CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review

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1 Review Accrding t the nd law f Thermdynamics, a prcess is spntaneus if S universe = S system + S surrundings > 0 Even thugh S system <0, if there is enugh changes t the surrundings such that S surrundings >0, such that Thermdynamics Change in Gibbs Free Energy, G S system + S surrundings >0 the prcess will be spntaneus. As chemists, we are interested in the criteria f spntaneity as they apply t chemical reactins. Deining S universe is nt always easy. We need t mdify the methd f deining the spntaneity f a prcess by nly lking within the system! Review Recall hw H and S favur spntaneity. 1. Spntaneus prcesses are: favured by a decrease in H (exthermic) favured by an increase in S Gibbs Free Energy, G At this pint, we intrduce a new functin, G, such that G = H TS where G is called the Gibbs free energy. Fr a prcess that is carried ut under cnstant T and P,. Nnspntaneus prcess, favured by an increase in H (endthermic) favured by an decrease in S We will expand n this. G = (H TS) = H (TS) at cnstant T G is the change in Gibbs free energy. G is a state functin and has the prperties f a state functin. The Gibbs energy is the maximum useful wrk that a system can d n the surrundings when the prcess is carried ut under cnstant temperature and pressure. At cnstant P, q surr = -q sys = - H sys T S surr = -q surr = - H sys T S univ = T S surr + T S sys = - H sys + T S sys G = T S univ fr cnstant T and P 3 4 Prcess carried ut under cnstant T and P G Irreversible and Spntaneus < 0 Reversible = 0 Irreversible and Nnspntaneus > 0 The change in the free energy f a system that ccurs during a reactin can be measured under any set f cnditins. If the data are cllected under standard-state cnditins, the result is the standard-state free energy f reactin (G ). G = H -T S H S G Case I < 0 > 0 Always < 0 Spntaneus at all T Case II > 0 < 0 Always > 0 Nnspntaneus at all T Case III > 0 > 0 > 0 at lw T < 0 at high T Nnspntaneus at lw T and becmes spntaneus as T is raised Case IV Factrs < 0 favuring < 0 a < spntaneus 0 at lw T prcess, Spntaneus making at lw G T < and 0 becmes > 0 at high T nnspntaneus as T is raised 5 C is cnsidered lw T. As a result, exthermic reactins are in general usually spntaneus at rm temperature and 1 atmsphere. [Nte: H is a fairly reliable indicatr f a reactin s spntaneity at rm temperature, but it is NOT a general criterin.] 5 6 1

2 H S G Case I < 0 > 0 Always < 0 Spntaneus at all T Case II > 0 < 0 Aways > 0 Nnspntaneus at all T Case III > 0 > 0 > 0 at lw T Nnspntaneus at lw < 0 at high T T and becmes spntaneus as T is raised Case IV < 0 < 0 < 0 at lw T Spntaneus at lw T > 0 at high T and becmes nnspntaneus as T is raised G versus T graphs at cnstant pressure (a) Sketch G versus T fr the fllwing reactin. (b) Calculate the crss-ver temperature. H S (g) + SO (g) 3 S (s) + H O (g) H = kcal G = kcal Answers: (a) Expect S < 0 G = H T S = ( S ) S = -44 cal K -1 (b) G = H T S 0 = T(-0.044) T = 795 K = 5 C H < 0 and S < 0 Prcess is spntaneus belw 5 C. 7 8 Standard Gibbs Free Energy f frmatin, G f Wuld yu invest yur mney and time in develping a glue which is endthermic as it sets? G f f a substance refers t the reactin in which that substance is frmed frm the elements as they exist in their mst stable frms at 1 atm pressure and (usually) 98K. Standard free energies f frmatin, G f are analgus t standard enthalpies f frmatin, H f. Answer: Glue (l) Glue (s) Like the enthalpies f elements in their standard states (1 atm, 5 C) is assigned zer, we assign a free energy f zer t each free element in its standard state. H > 0 S < 0 Expect G > 0 ver all temperature. This is a nnspntaneus prcess. The glue will never set. G f can be lked up in tables r calculated frm S and H f 9 10 Standard Gibbs Free Energy f frmatin, G f What is the standard free energy fr the burning f n-pentane? C 5 H 1 (g) + 8 O (g) 5 CO (g) + 6 H O (g) Given G f (CO (g)) = kj mle -1 G f (H O (g)) = -8.6 kj mle -1 G f (C 5 H 1 (g)) = kj mle -1 G reactin = 5 G f (CO (g)) + 6 G f (H O (g ) - G f (C 5 H 1 (g)) 8 G f (O (g)) 5(-394.4) + 6(-8.6) (-8.37) 8(0) = kj G < 0 Prcess is spntaneus. 11 G Examples Given that at 400 C the reactin 4Cu(s) + O (g) Cu O(s) G = -50 kj/ml 1. What is G at 400 C fr the fllwing reactins a) Cu(s) + ½O (g) Cu O(s) G = ½(-50 kj/ml) = -15 kj/ml b) Cu O(s) Cu(s) + ½O (g) G = -½(-50 kj/ml) = 15 kj/ml. Will Cu O(s) spntaneusly decmpse at 400 C with the cmpunds in their standard states? G = >0, n 3. What happens if the Cu O(s) decmpsitin reactin is made t run simultaneusly with C(s) + ½O (g) CO(g) G = -175 kj/ml C(s) + ½O (g) CO(g) G = -175 kj/ml Cu O(s) Cu(s)+ ½O (g) G = 15 kj/ml Cu O(s) + C(s) Cu(s) + CO(g) G = -50 kj/ml G < 0, reactin is spntaneus

3 G Examples Given fr water at 5 C H vap = 4 kj/ml S vap = 119 J ml -1 K -1 What is G vap at 5 C? Will water spntaneusly evaprate at 5 C? G vap = H - T S = 4 kj/ml (98 K)(0.119 kj ml -1 K -1 ) G vap = 7 kj/ml Since G vap > 0, 5 C will nt evaprate under standard cnditins Des this agree with yur intuitin? Fr many reactins, the rates at 5 C are t lw t be useful. We need t find G at ther temperatures. We will assume that H f and S t be cnstant and use them t estimate G at temperatures ther then 5 C. Estimate the G f the frmatin and decmpsitin f ammnia at 5 C and at 47 C. G therefre describes this N (g) + 3 H (g) NH 3 (g) reactin when all three Given H = -9. kj and S = kj K -1 cmpnents are present at 1 atm pressure. G < 0, therefre, expect the prcess Answer: t be spntaneus. G Under standard cnditin, the = -9. (98)( ) = kj equilibrium favurs the prducts. G 47 C -9. (47+73)( ) = 47 kj G 47 C > 0 Expect that the prcess is nnspntaneus. Therefre, we can prduce ammnia at 5 C, but nt at 47 C spntaneusly. But cmmercial prcesses synthesize ammnia at near 500 C. Why? 1 N (g) + 3 H (g) NH 3 (g) N (g) + 3 H (g) NH 3 (g) G = H - T S G = H - T S Fr ideal gases H = H S N = S N Rln P N S H = S H Rln P H S = S f S i = R ln P i = -R ln P f S NH3 = S NH3 Rln P NH3 Let i be the standard state S f = S - R ln P f P = S - R ln P P i P = S - R ln P 1 13 S rxn = (S NH3 Rln P NH3 ) (S N Rln P N ) 3(S H Rln P H ) S rxn = S NH3 S N 3S H + Rln P N P 3 H S rxn = S rxn + Rln 14 N (g) + 3 H (g) NH 3 (g) rxn TR ln S rxn = S rxn + Rln Remember if G = 0, the system is at equilibrium. S G must be related t the equilibrium cnstant, K eq. Standard free energy, G, is directly related t K eq G appraches zer, and Q = K eq. G = G + RT ln P NH3 0 G + RT ln K eq G = G + RT ln Q where G is the standard free energy change R is the gas cnstant T is the temperature in Kelvin K eq is the equilibrium cnstant Q is the reactin qutient

4 Cnsider the equilibrium reactin at 47 C N (g) + 3 H (g) NH 3 (g) + heat Accrding t Le Chatelier s Principle there are tw ways t increase the yield f ammnia. 1. Increase the pressure causes the equilibrium psitin t shift t the right resulting in a higher yield f ammnia (since there are mre gas mlecules n the left hand side f the equatin).. Decrease the temperature causes the equilibrium psitin t shift t the right resulting in a higher yield f ammnia since the reactin is exthermic (releases heat). The rate f the reactin at lwer temperatures is extremely slw making the prductin f ammnia is negligible. Even thugh increasing the temperature shifts the equilibrium t the left, a higher temperature must be used t speed up the reactin. At the same time, pressure is als increased in rder t increase the yield f ammnia. 17 Cnsider the equilibrium reactin at 47 C N (g) + 3 H (g) NH 3 (g) If p(n ) = 33.0 atm, p(h ) = 99.0 atm, p(nh 3 ) =.0 atm, calculate G. Evaluate Q, G G 47 C + RT ln Q G (700K) ln (1.3x10-7 ) = - 45 kj P ttal = 134 atm Recall G 47 C = H T S G 47 C -9. (47+73)( ) 47 kj As a result f the new pressure cnditins at 47 C, G 47 C < 0. Usually, a catalyst (eg Fe) is used t speed up the reactin by lwering the activatin energy s that N bnds and H bnds can be brken mre readily. 18 Cnsider the equilibrium reactin at 47 C N (g) + 3 H (g) NH 3 (g) P ttal = 134 atm Synthesis f Ammnia: N (g) + 3H (g) NH 3(g) Under this cnditin, p(n ) = 33.0 atm, p(h ) = 99.0 atm, p(nh 3 ) =.0 atm, G < C and 750 atm gives near 100% yield. But there are safety cncerns f equipment running at high pressure. In practice, cmmercial prcess is carried ut arund 00 atm and 500 C t yield 10-0% t minimize n the cst and safety cncerns f the plant Nnspntaneus changes can be made t ccur if they are cupled with spntaneus changes. Spntaneus Irn rusts 4 Fe (s) + 3 O (g) Fe O 3 (s) G = kj The reverse f this reactin is NOT spntaneus. Fe O 3 (s) 4 Fe (s) + 3 O (g) G = kj 6 CO (g) + 3 O (g) 6 CO (g) G = kj Cmmercially, many tns f Fe O 3 are used t prduce irn. Hw d they d that? Gibbs energy allws us t make predictins based n the thermdynamic prperties f the reactants and prducts themselves, eliminating the need t carry ut the experiment. But while thermdynamics always crrectly predicts whether a given prcess is spntaneus, it is unable t tell whether it will take place at an bservable rate. Fe O 3 (s) + 6 CO (g) 4 Fe (s) + 6 CO (g) G = kj Spntaneus! Irn re (primarily rust) is reduced t irn metal in a blast furnace by a reactin with carbn mnxide, r carbn. 1 4

5 G = - nfe G = - nfe G is the free energy that is available t d useful wrk such as electrchemical wrk. G = - nfe An electrchemical cell is able d wrk if its cell ptential, E, is psitive because G < 0. where n is the number f mles f electrns that flw in the cell, F is Faraday s cnstant, culmbs / mle f electrns, E is the standard vltage measured f the electrchemical cell. Hw much useful wrk can be btained frm a Zn and Cu battery perating at standard cnditins when 0.15 g Zn react? , Zinc and Cpper vltaic cell E =1.10 V Therefre, 490 J f useful wrk can be btained 3 4 G = - nfe Hydrgen fuel cell Use slar energy t break apart water t generate H and O as fuel fr the fuel cell. The spntaneity f a prcess that is carried ut under cnstant T and P can be predicted by G G = H - T S G < 0 is a spntaneus prcess. G = 0 is when the prcess is in equilibrium. G > 0 is a nnspntaneus prcess. E =1.3 V Ande: H (g) 4H + (g) + 4 e - Cathde: O (g) + 4 H + (g) + 4 e - H O (l) H (g) + O (g) H O (l) E = 1.3 V 5 G = - RT ln K eq G = - nfe 6 5