Thermohemija. Energija. Prvi zakon termodinamike. Entalpija / kalorimetrija
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1 Thermohemija Energija Prvi zakon termodinamike Entalpija / kalorimetrija Hess-ov zakon Kirchoff-ov zakon
2 Prenos toplote U endotermalnom procesu, toplota se uvek prenosi sa toplijeg objekta (okruženja) na hladniji sistem. T(sistema) raste T (okruženja) opada
3 Prenos toplote U egzotermalnom procesu, toplota se prenosi iz toplijeg SISTEMA u hladnije OKRUŽENJE. T(sistema) opada T(okruženja) raste
4 U toku hemijske reakcije se može apsorbovati (trošiti) ili oslobadjati (davati) tolota. Medjutim u toku hemijskih reakcija se takodje može vršiti rad. Zn(s) + 2H + (aq) Zn 2+ (aq) + H 2 (g) Ako je pritisak konstantan w P V Sistem vrši rad!
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7 Entalpija je tolota razmenjena izmedju sistema i okoline pri konstantnom pritisku Entalpija je funkcija stanja definisana sa: H = U + PV Na konstantnom pritisku promena entalpije je: H = H 2 - H 1 = U 2 + PV 2 - (U 1 + PV 1 ) H = U + P V Iz prvog zakona : U = Q - P V H = Q p (konstantnom P)
8 Veza U i H H U P V V H nrt P U n g n g RT RT Prethodna jednačina znači da se promena unutrašnje energije i promena entalpije sistema razlikuju značajno samo kada u reakciji dolazi do promene broja molova kada u reakciji učestvuju gasovi bilo kao reaktanti, ili kao produkti. P
9 CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O (g) + Q (-802,3 kj/mol) Q + H 2 O (s) H 2 O (l). Većina hemijskih reakcija odvija se na konstantnom pritisku, pa se toplota razmenjena u tim uslovima može izraziti termodinamičkom funkcijom stanja - entalpijom. Kako je entalpija funkcija stanja, H zavisi samo od početnog i krajnjeg stanja. Takodje, treba napomenuti da je kod mnogih reakcija na P = const, i V = 0, tako da je H = U. 1. u reakciji nema gasova: 2 KOH (aq) + H 2 SO 4 (aq) K 2 SO 4 (aq) + 2H 2 O (l) Kako se radi o tečnostima, promena V = U. 0, pa je P V = 0, i H
10 2. broj molova reaktanata jednak je broju molova produkata u gasnoj reakciji: N 2 (g) + O 2 (g) 2 NO (g) V = 0, P V = 0 H = U. 3. broj molova gasa menja se tokom reakcije, pa P V 0. Medjutim, kako je najčešće q P >> P V H U. 2 H 2 (g) + O 2 (g) 2H 2 O (g); H kj; P V = kj U = H - P V = - 481,1 kj H Za najveći broj hemijskih reakcija može se pouzdano uzeti vrednost H kao jednaka ili približno jednaka U.
11 Promena entalpije hemijske reakcije prikazuje se entalpijskim dijagramom CaO(s) + CO 2 (g) CaCO 3 (s) rxnh o = kJmol -1 Entalpija je ekstenzivna veličina, pa je promena entalpije H direktno proporcionalna količini supstance!
12 Promena Standardne Enthalpije, odnosno, H 0, je promena entalpije za proces u kome se i reaktanti i produkti nalaze u svojim standardnim stanjima (STP). Standardne entalpije nastajanja supstanci: toplota nastanka jednog jedinjenja jeste povećanje toplotnog sadržaja H kada se nagradi 1 mol supstance iz njenih elemenata. Dogovorno je uzeto da je: Toplotni sadržaj jednog jedinjenja jednak njegovoj toploti postojanja. C (s) + O 2 (g) = CO 2 (g) H = kj/mol=-94 kcal/mol H = U + P V U = H - P V; U = H - PV = H - RT n g. Ovde je - promena broja molova, a n g - promena broja molova gasnih učesnika reakcije. Kako je u ovoj reakciji, n g = 0 H = U.
13 izračunavanje toplote nastanka metana iz ugljenika i vodonika: C grafit + 2 H 2 (g) CH 4 (g) H = kj Znamo toplote nastanka CO 2 i H 2 O (toplotni sadržaji elemenata - C, O i H su jednaki 0!), a takodje i toplotu rekacije izmedju metana i kiseonika: a) C grafit + O 2 (g) CO 2 (g) H = kj b) H 2 (g) + 1/2 O 2 (g) H 2 O (l) H = kj c) CH 4 (g) + 2O 2 (g) CO 2 (g) + 2 H 2 O (l) H = kj Kombinacijom: (a) + 2(b) - (c),dobija se da je H za traženu reakciju kj.
14 Standardne Entalpije Formiranja C(s, graphite) + O 2 (g) CO 2 (g) H f 0 = kj mol -1 C(s, graphite) + 2H 2 (g) CH 4 (g) H f 0 = kj mol -1 ½ N 2 (g) + 3/2 H 2 (g) NH 3 (g) H f 0 = kj mol -1 (1/2) N 2 (g) + (1/2) O 2 (g) NO(g) H f 0 = kj mol -1
15 2. Toplota sagorevanja: promena toplotnog sadržaja koja prati potpuno sagrevanje jednog mola nekog jedinjenja: 3. Toplota topljenja: promena toplotnog sadržaja koja prati topljenje jednog mola čvrste supstance. 4. Toplota isparavanja: promena toplotnog sadržaja koja prati isparavanje jednog mola čvrste supstance. 5. Toplota hidrogenacije: promena toplotnog sadržaja koja prati prevodjenje jednog mola nezasiććenog ugljovodonika u odgovarajuće zasićeno jedinjenje pomoću gasovitog vodonika:
16 Energy H 2 O(l) H final > H initial H = H final H initial Endothermic H 2 O(s) H 2 O(s) H 2 O(l) ΔH = kj mol -1
17 Energy H 2 O(l) H final < H initial H = H final H initial Exothermic H 2 O(s) H 2 O(l) H 2 O(s) ΔH = kj mol -1
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19 Odakle potiče toplota reakcije? Kada raguje 2 g H 2 i 16 g O 2 nastaje 18 g vode i oslobadja se 242 kj toplote. Pitanje je: odakle potiče ova energija? Energija nastala tokom stvaranja jednog mola vode nastala je kao transformisana energija iz unutrašnje energije reaktanata, vodonika i kiseonika. Unutrašnja energija čestica: U = E kinetičko + E potencijalno Dakle, kada se stvara voda u reakciji: H 2 + 1/2 O 2 H 2 O E 1 E 2 i E 1 > E 2 Pitanje je: koji delovi unutrašnje energije sistema su odgovorni za promenu u energiji koja se javlja tokom hemijske reakcije (odnosno za Q P = H)?
20 U = E K trans + E K rot + E K vib + E P(atom) + E P(nukleus) + E P (bond) Kako je kineticka energija E K T, na T = const E K 0. Energija koja se oslobadja ili apsorbuje tokom hemijske promene potiče od razlike potencijalne energije E P medjuatomskih (unutarmolekulskih) veza reaktanata i produkata. Hemijska reakcija je proces u kome se hemijske veze u molekulima reaktanata kidaju (potrebna je energija E 1 ), a veze u novim molekulima, produktima nastaju (oslobadja se energija E 2 ). Kada je E 2 >E 1, reakcija je egzotermna.
21 Stehiometrija termohemijske jednačine Pomoću termohemijskih jednačina se predstavljaju termohemijske promene. Postoje 2 pravila: 1. H je iste vrednosti ali suprotnog znaka za reakciju u suprotnom smeru. 2. Veličina H je proporcionalna količini susptance. H 2 (g) + 1/2O 2 (g) H 2 O (l) H = kj/mol 2H 2 (g) + O 2 (g) 2H 2 O (l) H = kj/mol 2H 2 O (l) 2H 2 (g) + O 2 (g) H = kj/mol
22 H isparavanja: Promena entalpije potrebna da jedan mol supstance ispari na tački ključanja pri pritisku od 1 atm. ( kjmol-1 at 373K za vodu ) endoterman H topljenja : Promena entalpije potrebna da se jedan mol čiste supstance istopi na tački topljena pri pritisku od 1 atm. (+6.01 kjmol-1 at 273K za led) endoterman
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24 Hesov zakon sumiranja toplota Promena entalpije ukupnog hemijskog procesa je suma entalpija u pojedinačnim procesima. Promena toplote u hemijskoj reakciji je ista nezavisno od toga da li se reakcija dešava u jednom ili više stupnjeva.
25 Posledica činjenice da je entalpija funkcija stanja (ne zavisi od prethodnog stanja sistema) je da se: može se odrediti promena entalpije H za složenu reakciju bez razmatranja kako se reakcija odvila u realnosti. Termohemijske jednačine se mogu sabirati i oduzimati, pa se entalpija velikog broja hemijskih reakcija može izračunati bez meranja. Ovo je posebno važno kada su u pitanju složene reakcije, kao i reakcije koje se odvijaju u nepovoljnim eksperimentalnim uslovima.
26 C + O 2 CO 2 H = kj C+ ½ O 2 CO H = kj CO (g) ½ O 2 CO 2 H = kj (-283) = kj/mol
27 Izračunavanje H za reakciju kod koje se ta vrednost teško meri, iz poznatih vrednosti, kada znamo H za neke druge reakcije.
28 Hess-ov zakon Promena entalpije ukupnog hemijskog procesa je suma entalpija u pojedinačnim procesima. rxnh o = H o products - H o reactants
29 A C Hess-ov zakon Ako je reakcija zbir dve reakcije tada je promena entalpije u toku zbirne reakcije zbir promena entalpije komponenata reakcije. -C = H A-B + H B-C products C A B C reactants A
30 Kalorimetrija - Promena unutrašnje energije se može meriti kalorimetrijski U = q - P V pri konstantnoj zapremini V=0! Tako da je U = q
31 H = ΔQ ΔT H = ΔQ = const ΔT H =ΔQ = C ΔT Promena enthalpije Promena temperature Tolotni kapacitet C = m C spec C spec = specifični toplotni kapacitet H = ΔQ = m C spec ΔT
32 Specifični tolotni kapacitet Sposobnost supstance da primi toplotu C spec = J g -1 K -1 Količina energije (J) potrebna da se 1g supstance zagreje za 1 K
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34 Al C Fe Cu Au NH 3 H 2 O C 2 H 5 OH drvo staklo
35 Kirhoff-ov zakon - uticaj temperature na toplotu reakcije I H 1 + T C PB II H 2 + T C PA H 1 = (H B - H A ) T1 ; H 2 = (H B - H A ) T2 H 1 + T C PB = H 2 + T C PA H T H 2 1 H T T H P C C C P PB PA P H produkata T P H H T reak tanata T ( C ) ( C ) C tan P produkti P reak ti P P P C P
36 Za reakciju: n A A + n B B + n C C + = n L L + n M M + n N N + C [ n ( C ) n ( C )...] [ n ( C ) n ( C )...] P L P L M P M A P A B P B H H C dt 0 T 0 P 2 3 C a bt ct dt P C na nb T nc T... T T T... P T 2 H H T T... dt H T 1/ 2 T 1/ 3 T H H T 0
37 rh(t 2 ) Reactants(T 2 ) Products(T 2 ) H R Reactants(T 1 ) Products(T 1 ) H P rh(t 1 ) rh(t 2 ) = H R + rh(t 1 ) + H P
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39 Foods: Fuels for the Body Three classes of food are carbohydrates, proteins, and fat. Carbohydrates include starches and sugars. During digestion, they are converted to simple sugar glucose C 6 H 12 O 6. C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l) H = kj Fats belong to the ester family, which is derived from a carboxylic acid RCOOH and an alcohol ROH. An OH group of the acid is replaced by a OR group of the alcohol. On a mass basis, the heat of combustion of fats (-37.4 kj/g C39H74O6) is almost twice that of glucose ( kj/g). Therefore, fats are the perfect material for energy storage. Proteins are polymers of amino acids, and have about the same energy values as O carbohydrates. Food calories (Cal) 1 Kcal = 1 Cal = 1000 cal = 4.18 kj H H C C NH 2 OH
40 Energy Content of Foods Chemical energy in animals is derived from carbohydrates, fats, and proteins. Carbohydrates 17 Fat 38 Protein 17 Fuel value (kj/g) Fuel value is usually expressed in kcal or Cal per serving. 1 Cal = 1 kcal = kj
41 Energy Content of Foods If a person uses about 420 kj/mi when running, how many candy bars are required to run three miles? 1 Butterfinger : 42 g carbohydrates 11 g fat 3 g protein 56 g 710 kj 420 kj 50 kj 1180 kj
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43 Energy 2 H 2 (g) + O 2 (g) NH 4 NO 3 (aq) H final < H initial H final > H initial Exothermic Endothermic 2 H 2 O(l) NH 4 NO 3 (s) + H 2 O(l)
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