All Excuses must be taken to 233 Loomis before 4:15, Monday, April 30.

Transcription

1 Miscellaeous Notes The ed is ear do t get behid. All Excuses must be take to 233 Loomis before 4:15, Moday, April 30. The PYS 213 fial exam times are * 8-10 AM, Moday, May 7 * 8-10 AM, Tuesday, May 8 ad * 1:30-3:30 PM, Friday, May 11. The deadlie for chagig your fial exam time is 10pm, Moday, April 30. omework 6 is due Tuesday, May 1 at 8 am. (NO late tur-i). Course Survey = 2 bous poits (ow o SmartPhysics due Wed. May 2) Lecture 18, p 1

2 Lecture 19: Chemical Equilibria, Surfaces, ad Phase Trasitios Chemical equilibria - Law of mass actio Surface chemistry Phase equilibria ad chemical potetials Vapor pressure of a solid Readig for this Lecture: Elemets Ch 13 Lecture 19, p 2

3 Chemical Equilibrium Chemical is a bit of a misomer. We re describig ay process i which thigs combie (or rearrage) to form ew thigs. These problems ivolve reactios like: aa + bb cc, where A, B, ad C are the particle types ad a, b, ad c are itegers. I equilibrium the total free eergy, F, is a miimum. We must have DF = 0 whe the reactio is i equilibrium, for ay reactio that takes us away from equilibrium: df F F dnb F dn dn N N dn N dn A A B A C A F b F c F dna dnb dn usig N a N a N a b c 0 A Therefore: am A + bm B = cm C B C C C Lecture 19, p 3

4 Chemical Equilibrium (2) Treatig the compoets as ideal gases or solutes: m i kt l i Q i D Plug these chemical potetials ito the equilibrium coditio, am A + bm B = cm C, ad solve for the desity ratios: c c D C QC K( T ), where K( T ) e kt D cd ad bd a b a b A B Q Q i Iteral eergy per molecule A B C A B K(T) is called the equilibrium costat. It depeds o D s ad T, but ot o desities. This equilibrium coditio is a more geeral versio of the law of mass actio that you saw before for electros ad holes. The exact form of the equilibrium coditio (how may thigs i the umerator ad deomiator, ad the expoets) depeds o the reactio formula: aa + bb cc RS umerator LS deomiator Lecture 19, p 4

5 Examples of Chemical Equilibrium Process Reactio Equilibrium coditio Dissociatio of 2 molecules 2 2 m 2 = 2m Ioizatio of atoms e + p m = m e + m p Sythesize ammoia N N 3 m N2 + 3m 2 = m N3 Geeral reactio aa + bb cc + dd am A + bm B = cm C + dm D For the moatomic gases (circled) you ca use T = Q. The others are more complicated, ad we wo t deal with it here. owever, remember that T ofte cacels, so it wo t be a problem. Ideal solutios follow the same geeral form, but m is t close to the ideal moatomic gas value, because iteractios i a liquid ca be strog, modifyig both U ad S. Uits ad otatio: Chemists measure desity usig uits of moles per liter, ad write the law of mass actio like this: c [ C] ( ) a b [ A] [ B] KT Lecture 19, p 5

6 Chemical Equilibrium (3) Iteractios betwee the particles (e.g., molecules): I additio to simple PE terms from exteral fields, there are usually PE terms from iteractios betwee particles (which are ot usually ideal gases). Iteractios betwee the molecules ca ofte be eglected. That is, we ll treat the molecules as ideal gases. Iteral eergy of each particle (e.g., molecule): Atoms ca combie i ay of several molecular forms, each of which has a differet bidig eergy. The U term i F icludes all those bidig eergies (which we ll call D s), so they must be icluded i the m s. (df/dn) The reactio will NOT proceed to completio i either directio, because m depeds o for each type of molecule. As ay oe type becomes rare, its m drops util equilibrium is reached, with some of each type preset. (Just as ot all air molecules settle ito the lower atmosphere.) Lecture 19, p 6

7 ACT 1: Equilibrium i the Ammoia Reactio Cosider a reactio that is essetial to agriculture: the sythesis of ammoia from itroge ad hydroge: N N 3 1) Isert the correct superscripts ad subscripts i the equilibrium equatio: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) KT ( ) 2) Suppose the reactio is i equilibrium. Now double the umber of N 2 molecules. What will happe? A) Make more N 3. B) Dissociate more N 3. C) Nothig. Lecture 19, p 7

8 Solutio Cosider a reactio that is essetial to agriculture: the sythesis of ammoia from itroge ad hydroge: N N 3 1) Isert the correct superscripts ad subscripts i the equilibrium equatio: X ( 2 ) ( ) ( N 3 ) ( ) ( 1 ) ( 3 ) ( ) N 2 ( 2 ) KT ( ) Of course, you could write the whole thig upside dow, with K (T) = 1/K(T). 2) Suppose the reactio is i equilibrium. Now double the umber of N 2 molecules. What will happe? A) Make more N 3. B) Dissociate more N 3. C) Nothig. Lecture 19, p 8

9 Solutio Cosider a reactio that is essetial to agriculture: the sythesis of ammoia from itroge ad hydroge: N N 3 1) Isert the correct superscripts ad subscripts i the equilibrium equatio: X ( 2 ) ( ) ( N 3 ) ( ) ( 1 ) ( 3 ) ( ) N 2 ( 2 ) KT ( ) Of course, you could write the whole thig upside dow, with K (T) = 1/K(T). 2) Suppose the reactio is i equilibrium. Now double the umber of N 2 molecules. What will happe? A) Make more N 3. B) Dissociate more N 3. C) Nothig. You ve decreased the desity ratio. To restore it, N3 must icrease ad/or 2 must decrease. Some of the ew N 2 reacts with some of the 2, (decreasig 2 ), producig more N 3 (icreasig N3 ). There s still some of the ew N 2, i.e., N2 still icreases somewhat. Lecture 19, p 9

10 No-moatomic Gases Formatio of 2 from hydroge atoms: 2 2, so m 2 = 2m. Equilibrium coditio: D Q K( T ), where K( T ) e kt Q We ca use Q, because it s moatomic. D = 2 bidig eergy = 4 ev T = 1000 K We do t kow how to calculate Q, because it is diatomic ad has extra 2 U ad S. owever, we saw last time the effect that 2 beig diatomic has o m 2, amely, it reduces the chemical potetial. Lecture 19, p 10

11 Act 3: Formatio of 2 We have: 2 KT ( ) 2 We ca also write it like this: 2 K( T ) 2 1) What happes to if we decrease 2? A) Decrease B) Icrease C) Icrease, the decrease 2) What happes to / 2 if we decrease 2? A) Decrease B) Icrease C) Icrease, the decrease Lecture 19, p 11

12 Solutio We have: 2 KT ( ) 2 2 K( T ) 2 1) What happes to if we decrease 2? A) Decrease B) Icrease C) Icrease, the decrease Sice 2, decreasig 2 decreases. Makes sese: The overall desity is reduced. 2) What happes to / 2 if we decrease 2? A) Decrease B) Icrease C) Icrease, the decrease Lecture 19, p 12

13 Solutio We have: 2 KT ( ) 2 2 K( T ) 2 1) What happes to if we decrease 2? A) Decrease B) Icrease C) Icrease, the decrease Sice 2, decreasig 2 decreases. Makes sese: The overall desity is reduced. 2) What happes to / 2 if we decrease 2? A) Decrease B) Icrease C) Icrease, the decrease / 2 1/ 2, so decreasig 2 icreases the fractio of free atoms. 2 2 requires that two atoms meet, while 2 2 oly requires a sigle molecule. At low desity, the rate of the secod process is higher, shiftig equilibrium to more. At a give T, the fractio of atoms icreases at lower molecule desity! There are more atoms i outer space tha 2 molecules. Why? Two particles ( + ) have more etropy tha oe particle ( 2 ). Etropy maximizatio domiates the tedecy of atoms to bid!!! Lecture 19, p 13

14 Phase Trasitios Roadmap: We ll start by lookig at a simple model of atoms o surfaces, ad discover that, depedig o the temperature, the atoms prefer to be boud or to be flyig free. This is related to the commo observatio that materials ca be foud i distict phases: E.g., solid, liquid, gas. We ll lear how equilibria betwee these phases work. The we ll go back ad try to uderstad why distict phases exist i the first place. Lecture 19, p 14

15 Adsorptio of Atoms o a Surface M = # (sigle occupacy) bidig sites o the surface F U TS N D kt l s s s s M! ( M N )! N! m m s g s s Calculate the chemical potetials: Boud atoms: l l! d M Ns d N l, usig ln dns Ns dn df s M N s ms D kt l dns Ns Equilibrium: M N N s s p Q p e D/ kt D = bidig eergy of a atom o site N s = umber of boud atoms F s = Free eergy of boud atoms Atoms i the gas: p mg kt l, where pq QkT p Q We could solve this equatio for N s, but Lecture 19, p 15

16 Adsorptio of Atoms (2) usually we wat to kow what fractio of the surface sites are occupied, for a give gas pressure p ad temperature T: Usig our result: M N p s Q p e, where p p e N p p s D / kt 0 D / kt 0 Q We obtai a simple relatio for the fractio of occupied sites: f Ns p M p p o More atoms go oto the surface at high pressure, because m gas icreases with pressure. p o (T) is the characteristic pressure at which half the surface sites are occupied. It icreases with temperature due to the Boltzma factor. f T 1 p 0 (T 2 ) T 2 T 3 T 3 > T 2 > T 1 p Lecture 19, p 16

17 Act 4: Adsorptio 1) At 10 atm, half the sites are occupied. What fractio are occupied at 0.1 atm? A) 1% B) 11% C) 90% 2) Keep the pressure costat, but icrease T. What happes to f? A) Decrease B) No effect C) Icrease Lecture 19, p 17

18 Solutio 1) At 10 atm, half the sites are occupied. What fractio are occupied at 0.1 atm? A) 1% B) 11% C) 90% p p p 3 f % At lower pressure, gas atoms hit the surface less ofte. 2) Keep the pressure costat, but icrease T. What happes to f? A) Decrease B) No effect C) Icrease Lecture 19, p 18

19 Solutio 1) At 10 atm, half the sites are occupied. What fractio are occupied at 0.1 atm? A) 1% B) 11% C) 90% p p p 3 f % At lower pressure, gas atoms hit the surface less ofte. 2) Keep the pressure costat, but icrease T. What happes to f? A) Decrease B) No effect C) Icrease igher T: higher p Q ad e -D/kT / kt p 0 icreases p0 pqe D f decreases Makes sese? More atoms have eough thermal eergy to leave. Lecture 19, p 19

20 Solids, Liquids ad Gases Solid Liquid Gas Solids have fixed relatioships amog the atoms (or molecules) Liquids have looser relatioships amog atoms. The liquid has more etropy, because there are more ways to arrage atoms i the liquid. I liquids there are still some correlatios betwee atoms, but i gases there are essetially oe. I most situatios the etropy of the gas is vastly larger tha that of the liquid or solid. Lecture 19, p 20

21 Phase Trasitios ad Etropy If S gas > S liguid > S solid, why are differet phases stable at differet temperatures? The aswer is that we must also cosider the etropy of the eviromet. That s what free eergy ad chemical potetial do for us. For example: At low eough temperatures a substace like water is a solid. Its etropy is lower tha that of the liquid so it must give up eough eergy to its eviromet to make the total etropy icrease whe ice forms: ice Liquid 2 O Solid: DS tot = DS L +DS S + DS ev 0 < 0 > 0 I order for this to work, eough heat must be give to the eviromet to make DS tot 0. OK. So, why is liquid 2O favored at higher temperatures? The relative sizes of the DS terms must be differet. water Let s look at the problem more quatitatively. Lecture 19, p 21

22 Solid-gas equilibrium: vapor pressure Cosider solid-gas equilibrium at costat volume ad temperature. Some substaces (e.g., CO 2 ) do t exist as liquids at atmospheric pressure. The solid has egligible etropy (compared to the gas), so we ll igore it. I that case: F s = U s -TS s = -ND m s = -D The gas: m g = kt l(/ Q ) Bidig eergy of a atom i the solid Set them equal ad solve for the equilibrium gas pressure: p kt l = kt l = D p pvapor pqe Q p Q The equilibrium pressure is called the vapor pressure of the solid at temperature T. If p < p vapor, atoms will leave the solid util p gas = p vapor. This is called sublimatio. For liquids, it s called evaporatio. - D/kT V,T Examples: Si (28 g/mol): p vapor = ( atm)(28) 3/2 e -3eV/.026eV = atm CO 2 (44 g/mol): p vapor = ( atm)(44) 3/2 e -0.26eV/.026eV = 535 atm Some solids do t sublimate. Some do. Questio: Does water ice sublimate? Note the differet D s Lecture 19, p 22

23 Next time Phase diagrams Latet heats Phase-trasitio fu Freezig poit depressio/boilig poit elevatio Superheated/cooled water Lecture 19, p 23