Chapter 17 Free Energy and Thermodynamics

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1 Chemistry: A Mlecular Apprach, 1 st Ed. Nivald Tr Chapter 17 Free Energy and Thermdynamics Ry Kennedy Massachusetts Bay Cmmunity Cllege Wellesley Hills, MA 2008, Prentice Hall

2 First Law f Thermdynamics yu can t win! First Law f Thermdynamics: Energy cannt be Created r Destryed the ttal energy f the universe cannt change thugh yu can transfer it frm ne place t anther E universe = 0 = E system + E surrundings Tr, Chemistry: A Mlecular Apprach 2

3 First Law f Thermdynamics Cnservatin f Energy Fr an exthermic reactin, lst heat frm the system ges int the surrundings tw ways energy lst frm a system, cnverted t heat, q used t d wrk, w Energy cnservatin requires that the energy change in the system equal the heat released + wrk dne E = q + w E = H + P V E is a state functin internal energy change independent f hw dne Tr, Chemistry: A Mlecular Apprach 3

4 Energy Tax yu can t break even! t recharge a battery with 100 kj f useful energy will require mre than 100 kj every energy transitin results in a lss f energy cnversin f energy t heat which is lst by heating up the surrundings Tr, Chemistry: A Mlecular Apprach 4

5 Heat Tax fewer steps generally results in a lwer ttal heat tax Tr, Chemistry: A Mlecular Apprach 5

6 Thermdynamics and Spntaneity thermdynamics predicts whether a prcess will prceed under the given cnditins spntaneus prcess nnspntaneus prcesses require energy input t g spntaneity is determined by cmparing the free energy f the system befre the reactin with the free energy f the system after reactin. if the system after reactin has less free energy than befre the reactin, the reactin is thermdynamically favrable. spntaneity fast r slw Tr, Chemistry: A Mlecular Apprach 6

7 Cmparing Ptential Energy The directin f spntaneity can be determined by cmparing the ptential energy f the system at the start and the end. Tr, Chemistry: A Mlecular Apprach 7

8 Reversibility f Prcess any spntaneus prcess is irreversible it will prceed in nly ne directin a reversible prcess will prceed back and frth between the tw end cnditins equilibrium results in n change in free energy if a prcess is spntaneus in ne directin, it must be nnspntaneus in the ppsite directin Tr, Chemistry: A Mlecular Apprach 8

9 Thermdynamics vs. Kinetics Tr, Chemistry: A Mlecular Apprach 9

10 Diamnd Graphite Graphite is mre stable than diamnd, s the cnversin f diamnd int graphite is spntaneus but dn t wrry, it s s slw that yur ring wn t turn int pencil lead in yur lifetime (r thrugh many f yur generatins). Tr, Chemistry: A Mlecular Apprach 10

11 Factrs Affecting Whether a Reactin Is Spntaneus The tw factrs that determine the thermdynamic favrability are the enthalpy and the entrpy. The enthalpy is a cmparisn f the bnd energy f the reactants t the prducts. bnd energy = amunt needed t break a bnd. H The entrpy factrs relates t the randmness/rderliness f a system S The enthalpy factr is generally mre imprtant than the entrpy factr Tr, Chemistry: A Mlecular Apprach 11

12 Enthalpy related t the internal energy H generally kj/ml strnger bnds = mre stable mlecules if prducts mre stable than reactants, energy released exthermic H = negative if reactants mre stable than prducts, energy absrbed endthermic H = psitive The enthalpy is favrable fr exthermic reactins and unfavrable fr endthermic reactins. Hess Law H rxn = Σ( H prd ) - Σ( H react ) Tr, Chemistry: A Mlecular Apprach 12

13 Substance H kj/ml Substance H kj/ml Al(s) 0 Al 2 O Br 2 (l) 0 Br 2 (g) C(diamnd) C(graphite) 0 CO(g) CO 2 (g) Ca(s) 0 CaO(s) Cu(s) 0 CuO(s) Fe(s) 0 Fe 2 O 3 (s) H 2 (g) 0 H 2 O 2 (l) H 2 O(g) H 2 O(l) HF(g) HCl(g) HBr(g) HI(g) I 2 (s) 0 I 2 (g) N 2 (g) 0 NH 3 (g) NO(g) NO 2 (g) Na(s) 0 O 2 (g) 0 S(s) 0 SO 2 (g)

14 Entrpy entrpy is a thermdynamic functin that increases as the number f energetically equivalent ways f arranging the cmpnents increases, S S generally J/ml S = k ln W k = Bltzmann Cnstant = 1.38 x J/K W is the number f energetically equivalent ways, unitless Randm systems require less energy than rdered systems Tr, Chemistry: A Mlecular Apprach 14

15 W Energetically Equivalent States fr the Expansin f a Gas Tr, Chemistry: A Mlecular Apprach 15

16 Macrstates Micrstates These micrstates all have the same macrstate S there are 6 different This particle macrstate can be achieved thrugh arrangements several different that arrangements f the particles result in the same macrstate Tr, Chemistry: A Mlecular Apprach 16

17 Macrstates and Prbability There is nly ne pssible arrangement that gives State A and ne that gives State C There are 6 pssible arrangements that give State B Therefre State B has higher entrpy than either State A r State B The macrstate with the highest entrpy als has the greatest dispersal f energy Tr, Chemistry: A Mlecular Apprach 17

18 Changes in Entrpy, S entrpy change is favrable when the result is a mre randm system. S is psitive Sme changes that increase the entrpy are: reactins whse prducts are in a mre disrdered state. (slid > liquid > gas) reactins which have larger numbers f prduct mlecules than reactant mlecules. increase in temperature slids dissciating int ins upn disslving Tr, Chemistry: A Mlecular Apprach 18

19 Increases in Entrpy Tr, Chemistry: A Mlecular Apprach 19

20 The 2 nd Law f Thermdynamics the ttal entrpy change f the universe must be psitive fr a prcess t be spntaneus fr reversible prcess S univ = 0, fr irreversible (spntaneus) prcess S univ > 0 S universe = S system + S surrundings if the entrpy f the system decreases, then the entrpy f the surrundings must increase by a larger amunt when S system is negative, S surrundings is psitive the increase in S surrundings ften cmes frm the heat released in an exthermic reactin Tr, Chemistry: A Mlecular Apprach 20

21 Entrpy Change in State Change when materials change state, the number f macrstates it can have changes as well fr entrpy: slid < liquid < gas because the degrees f freedm f mtin increases slid liquid gas Tr, Chemistry: A Mlecular Apprach 21

22 Entrpy Change and State Change Tr, Chemistry: A Mlecular Apprach 22

23 Heat Flw, Entrpy, and the 2 nd Law Heat must flw frm water t ice in rder fr the entrpy f the universe t increase Tr, Chemistry: A Mlecular Apprach 23

24 Temperature Dependence f S surrundings when a system prcess is exthermic, it adds heat t the surrundings, increasing the entrpy f the surrundings when a system prcess is endthermic, it takes heat frm the surrundings, decreasing the entrpy f the surrundings the amunt the entrpy f the surrundings changes depends n the temperature it is at riginally the higher the riginal temperature, the less effect additin r remval f heat has S surrundings H Tr, Chemistry: A Mlecular Apprach 24 = T system

25 Gibbs Free Energy, G maximum amunt f energy frm the system available t d wrk n the surrundings G = H T S G sys = H sys T S sys G sys = T S universe G reactin = Σ n G prd Σ n G react when G < 0, there is a decrease in free energy f the system that is released int the surrundings; therefre a prcess will be spntaneus when G is negative Tr, Chemistry: A Mlecular Apprach 25

26 Ex. 17.2a The reactin C 3 H 8(g) + 5 O 2(g) 3 CO 2(g) + 4 H 2 O (g) has H rxn = kj at 25 C. Calculate the entrpy change f the surrundings. Given: Find: Cncept Plan: Relatinships: H system = kj, T = 298 K S surrundings, J/K T, H S surr = H T sys S Slutin: S S surr surr = = H T 6.86 sys kj K ( 2044 kj) = 298 K 3 = J K Check: cmbustin is largely exthermic, s the entrpy f the surrunding shuld increase significantly

27 Free Energy Change and Spntaneity Tr, Chemistry: A Mlecular Apprach 27

28 Gibbs Free Energy, G prcess will be spntaneus when G is negative G will be negative when H is negative and S is psitive exthermic and mre randm H is negative and large and S is negative but small H is psitive but small and S is psitive and large r high temperature G will be psitive when H is + and S is never spntaneus at any temperature when G = 0 the reactin is at equilibrium Tr, Chemistry: A Mlecular Apprach 28

29 G, H, and S Tr, Chemistry: A Mlecular Apprach 29

30 Ex. 17.3a The reactin CCl 4(g) C (s, graphite) + 2 Cl 2(g) has H = kj and S = J/K at 25 C. Calculate G and determine if it is spntaneus. Given: H = kj, S = J/K, T = 298 K Find: G, kj Cncept Plan: T, H, S G Relatinships: G = H T S Slutin: Answer: G = H T S = ( 3 ) ( )( J J 298 K ) = J Since G is +, the reactin is nt spntaneus at this temperature. T make it spntaneus, we need t increase the temperature. K

31 Ex. 17.3a The reactin CCl 4(g) C (s, graphite) + 2 Cl 2(g) has H = kj and S = J/K. Calculate the minimum temperature it will be spntaneus. Given: H = kj, S = J/K, G < 0 Find: Τ, Κ Cncept Plan: G, H, S T Relatinships: Slutin: Answer: ( H T S) G = H T S G = < 0 ( ) ( J ) 3 ( )( J J T ) < 0 ( J K ) K 3 ( ) ( )( J J < T ) The temperature must be higher than 673K fr the reactin t be spntaneus K 673 K < T < T

32 The 3 rd Law f Thermdynamics Abslute Entrpy the abslute entrpy f a substance is the amunt f energy it has due t dispersin f energy thrugh its particles the 3 rd Law states that fr a perfect crystal at abslute zer, the abslute entrpy = 0 J/ml K therefre, every substance that is nt a perfect crystal at abslute zer has sme energy frm entrpy therefre, the abslute entrpy f substances is always + Tr, Chemistry: A Mlecular Apprach 32

33 Standard Entrpies S extensive entrpies fr 1 mle at 298 K fr a particular state, a particular alltrpe, particular mlecular cmplexity, a particular mlar mass, and a particular degree f disslutin Tr, Chemistry: A Mlecular Apprach 33

34 Substance S J/ml-K Substance S J/ml-K Al(s) 28.3 Al 2 O 3 (s) Br 2 (l) Br 2 (g) C(diamnd) 2.43 C(graphite) 5.69 CO(g) CO 2 (g) Ca(s) 41.4 CaO(s) Cu(s) CuO(s) Fe(s) Fe 2 O 3 (s) H 2 (g) H 2 O 2 (l) H 2 O(g) H 2 O(l) HF(g) HCl(g) HBr(g) HI(g) I 2 (s) I 2 (g) N 2 (g) NH 3 (g) NO(g) NO 2 (g) Na(s) O 2 (g) S(s) SO 2 (g) 248.5

35 Relative Standard Entrpies States the gas state has a larger entrpy than the liquid state at a particular temperature the liquid state has a larger entrpy than the slid state at a particular temperature Substance H 2 O (g) H 2 O (l) S, (J/ml K) Tr, Chemistry: A Mlecular Apprach 35

36 Relative Standard Entrpies Mlar Mass the larger the mlar mass, the larger the entrpy available energy states mre clsely spaced, allwing mre dispersal f energy thrugh the states Tr, Chemistry: A Mlecular Apprach 36

37 Relative Standard Entrpies the less cnstrained the structure f an alltrpe is, the larger its entrpy Alltrpes Tr, Chemistry: A Mlecular Apprach 37

38 Relative Standard Entrpies Mlecular Cmplexity larger, mre cmplex mlecules generally have larger entrpy Substance Ar (g) Mlar Mass S, (J/ml K) mre available energy states, allwing mre dispersal f energy thrugh the states NO (g) Tr, Chemistry: A Mlecular Apprach 38

39 Relative Standard Entrpies Disslutin disslved slids generally have larger entrpy distributing particles thrughut the mixture Substance KClO 3 (s) KClO 3 (aq) S, (J/ml K) Tr, Chemistry: A Mlecular Apprach 39

40 Ex Calculate S fr the reactin 4 NH 3(g) + 5 O 2(g) 4 NO (g) + 6 H 2 O (l) Given: Find: Cncept Plan: Relatinships: Slutin: Check: standard entrpies frm Appendix IIB S, J/K S NH3, S O2, S NO, S H2O, S = = [4(S = ( ) ( S = Σn S ) p prducts Σn r S reactants ( ) ( Σ S ) prducts Σ S reactants NO( g) = [4(210.8 n p n r J K J K ) + 6(S H 2 ) + 6(188.8 O( g) J K )] [4(S Substance NH 3 (g) O 2 (g) NO(g) H 2 O(g) NH )] [4(192.8 S S is +, as yu wuld expect fr a reactin with mre gas prduct mlecules than reactant mlecules 3 ( g) J K ) + 5(S O 2 ) + 5(205.2 S, J/ml K ( g) J K )] )]

41 Calculating G at 25 C: G reactin = ΣnG f (prducts) - ΣnG f (reactants) at temperatures ther than 25 C: assuming the change in H reactin and S reactin is negligible G reactin = H reactin T S reactin Tr, Chemistry: A Mlecular Apprach 41

42 Substance G f kj/ml Substance G f kj/ml Al(s) 0 Al 2 O Br 2 (l) 0 Br 2 (g) C(diamnd) C(graphite) 0 CO(g) CO 2 (g) Ca(s) 0 CaO(s) Cu(s) 0 CuO(s) Fe(s) 0 Fe 2 O 3 (s) H 2 (g) 0 H 2 O 2 (l) H 2 O(g) H 2 O(l) HF(g) HCl(g) HBr(g) HI(g) I 2 (s) 0 I 2 (g) N 2 (g) 0 NH 3 (g) NO(g) NO 2 (g) Na(s) 0 O 2 (g) 0 S(s) 0 SO 2 (g)

43 Ex Calculate G at 25 C fr the reactin CH 4(g) + 8 O 2(g) CO 2(g) + 2 H 2 O (g) + 4 O 3(g) Given: Find: Cncept Plan: Relatinships: Slutin: G = = [( G = kj standard free energies f frmatin frm Appendix IIB G, kj G f f prd & react G = ( ) ( Σn G ) prducts Σn G reactants ( ) ( Σn G G ) p prducts Σn reactants f CO ) + = [( kj) + f 2( G f r p ) + ( G f f f r )] [( G Substance CH 4 (g) O 2 (g) CO 2 (g) H 2 O(g) O 3 (g) f G ) + 8( G 2 H2O O3 CH4 O2 2( kj) + ( kj)] [( 50.5 kj) + 8(0.0 kj)] f G f, kj/ml f )]

44 Ex The reactin SO 2(g) + ½ O 2(g) SO 3(g) has H = kj and S = J/K at 25 C. Calculate G at 125 C and determine if it is spntaneus. Given: H = kj, S = J/K, T = 398 K Find: G, kj Cncept Plan: T, H, S G Relatinships: G = H T S Slutin: Answer: G = = H ( ) ( )( ) 3 J J 398 K 94.0 = T S 3 J = 61.5 kj Since G is -, the reactin is spntaneus at this temperature, thugh less s than at 25 C K

45 G Relatinships if a reactin can be expressed as a series f reactins, the sum f the G values f the individual reactin is the G f the ttal reactin G is a state functin if a reactin is reversed, the sign f its G value reverses if the amunts f materials is multiplied by a factr, the value f the G is multiplied by the same factr the value f G f a reactin is extensive Tr, Chemistry: A Mlecular Apprach 45

46 Free Energy and Reversible Reactins the change in free energy is a theretical limit as t the amunt f wrk that can be dne if the reactin achieves its theretical limit, it is a reversible reactin Tr, Chemistry: A Mlecular Apprach 46

47 Real Reactins in a real reactin, sme f the free energy is lst as heat if nt mst therefre, real reactins are irreversible Tr, Chemistry: A Mlecular Apprach 47

48 G under Nnstandard Cnditins G = G nly when the reactants and prducts are in their standard states there nrmal state at that temperature partial pressure f gas = 1 atm cncentratin = 1 M under nnstandard cnditins, G = G + RTlnQ Q is the reactin qutient at equilibrium G = 0 G = RTlnK Tr, Chemistry: A Mlecular Apprach 48

49 Tr, Chemistry: A Mlecular Apprach 49

50 Example - G Calculate G at 427 C fr the reactin belw if the P N2 = 33.0 atm, P H2 = 99.0 atm, and P NH3 = 2.0 atm N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Q = P NH3 2 P N21 x P H2 3 (2.0 atm) 2 = (33.0 atm) 1 (99.0) 3 = 1.2 x 10-7 H = [ 2(-46.19)] - [0 +3( 0)] = kj = J S = [2 (192.5)] - [(191.50) + 3(130.58)] = J/K G = J - (700 K)( J/K) G = J G = G + RTlnQ G = J + (8.314 J/K)(700 K)(ln 1.2 x 10-7 ) G = J = -46 kj 50

51 Tr, Chemistry: A Mlecular Apprach 51

52 Example - K Estimate the equilibrium cnstant and psitin f equilibrium fr the fllwing reactin at 427 C N 2 (g) + 3 H 2 (g) 2 NH 3 (g) H = [ 2(-46.19)] - [0 +3( 0)] = kj = J S = [2 (192.5)] - [(191.50) + 3(130.58)] = J/K G = J - (700 K)( J/K) G = J G = -RT lnk J = -(8.314 J/K)(700 K) lnk lnk = K = e = 3.45 x 10-4 since K is << 1, the psitin f equilibrium favrs reactants 52

53 Temperature Dependence f K fr an exthermic reactin, increasing the temperature decreases the value f the equilibrium cnstant fr an endthermic reactin, increasing the temperature increases the value f the equilibrium cnstant H ln K = R rxn 1 T + S R rxn Tr, Chemistry: A Mlecular Apprach 53

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