The Gibbs Free Energy

Transcription

1 he Gibbs Free Energy Using the entrpy criterin t decide whether a prcess can ccur we have t calculate the entrpy change f the system and f the surrundings We have cncluded that, S univ > 0 S univ 0 S univ < 0 Spntaneus Equilibrium Impssible Very impractical! We need a state prperty t give a feasibility f the prcess withut reference t surrundings Gibbs Free Energy (G): a state functin (cnstant and P) G H S (in Jules) Since is cnstant, G system system - S system system Recall, Ssurrundings If the prcess is spntaneus: S universe S system + S surrundings > 0 system herefre, ( Ssystem + ) > 0 We multiply bth sides f the inequality by, and ( S system - system ) > 0 and system - S system <0 G system system - S system <0 (at cnstant and P) If the Gibbs free energy is frecast t decrease during a prcess at cnstant and P the prcess is spntaneus. If the Gibbs free energy is frecast t increase the prcess cannt ccur under the cnditins specified ( S univ < 0) but the reverse f the prcess under the specified cnditins is spntaneus. If the Gibbs free energy is frecast t stay cnstant, S univ 0 fr the prcess and n change in either directin can ccur under the specified cnditins and this is synnymus t EQUILIBRIUM.

2 Fr a change at cnstant and P: G system < 0 Spntaneus G system 0 Equilibrium G system >0 Nn-spntaneus but reverse is he Gibbs Energy and Phase ransitins H O(l) H O(s) At the freezing pint (73.5K) and atm pressure, J ml - and Jml S JK ml freeze 73.5K G S 73.5 (-6007 J ml - ) (73.5 K) (-.99 J K - ml - ) 0 J ml - - N surprise, equilibrium! What happens if water is cled dwn t 63.5 K (- 0 C)? We assume that bth and S d nt change significantly with the drp in temperature, and therefre, G S 63.5 (-6007 J ml - ) (63.5 K) (-.99 J K - ml - ) - 0 J ml - - water freezes spntaneusly (n external interventin) What happens if water is heated up t 83.5 K (+ 0 C)? Again we assume that bth and S d nt change significantly with the rise in temperature, and therefre, G S 83.5 (-6007 J ml - ) (83.5 K) (-.99 J K - ml - ) + 0 J ml - - freezing f water is nn-spntaneus S, liquid water at atmspheric pressure never freezes when the temperature is greater than 73.5 K. (Rather the reverse prcess ccurs and slid water melts). At 73.5 K the tw curves crss, G 0 (ice and water cexist) At < 73.5 K water spntaneusly freezes t ice At > 73.5 K the reverse prcess, the melting f ice is spntaneus.

3 Example CS is a liquid at rm temperature. Calculate G vap at 5 C given that vap 7.66 kj ml - and S vap J K - ml -. Calculate als the nrmal biling pint f CS. G vap vap - S vap (7.66 kj ml - ) (98.5 K) (86.39 J K - ml - ) ( kj/000j).90 kj ml - In rder t calculate the nrmal biling pint f CS we cnsider G vap 0, because it is assciated with the prcess CS (l) CS (g) (equilibrium) vap G vap vap - S vap 0 b 30. K S rutn s Rule vap Mst liquids have the apprximately the same mlar entrpy f vaprizatin at their nrmal biling pints S vap 88 ± 5 J K - ml - rutn s rule was derived because vaprizatin ccurs within an increase in mlar vlume that is very large and nearly the same frm substance t substance. he vlume cntributin t entrpy is s large that we neglect any ther. vap Svap (get vap if we knw bp ) bp rutn s rule breaks when liquids have large amunts f rder. Fr water, S vap 09 J K - ml - - extensive H-bnding Gibbs Free Energy and Chemical Reactins G G prducts G reactants A chemical can ccur under cnstant and P nly if it lses Gibbs free energy t the surrundings he negative f the Gibbs free energy (- G ) is a measure f the driving frce f a reactin under the cnditins f cnstant and P. he units are J ml -. he ml -, per mle refers t the reactin as written. G - S 3

4 Standard Gibbs Free Energies f Frmatin G - S Reactants at standard states give prducts at standard states. Standard Mlar Gibbs Free Energies f Frmatin he standard mlar Gibbs free energy f frmatin is the change in Gibbs free energy when ml f substance frms in a standard state at a specified temperature frm the mst stable frms f its cnstituent elements in standard states at the same temperature. We use G f f - S f t cmpute G f f any substance at any temperature Cnsider C (graphite) + O (g) CO (g) at 5 C. he G f (CO (g)) equals G We calculate f f (CO (g)) kj ml - S x S (CO (g)) [ x S (O (g)) + x S (C (graphite))].86 J K - ml - G f (CO (g)) f (CO (g)) - S f (CO (g)) kj ml - Fr: aa + bb cc + dd, G [c G f (C) + d G f (D)] - [a G f (A) + b G f (B)] In general, G Σm G f (prducts) - Σn G f (reactants) Example Calculate G frm enthalpy and entrpy values r the reactin: 4KClO 3 (s) 3KClO 4 (s) + KCl(s) [(3 ml)(-43.8 kj/ml) + ( ml)( kj/ml)] - (4 ml)( kj/ml) - 44 kj S [(3 ml) (5 J ml - K - ) + ( ml) (8.6 J ml - K - )] - (4 ml)(43. J ml - K - ) J K - G - S (- 44 kj) (98.5K) ( J K - ) ( kj/000j) - 33 kj G <0 is spntaneus! 4

5 We culd the same calculatin if we had in ur dispsal G f values. G [(3 ml)(-303. kj/ml) + ( ml)(-409. kj/ml)] - (4 ml)(-96.3 kj/ml) -34 kj <0 is spntaneus! Effect f emperature n G G - S and S change nly slightly as the temperature changes but G changes cnsiderably. he spntaneity f a reactin at different temperatures depends n the signs f and S. When bth signs are the same, the temperature determines the spntaneity f the reactin We calculate G frm G f nly at 98.5 K. At any ther temperature G - S. We assume that and S are independent f temperature in the temperature range we cnsider!!!! In the yellw quarters there is a temperature at which G 0 and in this case G - S 0 S and. At this reactants and prducts S 5

6 are at their standard states (gases at partial pressure bar and slutins at M) his temperature is usually called crssver temperature. Example Using Mlecular Scenes t Examine the Signs f, S, and G he fllwing scenes represent a familiar phase change fr water (blue spheres): (a) What are the signs f and S fr this prcess? Explain. he scenes represent cndensatin f a gas, S < 0 (mre rder) and < 0 (exthermic) (b) Is the prcess spntaneus at all, n, lw, r high? Explain. G - S. With a negative S and negative, G will be negative. herefre, the prcess is spntaneus at lw. Example Determining the Effect f emperature n G A key step in the prductin f sulfuric acid is the xidatin f SO (g) t SO 3 (g): SO (g) + O (g) SO 3 (g) At 98 K, G -4.6 kj; kj; and S J K - (a) Use the data t decide if this reactin is spntaneus at 5 C, and predict hw G will change with increasing. he reactin is spntaneus at 5 C because G is (-). Since is (-) but S is als (-), G will becme less negative, and the reactin less spntaneus as the temperature increases. (b) Assuming and S are cnstant with, is the reactin spntaneus at 900 C? G - S G kj [(73 K)(-87.9 J K - )( kj/000 J).0 kj he reactin is nt spntaneus at 900 C. 6

7 Example At which temperature is the fllwing prcess Br (l) Br (g) spntaneus at atm? 3.0 kj ml - and S 93.0 J K - ml - A bit deep thught he vaprizatin prcess is spntaneus at all temperatures at which G < 0. S favrs vaprizatin and favrs cndensatin (exthermic). hese tendencies balance at the biling pint. At the biling pint: Br (l) Br (g) G 0 (equilibrium) and At > 333 K, S favrs vaprizatin At < 333 K, favrs an exthermic prcess At 333 K, G 0 nrmal biling pint 333 K S Gibbs Free Energy and the Equilibrium Cnstant G G + R ln Q Q is the reactin qutient. If all reactants and prducts are at their standard states, then Q, ln Q and G G he reactin qutient gives the prgress f the reactin. It varies frm 0 (nly pure reactants) t infinity (nly pure prducts) and it becmes equal t the equilibrium cnstant, K at equilibrium. he relative values f Q and K establish the directin that the reactin takes in cming t equilibrium. Q < K reactin prceeds as written t the right Q > K reactin prceeds as written t the left Q K equilibrium, G 0 G - R ln K G G + R ln Q - R ln K + R ln Q R ln (Q/K) 7

8 Criteria fr Spntaneity in a Chemical Reactin Spntaneus Prcesses Equilibrium Prcesses Nn-spntaneus prcesses Cnditins S univ > 0 S univ 0 S univ < 0 all G < 0 G 0 G > 0 Cnst., P Q < K Q K Q > K Cnst., P he relatin between Gibbs free energy and the prfess f reactin. Keep in mind! At equilibrium G 0 G - R ln K Example G ln K and R K e G R Cnsider N (g) + 3H (g) NH 3 (g). G kj ml - at 98.5 K. Predict the directin in which the system will shift t reach equilibrium when a) P f NH 3 is atm, P f N is. 47 atm and P f H is. 0 x 0 - atm We shuld calculate G, P G G NH3 + R ln Q 33.3kJ + (8.34JK ml )(98.5K) ln 3 PH P N G 33.3kJ + (.48kJml ln 33.3kJ kJ 0 3 EQUILIBRIUM (.47)(.0x0 ) NO SHIF OCCURS! b) P f NH 3 is.0 atm, P f N is.0 atm and P f H is. 0 atm Q, ln 0, s G G Rxn shifts t right t reach equilibrium 8

9 Example Cnsider the fllwing reactin: CO (g) + H (g) CH 3 OH (l) G f (kj ml - ) Calculate G at 5 C when the partial pressure f CO is 5.0 atm and the partial pressure f hydrgen is 3.0 atm. G G + R ln Q G - 66 (-37) 0-9 kj ml - Q.x0 PCOPH (5.0)(3.0) G G + R ln Q (-.9x0 4 J ml - ) + (8.34 J K - ml - )(98.5K)ln(.x0 - ) G - 38 kj ml - We bserve that G is mre negative than G. his implies that the reactin is mre spntaneus at reactin pressures greater than.0 atm. he emperature Dependence f Equilibrium Cnstants G S ln K + R R R the extent that the temperature dependence f and S can be neglected then ln K is a linear functin f /. A graph f ln K vs. / is apprximately a straight line with a slpe f and R S intercept R ln K R ln K K ln K S + R ln K R R ln K R van t Hff Equatin S + R 9

10 he Variatin f Vapr Pressure with emperature liquid (l) vapr (g) P vapr K Let s write van t Hff equatin fr vaprizatin at tw different temperatures and K P vap, H vap ln ln We assume that and S fr vaprizatin are K Pvap, R independent f temperature. Claussius-Clapeyrn Equatin At the nrmal biling pint f a substance, b the vapr pressure is atm. aking t crrespnd t b and sme ther temperature, then ln Pvapr, R vap b 0