Chapter 20 in Silberberg. 1,2,12,15,19,25,34,36,46,51,55,57,62,69,83,86,90. Chem 45.5 Firday, October 16, 930:11:30AM

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1 Your Projects ---Make sure you have all your materials or next week s lab ---It s basically a live exam that you are responsible or. Think Apprentice or Next Top Designer ---I your group is ound to be unprepared your group will be asked to leave the lab and a grade o either F will have been earned. ---No excuses so please make sure all gears are untioning beore you walk into the lab. Chapter 20 in Silberberg. 1,2,12,15,19,25,34,36,46,51,55,57,62,69,83,86,90 Chem 45.5 Firday, October 16, 930:11:30AM Why Does Copyright The This McGraw-Hill Happen Companies, Inc. Permission With required Most or reproduction Mixtures? or display.

2 Standard Molar Entropy, S rxn: The standard molar entropy o a chemical reaction can be determined rom the standard molar entropies o reactants and products. aa + bb cc + dd ΔS 0 rxn =? ΔS 0 rxn = [ cs (C) + ds (D) ] - [ as (A) + bs (B)] Ssys = ni Si (products) nisi (reactants) S or each component is ound in a table. Also note that S = 0 or elements (unlike H).

3 Because its diicult to measure ΔS univ, ΔS sys and ΔS surr we recast the 2nd Law equation into something more useul called the Gibb s Free Energy, ΔG. ΔS univ = ΔS sys + ΔS surr 1. substitute ΔS surr = -q sys /T = -ΔH sys /T ΔS univ = ΔS sys - ΔH sys /T 2. multiply through by (-T) -TΔS univ = - TΔS sys + ΔH sys 3. Deine ΔG universe = -TΔS univ ΔG universe = ΔH sys - TΔS sys Criteria or spontaneity!

4 The Gibb s Free Energy, ΔG universe is the unction that indicates whether a reaction will occur spontaneously. The sign o ΔG determines the reaction direction, the magnitude determines the maximum amount o work the system can do (i negative). Temperature in Kelvin ΔGuniverse = ΔHsys - T ΔSsys Gibb s Free Energy Change Enthalpy term Useless or dispersed energy ΔGuniverse = ΔHsys - T } ΔSsys ΔGsys = ΔHsys - T ΔSsys ΔGrxn = ΔHrxn - T ΔSrxn Don t be conused by symbols. These all mean the same thing and pertain to the system!

5 There are 3 Cases or the Sign o ΔG sys : 1. ΔG sys < 0 (ΔS univ > 0) or or a spontaneous process 2. ΔG sys > 0 (ΔS univ < 0) or a non-spontaneous process 3. ΔG sys = 0 (ΔS univ = 0) or or a process at equilibrium -ΔG < 0 is the maximum useul work that can be produced by a chemical reaction. ΔG = work max - For ΔG > 0 (non-spontaneous process) ΔG is the minimum work that must be done on the system to make the process take place. What is Free Energy physically? ΔGuniverse = ΔHsys - T ΔSsys ree energy Total energy available as heat energy dispersed or spread as it occupies more microstates

6 There are 3 Cases or the Sign o ΔG sys : 1. ΔG sys < 0 (ΔS univ > 0) or or a spontaneous process 2. ΔG sys > 0 (ΔS univ < 0) or a non-spontaneous process 3. ΔG sys = 0 (ΔS univ = 0) or or a process at equilibrium Any process where the system looses energy ΔE, and its entropy alls by ΔS, a quantity at least TR ΔS o that energy must be given up to the system's surroundings as unusable heat (TR is the temperature o the system's external surroundings). Otherwise the process will not go orward.

7 Using Hess s Law and thermodynamic data, we can compute ΔG o rxn under standard state conditions in two dierent ways as well. ΔGsys 1 2 Use Table o ΔH o and ΔS o Use Table o ΔG o ΔG o rxn = ΔH o rxn - TΔS o rxn ΔH o rxn = Σ miδh o i (products) - Σ niδh o i(reactants) ΔS o rxn = Σ mis o i (products) - Σ nis o i(reactants) ΔG rxn = [ cδg (C) dδg (D) + ] - [ aδg (A) + bδg (B) ]

8 Chemists have a deined reerence reaction condition called the standard state (below). We denote this standard state symbolically with a superscript degree sign above the thermodynamic unctions: ΔH, ΔG, ΔS Standand State Reaction Conditions Standard state ΔH o rxn ΔG o rxn Non-standard ΔHrxn ΔGrxn H = 0 1M concentration Think About It: It s great to have thermodynamic data or standard conditions but what about non-standard conditions that exist in the real-world?

9 ΔH rxn or any reaction can be calculated using ΔH data rom thermodynamic tables. aa + bb cc + dd ΔH rxn =? H rxn = all products n i H i all reactants n i H i i=1 i=1 When applied the equation above translates to this: H rxn = [c H C + d H D] [b H B + a H A]

10 Recall: The Standard Enthalpy o Formation, ΔH, is the enthalpy change associated with the ormation o 1 mole o compound rom its naturally occurring elements under standard state conditions or both. Examples: Note: 1 mol product! H 2 (g) + O 2 (g) H 2 O(l ) ΔH = kj/mol Ag(s) + 1/2Cl 2 (g) AgCl(s) ΔH = kj/mol Elements as they exist in elemental orm Oxygen exists as O 2 gas at 25 C ΔH = 0 Carbon exists as solid graphite (C) at 25 C. ΔH = 0 Sulur exits as S8 as a solid at 25 C Note: ractional coeicients allowed Water is H 2 O(l ) in its standard state (not ice or water vapor). We can measure ΔH or all compounds and put them into a table.

11 Thermodynamic Tables Substance ΔH ΔG S ΔH and ΔG have units o kj/ mol (energy/mol) S has units o J/mol K

12 Energy level diagrams are useul pictorials o the thermodynamic transitions that take place during a chemical reaction. Elements Enthalpy, H Reactants decomposition -ΔH 0 ormation ΔH 0 Products ΔH 0 rxn ΔH rxn = Σ mδh (products) - Σ nδh (reactants)

13 It is also not possible to measure the absolute enthalpy. We deine an arbitrary scale with the standard enthalpy o ormation (ΔH ) as a reerence point. 1) The standard enthalpy o ormation, ΔH o any element in its most stable orm is zero. ΔH 0 (O 2 ) = 0 ΔH 0 (O 3 ) = 142 kj/mol ΔH 0 (C, graphite) = 0 ΔH 0 (C, diamond) = 1.90 kj/mol ΔH 0 (S8, rhombic) = 0 kj/mol 2) Computing ΔHrxn works because it is a state unction. We only care about the initial and inal states under deined set o reaction conditions.

14 Calculating ΔH rom tabulated data Calculate the ΔH comb or the combustion o 2 moles o C 6 H 6 (l ) rom enthalpies o ormation data table. ΔH Data From Table In Back o the Textbook (or Handbook o Physics and Chemistry C(s) + O 2 (g) CO 2 (g) ΔH = H 2 (g) + 1/2O 2 (g) H 2 O(l ) 6C(s) + 3H 2 (g) C 6 H 6 (l ) ΔH = kj ΔH = kj

15 Calculating ΔH rom tabulated data 6C(s) + 6O 2 (g) 6CO 2 (g) 3H 2 (g) + O 2 (g) 3H 2 O(l ) C 6 H 6 (l ) 6C(s) + 3H 2 (g) ΔH = 6( kj) ΔH = 3( kj) ΔH = ( kj) C 6 H 6 (l ) + O 2 (g) 6CO 2 (g) + 3H 2 O(l) ΔH comb =? ΔH comb = [6( kj) + 3( kj)] - [( kj)] = -3,267.4 kj ΔH rxn = Σ n i ΔHi (products) - Σ mi ΔHj (reactants) that s or one mole o benzene only! 2 C 6 H 6 (l ) +15 O 2 (g) 12 CO 2 (g) + 6 H 2 O(l ) ΔH comb = 2 X kj/mol

16 Calculate the heat o decomposition or the ollowing process using standard enthalpies o ormation ound in Appendix. What kind o reaction is this? Is energy given o or needed to make this reaction proceed? CaCO 3 (s) CaO (s) + CO 2 (g) ΔH rxn = Σ ni ΔHi (products) - Σ mi ΔHi (reactants)

17 Calculate the heat o decomposition or the ollowing process using standard enthalpies o ormation ound in Appendix. What kind o reaction is this? Is energy given o or needed to make this reaction proceed? CaCO 3 (s) CaO (s) + CO 2 (g) ΔH rxn = Σ ni ΔHi (products) - Σ mi ΔHi (reactants) ΔH 0 rxn = [ ΔH 0 (CaO) + ΔH 0 (CO 2 )] - [ ΔH 0 (CaCO 3 )] ΔH 0 rxn = [ ] [ ] = 179 kj Reaction is decomposition and endothermic. Energy must be absorbed to proceed.

18 Using Hess s Law ormation data we can easily compute ΔG o rxn under standard state conditions in two dierent ways as well. ΔGsys 1 2 Use Table o ΔH o and ΔS o Use Table o ΔG o ΔG o rxn = ΔH o rxn - TΔS o rxn ΔH o rxn = Σ miδh o i (products) - Σ niδh o i(reactants) ΔS o rxn = Σ mis o i (products) - Σ nis o i(reactants) ΔG rxn = [ cδg (C) dδg (D) + ] - [ aδg (A) + bδg (B) ]

19 Recall ΔH o and ΔG o o any element in its most stable orm is deined as zero. ΔH 0 (O 2 ) = 0 ΔH 0 (O 3 ) = 142 kj/mol ΔH 0 (N 2 ) = 0 ΔH 0 (C, graphite) = 0 ΔH 0 (C, diamond) = 1.90 kj/mol ΔH 0 (S8, rhombic) = 0 kj/mol ΔH 0 (Na(s) = 0 Both ΔH o and ΔG o o many compounds are tabulated in handbooks. This gives us predictive capability and is someone useul in the real world.

20 Thermodynamic Tables Substance ΔH ΔG S ΔH and ΔG have units o kj/ mol (energy/mol) S has units o J/mol K

21 ΔGrxn, the standard ree energy o a reaction can be computed directly rom its ormation data, or rom ΔH o and S o ormation data, ΔG aa + bb cc + dd ΔG rxn =? ΔG rxn = Σ ni ΔGi (products) - Σ mi ΔGj (reactants) = [cδg (C) + dδg (D)] [aδg (A) + bδg (B)] OR ΔG o rxn = ΔH o rxn - TΔS o rxn ΔH o rxn = Σ miδh o i (products) - Σ niδh o i(reactants) = [cδh (C) + dδh (D)] [aδh (A) + bδh (B)] ΔS o rxn = Σ mis o i (products) - Σ nis o i(reactants) = [cs (C) + ds (D)] [as (A) + bs (B)]

22 Calculating ΔG o rxn rom ΔG o values Use ΔG o values to calculate ΔG o rxn or the ollowing reaction: 4KClO 3 (s) Δ 3KClO 4 (s) + KCl(s) PLAN: Look up the values o products and reactants in a thermodynamic table (appendix o your book!) Use the ΔG o rxn summation equation! ΔG o KClO 3 (s) KClO 4 (s) + KCl(s) kj/mol kj/mol kj/mol

23 Calculating ΔG o rxn rom ΔG o values Use ΔG o values to calculate ΔG o rxn or the ollowing reaction: 4KClO 3 (s) Δ 3KClO 4 (s) + KCl(s) PLAN: Use the ΔG o rxn summation equation and look up the values o products and reactants in a thermodynamic table. ΔG o 4KClO 3 (s) Δ 3KClO 4 (s) + KCl(s) kj/mol kj/mol kj/mol ΔG o rxn ΔG o rxn = Σ miδg o i (products) - Σ niδg o i (reactants) = [(3 mol)( kj/mol) + (1 mol)( kj/mol)] - [(4 mol)( kj/mol)] ΔG o rxn = -134 kj (reaction is spontaneous as written)

24 What is the standard ree-energy change or the ollowing reaction at 25 0 C? 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) ΔG o rxn = Σ miδg o i (products) - Σ niδg o i (reactants) ΔGrxn 0 = [ 12ΔG 0 (CO 6ΔG 0 2 ) + (H 2 O) ] - [ 2ΔG 0 (C 6 H 6 )] Is the reaction spontaneous at 25 0 C and what does the ree energy tell us? ΔG 0 rxn = [ 12(-394.4) + 6(-237.2) ] [ 2 x ] = kj ΔG 0 = kj < 0 and thereore a spontaneous reaction It tells us that under standard state conditions the combustion o propane is product avored and that -6405kJ is the maxium work that could be done.

25 What is the standard ree-energy change or the ollowing reaction at 25 0 C? 2C 6 H 6 (l) + 15O 2 (g) 12CO 2 (g) + 6H 2 O (l) ΔG o rxn = Σ miδg o i (products) - Σ niδg o i (reactants) ΔG 0 ] rxn = [ 12ΔG 0 (CO 6ΔG 0 2 ) + (H 2 O) ] - [ 2ΔG 0 (C 6 H 6 ) Is the reaction spontaneous at 25 0 C and what does the ree energy tell us? ΔG 0 rxn = [ 12(-394.4) + 6(-237.2) ] [ 2 x ] = kj ΔG 0 = kj < 0 and thereore a spontaneous reaction It tells us that under standard state conditions the combustion o propane is product avored and that -6405kJ is the maxium work that could be done.

26 Using thermodynamic tables o data we can easily compute ΔG o rxn under standard state conditions in two dierent ways. ΔGsys 1 2 Use Table o ΔH o and ΔS o Use Table o ΔG o ΔG o rxn = ΔH o rxn - TΔS o rxn ΔH o rxn = Σ miδh o i (products) - Σ niδh o i(reactants) ΔS o rxn = Σ mis o i (products) - Σ nis o i(reactants) ΔG rxn = [ cδg (C) dδg (D) + ] - [ aδg (A) + bδg (B) ]

27 The standard ree-energy (ΔG 0 rxn) is the ree-energy change or the complete reaction o reactants to products with both reactants and products under standard-state conditions. aa + bb cc + dd ΔG rxn =? G rxn = H rxn T S rxn H rxn = [d H D + c H C] [b H B + a H A] S rxn = [ds D + cs C] [bs B + as A] ΔG 0 values are a special hypothetical case o 100% conversion. We are usually more interested in real world non-standard conditions and equilibrium values or ΔG.

28 Calculating ΔG o rom standard enthalpy and entropy values Potassium chlorate, one o the common oxidizing agents in explosives, ireworks and match heads, undergoes a solidstate redox reaction when heated. In this reaction, the oxidation number o Cl in the reactant is higher in one o the products and lower in the other (a disproportionation reaction). +5 4KClO 3 (s) Δ KClO 4 (s) + KCl(s) Use ΔH o and So values to calculate ΔG o sys (ΔGo rxn ) at 25 o C or this reaction. PLAN: Obtain appropriate thermodynamic data rom a data table; insert them into the Gibbs ree energy equation and solve. ΔH rxn ΔH o rxn = Σ ni ΔHi (products) - Σ mi ΔHj (reactants) ΔH = [cδh (C) + dδh (D)] [aδh (A) + bδh (B)] = [(3 mol)( kj/mol) + (1 mol)( kj/mol)] - [(4 mol)( kj/mol)] ΔH o rxn = -144 kj

29 ΔS o rxn = Σ mis o i (products) - Σ nis o i(reactants) = [cs (C) + ds (D)] [as (A) + bs (B)] ΔS o rxn = [(3 mol)(151 J/mol. K) + (1 mol)(82.6 J/mol. K)] - [(4 mol)(143.1 J/mol. K)] ΔS o rxn = J/K ΔH o rxn = -144 kj ΔG o rxn = ΔH o rxn - TΔS o rxn = -144 kj - (298 K)(-36.8 J/K)(kJ/10 3 J) = -133 kj The reaction is spontaneous or the reaction goes to proceeds to the right.

30 ΔGrxn is the ree energy unction used or nonstandard thermodynamic reaction conditions It connects the ree energy to the equilibrium reaction quotient, Q. It can be shown that: G = G + RT ln Q Non-standard Free Energy Standard Free Energy Reaction Quotient R is the gas constant (8.314 J/K mol) T is the absolute temperature (K)

31 Recall our expressions or the equilibrium constants (all o them Kc, Kp, Ka, Ksp) aa(g) + bb(g) cc(g) + dd(g) Concentrations or partial pressures at point beore equilibrium Q = [C]c [D] d [A] a [B] b Q = (P c) c (Pd) d (Pa) a (Pb) b Concentrations or partial pressures at equilibrium Kc = [C]c [D] d [A] a [B] b Kp = (P c) c (Pd) d (Pa) a (Pb) b

32 Q is related to non-standard state ΔGrxn and like Q tell us the direction o a chemical reaction! aa(g) + bb(g) cc(g) + dd(g) Q < K Q = Kc Q > K Product-Favored ΔG sys < 0 Equilibrium Reactant-Favored ΔG sys = 0 ΔG sys > 0

33 G rxn o a chemical reaction and the equilibrium constant K are related. aa + bb cc + dd Kc = [C]c [D] d [A] a [B] b Grxn = G rxn + RT ln Q R = J/K mol T in Kelvin 0 = G + RT ln Kc G rxn = - RT ln K K = exp (- G rxn/rt) At equilibrium: G = 0 and Q = Kc Solving or G rxn R = J/K mol T in Kelvin Taking antiloge or exp o both sides R = J/K mol T in Kelvin

34 ΔG o and Kc at 25 o C are inversely related...big K means smaller G and vis versa. ΔG o (kj) K Signiicance x x x x x x x x x Essentially no orward reaction; reverse reaction goes to completion Forward and reverse reactions proceed to the same extent Forward reaction goes to completion; essentially no reverse reaction FORWARD REACTION REVERSE REACTION

35 Free energy diagrams show how the ree energy changes (drive a reaction) towards equilibrium Q = K, ΔG o =0. ΔG o (reactants) Total ree energy ΔG o Q < K Spontaneous Direction Spontaneous Direction Q = K Q > K ΔG o (rxn) (rxn) ΔG o (products) 100% Reactants Equilibrium Mixture Reaction Progress ===> 100% Products

36 Calculating ΔG at non-standard conditions The oxidation o SO 2 to SO3 is too slow at 298 K to be useul in the manuacture o suluric acid. To overcome this low rate, the process is conducted at an elevated temperature. 2SO 2 (g) + O 2 (g) 2SO 3 (g) (a) Calculate K at 298 K and at 973 K. (ΔG o 298 = kj/mol or the reaction as written using ΔH o and ΔS o values. At 973 K, ΔG o 973 = kj/mol or the reaction as written.) (b) In experiments to determine the eect o temperature on reaction spontaneity, two sealed containers are illed with atm o SO 2, atm o O 2, and atm o SO 3 and kept at 25 o C and at 700. o C. In which direction, i any, will the reaction proceed to reach equilibrium at each temperature? (c) Calculate ΔG or the system in part (b) at each temperature.

37 SOLUTION: (a) Calculating K ΔG o 298 and ΔG o 973 rom at the two temperatures: ΔG o = -RT ln K thus: At 298 K: ΔG o /RT = - ( kj/mol)(103 J/kJ) (8.314 J/mol. K)(298 K) = 57.2 = e 57.2 = 7 x ( kj/mol)(10 At 973 K: -ΔG o /RT = - 3 J/kJ) = 1.50 (8.314 J/mol. K)(973 K) = e 1.50 = 4.5

38 (b) Q = pso 3 2 (pso 2 ) 2 (po 2 ) = (0.100) 2 (0.500) 2 (0.0100) = 4.00 Since Q is < K at both temperatures the reaction equilibrium will shit right and be product avored; or 298 K there will be a dramatic shit while at 973 K the shit will be slight. (c) The non-standard ΔG is calculated using: ΔG = ΔG o + RT ln Q ΔG 298 = kj/mol + (8.314 J/mol. K)(kJ/10 3 J)(298 K)(ln 4.00) ΔG 298 = kj/mol ΔG 973 = kj/mol + (8.314 J/mol. K)(kJ/10 3 J)(973 K)(ln 4.00) ΔG 298 = -0.9 kj/mol

39 Summary o Key Points: 1. ΔG o tells us the maximum amount o work a spontaneous reaction can do (when negative) ΔG o sys = ΔHo sys - TΔSo sys 2. Q and ΔGrxn are linked Q = Kc, ΔGrxn = 0 and ΔSuniverse = 0 ====> Equilibrium. Q < Kc, ΔGrxn < 0 and ΔSuniverse > 0 ====> Product-Favored Q > Kc, ΔGrxn > 0 and ΔSuniverse < 0 ====> Reactant-Favored 3. ΔG o rxn gives the position o equilibrium and can be calculated three ways: ΔG o rxn = Σ miδg o i (products) - Σ niδg o i (reactants) ΔG o rxn = ΔH o rxn - TΔS o rxn ΔG o rxn = -RT ln K 4. ΔG rxn describes the direction in which a non-standard reaction proceeds to reach equilibrium calculated by: ΔGrxn = ΔG o + RT ln Q

40 Plots o ree energy diagrams or ΔG o > 0 (let) and ΔG o < 0 (right). Total ree energy ΔG Q < K 100% Reactants Reactant Favored ΔG o reactant ΔG o (rxn) > 0 Q = K Equilibrium Mixture ΔG o product Q > K Reaction Progress ===> 100% Products Total ree energy ΔG Q < K Pure Reactants Product Favored ΔG o reactant ΔG o (rxn)< 0 Q = K Equilibrium Mixture ΔG o product Q > K Reaction Progress ===> Pure Products ΔG o (rxn) > 0, K<1 ΔG o (rxn)<0, K>1

41 Temperature plays an important role in determining whether a reaction is spontaneous. It determines the spontaneity in 2 o 4 cases. ΔG o sys = ΔHo sys - TΔSo sys Case ΔH o rxn ΔS o sys TΔS o ΔG o Process Description spontaneous at all T nonspontaneous at all T spontaneous at high T; nonspontaneous at low T spontaneous at low T; nonspontaneous at high T

42 When ΔH o rxn and ΔSo rxn have the same sign, then the reaction temperature will determine spontaneity. We can compute the cross-over temperature that makes the reaction spontaneous Cross-over logic: G o sys > 0 = not spontaneous AND Go sys < 0 = spontaneous. We can set G o sys = 0 to ind that temperature where G changes over. The approach assumes that H and S are constant over temperature. ΔG o rxn = ΔHo rxn - TΔSo rxn 0 = ΔH o rxn - TΔSo rxn At cross-over T ΔG o rxn = 0 T = ΔH o rxn ΔS o rxn Solve or T Watch your units! S is usually J and H is usually kj

43 Consider the reaction o copper(i) oxide with carbon with the ollowing thermodynamic data.. kj/mol kj/mol J/mol K Cu2O(s) + C(s) ===> 2Cu(s) + CO(g) ΔH ΔG S By inspection only o the chemical equation determine i entropy o the system increases or decreases? 2. Calculate ΔH rxn 3. Calculate ΔS rxn 4. Calculate ΔG rxn, using ΔG rxn = ΔH rxn TΔS rxn 5. Caculate ΔG rxn using the thermodynamic ΔG data 6. Determine i the reaction is spontaneous under standard state conditions (298K, 1M, 1atm)? 7. What impact would temperature have on the spontaneity o the reaction assuming constant enthalpy and entropy vs T? 8. Calculate the cross-over temperature which must be exceeded to make this reaction spontaneous. 9. Determine Kp or this reaction.

44 kj/mol kj/mol J/mol K Cu2O(s) + C(s) ===> 2Cu(s) + CO(g) ΔH ΔG S The entropy o the system increases (>0) as we move rom solid reactants to gaseous products. 2. ΔH rxn = [1ΔH (CO(g)) + 2ΔH (Cu(s))] [1ΔH (Cu2O(s) ) + 1ΔH (C(s))] = [1(-110.5) + 2(0)] [1(-168.6) + 1(0)] = 58.1 kj 3. ΔS rxn = [1S (CO(g)) + 2S (Cu(s))] [1S (Cu2O(s) ) + 1S (C(s))] = [1(197.5) + 2(33.1)] [1(93.1) + 1(5.68)] = J/K 4. ΔG o rxn = ΔHo rxn - =.1649 kj/ K TΔS rxn = (58.1kJ) - (298K) ( kj/k) = +8.9 kj

45 kj/mol kj/mol J/mol K Cu2O(s) + C(s) ===> 2Cu(s) + CO(g) ΔH ΔG S ΔG rxn = [1ΔG (CO(g)) + 2ΔG (Cu(s))] [1ΔG (Cu2O(s) ) + 1ΔG (C(s))] = [1(-137.2) + 2(0)] [1(-146.0) + 1(0)] = 8.8kJ 6. ΔG o rxn > 0. Thereore the reaction is not spontaneous under standard state conditions at 298K. 7. ΔH o rxn > 0 and ΔS rxn > 0. Analysis o the Gibb s equation: ΔG o rxn = ΔHo rxn - TΔSo rxn will show that increasing T will increase TΔS rxn and drive ΔG rxn <0 as ΔH o rxn remains unchanged (which is assumed).

46 8. The cross-over temperature that must be exceeded is ound by setting ΔG o rxn = 0. 0 = ΔH o rxn - TΔSo rxn T = ΔH o rxn ΔS o rxn Watch the units o J/K and kj! T = 58.1 kj.1649 kj/k > 352.K 9. K298 = exp (- G rxn/rt) kj 1000J/1kJ J/mol K (298K) = 2.9 x 10-2 = exp ( ) This temperature has to exceeded or the reaction to be spontaneous!

47 Determining the eect o temperature on ΔG o An important reaction in the production o suluric acid is the oxidation o SO 2 (g) to SO 3 (g): 2SO 2 (g) + O 2 (g) 2SO 3 (g) ΔG o rxn = kj, ΔH o rxn = kj, and ΔS o rxn = J/K. The BOOK OMITS THE ΔG o rxn sign! (a) Is this reaction spontaneous at 25 o C, and predict how ΔG o will change with increasing T. (b) Assuming ΔH o and ΔS o are constant with increasing T, is the reaction spontaneous at 900. o C? PLAN: The sign o ΔG o tells us whether the reaction is spontaneous and the signs o ΔH o and ΔS o will be indicative o the T eect. Analyze the Gibbs ree equation to see the aect o temperature.

48 SOLUTION: The Given Data:At 298 K, ΔG o = kj, ΔH o = kj, and ΔS o = J/K. (a) ΔG o can be ound two ways, rom the data ΔG o = kj or via the Gibb s equation: ΔG o rxn = ΔH o rxn - TΔS o rxn ΔG o rxn = kj - [(298 K)( J/mol. K)(kJ/10 3 J)] = kj The reaction is spontaneous at 25 o C because ΔG o rxn < 0! Since ΔH o is (-) and ΔS o are both (-), ΔG o will become less negative and thereore the reaction less spontaneous as temperature increases. (b) ΔG o rxn = ΔH o rxn - TΔS o rxn ΔG o rxn = kj - [(1173 K)( J/mol. K)(kJ/10 3 J)] = 22 kj ΔG o rxn = kj; the reaction will not be spontaneous at 900. o C

49 Temperature and Spontaneity o Chemical Reactions CaCO 3 (s) CaO (s) + CO 2 (g) G (kj/mol) H (kj/mol) S (J/K) mol ΔH rxn = kj At 25 0 C ΔS rxn = J/K ΔG rxn = ΔH TΔS ΔG rxn = K (.1605) ΔG 0 = kj not spontaneous! What temperature can we make this reaction go?

50 Coupled Reactions In order to drive a non-spontaneous reactions we changed the conditions (i.e. temperature or electrolysis) Another method is to couple two reactions. One with a positive ΔG and one with a negative ΔG. Overall spontaneous process. Prentice-Hall 2002 General Chemistry: Chapter 20 Slide o 37

51 Smelting Copper Ore Cu 2 O(s) Δ 2 Cu(s) + ½ O 2 (g) ΔG 673K = +125 kj Non-spontaneous reaction: Cu 2 O(s) 2 Cu(s) + ½ O 2 (g) +125 kj Spontaneous reaction: C(s) + ½ O 2 (g) CO(g) -175 kj Cu 2 O(s) + C(s) 2 Cu(s) + CO(g) Spontaneous reaction! -50 kj Prentice-Hall 2002 General Chemistry: Chapter 20 Slide o 37