Miscellaneous Notes. Lecture 19, p 1

Transcription

1 Miscellaeous Notes The ed is ear do t get behid. All Excuses must be take to 233 Loomis before oo, Thur, Apr. 25. The PHYS 213 fial exam times are * 8-10 AM, Moday, May 6 * 1:30-3:30 PM, Wed, May 8 The deadlie for chagig your fial exam time is 10 pm, Tuesday, April 30 Homework 6 is due Tuesday, Apr. 30 at 8 am. (NO late tur-i). Course Survey = 2 bous poits (accessible i SmartPhysics, but ot util Apr. 24) Lecture 19, p 1

2 Lecture 19: Chemical Equilibria, Surfaces, ad Phase Trasitios Chemical equilibria - Law of mass actio Surface chemistry 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 F = 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: aµ A + bµ B = cµ C B C C C *Obviously we ca have more or fewer species. Lecture 19, p 3

4 Chemical Equilibrium (2) Treatig the compoets as ideal gases or solutes: µ = kt l i i Q i i Iteral eergy per molecule Plug these chemical potetials ito the equilibrium coditio, aµ A + bµ B = cµ C, ad solve for the desity ratios: c C a b AB c + QC = K( T ), where K( T ) = e kt c a b a b Q A Q C C A B K(T) is called the equilibrium costat. It depeds o 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 RHS umerator LHS deomiator Lecture 19, p 4

5 Examples of Chemical Equilibrium Process Reactio Equilibrium coditio Dissociatio of H 2 molecules H 2 2H µ H2 = 2µ H Ioizatio of H atoms H e + p µ H = µ e + µ p Sythesize ammoia N 2 + 3H 2 2NH 3 µ N2 + 3µ H2 = µ NH3 Geeral reactio aa + bb cc + dd aµ A + bµ B = cµ C + dµ D For the moatomic gases (circled) you ca use T = Q. The others are more complicated, ad we wo t deal with it here. However, remember that T ofte cacels, so it wo t be a problem. Ideal solutios follow the same geeral form, but µ 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 ] = K T 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 s), so they must be icluded i the µ s. (df/dn) The reactio will NOT proceed to completio i either directio, because µ depeds o for each type of molecule. As ay oe type becomes rare, its µ 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 H 2 2 NH 3 1) Isert the correct superscripts ad subscripts i the equilibrium equatio: ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) = K( T) 2) Suppose the reactio is i equilibrium. Now double the umber of N 2 molecules. What will happe? A) Make more NH 3. B) Dissociate more NH 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 H 2 2 NH 3 1) Isert the correct superscripts ad subscripts i the equilibrium equatio: X ( 2 ) ( ) ( NH 3 ) ( ) ( 1 ) ( 3 ) ( ) ( ) N 2 H 2 = K( T) 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 NH 3. B) Dissociate more NH 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 H 2 2 NH 3 1) Isert the correct superscripts ad subscripts i the equilibrium equatio: X ( 2 ) ( ) ( NH 3 ) ( ) ( 1 ) ( 3 ) ( ) ( ) N 2 H 2 = K( T) 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 NH 3. B) Dissociate more NH 3. C) Nothig. You ve decreased the desity ratio. To restore it, NH3 must icrease ad/or H2 must decrease. Some of the ew N 2 reacts with some of the H 2, (decreasig H2 ), producig more NH 3 (icreasig NH3 ). There s still some of the ew N 2, i.e., N2 still icreases somewhat. Lecture 19, p 9

10 Example: p + e H at 6000K We pick this reactio because each compoet (proto, electro, hydroge atom) has o iteral modes (except spi, ot importat here) so we ca write dow the compoet µ s easily. What is the relatio betwee p ad H at T = 6000 K? Lecture 19, p 10

11 Solutio We pick this reactio because each compoet (proto, electro, hydroge atom) has o iteral modes (except spi, ot importat here) so we ca write dow the compoet µ s easily. What is the relatio betwee p ad H at T = 6000 K? Start with the exact equilibrium rule: µ p + µ e = µ H Now use that these are almost ideal moatomic compoets, (as log as the desity is low). So: p e H kt = e pq eq HQ 2 H p = ( pqeq ) eqhe HQ because pq HQ p e H kt l + kt l = kt l pq eq HQ Also: = because they are kt Now use: = 13.6eV ad eq = 2x(1.2x10 27 m -3 ) at 6000K 10 m at 6000 K p H p e produced i pairs. O the Su: ρ ~ ρ H2O. So, H ~ /m 3. p ~ /m 3, ot very large. Oly 1 i 10 7 ioized. Lecture 19, p 11

12 ACT 2 The temperature of the Su actually varies a great deal, from ~5700K at the surface, to ~10 7 K i the iterior. 1) For what approximate temperature will we have p ~ H? A) 10,000 K B) 60,000 K C) 160,000 K D) 6 x 10 6 K Lecture 19, p 12

13 Solutio The temperature of the Su actually varies a great deal, from ~5700K at the surface, to ~10 7 K i the iterior. 1) For what approximate temperature will we have p ~ H? A) 10,000 K B) 60,000 K C) 160,000 K D) 6 x 10 6 K As a rough estimate, we set kt = 13.6 ev T = 160,000 K Lecture 19, p 13

14 No-moatomic Gases Formatio of H 2 from hydroge atoms: H 2 2H, so µ H2 = 2µ H. Equilibrium coditio: H + QH = K( T ), where K( T ) = e kt H Q We ca use QH, because it s moatomic. H = H 2 bidig eergy = 4 ev T = 1000 K We do t kow how to calculate QH2, because it is diatomic ad has extra U ad S. However, we saw last week that we ca estimate the effect that H 2 beig diatomic has o µ H2, amely it reduces it. Lecture 19, p 14

15 We have: Act 2: Formatio of H 2 H2 = K( T ) 2 H 1) What happes to H if we decrease H2? A) Decrease B) Icrease C) Icrease, the decrease 2) What happes to H / H2 if we decrease H2? A) Decrease B) Icrease C) Icrease, the decrease Lecture 19, p 15

16 We have: H2 = K( T ) 2 H Solutio 1) What happes to H if we decrease H2? A) Decrease B) Icrease C) Icrease, the decrease Sice H H2, decreasig H2 decreases H. Makes sese: The overall desity is reduced. 2) What happes to H / H2 if we decrease H2? A) Decrease B) Icrease C) Icrease, the decrease Lecture 19, p 16

17 We have: H2 = K( T ) 2 H Solutio 1) What happes to H if we decrease H2? A) Decrease B) Icrease C) Icrease, the decrease Sice H H2, decreasig H2 decreases H. Makes sese: The overall desity is reduced. 2) What happes to H / H2 if we decrease H2? We ca also write it like this: A) Decrease B) Icrease C) Icrease, the decrease 2 H 2 H = 2 H 1 ( ) = ( ) 2 H K T 2 H K T 2 H H H2 H2 Thus, decreasig H2 icreases the fractio of free atoms. 2H H 2 requires that two atoms meet, while H 2 2H oly requires a sigle molecule. At low desity, the rate of the secod process is higher, shiftig equilibrium to more H. At a give T, the fractio of atoms icreases at lower molecule desity! There are more H atoms i outer space tha H 2 molecules. Why? Two particles (H + H) have more etropy tha oe particle (H 2 ). Etropy maximizatio domiates the tedecy of atoms to bid!!! Lecture 19, p 17

18 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 18

19 Applicatios of Surface Chemistry Catalysis -- purify egie exhaust Oxidatio of surfaces Fabricatio of high quality films Bidig of O 2 gas to hemoglobi ad myoglobi i your body... Adsorbed Atoms Gas at some pressure p = kt I equilibriuim: Chemical potetial of the gas: kt p µ g = kt l = kt l = kt l Q QkT p ɶ ɶ Q µ g = µ s where we ve defied p ɶ kt Q Q p Q, the quatum pressure, is the pressure it would take (hypothetically) to compress a ideal gas to the quatum desity, so that there was 1 particle per quatum cell. Chemical potetial of the surface: dfs µ s = dn s with Fs = Us TSs =? Lecture 19, p 19

20 Adsorptio of Atoms o a Surface M = # (sigle occupacy) bidig sites o the surface F = U TS = N kt l( Ω) s s s s M! Ω = ( M N )! N! µ = µ s g s s Calculate the chemical potetials: Boud atoms: ( l Ω) ( l!) d M Ns d N = l, usig = ln dns Ns dn df s M N s µ s = = kt l dns Ns Equilibrium: M N N s s = p Q p e / kt = 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 µ g = kt l, where pq = QkT p Q We could solve this equatio for N s, but Lecture 19, p 20

21 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 / kt 0 / kt 0 Q We obtai a simple relatio for the fractio of occupied sites: f N p M p + p s = o More atoms go oto the surface at high pressure, because µ 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 T 2 T 3 p 0 (T 2 ) T 3 > T 2 > T 1 p Lecture 19, p 21

22 Example: Adsorptio of Atoms At T = 300 K ad p = 1 atm, it is observed that 50% of the bidig sites o a particular metal surface are occupied. Whe the temperature is raised to 320 K at costat pressure, oly 25% of the sites are occupied. (You may assume that p Q is costat over this small temperature rage.) What is the bidig eergy of a site o the surface? k = ev/k Lecture 19, p 22

23 Solutio At T = 300 K ad p = 1 atm, it is observed that 50% of the bidig sites o a particular metal surface are occupied. Whe the temperature is raised to 320 K at costat pressure, oly 25% of the sites are occupied. (You may assume that p Q is costat over this small temperature rage.) What is the bidig eergy of a site o the surface? k = ev/k 1 1 Atm f (300 K) = = p0(300 K) = 1 Atm = pqe 2 1 Atm + p (300 K) 0 What is p (320K)? p (320 K) = 3 Atm = p e 0 0 ( / k )( 1/ 320 1/ 300) / k(300 K ) / k(320 K ) So : 3 = e = k l( 3 )/ ( 1/ 300 1/ 320) = ev Q Lecture 19, p 23

24 Act 3: 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 24

25 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 0.1 p + p f = = = % 0 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 25

26 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 0.1 p + p f = = = % 0 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 Higher T: higher p Q ad e - /kt / kt p 0 icreases p0 pqe f decreases Makes sese? More atoms have eough thermal eergy to leave. Lecture 19, p 26

27 Example: Oxyge i blood Your body eeds to carry O 2 from the lugs out to tissues (called T i the reactios below), usig some carrier molecule. Suppose it s myoglobi. If there are M myoglobi bidig sites ad N B boud oxyge molecules, (ad is the bidig eergy of oxyge to myoglobi), we ca aalyze the trasport of oxyge from lugs to tissues. Lecture 19, p 27

28 Solutio First look at the lugs. Igore the effect of the oxyge o the myoglobi to which it bids, so it s just like the problem of adsorptio by surface sites. Equilibrium is described by: µ gas = µ boud. N 2 The fractio of occupied myoglobi sites is: B p f = = O M p + p We ca t calculate p 0, but we kow empirically that f 2/3 i the lugs (where 1 p O2 = p lugs 0.19 atm). That is, p p 0 2 lugs What happes i the tissue? p 0 is the same. 1 1 Empirically, f 1/3, so p p p. tissue lugs 02 0 About 1/3 of the myoglobi sites are used to trasport oxyge to the tissues. Trasport will occur as log as the O 2 pressure i the tissue is less tha that i the lugs. Lecture 19, p 28

29 Next time Phase diagrams Latet heats Phase-trasitio fu Lecture 19, p 29