample reaction before studying the chemical reactions that produce biodiesel: |
A common experimental approach to the determination of rate laws is the method of initial rates. This method involves measuring reaction rates for multiple experimental trials carried out using different initial reactant concentrations. Comparing the measured rates for these trials permits determination of the reaction orders and, subsequently, the rate constant, which together are used to formulate a rate law. This approach is illustrated in the next two example exercises.
Determining a Rate Law from Initial Rates
Ozone in the upper atmosphere is depleted when it reacts with nitrogen oxides. The rates of the reactions of nitrogen oxides with ozone are important factors in deciding how significant these reactions are in the formation of the ozone hole over Antarctica (). One such reaction is the combination of nitric oxide, NO, with ozone, O3: |
Determine the values of m, n, and k from the experimental data using the following three-part process: Step 1.
Determine the value of m from the data in which [NO] varies and [O3] is constant. In the last three
experiments, [NO] varies while [O3] remains constant. When [NO] doubles from trial 3 to 4, the rate doubles, and when [NO] triples from trial 3 to 5, the rate also triples. Thus, the rate is also directly proportional to [NO], and m in the rate law is equal to 1.
Step 2.
Determine the value of n from data in which [O3] varies and [NO] is constant. In the first three experiments, [NO] is constant and [O3] varies. The reaction rate changes in direct proportion to the change in [O3]. When [O3] doubles from trial 1 to 2, the rate doubles; when [O3] triples from trial 1 to 3, the rate increases also triples. Thus, the rate is directly proportional to [O3], and n is equal to 1.The rate law is thus: |
As in , approach this problem in a stepwise fashion, determining the values of m and n from the experimental data and then using these values to determine the value of k. In this example, however, an explicit algebraic approach (vs. the implicit approach of the previous example) will be used to determine the values of m and n:
Step 1.
Determine the value of m from the data in which [NO] varies and [Cl2] is constant. Write the ratios with the subscripts x and y to indicate data from two different trials: |
Step 3.
Determine the numerical value of the rate constant k with appropriate units. The units for the rate of a reaction are mol/L/s. The units for k are whatever is needed so that substituting into the rate law expression affords the appropriate units for the rate. In this example, the concentration units are mol3/L3. The units for k should be mol−2 L2/s so that the rate is in terms of mol/L/s.
To determine the value of k once the rate law expression has been solved, simply plug in values from the first experimental trial and solve for k: |
chemical equation for the reaction. This is merely a coincidence and very often not the case.
Rate laws may exhibit fractional orders for some reactants, and negative reaction orders are sometimes observed when an increase in the concentration of one reactant causes a decrease in reaction rate. A few examples illustrating these points are provided:
It is important to note that rate laws are determined by experiment only and are not reliably predicted by reaction stoichiometry.
The units for a rate constant will vary as appropriate to accommodate the overall order of the reaction. The unit of the rate constant for the second-order reaction described in was determined to be
For the third-order reaction described in , the unit for k was derived to be Dimensional analysis requires the rate constant unit for a reaction whose overall order is x to be
summarizes the rate constant units for common reaction orders. |
Note that the units in this table were derived using specific units for concentration (mol/L) and time (s), though any valid units for these two properties may be used.
Integrated Rate Laws
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Explain the form and function of an integrated rate law
Perform integrated rate law calculations for zero-, first-, and second-order reactions
Define half-life and carry out related calculations
Identify the order of a reaction from concentration/time data
The rate laws discussed thus far relate the rate and the concentrations of reactants. We can also determine a second form of each rate law that relates the concentrations of reactants and time. These are called integrated rate laws. We can use an integrated rate law to determine the amount of reactant or product present after a period of time or to estimate the time required for a reaction to proceed to a certain extent. For example, an integrated rate law is used to determine the length of time a radioactive material must be stored for its radioactivity to decay to a safe level.
Using calculus, the differential rate law for a chemical reaction can be integrated with respect to time to give an equation that relates the amount of reactant or product present in a reaction mixture to the elapsed time of the reaction. This process can either be very straightforward or very complex, depending on the complexity of the differential rate law. For purposes of discussion, we will focus on the resulting integrated rate laws for first-
, second-, and zero-order reactions.
First-Order Reactions
Integration of the rate law for a simple first-order reaction (rate = k[A]) results in an equation describing how the reactant concentration varies with time:
where [A]t is the concentration of A at any time t, [A]0 is the initial concentration of A, and k is the first-order rate constant.
For mathematical convenience, this equation may be rearranged to other formats, including direct and indirect proportionalities: |
The initial concentration of C4H8, [A]0, is not provided, but the provision that 80.0% of the sample has decomposed is enough information to solve this problem. Let x be the initial concentration, in which case the concentration after 80.0% decomposition is 20.0% of x or 0.200x. Rearranging the rate law to isolate t and substituting the provided quantities yields: |
A plot of ln[A]t versus t for a first-order reaction is a straight line with a slope of −k and a y-intercept of ln[A]0. If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in A.
Graphical Determination of Reaction Order and Rate Constant
Show that the data in can be represented by a first-order rate law by graphing ln[H2O2] versus time. Determine the rate constant for the decomposition of H2O2 from these data.
Solution
The data from are tabulated below, and a plot of ln[H2O2] is shown in . |
The plot of ln[H2O2] versus time is linear, indicating that the reaction may be described by a first-order rate law.
According to the linear format of the first-order integrated rate law, the rate constant is given by the negative of this plot’s slope.
The slope of this line may be derived from two values of ln[H2O2] at different values of t (one near each end of the line is preferable). For example, the value of ln[H2O2] when t is 0.00 h is 0.000; the value when t = 24.00 h is
−2.772 |
Second-Order Reactions
The equations that relate the concentrations of reactants and the rate constant of second-order reactions can be fairly complicated. To illustrate the point with minimal complexity, only the simplest second-order reactions will be described here, namely, those whose rates depend on the concentration of just one reactant. For these types of reactions, the differential rate law is written as: |
This “dimerization” reaction is second order with a rate constant equal to 5.76 10−2 L mol−1 min−1 under certain conditions. If the initial concentration of butadiene is 0.200 M, what is the concentration after 10.0 min?
Solution
For a second-order reaction, the integrated rate law is written |
Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L that was originally present.
Check Your Learning
If the initial concentration of butadiene is 0.0200 M, what is the concentration remaining after 20.0 min?
Answer:
0.0195 mol/L |
Graphical Determination of Reaction Order and Rate Constant
The data below are for the same reaction described in . Prepare and compare two appropriate data plots to identify the reaction as being either first or second order. After identifying the reaction order, estimate a value for the rate constant.
Solution |
The plots are shown in , which clearly shows the plot of ln[C4H6]t versus t is not linear, therefore the reaction is not first order. The plot of versus t is linear, indicating that the reaction is second order.
FIGURE 12.10 These two graphs show first- and second-order plots for the dimerization of C4H6. The linear trend in the second-order plot (right) indicates that the reaction follows second-order kinetics.
According to the second-order integrated rate law, the rate constant is equal to the slope of the versus t
plot. Using the data for t = 0 s and t = 6200 s, the rate constant is estimated as follows: |
A zero-order reaction thus exhibits a constant reaction rate, regardless of the concentration of its reactant(s). This may seem counterintuitive, since the reaction rate certainly can’t be finite when the reactant concentration is zero. For purposes of this introductory text, it will suffice to note that zero-order kinetics are observed for some reactions only under certain specific conditions. These same reactions exhibit different kinetic behaviors when the specific conditions aren’t met, and for this reason the more prudent term pseudo- zero-order is sometimes used.
The integrated rate law for a zero-order reaction is a linear function:
A plot of [A] versus t for a zero-order reaction is a straight line with a slope of −k and a y-intercept of [A]0. shows a plot of [NH3] versus t for the thermal decomposition of ammonia at the surface of two different heated solids. The decomposition reaction exhibits first-order behavior at a quartz (SiO2) surface, as suggested by the exponentially decaying plot of concentration versus time. On a tungsten surface, however, the plot is linear, indicating zero-order kinetics.
Graphical Determination of Zero-Order Rate Constant
Use the data plot in to graphically estimate the zero-order rate constant for ammonia decomposition at a tungsten surface.
Solution
The integrated rate law for zero-order kinetics describes a linear plot of reactant concentration, [A]t, versus time, t, with a slope equal to the negative of the rate constant, −k. Following the mathematical approach of previous examples, the slope of the linear data plot (for decomposition on W) is estimated from the graph. Using the ammonia concentrations at t = 0 and t = 1000 s:
Check Your Learning
The zero-order plot in shows an initial ammonia concentration of 0.0028 mol L−1 decreasing linearly with time for 1000 s. Assuming no change in this zero-order behavior, at what time (min) will the concentration reach 0.0001 mol L−1?
Answer:
35 min
FIGURE 12.11 The decomposition of NH3 on a tungsten (W) surface is a zero-order reaction, whereas on a quartz (SiO2) surface, the reaction is first order.
The Half-Life of a Reaction
The half-life of a reaction (t1/2) is the time required for one-half of a given amount of reactant to be consumed. In each succeeding half-life, half of the remaining concentration of the reactant is consumed. Using the decomposition of hydrogen peroxide () as an example, we find that during the first half-life (from 0.00 hours to 6.00 hours), the concentration of H2O2 decreases from 1.000 M to 0.500 M. During the second half-life (from 6.00 hours to 12.00 hours), it decreases from 0.500 M to 0.250 M; during the third half-life, it decreases from 0.250 M to 0.125 M. The concentration of H2O2 decreases by half during each successive period of 6.00 hours. The decomposition of hydrogen peroxide is a first-order reaction, and, as can be shown, the half-life of a first-order reaction is independent of the concentration of the reactant. However, half-lives of reactions with other orders depend on the concentrations of the reactants.
First-Order Reactions
An equation relating the half-life of a first-order reaction to its rate constant may be derived from the integrated rate law as follows: |
This equation describes an expected inverse relation between the half-life of the reaction and its rate constant,
k. Faster reactions exhibit larger rate constants and correspondingly shorter half-lives. Slower reactions exhibit smaller rate constants and longer half-lives. |
FIGURE 12.12 The decomposition of H2O2 at 40 °C is illustrated. The intensity of the color symbolizes the concentration of H2O2 at the indicated times; H2O2 is actually colorless.
Solution
Inspecting the concentration/time data in shows the half-life for the decomposition of H2O2 is
2.16 104 s: |
Second-Order Reactions
Following the same approach as used for first-order reactions, an equation relating the half-life of a second- order reaction to its rate constant and initial concentration may be derived from its integrated rate law: |
For a second-order reaction, is inversely proportional to the concentration of the reactant, and the half-life increases as the reaction proceeds because the concentration of reactant decreases. Unlike with first-order reactions, the rate constant of a second-order reaction cannot be calculated directly from the half-life unless the initial concentration is known.
Zero-Order Reactions
As for other reaction orders, an equation for zero-order half-life may be derived from the integrated rate law: |
As for all reaction orders, the half-life for a zero-order reaction is inversely proportional to its rate constant. However, the half-life of a zero-order reaction increases as the initial concentration increases.
Equations for both differential and integrated rate laws and the corresponding half-lives for zero-, first-, and second-order reactions are summarized in . |
Half-Life for Zero-Order and Second-Order Reactions
What is the half-life for the butadiene dimerization reaction described in ?
Solution
The reaction in question is second order, is initiated with a 0.200 mol L−1 reactant solution, and exhibits a rate constant of 0.0576 L mol−1 min−1. Substituting these quantities into the second-order half-life equation: |
Collision Theory
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Use the postulates of collision theory to explain the effects of physical state, temperature, and concentration on reaction rates
Define the concepts of activation energy and transition state
Use the Arrhenius equation in calculations relating rate constants to temperature
We should not be surprised that atoms, molecules, or ions must collide before they can react with each other. Atoms must be close together to form chemical bonds. This simple premise is the basis for a very powerful theory that explains many observations regarding chemical kinetics, including factors affecting reaction rates.
Collision theory is based on the following postulates:
The rate of a reaction is proportional to the rate of reactant collisions:
The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product.
The collision must occur with adequate energy to permit mutual penetration of the reacting species’ valence shells so that the electrons can rearrange and form new bonds (and new chemical species).
We can see the importance of the two physical factors noted in postulates 2 and 3, the orientation and energy |
Carbon monoxide is a pollutant produced by the combustion of hydrocarbon fuels. To reduce this pollutant, automobiles have catalytic converters that use a catalyst to carry out this reaction. It is also a side reaction of the combustion of gunpowder that results in muzzle flash for many firearms. If carbon monoxide and oxygen are present in sufficient amounts, the reaction will occur at high temperature and pressure.
The first step in the gas-phase reaction between carbon monoxide and oxygen is a collision between the two molecules:
Although there are many different possible orientations the two molecules can have relative to each other, consider the two presented in . In the first case, the oxygen side of the carbon monoxide molecule collides with the oxygen molecule. In the second case, the carbon side of the carbon monoxide molecule collides with the oxygen molecule. The second case is clearly more likely to result in the formation of carbon dioxide, which has a central carbon atom bonded to two oxygen atoms This is a rather simple example of how important the orientation of the collision is in terms of creating the desired product of the reaction.
FIGURE 12.13 Illustrated are two collisions that might take place between carbon monoxide and oxygen molecules. The orientation of the colliding molecules partially determines whether a reaction between the two molecules will occur.
If the collision does take place with the correct orientation, there is still no guarantee that the reaction will proceed to form carbon dioxide. In addition to a proper orientation, the collision must also occur with sufficient energy to result in product formation. When reactant species collide with both proper orientation and adequate energy, they combine to form an unstable species called an activated complex or a transition state. These species are very short lived and usually undetectable by most analytical instruments. In some cases, sophisticated spectral measurements have been used to observe transition states.
Collision theory explains why most reaction rates increase as concentrations increase. With an increase in the concentration of any reacting substance, the chances for collisions between molecules are increased because there are more molecules per unit of volume. More collisions mean a faster reaction rate, assuming the energy of the collisions is adequate.
Activation Energy and the Arrhenius Equation
The minimum energy necessary to form a product during a collision between reactants is called the activation energy (Ea). How this energy compares to the kinetic energy provided by colliding reactant molecules is a primary factor affecting the rate of a chemical reaction. If the activation energy is much larger than the average kinetic energy of the molecules, the reaction will occur slowly since only a few fast-moving molecules will have enough energy to react. If the activation energy is much smaller than the average kinetic energy of the molecules, a large fraction of molecules will be adequately energetic and the reaction will proceed rapidly.
shows how the energy of a chemical system changes as it undergoes a reaction converting reactants to products according to the equation
These reaction diagrams are widely used in chemical kinetics to illustrate various properties of the reaction of interest. Viewing the diagram from left to right, the system initially comprises reactants only, A + B. Reactant molecules with sufficient energy can collide to form a high-energy activated complex or transition state. The unstable transition state can then subsequently decay to yield stable products, C + D. The diagram depicts the reaction's activation energy, Ea, as the energy difference between the reactants and the transition state. Using a specific energy, the enthalpy (see chapter on thermochemistry), the enthalpy change of the reaction, ΔH, is estimated as the energy difference between the reactants and products. In this case, the reaction is exothermic (ΔH < 0) since it yields a decrease in system enthalpy. |
In this equation, R is the ideal gas constant, which has a value 8.314 J/mol/K, T is temperature on the Kelvin scale, Ea is the activation energy in joules per mole, e is the constant 2.7183, and A is a constant called the frequency factor, which is related to the frequency of collisions and the orientation of the reacting molecules.
Postulates of collision theory are nicely accommodated by the Arrhenius equation. The frequency factor, A, reflects how well the reaction conditions favor properly oriented collisions between reactant molecules. An increased probability of effectively oriented collisions results in larger values for A and faster reaction rates.
The exponential term, e−Ea/RT, describes the effect of activation energy on reaction rate. According to kinetic molecular theory (see chapter on gases), the temperature of matter is a measure of the average kinetic energy of its constituent atoms or molecules. The distribution of energies among the molecules composing a sample of matter at any given temperature is described by the plot shown in (a). Two shaded areas under the curve represent the numbers of molecules possessing adequate energy (RT) to overcome the activation barriers (Ea). A lower activation energy results in a greater fraction of adequately energized molecules and a faster reaction.
The exponential term also describes the effect of temperature on reaction rate. A higher temperature represents a correspondingly greater fraction of molecules possessing sufficient energy (RT) to overcome the activation barrier (Ea), as shown in (b). This yields a greater value for the rate constant and a correspondingly faster reaction rate.
FIGURE 12.15 Molecular energy distributions showing numbers of molecules with energies exceeding (a) two different activation energies at a given temperature, and (b) a given activation energy at two different temperatures.
A convenient approach for determining Ea for a reaction involves the measurement of k at two or more different temperatures and using an alternate version of the Arrhenius equation that takes the form of a linear equation |
is a graph of ln k versus In practice, the equation of the line (slope and y-intercept) that best fits these plotted data points would be derived using a statistical process called regression. This is helpful for most experimental data because a perfect fit of each data point with the line is rarely encountered. For the data here, the fit is nearly perfect and the slope may be estimated using any two of the provided data pairs. Using
the first and last data points permits estimation of the slope. |
Alternative approach: A more expedient approach involves deriving activation energy from measurements of the rate constant at just two temperatures. In this approach, the Arrhenius equation is rearranged to a convenient two-point form: |
and the result is Ea = 1.8 105 J mol−1 or 180 kJ mol−1
This approach yields the same result as the more rigorous graphical approach used above, as expected. In practice, the graphical approach typically provides more reliable results when working with actual experimental data.
Check Your Learning
The rate constant for the rate of decomposition of N2O5 to NO and O2 in the gas phase is 1.66 L/mol/s at 650 K and 7.39 L/mol/s at 700 K: |
Reaction Mechanisms
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Distinguish net reactions from elementary reactions (steps)
Identify the molecularity of elementary reactions
Write a balanced chemical equation for a process given its reaction mechanism
Derive the rate law consistent with a given reaction mechanism
Chemical reactions very often occur in a step-wise fashion, involving two or more distinct reactions taking place in sequence. A balanced equation indicates what is reacting and what is produced, but it reveals no details about how the reaction actually takes place. The reaction mechanism (or reaction path) provides details regarding the precise, step-by-step process by which a reaction occurs.
The decomposition of ozone, for example, appears to follow a mechanism with two steps:
Each of the steps in a reaction mechanism is an elementary reaction. These elementary reactions occur precisely as represented in the step equations, and they must sum to yield the balanced chemical equation representing the overall reaction:
Notice that the oxygen atom produced in the first step of this mechanism is consumed in the second step and therefore does not appear as a product in the overall reaction. Species that are produced in one step and consumed in a subsequent step are called intermediates.
While the overall reaction equation for the decomposition of ozone indicates that two molecules of ozone react
to give three molecules of oxygen, the mechanism of the reaction does not involve the direct collision and reaction of two ozone molecules. Instead, one O3 decomposes to yield O2 and an oxygen atom, and a second O3 molecule subsequently reacts with the oxygen atom to yield two additional O2 molecules.
Unlike balanced equations representing an overall reaction, the equations for elementary reactions are explicit representations of the chemical change taking place. The reactant(s) in an elementary reaction’s equation undergo only the bond-breaking and/or making events depicted to yield the product(s). For this reason, the rate law for an elementary reaction may be derived directly from the balanced chemical equation describing the reaction. This is not the case for typical chemical reactions, for which rate laws may be reliably determined only via experimentation.
Unimolecular Elementary Reactions
The molecularity of an elementary reaction is the number of reactant species (atoms, molecules, or ions). For example, a unimolecular reaction involves the reaction of a single reactant species to produce one or more molecules of product: |
illustrates a unimolecular elementary reaction that occurs as one part of a two-step reaction mechanism as described above. However, some unimolecular reactions may be the only step of a single-step reaction mechanism. (In other words, an “overall” reaction may also be an elementary reaction in some cases.) For example, the gas-phase decomposition of cyclobutane, C4H8, to ethylene, C2H4, is represented by the following chemical equation:
This equation represents the overall reaction observed, and it might also represent a legitimate unimolecular elementary reaction. The rate law predicted from this equation, assuming it is an elementary reaction, turns out to be the same as the rate law derived experimentally for the overall reaction, namely, one showing first- order behavior:
This agreement between observed and predicted rate laws is interpreted to mean that the proposed unimolecular, single-step process is a reasonable mechanism for the butadiene reaction.
Bimolecular Elementary Reactions
A bimolecular reaction involves two reactant species, for example: |
FIGURE 12.17 The probable mechanism for the reaction between NO2 and CO to yield NO and CO2.
Bimolecular elementary reactions may also be involved as steps in a multistep reaction mechanism. The reaction of atomic oxygen with ozone is the second step of the two-step ozone decomposition mechanism discussed earlier in this section:
Termolecular Elementary Reactions
An elementary termolecular reaction involves the simultaneous collision of three atoms, molecules, or ions. Termolecular elementary reactions are uncommon because the probability of three particles colliding simultaneously is less than one one-thousandth of the probability of two particles colliding. There are, however, a few established termolecular elementary reactions. The reaction of nitric oxide with oxygen appears to involve termolecular steps: |
Relating Reaction Mechanisms to Rate Laws
It’s often the case that one step in a multistep reaction mechanism is significantly slower than the others. Because a reaction cannot proceed faster than its slowest step, this step will limit the rate at which the overall reaction occurs. The slowest step is therefore called the rate-limiting step (or rate-determining step) of the reaction .
FIGURE 12.18 A cattle chute is a nonchemical example of a rate-determining step. Cattle can only be moved from one holding pen to another as quickly as one animal can make its way through the chute. (credit: Loren Kerns)
As described earlier, rate laws may be derived directly from the chemical equations for elementary reactions. This is not the case, however, for ordinary chemical reactions. The balanced equations most often encountered represent the overall change for some chemical system, and very often this is the result of some multistep reaction mechanisms. In every case, the rate law must be determined from experimental data and the reaction mechanism subsequently deduced from the rate law (and sometimes from other data). The reaction of NO2 and CO provides an illustrative example: |
The reaction is first order with respect to NO2 and first-order with respect to CO. This is consistent with a single-step bimolecular mechanism and it is possible that this is the mechanism for this reaction at high temperatures.
At temperatures below 225 °C, the reaction is described by a rate law that is second order with respect to NO2: |
The rate-determining (slower) step gives a rate law showing second-order dependence on the NO2 concentration, and the sum of the two equations gives the net overall reaction.
In general, when the rate-determining (slower) step is the first step in a mechanism, the rate law for the overall reaction is the same as the rate law for this step. However, when the rate-determining step is preceded by a step involving a rapidly reversible reaction the rate law for the overall reaction may be more difficult to derive.
As discussed in several chapters of this text, a reversible reaction is at equilibrium when the rates of the forward and reverse processes are equal. Consider the reversible elementary reaction in which NO dimerizes to yield an intermediate species N2O2. When this reaction is at equilibrium: |
Step 2 is the rate-determining step, and so the rate law for the overall reaction should be the same as for this step. However, the step 2 rate law, as written, contains an intermediate species concentration, [NOCl2]. To remedy this, use the first step’s rate laws to derive an expression for the intermediate concentration in terms of the reactant concentrations.
Assuming step 1 is at equilibrium: |
Catalysis
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Explain the function of a catalyst in terms of reaction mechanisms and potential energy diagrams
List examples of catalysis in natural and industrial processes
Among the factors affecting chemical reaction rates discussed earlier in this chapter was the presence of a catalyst, a substance that can increase the reaction rate without being consumed in the reaction. The concepts introduced in the previous section on reaction mechanisms provide the basis for understanding how catalysts are able to accomplish this very important function.
shows reaction diagrams for a chemical process in the absence and presence of a catalyst. Inspection of the diagrams reveals several traits of these reactions. Consistent with the fact that the two diagrams represent the same overall reaction, both curves begin and end at the same energies (in this case, because products are more energetic than reactants, the reaction is endothermic). The reaction mechanisms, however, are clearly different. The uncatalyzed reaction proceeds via a one-step mechanism (one transition state observed), whereas the catalyzed reaction follows a two-step mechanism (two transition states observed) with a notably lesser activation energy. This difference illustrates the means by which a catalyst functions to accelerate reactions, namely, by providing an alternative reaction mechanism with a lower activation energy. Although the catalyzed reaction mechanism for a reaction needn’t necessarily involve a different number of steps than the uncatalyzed mechanism, it must provide a reaction path whose rate determining step is faster (lower Ea).
FIGURE 12.19 Reaction diagrams for an endothermic process in the absence (red curve) and presence (blue curve) of a catalyst. The catalyzed pathway involves a two-step mechanism (note the presence of two transition states) and an intermediate species (represented by the valley between the two transitions states).
Reaction Diagrams for Catalyzed Reactions
The two reaction diagrams here represent the same reaction: one without a catalyst and one with a catalyst. Estimate the activation energy for each process, and identify which one involves a catalyst. |
The catalyzed reaction is the one with lesser activation energy, in this case represented by diagram b.
Check Your Learning
Reaction diagrams for a chemical process with and without a catalyst are shown below. Both reactions involve a two-step mechanism with a rate-determining first step. Compute activation energies for the first step of each mechanism, and identify which corresponds to the catalyzed reaction. How do the second steps of these two mechanisms compare? |
Homogeneous Catalysts
A homogeneous catalyst is present in the same phase as the reactants. It interacts with a reactant to form an intermediate substance, which then decomposes or reacts with another reactant in one or more steps to regenerate the original catalyst and form product.
As an important illustration of homogeneous catalysis, consider the earth’s ozone layer. Ozone in the upper atmosphere, which protects the earth from ultraviolet radiation, is formed when oxygen molecules absorb ultraviolet light and undergo the reaction: |
Notice that NO is a reactant in the first step of the mechanism and a product in the last step. This is another characteristic trait of a catalyst: Though it participates in the chemical reaction, it is not consumed by the reaction. |
Glucose-6-Phosphate Dehydrogenase Deficiency
Enzymes in the human body act as catalysts for important chemical reactions in cellular metabolism. As such, a deficiency of a particular enzyme can translate to a life-threatening disease. G6PD (glucose-6-phosphate dehydrogenase) deficiency, a genetic condition that results in a shortage of the enzyme glucose-6-phosphate dehydrogenase, is the most common enzyme deficiency in humans. This enzyme, shown in , is the rate-limiting enzyme for the metabolic pathway that supplies NADPH to cells ().
FIGURE 12.21 Glucose-6-phosphate dehydrogenase is a rate-limiting enzyme for the metabolic pathway that supplies NADPH to cells.
A disruption in this pathway can lead to reduced glutathione in red blood cells; once all glutathione is consumed, enzymes and other proteins such as hemoglobin are susceptible to damage. For example, hemoglobin can be metabolized to bilirubin, which leads to jaundice, a condition that can become severe. People who suffer from G6PD deficiency must avoid certain foods and medicines containing chemicals that can trigger damage their glutathione-deficient red blood cells.
FIGURE 12.22 In the mechanism for the pentose phosphate pathway, G6PD catalyzes the reaction that regulates NADPH, a co-enzyme that regulates glutathione, an antioxidant that protects red blood cells and other cells from oxidative damage.
Heterogeneous Catalysts
A heterogeneous catalyst is a catalyst that is present in a different phase (usually a solid) than the reactants. Such catalysts generally function by furnishing an active surface upon which a reaction can occur. Gas and liquid phase reactions catalyzed by heterogeneous catalysts occur on the surface of the catalyst rather than within the gas or liquid phase.
Heterogeneous catalysis typically involves the following processes:
Adsorption of the reactant(s) onto the surface of the catalyst
Activation of the adsorbed reactant(s)
Reaction of the adsorbed reactant(s)
Desorption of product(s) from the surface of the catalyst
illustrates the steps of a mechanism for the reaction of compounds containing a carbon–carbon double bond with hydrogen on a nickel catalyst. Nickel is the catalyst used in the hydrogenation of polyunsaturated fats and oils (which contain several carbon–carbon double bonds) to produce saturated fats and oils (which contain only carbon–carbon single bonds).
FIGURE 12.23 Mechanism for the Ni-catalyzed reaction (a) Hydrogen is adsorbed on the surface, breaking the H–H bonds and forming Ni–H bonds. (b) Ethylene is adsorbed on the surface, breaking the C–C π-bond and forming Ni–C bonds. (c) Atoms diffuse across the surface and form new C–H bonds when they collide.
(d) C2H6 molecules desorb from the Ni surface.
Many important chemical products are prepared via industrial processes that use heterogeneous catalysts, including ammonia, nitric acid, sulfuric acid, and methanol. Heterogeneous catalysts are also used in the catalytic converters found on most gasoline-powered automobiles (). |
Enzyme Structure and Function
The study of enzymes is an important interconnection between biology and chemistry. Enzymes are usually proteins (polypeptides) that help to control the rate of chemical reactions between biologically important |
Enzyme molecules possess an active site, a part of the molecule with a shape that allows it to bond to a specific substrate (a reactant molecule), forming an enzyme-substrate complex as a reaction intermediate. There are two models that attempt to explain how this active site works. The most simplistic model is referred to as the lock-and-key hypothesis, which suggests that the molecular shapes of the active site and substrate are complementary, fitting together like a key in a lock. The induced fit hypothesis, on the other hand, suggests that the enzyme molecule is flexible and changes shape to accommodate a bond with the substrate. This is not to suggest that an enzyme’s active site is completely malleable, however. Both the lock-and-key model and the induced fit model account for the fact that enzymes can only bind with specific substrates, since in general a particular enzyme only catalyzes a particular reaction ().
FIGURE 12.25 (a) According to the lock-and-key model, the shape of an enzyme’s active site is a perfect fit for the substrate. (b) According to the induced fit model, the active site is somewhat flexible, and can change shape in order to bond with the substrate. |
Key Terms
activated complex (also, transition state) unstable combination of reactant species formed during a chemical reaction
activation energy (Ea) minimum energy necessary
in order for a reaction to take place
Arrhenius equation mathematical relationship between a reaction’s rate constant, activation energy, and temperature
average rate rate of a chemical reaction computed
as the ratio of a measured change in amount or concentration of substance to the time interval over which the change occurred
bimolecular reaction elementary reaction
involving two reactant species
catalyst substance that increases the rate of a reaction without itself being consumed by the reaction
collision theory model that emphasizes the energy
and orientation of molecular collisions to explain and predict reaction kinetics
elementary reaction reaction that takes place in a
single step, precisely as depicted in its chemical equation
frequency factor (A) proportionality constant in
the Arrhenius equation, related to the relative number of collisions having an orientation capable of leading to product formation
half-life of a reaction (tl/2) time required for half of
a given amount of reactant to be consumed
heterogeneous catalyst catalyst present in a different phase from the reactants, furnishing a surface at which a reaction can occur
homogeneous catalyst catalyst present in the
same phase as the reactants
initial rate instantaneous rate of a chemical reaction at t = 0 s (immediately after the reaction has begun)
instantaneous rate rate of a chemical reaction at any instant in time, determined by the slope of the line tangential to a graph of concentration as a function of time
integrated rate law equation that relates the |
concentration of a reactant to elapsed time of reaction
intermediate species produced in one step of a
reaction mechanism and consumed in a subsequent step
method of initial rates common experimental
approach to determining rate laws that involves measuring reaction rates at varying initial reactant concentrations
molecularity number of reactant species involved
in an elementary reaction
overall reaction order sum of the reaction orders for each substance represented in the rate law
rate constant (k) proportionality constant in a rate law
rate expression mathematical representation defining reaction rate as change in amount, concentration, or pressure of reactant or product species per unit time
rate law (also, rate equation) (also, differential rate
laws) mathematical equation showing the dependence of reaction rate on the rate constant and the concentration of one or more reactants
rate of reaction measure of the speed at which a
chemical reaction takes place
rate-determining step (also, rate-limiting step) slowest elementary reaction in a reaction mechanism; determines the rate of the overall reaction
reaction diagram used in chemical kinetics to
illustrate various properties of a reaction
reaction mechanism stepwise sequence of elementary reactions by which a chemical change takes place
reaction order value of an exponent in a rate law
(for example, zero order for 0, first order for 1, second order for 2, and so on)
termolecular reaction elementary reaction
involving three reactant species
unimolecular reaction elementary reaction involving a single reactant species |
The rate of a reaction can be expressed either in terms of the decrease in the amount of a reactant or the increase in the amount of a product per unit time. Relations between different rate expressions for a given reaction are derived directly from the stoichiometric coefficients of the equation representing the reaction.
The rate of a chemical reaction is affected by several parameters. Reactions involving two phases proceed more rapidly when there is greater surface area contact. If temperature or reactant concentration is increased, the rate of a given reaction generally increases as well. A catalyst can increase the rate of a reaction by providing an alternative pathway with a lower activation energy.
Rate laws (differential rate laws) provide a mathematical description of how changes in the concentration of a substance affect the rate of a chemical reaction. Rate laws are determined experimentally and cannot be predicted by reaction stoichiometry. The order of reaction describes how much a change in the concentration of each substance affects the overall rate, and the overall order of a reaction is the sum of the orders for each substance present in the reaction. Reaction orders are typically first order, second order, or zero order, but fractional and even negative orders are possible.
Integrated rate laws are mathematically derived from differential rate laws, and they describe the time dependence of reactant and product concentrations.
The half-life of a reaction is the time required to decrease the amount of a given reactant by one-half. A reaction’s half-life varies with rate constant and,
for some reaction orders, reactant concentration. The half-life of a zero-order reaction decreases as the initial concentration of the reactant in the reaction decreases. The half-life of a first-order reaction is independent of concentration, and the half-life of a second-order reaction decreases as the concentration increases.
Chemical reactions typically require collisions between reactant species. These reactant collisions must be of proper orientation and sufficient energy in order to result in product formation. Collision theory provides a simple but effective explanation for the effect of many experimental parameters on reaction rates. The Arrhenius equation describes the relation between a reaction’s rate constant, activation energy, temperature, and dependence on collision orientation.
The sequence of individual steps, or elementary reactions, by which reactants are converted into products during the course of a reaction is called the reaction mechanism. The molecularity of an elementary reaction is the number of reactant species involved, typically one (unimolecular), two (bimolecular), or, less commonly, three (termolecular). The overall rate of a reaction is determined by the rate of the slowest in its mechanism, called the rate-determining step.
Unimolecular elementary reactions have first-order rate laws, while bimolecular elementary reactions have second-order rate laws. By comparing the rate laws derived from a reaction mechanism to that determined experimentally, the mechanism may be deemed either incorrect or plausible. |
What is the difference between average rate, initial rate, and instantaneous rate?
Ozone decomposes to oxygen according to the equation Write the equation that relates the rate expressions for this reaction in terms of the disappearance of O3 and the formation of oxygen.
In the nuclear industry, chlorine trifluoride is used to prepare uranium hexafluoride, a volatile compound
of uranium used in the separation of uranium isotopes. Chlorine trifluoride is prepared by the reaction
Write the equation that relates the rate expressions for this reaction in terms of the disappearance of Cl2 and F2 and the formation of ClF3.
A study of the rate of dimerization of C4H6 gave the data shown in the table:
Determine the average rate of dimerization between 0 s and 1600 s, and between 1600 s and 3200 s.
Estimate the instantaneous rate of dimerization at 3200 s from a graph of time versus [C4H6]. What are the units of this rate?
Determine the average rate of formation of C8H12 at 1600 s and the instantaneous rate of formation at 3200 s from the rates found in parts (a) and (b).
A study of the rate of the reaction represented as gave the following data:
Determine the average rate of disappearance of A between 0.0 s and 10.0 s, and between 10.0 s and 20.0 s.
Estimate the instantaneous rate of disappearance of A at 15.0 s from a graph of time versus [A]. What are the units of this rate?
Use the rates found in parts (a) and (b) to determine the average rate of formation of B between 0.00 s and
10.0 s, and the instantaneous rate of formation of B at 15.0 s.
Consider the following reaction in aqueous solution:
If the rate of disappearance of Br–(aq) at a particular moment during the reaction is 3.5 10−4 mol L−1 s−1, what is the rate of appearance of Br2(aq) at that moment?
Describe the effect of each of the following on the rate of the reaction of magnesium metal with a solution of hydrochloric acid: the molarity of the hydrochloric acid, the temperature of the solution, and the size of the pieces of magnesium.
Explain why an egg cooks more slowly in boiling water in Denver than in New York City. (Hint: Consider the
effect of temperature on reaction rate and the effect of pressure on boiling point.)
Go to the interactive. Use the Single Collision tab to represent how the collision between monatomic oxygen (O) and carbon monoxide (CO) results in the breaking of one bond and the formation of another. Pull back on the red plunger to release the atom and observe the results. Then, click on “Reload Launcher” and change to “Angled shot” to see the difference.
What happens when the angle of the collision is changed?
Explain how this is relevant to rate of reaction.
In the interactive, use the “Many Collisions” tab to observe how multiple atoms and molecules interact under varying conditions. Select a molecule to pump into the chamber. Set the initial temperature and select the current amounts of each reactant. Select “Show bonds” under Options. How is the rate of the reaction affected by concentration and temperature?
In the interactive, on the Many Collisions
tab, set up a simulation with 15 molecules of A and 10 molecules of BC. Select “Show Bonds” under Options.
Leave the Initial Temperature at the default setting. Observe the reaction. Is the rate of reaction fast or slow?
Click “Pause” and then “Reset All,” and then enter 15 molecules of A and 10 molecules of BC once again. Select “Show Bonds” under Options. This time, increase the initial temperature until, on the graph, the total average energy line is completely above the potential energy curve. Describe what happens to the reaction.
How do the rate of a reaction and its rate constant differ?
Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:
What is the order of the reaction with respect to that reactant?
Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?
Tripling the concentration of a reactant increases the rate of a reaction nine-fold. With this knowledge,
answer the following questions:
What is the order of the reaction with respect to that reactant?
Increasing the concentration of a reactant by a factor of four increases the rate of a reaction four-fold. What is the order of the reaction with respect to that reactant?
How will the rate of reaction change for the process: if the rate law
for the reaction is
Decreasing the pressure of NO2 from 0.50 atm to 0.250 atm.
Increasing the concentration of CO from 0.01 M to 0.03 M.
How will each of the following affect the rate of the reaction: if the rate law for the reaction is ?
Increasing the pressure of NO2 from 0.1 atm to 0.3 atm
Increasing the concentration of CO from 0.02 M to 0.06 M.
Regular flights of supersonic aircraft in the stratosphere are of concern because such aircraft produce nitric oxide, NO, as a byproduct in the exhaust of their engines. Nitric oxide reacts with ozone, and it has been suggested that this could contribute to depletion of the ozone layer. The reaction
is first order with respect to both NO and O3 with a rate constant of 2.20 107
L/mol/s. What is the instantaneous rate of disappearance of NO when [NO] = 3.3 10−6 M and [O3] = 5.9 10−7 M?
Radioactive phosphorus is used in the study of biochemical reaction mechanisms because phosphorus atoms are components of many biochemical molecules. The location of the phosphorus (and the location of the molecule it is bound in) can be detected from the electrons (beta particles) it produces:
rate = 4.85 10−2
What is the instantaneous rate of production of electrons in a sample with a phosphorus concentration of 0.0033 M?
The rate constant for the radioactive decay of 14C is 1.21 10−4 year−1. The products of the decay are nitrogen atoms and electrons (beta particles):
What is the instantaneous rate of production of N atoms in a sample with a carbon-14 content of 6.5 10−9 M?
The decomposition of acetaldehyde is a second order reaction with a rate constant of 4.71 10−8 L mol−1 s−1. What is the instantaneous rate of decomposition of acetaldehyde in a solution with a concentration of
5.55 10−4 M?
Alcohol is removed from the bloodstream by a series of metabolic reactions. The first reaction produces acetaldehyde; then other products are formed. The following data have been determined for the rate at which alcohol is removed from the blood of an average male, although individual rates can vary by 25–30%. Women metabolize alcohol a little more slowly than men: |
Determine the rate law, the rate constant, and the overall order for this reaction.
Nitrosyl chloride, NOCl, decomposes to NO and Cl2.
Determine the rate law, the rate constant, and the overall order for this reaction from the following data: |
Nitrogen monoxide reacts with chlorine according to the equation:
The following initial rates of reaction have been observed for certain reactant concentrations:
[NO] (mol/L) [Cl2] (mol/L) Rate (mol L−1 h−1)
What is the rate law that describes the rate’s dependence on the concentrations of NO and Cl2? What is the rate constant? What are the orders with respect to each reactant?
Hydrogen reacts with nitrogen monoxide to form dinitrogen monoxide (laughing gas) according to the
equation:
Determine the rate law, the rate constant, and the orders with respect to each reactant from the following data: |
What is the order of the reaction with respect to [Q], and what is the rate law?
What is the rate constant?
The rate constant for the first-order decomposition at 45 °C of dinitrogen pentoxide, N2O5, dissolved in chloroform, CHCl3, is 6.2 10−4 min−1.
What is the rate of the reaction when [N2O5] = 0.40 M?
The annual production of HNO3 in 2013 was 60 million metric tons Most of that was prepared by the following sequence of reactions, each run in a separate reaction vessel.
(a)
(b)
(c)
The first reaction is run by burning ammonia in air over a platinum catalyst. This reaction is fast. The reaction in equation (c) is also fast. The second reaction limits the rate at which nitric acid can be prepared from ammonia. If equation (b) is second order in NO and first order in O2, what is the rate of formation of NO2 when the oxygen concentration is 0.50 M and the nitric oxide concentration is 0.75 M? The rate constant for the reaction is 5.8 10−6 L2 mol−2 s−1.
The following data have been determined for the reaction:
1 2 3 |
Describe how graphical methods can be used to determine the order of a reaction and its rate constant from a series of data that includes the concentration of A at varying times.
Use the data provided to graphically determine the order and rate constant of the following reaction: |
What is the half-life for the first-order decay of phosphorus-32? The rate constant for the decay is 4.85 10−2 day−1.
What is the half-life for the first-order decay of carbon-14? The rate constant for the decay is 1.21 10−4 year−1.
What is the half-life for the decomposition of NOCl when the concentration of NOCl is 0.15 M? The rate
constant for this second-order reaction is 8.0 10−8 L mol−1 s−1.
What is the half-life for the decomposition of O3 when the concentration of O3 is 2.35 10−6 M? The rate constant for this second-order reaction is 50.4 L mol−1 h−1.
The reaction of compound A to give compounds C and D was found to be second-order in A. The rate constant for the reaction was determined to be 2.42 L mol−1 s−1. If the initial concentration is 0.500 mol/L, what is the value of t1/2?
The half-life of a reaction of compound A to give compounds D and E is 8.50 min when the initial concentration of A is 0.150 M. How long will it take for the concentration to drop to 0.0300 M if the reaction is (a) first order with respect to A or (b) second order with respect to A?
Some bacteria are resistant to the antibiotic penicillin because they produce penicillinase, an enzyme with a molecular weight of 3 104 g/mol that converts penicillin into inactive molecules. Although the kinetics of enzyme-catalyzed reactions can be complex, at low concentrations this reaction can be described by a rate law that is first order in the catalyst (penicillinase) and that also involves the concentration of penicillin. From the following data: 1.0 L of a solution containing 0.15 µg (0.15 10−6 g) of penicillinase, determine the order of the reaction with respect to penicillin and the value of the rate constant.
[Penicillin] (M) Rate (mol L−1 min−1)
Both technetium-99 and thallium-201 are used to image heart muscle in patients with suspected heart problems. The half-lives are 6 h and 73 h, respectively. What percent of the radioactivity would remain for each of the isotopes after 2 days (48 h)?
There are two molecules with the formula C3H6. Propene, is the monomer of the polymer
polypropylene, which is used for indoor-outdoor carpets. Cyclopropane is used as an anesthetic:
When heated to 499 °C, cyclopropane rearranges (isomerizes) and forms propene with a rate constant of
5.95 10−4 s−1. What is the half-life of this reaction? What fraction of the cyclopropane remains after 0.75 h at 499 °C?
Fluorine-18 is a radioactive isotope that decays by positron emission to form oxygen-18 with a half-life of
109.7 min. (A positron is a particle with the mass of an electron and a single unit of positive charge; the equation is Physicians use 18F to study the brain by injecting a quantity of fluoro-
substituted glucose into the blood of a patient. The glucose accumulates in the regions where the brain is active and needs nourishment.
What is the rate constant for the decomposition of fluorine-18?
If a sample of glucose containing radioactive fluorine-18 is injected into the blood, what percent of the radioactivity will remain after 5.59 h?
How long does it take for 99.99% of the 18F to decay?
Suppose that the half-life of steroids taken by an athlete is 42 days. Assuming that the steroids biodegrade by a first-order process, how long would it take for of the initial dose to remain in the athlete’s body?
Recently, the skeleton of King Richard III was found under a parking lot in England. If tissue samples from the skeleton contain about 93.79% of the carbon-14 expected in living tissue, what year did King Richard III die? The half-life for carbon-14 is 5730 years.
Nitroglycerine is an extremely sensitive explosive. In a series of carefully controlled experiments, samples of
the explosive were heated to 160 °C and their first-order decomposition studied. Determine the average rate constants for each experiment using the following data:
For the past 10 years, the unsaturated hydrocarbon 1,3-butadiene has ranked 38th among the top 50 industrial chemicals. It is used primarily for the manufacture of synthetic rubber. An isomer exists also as cyclobutene:
The isomerization of cyclobutene to butadiene is first-order and the rate constant has been measured as
2.0 10−4 s−1 at 150 °C in a 0.53-L flask. Determine the partial pressure of cyclobutene and its concentration after 30.0 minutes if an isomerization reaction is carried out at 150 °C with an initial pressure of 55 torr.
Chemical reactions occur when reactants collide. What are two factors that may prevent a collision from producing a chemical reaction?
When every collision between reactants leads to a reaction, what determines the rate at which the reaction occurs?
What is the activation energy of a reaction, and how is this energy related to the activated complex of the reaction?
Account for the relationship between the rate of a reaction and its activation energy.
Describe how graphical methods can be used to determine the activation energy of a reaction from a series of data that includes the rate of reaction at varying temperatures.
How does an increase in temperature affect rate of reaction? Explain this effect in terms of the collision theory of the reaction rate.
The rate of a certain reaction doubles for every 10 °C rise in temperature.
How much faster does the reaction proceed at 45 °C than at 25 °C?
How much faster does the reaction proceed at 95 °C than at 25 °C?
In an experiment, a sample of NaClO3 was 90% decomposed in 48 min. Approximately how long would this decomposition have taken if the sample had been heated 20 °C higher? (Hint: Assume the rate doubles for each 10 °C rise in temperature.)
The rate constant at 325 °C for the decomposition reaction is 6.1 10−8 s−1, and the
activation energy is 261 kJ per mole of C4H8. Determine the frequency factor for the reaction.
The rate constant for the decomposition of acetaldehyde, CH3CHO, to methane, CH4, and carbon monoxide, CO, in the gas phase is 1.1 10−2 L mol−1 s−1 at 703 K and 4.95 L mol−1 s−1 at 865 K. Determine the activation energy for this decomposition.
An elevated level of the enzyme alkaline phosphatase (ALP) in human serum is an indication of possible liver or bone disorder. The level of serum ALP is so low that it is very difficult to measure directly. However, ALP catalyzes a number of reactions, and its relative concentration can be determined by measuring the rate of one of these reactions under controlled conditions. One such reaction is the conversion of p-nitrophenyl phosphate (PNPP) to p-nitrophenoxide ion (PNP) and phosphate ion. Control of temperature during the test is very important; the rate of the reaction increases 1.47 times if the temperature changes from 30 °C to 37 °C. What is the activation energy for the ALP–catalyzed conversion of PNPP to PNP and phosphate?
In terms of collision theory, to which of the following is the rate of a chemical reaction proportional?
the change in free energy per second
the change in temperature per second
the number of collisions per second
the number of product molecules
Hydrogen iodide, HI, decomposes in the gas phase to produce hydrogen, H2, and iodine, I2. The value of the rate constant, k, for the reaction was measured at several different temperatures and the data are shown here:
Temperature (K) k (L mol−1 s−1)
What is the value of the activation energy (in kJ/mol) for this reaction?
The element Co exists in two oxidation states, Co(II) and Co(III), and the ions form many complexes. The rate at which one of the complexes of Co(III) was reduced by Fe(II) in water was measured. Determine the activation energy of the reaction from the following data:
T (K) k (s−1)
The hydrolysis of the sugar sucrose to the sugars glucose and fructose,
follows a first-order rate law for the disappearance of sucrose: rate = k[C12H22O11] (The products of the reaction, glucose and fructose, have the same molecular formulas but differ in the arrangement of the atoms in their molecules.)
In neutral solution, k = 2.1 10−11 s−1 at 27 °C and 8.5 10−11 s−1 at 37 °C. Determine the activation
energy, the frequency factor, and the rate constant for this equation at 47 °C (assuming the kinetics remain consistent with the Arrhenius equation at this temperature).
When a solution of sucrose with an initial concentration of 0.150 M reaches equilibrium, the
concentration of sucrose is 1.65 10−7 M. How long will it take the solution to reach equilibrium at 27 °C in the absence of a catalyst? Because the concentration of sucrose at equilibrium is so low, assume that the reaction is irreversible.
Why does assuming that the reaction is irreversible simplify the calculation in part (b)?
Use the to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first reaction (A is yellow, B is purple, and C is navy blue). Using the “straight shot” default option, try launching the A atom with varying amounts of energy. What changes when the Total Energy line at launch is below the transition state of the Potential Energy line? Why? What happens when it is above the transition state? Why?
Use the to simulate a system. On the “Single collision” tab of the simulation applet, enable the “Energy view” by clicking the “+” icon. Select the first reaction (A is yellow, B is purple, and C is navy blue). Using the “angled shot” option, try launching the A atom with varying angles, but with more Total energy than the transition state. What happens when the A atom hits the BC molecule from different directions? Why?
Why are elementary reactions involving three or more reactants very uncommon?
In general, can we predict the effect of doubling the concentration of A on the rate of the overall reaction
? Can we predict the effect if the reaction is known to be an elementary reaction?
Define these terms:
unimolecular reaction
bimolecular reaction
elementary reaction
overall reaction
What is the rate law for the elementary termolecular reaction For
Given the following reactions and the corresponding rate laws, in which of the reactions might the elementary reaction and the overall reaction be the same? |
Consider the following questions:
Determine the order for each of the reactants, NO and H2, from the data given and show your reasoning.
Write the overall rate law for the reaction.
Calculate the value of the rate constant, k, for the reaction. Include units.
For experiment 2, calculate the concentration of NO remaining when exactly one-half of the original amount of H2 had been consumed.
The following sequence of elementary steps is a proposed mechanism for the reaction. Step 1:
Step 2:
Step 3:
Based on the data presented, which of these is the rate determining step? Show that the mechanism is consistent with the observed rate law for the reaction and the overall stoichiometry of the reaction.
The reaction of CO with Cl2 gives phosgene (COCl2), a nerve gas that was used in World War I. Use the
mechanism shown here to complete the following exercises:
(fast, k1 represents the forward rate constant, k−1 the reverse rate constant) (slow, k2 the rate constant)
(fast, k3 the rate constant)
Write the overall reaction.
Identify all intermediates.
Write the rate law for each elementary reaction.
Write the overall rate law expression. |
Compare the functions of homogeneous and heterogeneous catalysts.
Consider this scenario and answer the following questions: Chlorine atoms resulting from decomposition of chlorofluoromethanes, such as CCl2F2, catalyze the decomposition of ozone in the atmosphere. One simplified mechanism for the decomposition is: |
Assuming the diagrams in represent different mechanisms for the same reaction, which of the reactions has the faster rate?
Consider the similarities and differences in the two reaction diagrams shown in . Do these diagrams represent two different overall reactions, or do they represent the same overall reaction taking place by two different mechanisms? Explain your answer. |
INTRODUCTION Imagine a beach populated with sunbathers and swimmers. As those basking in the sun get too hot, they enter the surf to swim and cool off. As the swimmers tire, they return to the beach to rest. If the rate at which sunbathers enter the surf were to equal the rate at which swimmers return to the sand, then the numbers (though not the identities) of sunbathers and swimmers would remain constant. This scenario illustrates a dynamic phenomenon known as equilibrium, in which opposing processes occur at equal rates. Chemical and physical processes are subject to this phenomenon; these processes are at equilibrium when the forward and reverse reaction rates are equal. Equilibrium systems are pervasive in nature; the various reactions involving carbon dioxide dissolved in blood are examples (see ). This chapter provides a thorough introduction to the essential aspects of chemical equilibria.
Chemical Equilibria
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Describe the nature of equilibrium systems
Explain the dynamic nature of a chemical equilibrium
The convention for writing chemical equations involves placing reactant formulas on the left side of a reaction
arrow and product formulas on the right side. By this convention, and the definitions of “reactant” and “product,” a chemical equation represents the reaction in question as proceeding from left to right. Reversible reactions, however, may proceed in both forward (left to right) and reverse (right to left) directions. When the rates of the forward and reverse reactions are equal, the concentrations of the reactant and product species remain constant over time and the system is at equilibrium. The relative concentrations of reactants and products in equilibrium systems vary greatly; some systems contain mostly products at equilibrium, some contain mostly reactants, and some contain appreciable amounts of both.
illustrates fundamental equilibrium concepts using the reversible decomposition of colorless dinitrogen tetroxide to yield brown nitrogen dioxide, an elementary reaction described by the equation: |
FIGURE 13.2 (a) A sealed tube containing colorless N2O4 darkens as it decomposes to yield brown NO2. (b) Changes in concentration over time as the decomposition reaction achieves equilibrium. (c) At equilibrium, the forward and reverse reaction rates are equal. |
As the reaction begins (t = 0), the concentration of the N2O4 reactant is finite and that of the NO2 product is zero, so the forward reaction proceeds at a finite rate while the reverse reaction rate is zero. As time passes, N2O4 is consumed and its concentration falls, while NO2 is produced and its concentration increases ( b). The decreasing concentration of the reactant slows the forward reaction rate, and the increasing product concentration speeds the reverse reaction rate (c). This process continues until the forward and reverse reaction rates become equal, at which time the reaction has reached equilibrium, as characterized by constant concentrations of its reactants and products (shaded areas of b and c). It’s important to emphasize that chemical equilibria are dynamic; a reaction at equilibrium has not “stopped,” but is proceeding in the forward and reverse directions at the same rate. This dynamic nature is essential to understanding equilibrium behavior as discussed in this and subsequent chapters of the text.
FIGURE 13.3 A two-person juggling act illustrates the dynamic aspect of chemical equilibria. Each person is throwing and catching clubs at the same rate, and each holds a (approximately) constant number of clubs.
Physical changes, such as phase transitions, are also reversible and may establish equilibria. This concept was introduced in another chapter of this text through discussion of the vapor pressure of a condensed phase (liquid or solid). As one example, consider the vaporization of bromine:
When liquid bromine is added to an otherwise empty container and the container is sealed, the forward process depicted above (vaporization) will commence and continue at a roughly constant rate as long as the exposed surface area of the liquid and its temperature remain constant. As increasing amounts of gaseous bromine are produced, the rate of the reverse process (condensation) will increase until it equals the rate of vaporization and equilibrium is established. A photograph showing this phase transition equilibrium is provided in .
FIGURE 13.4 A sealed tube containing an equilibrium mixture of liquid and gaseous bromine. (credit:
Equilibrium Constants
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Derive reaction quotients from chemical equations representing homogeneous and heterogeneous reactions
Calculate values of reaction quotients and equilibrium constants, using concentrations and pressures
Relate the magnitude of an equilibrium constant to properties of the chemical system
The status of a reversible reaction is conveniently assessed by evaluating its reaction quotient (Q). For a reversible reaction described by |
Note that the reaction quotient equations above are a simplification of more rigorous expressions that use relative values for concentrations and pressures rather than absolute values. These relative concentration and pressure values are dimensionless (they have no units); consequently, so are the reaction quotients. For purposes of this introductory text, it will suffice to use the simplified equations and to disregard units when computing Q. In most cases, this will introduce only modest errors in calculations involving reaction quotients. |
FIGURE 13.5 Changes in concentrations and Qc for a chemical equilibrium achieved beginning with (a) a mixture of reactants only and (b) products only.
The numerical value of Q varies as a reaction proceeds towards equilibrium; therefore, it can serve as a useful indicator of the reaction’s status. To illustrate this point, consider the oxidation of sulfur dioxide:
Two different experimental scenarios are depicted in , one in which this reaction is initiated with a mixture of reactants only, SO2 and O2, and another that begins with only product, SO3. For the reaction that begins with a mixture of reactants only, Q is initially equal to zero:
As the reaction proceeds toward equilibrium in the forward direction, reactant concentrations decrease (as does the denominator of Qc), product concentration increases (as does the numerator of Qc), and the reaction quotient consequently increases. When equilibrium is achieved, the concentrations of reactants and product remain constant, as does the value of Qc.
If the reaction begins with only product present, the value of Qc is initially undefined (immeasurably large, or infinite):
In this case, the reaction proceeds toward equilibrium in the reverse direction. The product concentration and the numerator of Qc decrease with time, the reactant concentrations and the denominator of Qc increase, and the reaction quotient consequently decreases until it becomes constant at equilibrium.
The constant value of Q exhibited by a system at equilibrium is called the equilibrium constant, K:
Comparison of the data plots in shows that both experimental scenarios resulted in the same value for the equilibrium constant. This is a general observation for all equilibrium systems, known as the law of mass action: At a given temperature, the reaction quotient for a system at equilibrium is constant. |
When 0.10 mol NO2 is added to a 1.0-L flask at 25 °C, the concentration changes so that at equilibrium, [NO2] =
0.016 M and [N2O4] = 0.042 M.
What is the value of the reaction quotient before any reaction occurs?
What is the value of the equilibrium constant for the reaction?
Solution
As for all equilibrium calculations in this text, use the simplified equations for Q and K and disregard any concentration or pressure units, as noted previously in this section.
Before any product is formed, and [N2O4] = 0 M. Thus, |
By its definition, the magnitude of an equilibrium constant explicitly reflects the composition of a reaction mixture at equilibrium, and it may be interpreted with regard to the extent of the forward reaction. A reaction exhibiting a large K will reach equilibrium when most of the reactant has been converted to product, whereas a small K indicates the reaction achieves equilibrium after very little reactant has been converted. It’s important to keep in mind that the magnitude of K does not indicate how rapidly or slowly equilibrium will be reached.
Some equilibria are established so quickly as to be nearly instantaneous, and others so slowly that no perceptible change is observed over the course of days, years, or longer.
The equilibrium constant for a reaction can be used to predict the behavior of mixtures containing its reactants and/or products. As demonstrated by the sulfur dioxide oxidation process described above, a chemical reaction will proceed in whatever direction is necessary to achieve equilibrium. Comparing Q to K for
an equilibrium system of interest allows prediction of what reaction (forward or reverse), if any, will occur. To further illustrate this important point, consider the reversible reaction shown below:
The bar charts in represent changes in reactant and product concentrations for three different reaction mixtures. The reaction quotients for mixtures 1 and 3 are initially lesser than the reaction’s equilibrium constant, so each of these mixtures will experience a net forward reaction to achieve equilibrium. The reaction quotient for mixture 2 is initially greater than the equilibrium constant, so this mixture will proceed in the reverse direction until equilibrium is established. |
Qc < Kc (0.48 < 0.64)
The reaction will proceed in the forward direction.
Check Your Learning
Calculate the reaction quotient and determine the direction in which each of the following reactions will proceed to reach equilibrium.
A 1.00-L flask containing 0.0500 mol of NO(g), 0.0155 mol of Cl2(g), and 0.500 mol of NOCl: |
Homogeneous Equilibria
A homogeneous equilibrium is one in which all reactants and products (and any catalysts, if applicable) are present in the same phase. By this definition, homogeneous equilibria take place in solutions. These solutions are most commonly either liquid or gaseous phases, as shown by the examples below: |
examples, water also functions as a reactant, but its concentration is not included in the reaction quotient. The reason for this omission is related to the more rigorous form of the Q (or K) expression mentioned previously in this chapter, in which relative concentrations for liquids and solids are equal to 1 and needn’t be included. Consequently, reaction quotients include concentration or pressure terms only for gaseous and solute species.
The equilibria below all involve gas-phase solutions:
For gas-phase solutions, the equilibrium constant may be expressed in terms of either the molar concentrations (Kc) or partial pressures (Kp) of the reactants and products. A relation between these two K values may be simply derived from the ideal gas equation and the definition of molarity: |
Again, note that concentration terms are only included for gaseous and solute species, as discussed previously. Two of the above examples include terms for gaseous species only in their equilibrium constants, and so Kp
expressions may also be written:
Coupled Equilibria
The equilibrium systems discussed so far have all been relatively simple, involving just single reversible reactions. Many systems, however, involve two or more coupled equilibrium reactions, those which have in common one or more reactant or product species. Since the law of mass action allows for a straightforward derivation of equilibrium constant expressions from balanced chemical equations, the K value for a system involving coupled equilibria can be related to the K values of the individual reactions. Three basic manipulations are involved in this approach, as described below.
Changing the direction of a chemical equation essentially swaps the identities of “reactants” and “products,” and so the equilibrium constant for the reversed equation is simply the reciprocal of that for the forward equation. |
Shifting Equilibria: Le Châtelier’s Principle
LEARNING OBJECTIVES
By the end of this section, you will be able to:
Describe the ways in which an equilibrium system can be stressed
Predict the response of a stressed equilibrium using Le Châtelier’s principle
A system at equilibrium is in a state of dynamic balance, with forward and reverse reactions taking place at equal rates. If an equilibrium system is subjected to a change in conditions that affects these reaction rates differently (a stress), then the rates are no longer equal and the system is not at equilibrium. The system will subsequently experience a net reaction in the direction of greater rate (a shift) that will re-establish the equilibrium. This phenomenon is summarized by Le Châtelier’s principle: if an equilibrium system is stressed, the system will experience a shift in response to the stress that re-establishes equilibrium.
Reaction rates are affected primarily by concentrations, as described by the reaction’s rate law, and temperature, as described by the Arrhenius equation. Consequently, changes in concentration and temperature are the two stresses that can shift an equilibrium.
Effect of a Change in Concentration
If an equilibrium system is subjected to a change in the concentration of a reactant or product species, the rate of either the forward or the reverse reaction will change. As an example, consider the equilibrium reaction |
The system will experience a temporary net reaction in the forward direction to re-establish equilibrium (the equilibrium will shift right). This same shift will result if some product HI is removed from the system, which decreases the rate of the reverse reaction, again resulting in the same imbalance in rates.
The same logic can be used to explain the left shift that results from either removing reactant or adding product to an equilibrium system. These stresses both result in an increased rate for the reverse reaction
and a temporary net reaction in the reverse direction to re-establish equilibrium.
As an alternative to this kinetic interpretation, the effect of changes in concentration on equilibria can be rationalized in terms of reaction quotients. When the system is at equilibrium,
If reactant is added (increasing the denominator of the reaction quotient) or product is removed (decreasing the numerator), then Qc < Kc and the equilibrium will shift right. Note that the three different ways of inducing this stress result in three different changes in the composition of the equilibrium mixture. If H2 is added, the right shift will consume I2 and produce HI as equilibrium is re-established, yielding a mixture with a greater concentrations of H2 and HI and a lesser concentration of I2 than was present before. If I2 is added, the new equilibrium mixture will have greater concentrations of I2 and HI and a lesser concentration of H2. Finally, if HI is removed, the concentrations of all three species will be lower when equilibrium is reestablished. Despite these differences in composition, the value of the equilibrium constant will be the same after the stress as it was before (per the law of mass action). The same logic may be applied for stresses involving removing reactants or adding product, in which case Qc > Kc and the equilibrium will shift left.
For gas-phase equilibria such as this one, some additional perspectives on changing the concentrations of reactants and products are worthy of mention. The partial pressure P of an ideal gas is proportional to its molar concentration M,
and so changes in the partial pressures of any reactant or product are essentially changes in concentrations and thus yield the same effects on equilibria. Aside from adding or removing reactant or product, the
pressures (concentrations) of species in a gas-phase equilibrium can also be changed by changing the volume occupied by the system. Since all species of a gas-phase equilibrium occupy the same volume, a given change in volume will cause the same change in concentration for both reactants and products. In order to discern what shift, if any, this type of stress will induce the stoichiometry of the reaction must be considered.
At equilibrium, the reaction H2(g) + I2(g) ⇌ 2HI(g) is described by the reaction quotient |
And so, changing the volume of this gas-phase equilibrium mixture does not result in a shift of the equilibrium.
A similar treatment of a different system, 2NO2(g) ⇌ 2 NO(g) + O2(g), however, yields a different result:
In this case, the change in volume results in a reaction quotient greater than the equilibrium constant, and so the equilibrium will shift left.
These results illustrate the relationship between the stoichiometry of a gas-phase equilibrium and the effect of a volume-induced pressure (concentration) change. If the total molar amounts of reactants and products are equal, as in the first example, a change in volume does not shift the equilibrium. If the molar amounts of reactants and products are different, a change in volume will shift the equilibrium in a direction that better “accommodates” the volume change. In the second example, two moles of reactant (NO2) yield three moles of product (2NO + O2), and so decreasing the system volume causes the equilibrium to shift left since the reverse reaction produces less gas (2 mol) than the forward reaction (3 mol). Conversely, increasing the volume of this equilibrium system would result in a shift towards products.
LINK TO LEARNING
Check out this to see a dramatic visual demonstration of how equilibrium changes with pressure changes. |
These same equilibrium reactions are the basis of today’s soft-drink carbonation process. Beverages are exposed to a high pressure of gaseous carbon dioxide during the process to shift the first equilibrium above to the right, resulting in desirably high concentrations of dissolved carbon dioxide and, per similar shifts in the other two equilibria, its hydrolysis and ionization products. A bottle or can is then nearly filled with the carbonated beverage, leaving a relatively small volume of air in the container above the beverage surface (the headspace) before it is sealed. The pressure of carbon dioxide in the container headspace is very low immediately after sealing, but it rises as the dissolution equilibrium is re-established by shifting to the left. Since the volume of the beverage is significantly greater than the volume of the headspace, only a relatively small amount of dissolved carbon dioxide is lost to the headspace.
When a carbonated beverage container is opened, a hissing sound is heard as pressurized CO2 escapes from the headspace. This causes the dissolution equilibrium to shift left, resulting in a decrease in the concentration of dissolved CO2 and subsequent left-shifts of the hydrolysis and ionization equilibria.
Fortunately for the consumer, the dissolution equilibrium is usually re-established slowly, and so the beverage may be enjoyed while its dissolved carbon dioxide concentration remains palatably high. Once the equilibria are re-established, the CO2(aq) concentration will be significantly lowered, and the beverage acquires a characteristic taste referred to as “flat.”
FIGURE 13.7 Opening a soft-drink bottle lowers the CO2 pressure above the beverage, shifting the dissolution equilibrium and releasing dissolved CO2 from the beverage. (credit: modification of work by “D Coetzee”/Flickr)
Effect of a Change in Temperature
Consistent with the law of mass action, an equilibrium stressed by a change in concentration will shift to re- establish equilibrium without any change in the value of the equilibrium constant, K. When an equilibrium shifts in response to a temperature change, however, it is re-established with a different relative composition |
The equilibrium constant is seen to be a mathematical function of the rate constants for the forward and reverse reactions. Since the rate constants vary with temperature as described by the Arrhenius equation, is stands to reason that the equilibrium constant will likewise vary with temperature (assuming the rate constants are affected to different extents by the temperature change). For more complex reactions involving multistep reaction mechanisms, a similar but more complex mathematical relation exists between the equilibrium constant and the rate constants of the steps in the mechanism. Regardless of how complex the reaction may be, the temperature-dependence of its equilibrium constant persists.
Predicting the shift an equilibrium will experience in response to a change in temperature is most conveniently accomplished by considering the enthalpy change of the reaction. For example, the decomposition of dinitrogen tetroxide is an endothermic (heat-consuming) process: |