FUNDAMENTALS
`OF
`
`CHEMISTRY
`
`WITH QUALITATIVE ANALYSIS
`
`Third Edi’rion
`
`Jomes E. Brody
`
`S1. John's University. New York
`
`A John R. Holum
`Augsburg College, Minnesofa
`
`WILEY
`
`John Wiley 84 Sons
`
`New York
`
`Chiohesfer
`
`Brisbane
`
`Toronto
`
`Singopore
`
`oooo01
`
`IPR2015-00171
`|PR2015-00171
`Exhibit 1032
`Exhibit ‘I032
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`000001
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`

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`“-'"-‘?f-‘-?‘f5.’-"-'-*”.‘:'°n":§-'‘C.
`
`Book production supervised by
`Lucille Buonooore ond Down Reiiz.
`
`Cover and text designed by Kevin Murphy.
`Cover orr by Roy Wiemonn.
`Phoio reseorched by Sofro Nimrod.
`Illustrations by John Bolbolis.
`Monusoripf editor wos Josephine Dello Perulo,
`under the supervision of Deboroh Herbert.
`
`Copyright © 1981, 1984, 1988, by John Wiley Sc Sons, Inc.
`
`All rights reserved. Published simultaneously in Canada.
`
`Reproduction or translation of any part of
`this work beyond that permitted by Sections
`107 and 108 of the 1976 United States Copyright
`Act without the permission of the copyright
`owner is unlawful. Requests for permission
`or further information should be addressed to
`the Permissions Department, john Wiley 86 Sons.
`
`ISBN 0-471-84491-8
`
`Printed in the United States of America
`
`10937654321
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`"sis-sew.‘ ror-Ic 10.1 sasszr-oariivo
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`Anyone who has watched much television has certainly _
`_ seen advertisements that praise-the virtues of freeze-'
`'
`- fdried instant coffee. "Campers routinely carry freeze-
`dried foods because they are lightweight. In addition.
`- bacteria cannot grow" and reproduce in the complete
`absence of moisture. so freeze-dried foods need. no
`refrigeration. Therefore, freeze-drying is clearly a
`- useful way of.preservi_ng foods. It is accomplished by
`"first freezing the food (or brewed coffee) and then
`placing it in a chamber that is connected to high4ca-
`pacity vacuum pumps. These pumps lower the pressure _
`in_the chamber below the vapor pressure of ice, which
`' causes the ice crystals to sI.Ib1i_me.'Drying foods in" this
`- way has one important advantage over other methods———
`
`the delicate molecules responsible for the flavor of
`
`
`'
`_
`.-
`_
`.
`‘s-..._
`-
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`It's common to find freeze-dried foodsarnong the supplies.
`-
`carried by campers."
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`foods are not destroyed. as they would be if the food
`were hfiflted. Freeze-dried foods |'l'I_3¥' 330 be @353?
`reconstituted simply by adding water.
`_
`Freeze-drying is also _used_ by biologists to preserve _
`tissue cultures and bacteria. ‘Removing moisture from
`tissue and storing it at low temperatures allows the
`cells to survive in what amounts to a state of sus-
`- pended animation for periods of at least ‘several -years.
`
`'_
`
`
`
`Vapor Pressures of Solids
`
`Solids have vapor pressures just as liquids do. In a crystal, the particles are not stationary.
`They vibrate back and forth about their equilibrium positions. At a given temperature
`there is a distribution of kinetic energies, so some particles vibrate slowly while others
`vibrate with a great deal of energy. Some particles at the surface have high enough
`energies to break away from their neighbors and enter the vapor state. When particles in
`the vapor collide with the crystal, they can be recaptured, so condensation can occur
`too. Eventually, the concentration of particles in the vapor reaches a point where the rate
`of sublimation from the solid is the same as the rate of condensation, and a dynamic
`equilibrium is established. The pressure of the vapor that is in equilibrium with the solid
`is called the f!t'fl.illll1l'l‘..lI’il vapor presstire of the solid. A5 with liquids, the equilibrium
`vapor pressure is usually referred to simply as the vapor pressure. Like that of a liquid, the
`vapor pressure of a solid is determined by the strengths of the attractive forces between
`its particles and by the temperature.
`
`‘l 0.8 DYNAMIC EQUILIBRIUM
`AND LE Cl-lATELIER'S PRINCIPLE
`
`Disturbing a system at equilibrium causes it to change in a way that counteracts the
`disturbing infiu-ante and brings the system to equilibrium again.
`
`In the last section, you learned that when the temperature of a liquid is raised, its vapor
`pressure increases. We could have reached this conclusion by analyzing what happens to
`the system in the following way. Initially, the liquid is in equilibrium with its vapor, which
`exerts a certain pressure. When the temperature is increased, equilibrium no longer exists
`because evaporation occurs more rapidly than condensation. Eventually, as the concen-
`tration of molecules in the vapor increases, the system reaches a new equilibrium in
`which there is more vapor. This greater amount of vapor exerts a larger pressure.
`What happens to the vapor pressure of a liquid when the temperature is raised is an
`
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`10.8
`
`_,
`
`Dynamic equiiibrium is indi-
`cated in a chemical equation by
`a pair of arrows pointing in
`opposite directions, =, which
`implies opposing changes
`happening at equal rates.
`
`(10.1)
`
`A rise in temperature moves an
`equilibrium in the direction of
`3" Endmhermic Change-
`
`A decrease in temperature
`f3V0fS 3 net Chafige that 35 EXO-
`"mimic-
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`heat + liquid : vapor
`
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`
`j'_ Chatelier’s Principie. When a system in equilibrium is subjected to a distur-
`bance that upsets the equilibrium, the system responds in a direction that tends to
`ounteract the disturbance and restore equilibrium in the system.
`
`
`
`DYNAMIC EQUILIBRIUM AND LE CI-lATELlER‘S PRINCIPLE
`367
`
` ple of a general phenomenon. Whenever a system at equilibrium is subjected to a
`
`
`bance that upsets the equilibrium, the system changes in a way that will return it to.
`"ilibrium again. For a liquid-vapor equilibrium, such a disturbance is a change of
`
`
`eratute, as we saw in the preceding paragraph.
`We will deal with many kinds of equiiibria, both chemical and physical, from time
`
`
`c. It would be very time consuming and sometimes very diflicult to carry out a
`"ailed analysis each time we wish to know the effects of some disturbance on the
`riutn system. Fortunately, there is a relatively simple and fast method of predicting
`
`
`é.-effect of a disturbance. It is based on a principle proposed in 1388 by a brilliant
`
`
`I;
`-chemist, Henry Le Chatelier (1850 - 1936).
`
`
`
`
`
`-Let's see how we can apply Le Chatelier’s principle to a liquid — vapor equilibrium
`Subjected to a temperature increase. To do this we have to ask ourselves, how do
`go,about increasing the temperature of a system? The answer, of course, is that we
`
`
`' eat to it. When the temperature is increased, it is the addition of heat that is really
`
`
`ilisturbing influence.
`"
`__we write the liquid — vapor equilibrium in the form ofa chemical equation, we can
`tide the energy change as follows.
`
`
`-'
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`-
`
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`
` __
`_' hen heat is added to a liquid— vapor system that is at equilibrium, Le Chatelier’s
`nu 'ple tells us that the system will try to adjust in a way that counteracts the distui-
`
`ce. The system will attempt to change in a way that absorbs some of the heat that is
`_. This can happen if some liquid evaporates, because vaporization is an endother-
`
`
`_
`laange. When liquid evaporates, the amount of vapor increases and the pressure
`
`
`Thus, we have reached the correct conclusion in a very simple way.
`We often use the term position of equilibrium to refer to the relative amounts of the
`
`
`__ stances on both sides of the double arrows in an equlibtium equation such as Equation
`
`
`.
`l_il.1. Then, we think of how a disturbance affects the position of equilibrium. For
`ple, increasing the temperature increases the amount of vapor and decreases the
`
`
`aunt of liquid, and we say that the position of equilibrium has shifted— in this case, it
`-
`ifted in the direction of the vapor, or it has shifted to the right. In using Le
`
`
`Ehfiteliefs principle, it is convenient to think of a disturbance as “shifting the position of
`btium" in one direction or another in the equilibrium equation.
`‘
`1 Now, let's use Le Chateiier’s principle to analyze what happens when the tempera-
`of a liquid —vapor equilibrium is lowered. To lower the temperature, heat must be
`_ovetl. The system responds by undergoing a change in a direction that tends to
`"pi re the lost heat—the position of equilibrium shifts to the left because as some
`=:
`condenses, some heat is evolved. At the new position of equilibrium there is more
`quidand less vapor. Since there is less vapor, the pressure is lower. Once again, we come
`the proper conclusion — that the vapor pressure is lowered by lowering the tempera-
`
`
`
`PRACTICE EXERCISE 2 Use Le Chételier’s principle to predict how a temperature increase will affect the vapor
`pressure of a solid. (Hint: solid + heat : vapor.)
`
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`324
`
`13 / Solutions and Colloids
`
`'
`
`-_ And in 1000 mL solution we have -the following mass of "H
`
`'1.54 _g solution)" - ' I
`
`I
`_
`I
`10.00-mEs.eIution'>< ,_><__
`Sothe ratio of grams of solute to grams of solut_ion,-taken as a" percent," is-'
`}
`_
`gzsgwcio‘)
`.
`,.
`..
`._
`_
`1.54 x 10? 9 (solution) X 19° = 5°‘_W°'_l”‘=’/‘fvi
`H-Thus, é.2o M Hero, is so.'o*i/s (w/w) i_-icIo,.
`__
`
`PRACTICE EXERCISE 14
`I-Iydriodic acid can be purchased as 5.51 M H1. The density of this solution is 1.50
`g/ml... What is the percent concentration (w/w) of Hi in this solution?
`
`
`_- -.
`
`LOWERING OF THE VAPOR PRESSURE
`
`The vapor pressure of any component in a mixture is lowered by the presence of
`the other components-
`The physical properties of solutions to be studied in this and succeeding sections are
`called colligative properties, because they depend mostly on the relative populations of
`articles in mixtures, not on their chemical identities. In this section we will look at the
`P
`.
`u
`c
`’
`|
`effect of a solute on the vapor pressure of a solvent in a liquid solution. The effect
`depends on whether the solute is volatile or nonvolatile, so we will study. these sepa-
`rarely.
`How a Nonvolatile Solute Affects the Vapor Pressure of a Solution
`All liquid solutions of nonvolatile solutes have lower vapor pressures than their pure
`solvents. The extent of the lowering of the vapor pressure of the solvent depends on
`whether the solute is molecular (and does not ionize) or is ionic. We will consider
`nonvolatile molecular solutes, like sugar, first, because there is a particularly simple law
`that correlates the solution’s vapor pressure with the mole fraction of the solvent. This is
`the vapor pressure—conceritration law, or Raoult’s law, and the best statement of it is
`just its equation.
`
`'
`
`Vapor Pressure — Concentration Law (Raoiilt’s Law}
`
`Psolutioli = solvent X
`
`P,,,.,,.;,,,, is the vapor pressure of the solution, P;,,,,,,,, is the vapor pressure of the pure
`solvent, and X,,,,,,,,, is the mole fraction of the solvent in this solution. Thus, the vapor.
`-
`-
`-
`-
`-
`pressure of the solution is a fraction — the mole fraction, X,,,1,,,,,‘—-of the pure solvents
`vapor pressure (at the same temperature).
`plot of P,,,u,i,,,, versus X,,,1,,,,,, _for a solution
`that obeyed Raoult’s law at all concentrations would be linear, as shown in the margin.
`The reason for Raoulfs law is easy to understand. The equilibrium vapor pressure
`of a liquid reflects how far the following equilibrium favors the vapor.
`solvent molecules in solution t————-‘“ solvent molecules in vapor state
`.
`_
`_
`_
`When the rate of evaporation—the forward change—ts relatively large, the vapor
`pressure is relatively high. When we add nonvolatile solute molecules to the solvent, they
`
`After the Greek kol.-‘igativ,
`depending on number and not
`on nature.
`
`Volatile means "can evapo-
`rate," has a low boiling point.
`Nonvolatile means "cannot
`evaporate." has a very high
`boiling point.
`
`we'll study solutions of ions in
`Section 1 3.1 1 .
`
`Francois Marie Raoult (1830 -
`1901) was a French scientist.
`
`Vapor pressure.
`IJUTE Sflllilfifll
`
`I,
`
`
`
`Psoln
`
`0
`
`Psoin =Xso1ventPsoIveni
`
`Xsolvent
`Raoulfs law plot
`
`1
`
`
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`000005
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`000005
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`13.7
`
`LOWERING OF THE VAPOR PRESSURE
`
`525
`
`FIGURE 13.23 Lowering of vapor
`pressure by a nonvolatile solute.
`Because particles of the solute are
`present. the opportunities for
`escape of the solvent particles are
`reduced. Therefore. the vapor
`pressure of this solution is less than
`the vapor pressure of the pure
`solvent {at the same temperature).
`
`take up spaces at the surface of the solution and leave the solvent’s molecules with a
`fraction of their former surface population. Thus, the solute molecules, which cannot
`evaporate, interfere with the escape of solvent molecules into the vapor state. They
`reduce the rate of the forward change. But they do not interfere with the return of
`solvent molecules to the solution. See Figure 13.23. The equilibrium thus shifts to the left
`and the vapor pressure decreases until the forward and reverse changes are again equal.
`The interference caused by the solute molecules and therefore the extent to which
`the vapor pressure is lowered depends on the fraction of molecules at the surface which
`are the solute. If 25% of these molecules are solute, then the vapor pressure is lowered by
`25%. Another way of looking at this, however, is that 75% of the molecules at the surface
`are those of the solvent, and the vapor pressure, having been lowered by 25%, is now 75%
`of its value for the pure soivent. This is exactly what Raoult’s law says, because if 75% of
`the molecules are those of the solvent, then the mole fraction of the solvent must be 0.75,
`and 0.75 multiplied by the vapor pressure of pure solvent gives a value that is 75% of
`olvent '
`
`Os
`
`
`
`
` -
`
`I
`
`;_.E_-X'AMP___LI_5 13,14. _RAOULT‘S LAW CALCULATION FOR NONVOLATILE sotures I
`-
`- Problem: _.Carbon tetrachloride has a vapor pressure of 100 torr-at'23.°C. _This_solvent'.can dissolve
`'
`candle wax, which is essentially nonvolatile. Although candie wax isa rnixture; we can take
`its moiecular formula to be CnH,,5 (formula weight, 310)..What will be the vapor pressure "at
`23 °C of a solution prepared by dissolving 10.0 g of wax in 40.0 g of CCI,?-The formula
`-weight of CCL, is 154.
`'
`'
`V -
`-
`‘Solution.’ We haveto find the mole fraction of the solvent before.w_e'can-"use_ R_aou|t's lawto answer
`'
`'
`'
`this question. So we have to calculate the rfrolgg of solute and _-solvent-firs-t.
`-
`-
`--
`-
`
`.
`
`For ccr,
`
`_
`
`_
`
`_
`
`_
`
`_'
`
`'
`
`-
`
`.[-
`
`.
`
`.. 4-
`
`5
`
`40.0 g ccI.._>< 15L2‘;—E% = 0.250 ‘moi ccl,
`‘
`'
`"
`'
`t
`i 7Lf¢’r.2H.s
`'= o.o'_323 rnol cu:-1...
`10.0 9 C22H4e >.< ;1"3°;
`_
`total number of moles is the sum: 0.292 mol. Now, we can find the mole-fraction of
`
`Xco-=3Z§§3§§i.=°-3905
`
`
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`000006
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`Dibutyl phthalate, C151-[1104 (formula weight 278), is an oil sometimes worked into "
`plastic articles to give them greater softness. It has a negligible vapor pressure at room
`temperature. (It has to be heated to 148 “C before its vapor pressure is even 1 torr.) What
`is the vapor pressure, at 20 °C, of a solution of 20.0 g of dibutyl phthalate in 50.0 g of
`oc.t__ane, Cs}-I1, (formula weight, 1 14)? The vapor pressure of pure octane at 20 °C is 10.5
`tort.’ '-
`
`The Effect of a Volatile Solute
`on the Vapor Pressure of a Solution
`
`When two (or more) components of a liquid solution can evaporate, the vapor contains
`molecules of each. Each volatile component contributes its own partial pressure to the
`total pressure and this partial pressure is directly proportional to the component’s mole
`fractions in the solution. This is reasonable because the rate of evaporation of each
`compound has to be lowered by molecules of the others in the same way we described
`just above for nonvolatile solutes. So, by Da1ton’s law, the total vapor pressure will be the
`sum of the partial pressures. To calculate these partial pressures, we use Raoult’s law for
`each component. For component A, present in a mole fraction X3, its partial pressure,
`PA, is found as a fraction of its vapor pressure when pure, P3,; thus,
`
`-PA = XAP;
`
`And the partial pressure of component B would be
`
`Pa = X3133
`
`Similar calculations would be done on other components, but we will stick to two-com-
`ponent systems.
`The total pressure of the solution of liquids A and B is then
`
`P,,,,,]=P,,+P5
`
`APPLYING RAOUI.T'S LAW T0 sotunoris or VOLHATILEHSOLUTES '
`
`F211
`
`1
`
`"l
`
`,’ fstilution:-':1ml('jul!¢sitl's
`
`The mole fraction of CCl,,, the solvent, is 0.890. The-vapor pressure-of the-solution will be '
`this fraction of the vapor pressure of pure CC!‘ (1 00 torr) as calculated by 'Raou'It's law.
`P,..,,,,,,. = 0.890" ><_'1oo torr-
`= 89.0 torr_'_
`'
`The presence of the wax has lowered the vapor pressure ‘of CCl. from 1OD'to 89.0 torr.
`_j
`
`'
`
`'
`
`
`
`
`
`'
`
`Acetone is a solvent for both water and molecularliquids that do not dissolve in "water, lik
`benzene. At 20 ‘C acetone has a vapor pressure ‘of 16'2'torr. The vapor pressure ,_of watera
`20 ‘C is 1 7.5 torr. what is the vapor pressure of a solution of acetone and water with 50'
`
`-
`mole percent of each? (\Ne assume that Raou|t's law applies.)
`-
`-
`-
`'
`To find Plm, we need to calculate the individual partial "pressures and then" add__ them."
`Pm,” = 162 torr ><--0.500 =' 81:.0 tort" 1
`= 17.5 torr X 0.500.-= 8.7-'3-tort
`PW!‘M
`-
`_
`Pm. = B9.8.t_orr _
`Thus the vapor pressure of the solution is much higherthan that of pure water but
`less than that of pure acetone.
`'
`-
`
`_
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`000007
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`l":'{.-'5..C'l'lf;'E-. F:?IEF?.{'.l.'-ii?
`
`"=93
`
`lle-meml".-er, PA and P3 here are
`HIV: partial giressums as calcu-
`late:-:i by Flac:u'2t’:; law.
`
`EXAMPLE 13,15"
`Problem:
`
`I Solution:
`
`000007
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`
`
`13.7
`
`LOWERING OF THE VAPOR PRESSURE
`
`527
`
`PRACTICE EXERCISE 16 At 20 °C, the vapor pressure of cyciohexarie, a hydrocarbon solvent, is 66.9 torr and that
`of toluene (another solvent) is 21.1 torr. What is the vapor pressure of a solution of these
`two at 20 °C when each is present at a mole fraction of 0.500?
`
`. Figure 13.24 shows how the total vapor pressure of a two-component solution of
`olatile liquids is related to composition and partial pressures according to Raoult’s law.
`Each of the steeper lower lines shows how the partial vapor pressure of one component
`_ nges as its own mole fraction changes. Any point on the top line represents the total
`ressure at a particular pair of values of mole fractions. (Try this out yourself with a
`millimeter ruler. The distance from a point on the baseline to one of the steeper lines plus
`he distance from the same base point to the other steeper line gives the distance from this
`base point to the top line.)
`
`FIGURE 13.24 The vapor pressure
`of an ideal, two-component
`solution of voiatile compounds.
`
`Vapor
`pressure
`of pure B
`
`Vapor
`pressure
`of pure A
`
`Vapor
`
`pressure—D~
`
`A 1
`B D
`
`.
`0.5
`Mole fractions of A and B
`
`,_.
`
`,
`
`_
`
`Ideal Solutions and Deviations from Raou|t's Law
`ilis we described in Section 13.3, an ideal solution is one for which AH,,,.,,,-M is zero, and
`therefore one in which forces of attraction between both like molecules and unlike
`. molecules are identical. Only an ideal solution would obey Raoult’s law exactly. Only an
`ideal solution would give plots like those in Figure 13.24, which we just discussed.
`' Not very many real, two-component solutions come close to being an ideal solu-
`_ on. Figure 13.25 shows two of the typical ways in which real two-component solutions
`
`Positive deviation
`
`Negative deviation”
`
`
`
`
`
`VaDO!|.'.|iB5SIJl'E-—-F
`
`C 1
`‘B O
`
`I 3°»‘uti‘o;,
`
`
`
`Mole fractions of
`C and D
`
`1 F 0
`
`Mole fractions of
`E and F
`
`1
`
`000008
`
`FIGURE 13.25 Typical deviations
`from ideal behavior of the total
`"'ap°' prefsures °f real’ tw°'c°mp°'
`nent solutions of volatile substances.
`
`000008
`
`

`
`528
`
`ll:
`
`_-’
`
`.‘it:luti<n:s -_m.l i_Ti-liiiitls
`
`Acetor-.e-— water solutions show
`negative deviations.
`
`Ethyl nlcol1ol— hexaim Si'..iIl.J'i.lUfi$
`show positive deviations.
`{rliexeiie is El nonpolar compo-
`men! of ga::_~-')|ine.}
`
`
`
`behave. The individual partial pressures are not linear with composition as Raoult’s law
`demands. So the sums of these partial pressures are seldom linear either. The sums given
`by the upper, total-pressure plots show either upward or positive deviations or they give
`downward or negative deviations from Raoult’s law behavior. What causes this?
`Let us identify the molecules of a two-component solution as A and B. Solutions
`with negative deviations generally form exothermically. In them, the attractive forces
`between A and B are stronger than the sum of the attractive forces between A and A and
`between B and B, as we analyzed on page 503. The extra attractive force between A and
`B, therefore, is an extra interference with the escape of either A orB into the vapor phase.
`50 the partial pressure plots (the lower, steeper plots) for A and B sag, which makes the
`plot for total pressure sag, too. Thus, solutions that form exotherrnically tend to show
`negative deviations.
`Solutions with positive deviations from Raoult’s law generally form endothermi-
`cally. In them the attractive forces between A and B are less than the sum of the attractive
`forces fromA to Aand fromBto B. In the solution ofA and B, it is as ifA and Bmolecules
`are not held back as much by their own kind. They are released to evaporate at faster
`rates. This. causes their individual partial pressure plots to curve upward. So the plot of
`total pressure must curve upward, too. Thus, solutions that form endothermically tend
`to show positive deviations. Figure 13.26 summarizes the relationships between Raoult's
`law deviations and the relative attractive forces—A to A, B to B, and A to B—in the
`solute, the solvent, and the solution.
`
`HGURIE 13.26 Deviations from Raou|t's law and attractions between solute and solvent in a
`
`.neiative_Auractivefg_rces_
`[A to A. B to B] > A to 3
`
`.
`
`_
`
`_
`
`I Bae0ll5’s'.lai'r.-iiIdls':=',.
`positive deviations
`
`I
`
`liquid-liquid two-component solution. Letting A represent one component and 13 another compo
`nent, we can represent the attractive forces between molecules within pure A as "A to A", and
`within pure B as "B to B". In the solution, these forces can still operate. but we have to add the
`intermolecular attractive forces of A to B. The net effect of these forces on AH,“ and Raou|t's law
`plots (vapor pressure versus composition) are as follows.
`
`
`- 5 AH'-''
`positive
`(endothermic)
`zero
`
`[A to A. B to B] = A to B
`(ideal solution)
`negative deviations
`negative
`[A to A. B to B] < A to B
`(exothermic)
`
`obeyed perfectly
`
`ELEVATION OF THE BOILING POINT
`
`‘Elie boiling point of :1 solution of a nonvolatile solute is higher than the boiling point
`of the solvent.
`
`When the only volatile component of a solution is the solvent, then the vapor pressure of
`the solution is less than the vapor pressure that the solvent has at any given temperature. _'
`See Figure 13.27. At the temperature at which the pure solvent normally boils, the vapor
`pressure of the solution is still not equal to the atmospheric pressure. 50 to make the
`vapor pressure of the solution come up to atmospheric pressure, we have to increase the
`temperature of the solution further. The presence of a nonvolatile solute thus eleva
`the boiling point of the solution.
`The most common application of this property of solutions occurs in the use
`perrnanent-type antifreezes. These protect the liquid in a vehicle's cooling system no‘
`just from freezing but also from boiling over. These products are based on eithe
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`EQUILIBRIUM LAWS FOR GASEOUS REACTIONS
`
`557
`
`14.4
`
`I here P stands for the partial pressure of the gas. Solving for P gives
`n
`P = — RT
`V
`
`__The quantity :1_/V has units of mol/L, and is simply the molar concentration, M, so we
`-"can write
`
`(14.3)
`P = (molar concentration) X RT
`,
`_
`_
`,
`_
`_-Thus the partial pressure of a gaseous reactant or product is directly proportional to its
`molar concentration at a given temperature.
`The relationship expressed in Equation 14.3 is what lets us write the mass action
`;-expression for reactions between gases either in terms of molarities or in terms of partial
`"pressures. Of course, when we make such a switch we can’t expect the numerical values
`of the equilibrium constants to be the same, so we use two difierent symbols for K.
`‘When molar concentrations are used, we use the symbol K‘. When partial pressures are
`_used, then K, is the symbol. For example, the equilibrium law for the reaction of nitrogen
`and hydrogen to form ammonia
`
`Nzlgl + 3142(8) *1‘ ZNHJUE)
`
`"can be written in either of the following ways.
`
`Doubling the molar concentra-
`tion in a gas without changing
`its volume or temperature
`doubres the p,e55u,e_
`
`lNH3(El]2 = K
`[Nz(3)liH2(g)l’
`
`‘
`
`because concentrations are
`used in the mass action
`expression
`
`
`
`
`
`
`
`
`because partial pressures
`are used in the mass action
`expression
`
`slit equilibrium, the molar concentrations can be used to calculate Kc, while the partial
`' pressures of the gases at equilibrium can be used to calculate K,. We will discuss how to
`convert from K, to K, and vice versa in the next section.
`
`
`
`PRACTICE EXERCISE 2 Using partial pressures, write the equilibrium law for the reaction
`Hzlg) 4' 12(3) :‘ 2H1(gl-
`
`
`__is:-the."expression for'!(,, for the reaction"-
`'-'
`:-' -'
`__ __
`-
`_
`.
`N20-4(9)’ «———-"-* '2No;(§) "j
`,
`‘we use_'par:'tial pressures in the mass action expre'ss_ion.. Therefore,
`
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`"Z _
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