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# Class Note for CHM 218 with Professor Berger at IPFW

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Date Created: 02/06/15

61 Chapter 6 Inorganic Thermodynamics Thermodynamics is the branch of chemistry and physics that deals with the relationship between heat and other types of energy during the course of chemical and physical processes Some definitions to keep in mind System Surroundings Universe State Phase Internal Energy U Heat q Work w State Function that collection of matter upon which we choose to focus our attention anything in thermal andor mechanical contact with the system This implies that energy in the form of heat or work may be transferred between the system and its surroundings the system the surroundings a description of a system in terms of properties which do not change with time Properties may include temperature volume pressure etc a description of the physical form of a substance as a solid liquid or gas The sum of the kinetic and potential energies of the particles making up a system a form of energy which ows spontaneously from a warmer object to a cooler one the energy exchange resulting when a force F moves an object through a distance d wFd A property whose value depends only on the current state of the system and not how the system got to that state U P V T are all state lnctions q and w are not path independent and are therefore not state lnctions They may be referred to as path functions The First Law of Thermodynamics The First Law of Thermodynamics may be stated in a number of ways Basically it is a statement that energy may be neither created nor destroyed but rather is conserved The total energy of the universe is conserved 62 Uuniv Usys USUIT a constant AUuniV 0 AUys AUsun AU iv 0 AUys AUsm Thermodynamic Sign Convention Surroundings qgt0 gt wgt0 System qltolt gt Wlto Heat q is positive when it ows from the surroundings into the system Work is positive when the surroundings do work on the system The change in intemal energy of the system AU is given by the sum of the heat absorbed by the system q and the work done on the system w AUqw This last equation is also a statement of the First Law Enthalpy and Enthalpy Changes The heat absorbed or evolved by a reaction depends on the conditions under which the reaction occurs If a reaction occurs under conditions of constant pressure the heat of the reaction is designated qp where the subscript p indicates constant pressure The heat of reaction at constant pressure qp is related to a property of the reactants and products called enthalpy H Like U P V and T H is a state lnction 63 Enthalpy H ls another thermodynamre pmme of a system H U PV vvhereUrs the rntemal energy ofthe system Prsrts pressure andVrs rts volume As was rndeated earlrer the enthalpy ehange for a Eamon occumng at eonstant pressure ls gven by the heat rt absorbs or evolves That ls AH tr Fortunately we do not have to earry out a reaeuon and measure the heat absorbed or evolved every me we vvantto know AH for a Eamon Beeause Hess39 Lavv ls sueh apowel39ful tool rt ls only neeessary to tabulate AH values for a lrmrtednumber ofreacuons These values along vvrth Hess39 Lavv wlll allow us to eompute AH values for a vvrde vanety ofreaeuons The reamons for vvhreh vve tabulate data are formauon reaeuons The standard enthalpy change for any arbrtrary Eamon ls then gven by A3 I n A pudum I m AE muunu Spontaneous Proeesses A spontaneous proeess ls one that gven sutHnent ume wlll oeeur by ltself vnthout any eonunurng external rntervenuon All proeesses vvhether physreal or ehemreal have aprmerred drreeuon For example aboulder rolls downhlll rather than up hlll and a preee oflron lat outsrde m morst arr wlll spontaneously reaet vvrth q atmosphere produee F2203 rust The rust does not spontaneously deeompose to rron and q 2 Fe 5 32 q 9 F220 5 spontaneous F220 5 e 2 Fe 5 32 q 9 nonrmontaneous Both ofthese proeesses eonserve energy so the Frrst Lavv does not determrne the spontanerty of a proeess 64 In order to determine whether a given process is spontaneous we must invoke the Second Law of Thermodynamics which is expressed in terms of the thermodynamic quantity entropy Entropy Entropy S is a measure of the randomness or disorder of a system or its surroundings Like internal energy U and enthalpy H entropy is a state function Entropy and the Third Law of Thermodynamics The Third Law of Thermodynamics states that the entropy of a perfect crystalline substance at 0 K is 0 In general as the temperature of a substance is increased its entropy increases The accompanying gure shows how the entropy of a substance varies with temperature Entropy increases gradually with temperature but jumps abruptly when a phase change occurs The standard or absolute entropy of a substance 8quot is the entropy of the species in its standard state The Standard entropy 8 JK 60 Gas 50 Entropy of vaporization 40 W 6 30 Entropy of fusion 20 10 0 50 100 150 200 Temperature K 250 300 350 standard state of a substance is the pure substance at a pressure of 1 bar approximately 1 atm For species in solution the concentration is 1 M The table on the next page gives the standard entropies of a variety of substances Note that unlike standard enthalpies of formation the standard entropy of an element in its reference state is nonzero Standard Entropies at 25 C 8 8 8 Formula Jmol oK Formula Jmol oK Formula Jmol oK Hydrogen Carbon continued Sulfur Haq 0 0520 1510 829 2281 H2g 1306 HCNg 2017 Srhombic 319 Sodium HCNI 1128 Smonoclinic 326 Naaq 602 0049 3097 8029 2481 Nas 514 00141 2144 81259 2056 NaCls 721 CHscHOg 266 Fluorine NaHCOss 102 cszoHa 161 F39aq 96 Na2003s 139 Silicon F2g 2027 Calcium 515 180 HFg 1737 Ca2aq 552 Si02s 415 We Cas 416 SiF4g 285 craq 551 05105 382 Lead 0129 2230 CaCOss 929 Pbs 648 HClg 1868 Carbon PbOs 665 Bromine Cgraphite 57 PbSs 913 Braq 807 Cdiamond 24 Nitrogen BFZU 1522 009 1975 N2g 1915 Iodine 0029 2137 NH3g 193 l39aq 1094 HCOs39aq 950 NOg 2106 125 1161 01149 1861 N02g 2399 Silver 04149 2192 HN03aq 146 Agaq 739 04169 2295 Oxygen Ags 427 061160 1728 029 2050 AgFs 84 HCHOg 219 039 2388 AgCls 961 CH30HI 127 OH39aq 105 AgBrs 1071 0829 2378 11209 1887 Agls 114 H200 699 Some General Observations l S solid lt S liquid lt S gas Ballpark molar entropy 8quot values Solid z 0 90 J K391 mol391 Liquid z 50 150 J K391 mol391 Gas z 130 a J K391 mol391 NOW H and AH Values 2 usually given in M or 1d mol l while s and As values are usually givenin J K3910r J K391 mol39l 2 S simple molecule lt S complex molecule S H2g s 1306 J K391 mol391 S CC4g s 3097 J K391 mol391 3 S increases as the temperature of a substance increases S Cu s 200K 237 J K391 mol391 S Cu s 300K 333 J K391 mol391 Tables of S0 values can be used to nd ASm AS Z S pr0ducts Z S reactants just as AH data is used to calculate AH r m Example What is the entropy change associated with the following reaction at a pressure of 1 atm and a temperature of 25 C 2 H2 g t 02 g a 2 H20 1 66 67 The Second Law of Thermodynamics The Second Law of Thermodynamics states that in any spontaneous process the entropy of the universe that is the entropy of the system plus the entropy of its surroundings increases Stated mathematically AS Assys Assun gt 0 Note that the Second Law states that the criterion for a spontaneous process is an increase in the entropy of the universe not of the system or surroundings There are many examples of spontaneous processes that involve a decrease in the entropy of the system but these are invariably accompanied by a larger increase in the entropy of the surroundings Similarly many spontaneous processes involve a decrease in the entropy of the surroundings but these must be accompanied by an even larger increase in the entropy of the system Any process leading to a decrease in the entropy of the universe is forbidden but its exact reverse will lead to an increase in the entropy of the universe and is therefore spontaneous What happens if for a particular process ASu 0 If ASu 0 there is no driving force for either the forward process or its reverse and the process is at equilibrium A process involves a change in the system AND its surroundings and can go in one direction only The sign of ASu detennines whether the process will go For aprocess A a B if ASu gt 0 the process will go if ASu lt 0 the process will not go but B a Awill go if ASu 0 the process is at equilibrium The sign of ASu tells us whether a process will go It says absolutely nothing about how fast it will go For example C diamond a C graphite is a spontaneous process at 25 C and 1 atm but occurs so slowly that we could never even begin to see a change Thermodynamics tells us whether a process will go and if so how far Kinetics tells us how fast 68 We already know how to calculate the entropy change for the system during the course of a chemical reaction but how do we determine the change in the entropy of the surroundings so we can calculate the entropy change of the universe The entropy change of the surroundings is related to the amount of heat that ows between the system and its surroundings In general AS 39 1 39139 where q is the amount of heat absorbed by the system from its surroundings and T is the absolute temperature For an exothermic process q lt 0 so the entropy of the surroundings increases while for an endothermic process q gt 0 so the entropy of the surroundings decreases Now we can restate the Second Law in terms of ASsys q and T We ll drop the sys subscript and assume that a quantity without an explicit subscript refers to the system A AS qgto 5 1 Keep in mind however that for a constant pressure process qp AH so An AS AS o If we multiply through by i T 39139AS 39139AS An lt 0 or AK 39TAS lt 0 Now we can see that the spontaneity criterion can be expressed exclusively in terms of the absolute temperature and changes in properties of the system AH and AS This suggests the invention of still another thermodynamic quantity 69 GIBBS FREE ENERGY G The Gibbs Free Energy G is de ned by the equation by the G H 7 TS The change in G is given by AG AH ATS However for constant temperature processes ATS TAS so at constant temperature AG AH 7 TAS Therefore the sign of AG serves as the criterion for the spontaneity of a reaction If ASu gt 0 AG lt 0 process is spontaneous ASu lt 0 AG gt 0 exact reverse process is spontaneous ASu 0 AG 0 process is at equilibrium Standard Free Energy Changes As we have seen before we de ne a standard state for the purpose of tabulating thermodynamic data For liquids and solids the standard state is the pure substance at a pressure of 1 bar 1 atm for gases the gas at a partial pressure of 1 bar and for solutes l M concentration under a pressure of 1 bar We can calculate standard free energy changes AGO for reactions in either of two ways If we know AHquot AS0 and the temperature we can use the relationship AGO AHO TASquot Example What is the standard free energy change for the reaction of 2 moles of H2 g with 1 mole of 02 g to form 2 moles of H20 1 at 25 C We have already calculated ASOfor this reaction ASO 3264 J 39K391 610 Standard Free Energy of Formation The standard free energy of formation AGO of a substance is de ned in a manner similar to the standard enthalpy of formation The standard free energy of formation is the free energy change associated with the reaction in which one mole of a substance in its standard state is formed from its elements in their reference states at a pressure of 1 bar 1 atrn and any speci ed temperature Standard free energy changes for reactions can be calculated from tabulated values of the standard free energies of formation of all the reactants and products in exactly the same manner as standard enthalpies of reactions are from standard enthalpies of formation As with standard enthalpies of formation the standard free energy of formation of an element in its reference state 1s zero AGO Z n AG products 2 m AG reactants Standard Free Energies of Formation at 25 C AG AG rs Formula kJmol Formula kJmol He dmgequot Carbon continued E2 8 CzH4g 684 20dium CzHeg 329 C H 1245 Naaq 261 9 Hagen 1 10 Nas O CH OH NaCKs 3840 0339 0 122 NaHCOgS 851 9 032 39 NaZCOSs 10481 chljeg 1236 Calcium Ca2aq 5530 33 113 7 C 4 0223 5035 COW 686 gacgas 11288 305329 ar on 2 5 Cgraphite 0 Silicon Cdiamond 29 31 852 5 c309 1372 I 2 S 0029 3944 SiF4g 1506 HCOsaQ 5871 CH4g 508 AG AG p Formula kJmol Formula kJmol Lead Fluorine Pbs 0 Nat 2765 PbOs 1892 59 0 PbSs 967 HFg 275 Nitrogen Chlorine N2g 0 C39aq 1312 NH3g 16 C2g 0 NOg 8660 HCg 953 N02g 51 Bromine HN03aq 1105 Br39aq 1028 Oxygen BrzU 0 029 0 Iodine 03g 163 I aq 517 OH39aq 1573 2s 0 H209 2285 Silver H200 2372 Agaq 771 Sulfur Ags 0 32g 801 AgFs 185 Srhombic 0 AgCIs 1097 Smonoclinic 010 AgBrs 959 3029 3002 Ags 663 H2 Sg 33 611 612 Free Energy Changes During A Reaction A chemical reaction proceeds until a state of equilibrium is reached The driving force for this or any other reaction is a decrease in free energy The equilibrium composition of the system is that with the minimum free energy Ifyou look at the gure to the right you will see that the reaction of gasoline with 02 to produce H20 and C02 is highly exergonic that is AG0 has a large negative value As the reaction proceeds the free energy of the Reactants Products system decreases until a state of QaSOIine and 02 CO2 and H20 equilibrium is reached Any further reaction would require the free energy of the system to increase In general the more negative the value of AG the greater the value of the equilibrium constant and the farther the reaction will proceed before equilibrium is achieved Spontaneous reaction AG Free energy gt Equilibrium Let s now consider a reaction with a large positive value of AG According to the gure we should expect this reaction to proceed only to a small degree before equilibrium is reached As you might guess N nsp mane US there is a relationship between the value of gt readlon Ago AG0 and the equilibrium constant which we E will investigate next g L Equilibrium Reactants Products Relating AG0 to the Equilibrium Constant The thermodynamic equilibrium constant K is the equilibrium constant expression in which concentrations of gases are expressed in terms of their partial pressures in atmospheres and the concentrations of solutes in liquid solutions are expressed in terms of molarities In reality the thermodynamic equilibrium constant is expressed in terms of the activities of the substances taking part in the reaction Partial pressures and concentrations are merely approximations to the activities of substances 613 Under standard conditions unit concentration or partial pressure AGO for a reaction can be calculated from data found in thermodynamic tables as was previously illustrated When reactants andor products are not at unit concentrationpressure the following equation allows us to calculate AG AG AGO RT 1n Q where Q is the thermodynamic reaction quotient R is the gas constant 8314 J 39 K391 39 mol39l and T is the absolute temperature When a reaction has reached equilibrium there is no driving force for the reaction in either direction so AG 0 and the concentrations and or pressures in Q are the equilibrium concentrations andor pressures so Q K We can substitute these values into the previous equation 0 AGO RT In K and rearrange to obtain the relationship between AG0 and K that we alluded to in the previous section AGO RTan Note that the units of R are J 39 K391 39 mol391 while AG0 is normally expressed in 1d Before using this equation is used we must express these both either in terms of J or 1d Changes in AG with Temperature The thermodynamic tables in your book as well as most thermodynamic tables you might expect to see apply to a temperature of 25 C Nothing in the de nition of standard state refers to this speci c temperature however It is just that most of the data have been tabulated at this temperature While the temperature dependence of G and AG are explicit in their definitions G H TS and AG AH TAS the temperature dependence of AH and AS are not so obvious There are methods for determining values of AH0 and AS0 from data tabulated at 25 C but we will not concem ourselves with them Instead we will note that as long as the temperature changes are not too large AH0 and AS0 may be assumed to be constant with respect to temperature so that AG T z AH398 TASg98 614 As we have previously noted a negative value of AG indicates a spontaneous reaction and AGT z AH298 T AS298 so the signs of AH and AS will provide insight as to whether a reaction will become more or less favorable with increasing temperature Since both AH and AS can be either positive or negative there are four possible combinations of the signs of AH and AS Sign of AH Sign of AS Sign ofAG negative positive negative at all temperatures negative at low temperatures ne ative ne ative g g pos1t1ve at high temperatures positive at low temperatures pos1t1ve pos1t1ve negative at high temperatures positive negative positive at all temperatures You can verify that these conclusions regarding the sign of AG are correct For example when AH is negative and AS is positive we are subtracting a positive value TA S from a negative value AH to obtain AG The diiTerence is always negative irrespective of the temperature However when both AH and AS are negative the results will diiTer depending on whether the temperature is relatively low or high At suf ciently low temperature lAHl gt lTAS l so we are subtracting a less negative value from a more negative value resulting in a diiTerence which is negative However at suf ciently high temperature lAHl lt lTAS l so we are subtracting a more negative value from a less negative value resulting in a diiTerence which is positive We should note that the terms low temperature and high temperature are relative descriptions Relating AG0 to the Equilibrium Constant The thermodynamic equilibrium constant K is the equilibrium constant expression in which concentrations of gases are expressed in terms of their partial pressures in atmospheres and the concentrations of solutes in liquid solutions are expressed in terms of molarities In reality the thermodynamic equilibrium constant is expressed in terms of the activities of the substances taking part in the reaction Partial pressures and concentrations are merely approximations to the activities of substances 615 Under standard conditions unit concentration or partial pressure AG for a reaction can be calculated from data found in thermodynamic tables as was previously illustrated When reactants andor products are not at unit concentrationpressure the following equation allows us to calculate AG AG AG RT 1n Q where Q is the thermodynamic reaction quotient R is the gas constant 8314 J 39 K391 39 mol39l and T is the absolute temperature When a reaction has reached equilibrium there is no driving force for the reaction in either direction so AG 0 and the concentrations and or pressures in Q are the equilibrium concentrations andor pressures so Q K We can substitute these values into the previous equation 0 AG RT1nK and rearrange to obtain the relationship between AG and K that we alluded to in the previous section AG 7 RT In K Note that the units of R are J 39 K391 39 mol391 while AG is normally expressed in 1d Before using this equation is used we must express these both either in terms of J or 1d Example Silver oxide decomposes to silver and dioxygen according to the following equation 2 Agzo S v 4 Ag S 02 g At 25 C AH 621kJ and AS 1327 J 39K39l Calculate AG and K for the reaction at 25 C and at 225 C 616 Notice that AGO went from a positive value at the lower temperature to a negative value at the higher temperature in accord with out prediction from the table above We can in fact predict the temperature at which AGO goes from positive to negative Of course the temperature below which AG0 is positive and above which AG0 is negative is the temperature at which AGOequals zero We can solve for this temperature by setting AGO equal to zero in the equation AGO AHO TASquot 0 We should note that this calculation is only approximate because of the assumptions we made regarding the constancy of the values of AH0 and AS In general these values will change with temperature Energetics of Ionic Bonding According to Coulomb s Law the electrostatic force between two charged pa1ticles is given by FL 41 where 60 8854 X 10 3912 C2 39 J391 39 m391 permittivity offree space q and q are the charges in C d is the distance between the particles in m The corresponding electrostatic energy is amp 4121 617 Thxs relanonshp does not apply smeuy 16 a crystallan 10m 1amee beeause we have 16 take mto account the meeracnons between many was simultaneously For Examplem me Nac11amee there are 6 cr was surroundmg a Na m at admance onquot 12 Na xons at adutance 6 2 Wequot 8 c139 xons at adutance are 6 Na xons at a dutance 6f 2139 etc 56 wehave anm mte senes 6f altemanng amaeave and repulsive terms Fortunately ths senes comages 6 a smgle value whmh depends my on me geometzy 6f the 1auaee g u m mm mum M 11mm mom Sumng m m mam m marked Kr m m quotemu nmg mfs 1 mm m ncuml nughbmw sew1 m1 my 1117ther 1 m m am nan quotcum mam The value 15 ea11ed me Madelung con ant A Lamee A Nac1 174756 CsCl 176276 Car2 uonte 2 5194 A120 4 1719 A few others are gven m Table 6 4 56we can gvethe attractive energy ofmeeramon 6 am61e 6 a subgance as NAVAx z39 7 Azend 5 Ifthxswerethe 6111y conmbunonJhougmno sable 1amees could Emstbecause me snag W6u1d continually decrease as d e 0 6 l 8 However ions are not point charges As d becomes smaller and smaller we must consider the fact that at very short distances their electron clouds interact repulsively This repulsive term operates only at extremely small distances Eu dl where n is referred to as the Born exponent values range from 5 to 12 and depend on the electron con gurations of the ions Taking into account the attractive and repulsive terms we have 5 1 1 4 I a z n Lattice Enthalpy U We de ne the lattice enthalpy of an ionic compound as the energy necessary to convert 1 mol of the solid into its gaseous ions MmXX s a m M g X X g AH U This is the opposite convention to what the text uses BomLande Equation UNhAz z39 2 1l 41 n e39 con g of ion n He 5 Ne 7 Ar 9 Kr 10 Xe 12 Rolling all the constants into a single constant 1339 2105 l A 1 xquot U 1 to all V 619 BomMayer Equation Both of these equations require knowledge regarding the crystal lattice However AF Kapustinskii discovered that if the Madelung Constant for a variety of crystal structures was divided by the number of ions in the formula unit a relatively constant value is obtained Lattice v A Av NaCl 2 174756 0873780 CaF2 3 25194 083980 A1203 5 41719 083438 He proposed a hypothetical rock salt N aCl lattice energetically equivalent to any ionic solid making it possible to calculate U without knowing the details of the lattice This equation is known as the Kapustinskii Equation 1202 1105 Ir11quotquot v z xquot U ml 1 345 E Il to The extent to which the equations provide agreement with experimental values tells us how appropriate the ionic model is for a particular compound There is o en a signi cant diiTerence between calculated and measured values of U when there is signi cant covalence in a compound Formation of Ionic Compounds Direct experimental measurement of lattice enthalpies is dif cult Some degree of ionpairing always occurs so it is not possible to vaporize the solid completely into isolated ions However whenever we have a process for which an enthalpy change is dif cult to measure it is o en possible to use other data in conjunction with Hess Law to determine the enthalpy change for the process of interest BomHaber Cycle 620 The BomHaber cycle is based on Hess Law We have a process for which it is difficult to directly measure AH However if we can write the process as a sum of other reactions we can calculate AH from the AH values of the other reactions In the formation of Li20s from its elements Lis and 02 g we consider 5 processes 1 Atomization of Lis AH for the atomization of one mole of Lis is 162 k 2 Dissociation of 02 3 Ionization of Li 4 Formation of 0239 2Lis a 2Lig AH324kJ The 02 molecules are separated into individual atoms This requires the breaking of the 00 bond AHOO 494 ldmol 12 02 g a 0 g AH 12 AHOO 247 1d An electron is removed from each Li atom resulting in the formation of a L ion AH for this process is essentially equal to the first ionization energy of Li 520 kJmol 2 Li g a 2 Lil g 2 e39 g AH IE1 1040 k An electron is added to each 0 atom to result in a negatively charged oxide ion AH for this process is the sum of the first and second electron attachment enthalpies of O 0 g t e39 g z 0 g AH AHEA 141kJ 0 g e39 a 02 g AH 744 k 0 g 2 e39 g a 02 g AH 603 k 5 Formation of Li20s from its ions The L and 0239 ions condense to form solid LiZO AH for this process is the negative of the lattice enthalpy for LiZO since this process is the exact reverse of the process whose AH is the lattice enthalpy of LiZO 2LNg 0239g Li20s AHU 621 These ve steps can be added together to give us the equation for the formation of LiZO s from its elements in their reference states The enthalpy change for the latter process is simply the standard enthalpy of formation of LiZO s AH f 2Lis 2Lig AH324kJ 12 02 g a 0 g AH 12 AHOO 247 1d 2 Li g a 2 L g 2 e39 g AH IE1 1040 k 0 g 2 e39 g a 0239 g AH 603 Id 2 L g 0239 g U20 8 AH U 2 Li s 12 02 g Li20 s AHfLi20 s 7596 1d Therefore we can say 324 k 247 1d 1040 k 603 k U 7596 k or U324kJ247kJ1040kJ603kJ 596 kJ2810 1d Since this is the enthalpy change per mole of LiZO s U 2810 ldmol Compa1ing this to calculated values BomeLande 1389 z 105 M 25194 1 2 U 1 11 29ox1o 201 pm 5 BomMayer 1389 x 105 M 25194 1 2 U 1 130m 296103 201 pm 201 pm and Kapustinskii 1202 z 105 gt 3 1 2 U 1345l 2 2Mz1o 201m and Graphically 2 Li9 029 622 AH603 kJ 2Li g 2equot O 9 A AH104O kJ 2 i9 09 A AH247 kJ 2Li 9 12 029 A AH324 kJ EU 5 12 02 g AHf 596 kJ39 V U 2810 kJ Lizo S 623 We can use these cycles to rationalize the existence of certain compounds e g NaCl and the non existence of others eg NaClZ Ifwe assume the radius of a Nazl ion is similar to that of a Mgzl ion U for NaClZ is about 2650 ldmol What adjustments must be made to our cycle We can rationalize the nonexistence of NaClZ to the exceptionally large second IE of Na 624 Ihermodynamres of Soluuon Just as we ran assemble a BomrHaber Cyele for the formauon of an lonre eompound from rts elemems we ran assemble slmllar eyeles for other physleal proeesses Conslder the dssoluuon of an lonre solldln water MK 5 M39quot 3 X 3 AH AHertn We ran wnte thls equauon as the sum ofthree proeesses MXs M39gxn39gg AHU M39quot 9 M39quot 3 AH AHMM39 Xquot 9 Xquot 3 AH AEWW39 What ls the nature ofthe hydrated lons 7 The eauons normally have a primazy hydration when of4 or 6 water moleeules along wrth varymg numbers of adduonal water moleeules ln the seeondary hydrauon sphere The total number of watermoleeules surroundng an lon ls the hydration number Ion radlus pm hydrated ron hydrated radlus pm Na 116 NaH20 276 K 152 K0120 232 Whlle a larger ron ean aeeommodate more watermoleeules ln rts pnmary hydrauon sphere the hydrauon number wlll be more ln uencedby the eharge denle of the ron Consequences are lmportant For example K rons pass througn blologeal membranes more easlly than Na lons K rons are retarnedmore strongly on see exelusron reslns Ions wrth hlgher eharge denslues may have larger hydrauon numbers than those wrth smaller eharge densrues Thus whlle a smaller eauon may have a smaller pnmary hydrauon mhere rt may well have a larger hydranon number 625 Ions with larger charge densities also tend to have more negative hydration enthalpies Enthalpy of Solution Ion AHHyd 1d 39 mol39 1 Charge Density C 39 mm N85r 406 24 Mgrr 1920 120 Al3 4610 364 NaCl s Na g or g AH U 788 1d Nal g Nal 211 Cl g a Cl 211 NaCl s a Na aq Cl39 aq Entropy of Solution NaCl s a Na g Cl39 g Na g Na 3 1 C17 9 a C17 3 1 NaCl s a Na aq Cl39 aq Therefore AG AH TAS 4 k e 13 1d 9 k Formation of Covalent Compounds Approximation AH AthdNa 7406 1d AH AthdC139 7378 kJ AH AHsolnNaC1 4 1d TAS 68 1d TAS 27 kJ TAS 28 kJ TAS 13 1d and the formation of a 1M aqueous solution of NaCl at SATP is spontaneous We can use bond energies to estimate enthalpy changes for chemical reactions that occur in the gas phase This method applies only to gas phase reactions however AHO z 2 bond energies bonds broken 2 bond energies bonds formed 626 Consider the reaction between CH4 g and 02 g to form C02 g and H20 g H H c H g 2 g gt c gt g 2 H Eg Hg H In this reaction for every mole of CH4g we break 4 moles of C H bonds single and 2 moles of 00 double bonds and form 2 moles of CO double bonds and 4 moles of 0 H single bonds For obvious reasons it is use ll to write Lewis structures for all the reactants and products We can of course calculate a more accurate value for AH0 by using standard enthalpies of formation AH 2 H101 AH H20 9 t 1 H101 AH 002 g 1 H101 AH CH4 g AH 2 mol 2418 ldmol 1 mol 3935 1dmol 1 mol 749 1dmol AH 78022 k It should be pointed out that such good agreement between the estimate and the accurate value are not usually encountered and bond energies should only be used to estimate enthalpy changes when heats of formation are not available The values agree so well in part because we used the CO bond energy derived from the C02 molecule Furthermore this method of estimation should only be used for gas phase reactions 627 Thermodynamicvs Kmenc Control ofReacuons When ammomaxs burned the marmodynammally favoredproohd 15 dmnxogen 4NH3g301 2N2g6H20 However m the presence of a suitable catalyst me kmeucally favored product 15 NO 4N39Hg501 4NOg6HZO AG 958kJ Unwxm 51 men m4 Rmcuun mum

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