# Thermodynamics 1/16/2013. Thermodynamics

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1 Thermdynamics 1 Thermdynamics Study f energy changes and flw f energy Answers several fundamental questins: Is it pssible fr reactin t ccur? Will reactin ccur spntaneusly (withut utside interference) at given T? Will reactin release r absrb heat? Tells us nthing abut time frame f reactin Kinetics Tw majr cnsideratins must be balanced Enthalpy changes, H(heats f reactin) Heat exchange between system and surrundings Nature's trend t randmness r disrder 2 Examples f Spntaneus Reactins 3 1

2 Cnsider this Reactin 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) Cncerning this reactin: 1. Des this reactin naturally ccur as written? 2. Will the reactin mixture cntain sufficient amunt f prduct at equilibrium? 4 We can answer these questins with heat measurements nly!!! 1. We can predict the natural directin. 2. We can determine the cmpsitin f the mixture at equilibrium. Hw? 5 Cnsider the Laws f Thermdynamics 1 st Law: The change in internal energy f a system U, equals q + w E = U = q + w Yu can t win 2 nd Law: The ttal entrpy f a system and its surrundings increases fr a spntaneus prcess. Yu cant break even. 3 rd Law: A substance that is perfectly crystalline at 0 K has an entrpy f zer. Yu can t quit the game 6 2

3 Review f First Law f Thermdynamics Internal energy, E r U System's ttal energy Sum f KEand PEf all particles in system E = ( KE ) + ( PE ) system system E = E final E initial r fr chemical reactin E = E prducts E reactants E+ energy int system E energy ut f system system 7 Tw Methds f Energy Exchange Between System and Surrundings Heat q = H Wrk w = P V U = E = q + w Cnventins f heat and wrk q + Heat absrbed by system E system q Heat released by system E system w + Wrk dne n system E system w Wrk dne by system E system 8 U(Internal Energy) is a state functin: Mst ften we are interested in the change: Internal Energy = Ε = U = U f - U i q = Energy that mves in and ut f a system w = Frce x distance = F x d = P V 9 3

4 First Law f Thermdynamics Energy can neither be created nr destryed It can nly be cnverted frm ne frm t anther KE PE Chemical Electrical Electrical Mechanical E and E are state functins Path independent E = - U = q + w E = ( KE ) + ( PE ) system system system 10 Wrk in Chemical Systems 1. Electrical 2. Pressure-vlume r P V w = P V Where P = external pressure If P V nly wrk in chemical system, then E = q + ( P V ) = q P V 11 Heat vs. Wrk 12 4

5 Shwing wrk is P V w = - F x h = - F x V/A = -F/A x V = -P V 13 Heat at Cnstant Vlume Reactin dne at cnstant V V= 0 P V= 0, s E= q V Entire Echange due t heat absrbed r lst 14 Heat at Cnstant Pressure Mre cmmn Reactins pen t atmsphere Cnstant P Enthalpy H = E + PV Enthalpy change H = E + P V Substituting in first law fr Egives H = (q P V) + P V = q P H = q P Heat f reactin at cnstant pressure 15 5

6 P V = -92 J q = 165 J U = q + w = (+165) + (-92) = +73 J 16 Heat f Reactin and Internal Energy Zn (s) + HCl (l) ZnCl 2 (aq) + H 2 (g) 17 Here the system expands and evlves heat frm A t B. Zn 2+ (aq) + 2Cl - (aq) + H 2 (g) V is psitive, s wrk is negative. 18 6

7 q = - w = - Zn (s) + HCl (l) ZnCl 2 (aq) + H 2 (g) w = -P V = -(1.01 x 10 5 pa)(24.5 l) = x 10 5 pa)(24.5 x 10-3 m 3 ) = x 10 3 J = kj q = -152 kj U = -152 kj + (-2.47 kj) = kj 20 Diagram and explain the change in internal energy fr the fllwing reactin. CH 4 (g) + 2 O 2 (g) CO 2 (g) + 2 H 2 O (l) q = kj w U = P V = +(1.01 x 10 5 pa)(24.5 l)(2) = +(1.01 x 10 5 pa)(24.5 x 10-3 m 3 )(2) = kj = kj + (+4.95 kj) = kj 21 7

8 Enthalpy and Enthalpy Change Enthalpy is defined as q p Enthalpy = H = U + PV All are state functins. H = H f - H i H = (U f + PV f ) (U i + PV i ) = (U f U i ) + P(V f V i ) U = q p - P V H = (q p - P V) + P V = q p H f = Σn H f (prducts) - Σm H f (reactants) H f = Standard enthalpy change (25 C) 22 H f = Σn H f (prducts) - Σm H f (reactants) Calculate H f fr the reactin in slide 15 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) H f = fr NH 3 (g) = kj = fr CO 2 (g) = kj = fr NH 2 CONH 2 = kj = fr H 2 O (l) = kj H = [( ) (-2 x )] kj = Since H has a negative sign, heat is evlved 23 Cnverting Between Eand H Fr Chemical Reactins H E Differ by H E = P V Only differ significantly when gases frmed r cnsumed Assume gases are ideal nrt V = nrt V = P P Since P and T are cnstant RT V = n P 24 8

9 Cnverting Between Eand H Fr Chemical Reactins When reactin ccurs Vcaused by nf gas Nt all reactants and prducts are gases S redefine as n gas Where n gas = (n gas ) prducts (n gas ) reactants Substituting int H = E + P V gives RT H = E + P ngas r P H = E + ngasrt 25 Ex. 1. What is the difference between H and E fr the fllwing reactin at 25 C? 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) What is the % difference between H and E? Step 1: Calculate H using data (Table 7.2) Recall H = ( H f ) prducts ( H f ) reactants H = 4 Hf (NO2) + Hf (O2) 2 Hf (N2O5 ) H = (4 ml)(33.8 kj/ml) + (1 ml)(0.0 kj/ml) (2 ml)(11 kj/ml) H = 113 kj 26 Step 2: Calculate n gas = (n gas ) prducts (n gas ) reactants n gas = ( ) ml = 3 ml Step 3: Calculate E using R = J/K ml Ex. 1. (cnt.) T = 298 K E = 113 kj (3 ml)( J/K ml)(298 K)(1 kj/1000 J) E = 113 kj 7.43 kj = 106 kj 27 9

10 Ex. 1. finish Step 4: Calculate % difference 7. 43kJ % difference = 100% = 6. 6% 113kJ Bigger than mst, but still small Nte: Assumes that vlumes f slids and liquids are negligible V slid V liquid << V gas 28 Is Assumptin that V slid V liquid << V gas Justified? Cnsider CaCO 3 (s)+ 2H + (aq) Ca 2+ (aq)+ H 2 O(l) + CO 2 (g) 37.0 ml 2*18.0 ml 18 ml 18 ml 24.4 L Vlumes assuming each cefficient equal number f mles S V= V prd V reac = L 24.4 L Yes, assumptin is justified Nte: If N gasesare present reduces t E H 29 Learning Check Cnsider the fllwing reactin fr picric acid: 8O 2 (g) + 2C 6 H 2 (NO 2 ) 3 OH(l) 3 N 2 (g) + 12CO 2 (g) + 6H 2 O(l) What type f reactin is it? Calculate ΔΗ, ΔΕ 8O 2 (g) + 2C 6 H 2 (NO 2 ) 3 OH(l) 3N 2 (g) + 12CO 2 (g) + 6H 2 O(l) ΔΗ f (kj/ml) ΔH 0 = 12ml( kj/ml) + 6ml( kJ/ml) + 6ml(0.00kJ/ml) 8ml(0.00kJ/ml) 2ml( kJ/ml) ΔH 0 = 13,898.9 kj ΔΕ = ΔH Δn gas RT = ΔH (15 8)ml*298K* kj/(ml K) Ε = 13,898.9 kj 29.0 kj = 13,927.9 kj 30 10

11 Yur Turn! Given the fllwing: 3H 2 (g) + N 2 (g) 2NH 3 (g) H = kj ml -1 Determine E fr the reactin. A kj ml -1 B kj ml -1 C kj ml -1 D kj ml -1 H = E + nrt E = H - nrt E = kj ml ( NRT) -(-2 ml)(8.314 J K -1 ml -1 )(298K)(1 kj/1000 J) E = kj ml H f = Σn H f (prducts) - Σm H f (reactants) Calculate H f fr the reactin in slide 15 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) H f = fr NH 3 (g) = kj = fr CO 2 (g) = kj = fr NH 2 CONH 2 = kj = fr H 2 O (l) = kj H = [( ) (-2 x )] kj = Since H has a negative sign, heat is evlved 32 Spntaneity and Entrpy Definitin f Spntaneus Prcess: Physical r chemical prcess that ccurs by itself. Why?? Give several spntaneus prcesses: Still cannt predict spntaneity

12 Enthalpy Changes and Spntaneity What are relatinships amng factrs that influence spntaneity? Spntaneus Change Occurs by itself Withut utside assistance until finished Ex. Water flwing ver waterfall Melting f ice cubes in glass n warm day 34 Nnspntaneus Change Occurs nly with utside assistance Never ccurs by itself: Rm gets straightened up Pile f bricks turns int a brick wall Decmpsitin f H 2 O by electrlysis Cntinues nly as lng as utside assistance ccurs: Persn des wrk t clean up rm Bricklayer layers mrtar and bricks Electric current passed thrugh H 2 O 35 Entrpy and the 2 nd Law f Thermdynamics 2 nd Law: The ttal entrpy f a system and its surrundings increases fr a spntaneus prcess. Entrpy = S = A thermdynamic quantity that is a measure f hw dispersed the energy is amng the different pssible ways that a system can cntain energy. Cnsider: 1. A ht cup f cffee n the table 2. Rck rlling dwn the side f a hill. 3. Gas expanding. 4. Stretching a rubber band

13 Reactin Rate and Spntaneity H indicates if reactin has tendency t ccur Rate f reactin als plays rle Sme very rapid: Neurns firing in nerves in respnse t pain Detnatin f stick f dynamite Sme gradual: Ersin f stne Ice melting Irn rusting Sme s slw, appear t be nnspntaneus: Gasline and O 2 at RT Many bichemical prcesses Flask cnnected t an evacuated flask by a valve r stpcck S = S f - S i H 2 O (s) H 2 O (l) S = (63 41) J/K = 22 J/K 38 Directin f Spntaneus Change Many reactins which ccur spntaneusly are exthermic: Irn rusting Fuel burning Hand Eare negative Heat given ff Energy leaving system Early thughts were that exthermic reactins were spntaneus) Thus, H is ne factr that influences spntaneity 39 13

14 Directin f Spntaneus Change Sme endthermicreactins ccur spntaneusly: Ice melting Evapratin f water frm lake Expansin f CO 2 gas int vacuum Hand Eare psitive Heat absrbed Energy entering system Clearly ther factrs influence spntaneity 40 Cncept Check: Yu have a sample f slid idine at rm temperature. Later yu ntice that the idine has sublimed. What can yu say abut the entrpy change f the idine? 41 There are hw many Laws f thermdynamics? a. 1 b. 2 c. 3 d. 4 e

15 2 nd Law: The ttal entrpy f a system and its surrundings increases fr a spntaneus prcess. Prcess ccurs naturally as a result f energy dispersal In the system. S = entrpy created + q/t S > q/t Fr a spntaneus prcess at a given temperature, the change in entrpy f the system is greater than the heat divided by the abslute temperature. 43 Entrpy and Mlecular Disrder 44 Thermdynamic vs. Kinetics Thermdynamics tells us: Directin f reactin Is it pssible fr reactin t ccur? Will reactin ccur spntaneusly at given T? Will reactin release r absrb heat? Kinetics tells us: Speed f reactin Pathway between reactants and prducts Energy Reactants Dmain f Thermdynamics (initial and final states) Dmain f Kinetics (reactin pathway) Reactin Prgress Prducts 45 15

16 Heat Transfer Between Ht and Cld Objects Cnsider system f tw bjects Initially ne ht and ne cld Ht = higher KE ave f mlecules = faster Cld = lwer KE ave f mlecules = slwer When they cllide, what is mst likely t ccur? Faster bjects bump int clder bjects and transfer energy s Ht bjects cl dwn and slw dwn Cld bjects warm up and speed up The reverse desn t ccur Heat Transfer Between Ht and Cld Objects Result: Heat flws spntaneusly frm ht t clder bject Heat flws because f prbable utcme f intermlecular cllisins Spntaneus prcesses tend t prceed frm states f lw prbability t states f higher prbability Spntaneus prcesses tend t disperse energy Entrpy (symbl S) Thermdynamic quantity Describes number f equivalent ways that energy can be distributed Quantity that describes randmness f system Greater statistical prbability f particular state means greater the entrpy! Larger S, means mre randm and mre prbable 48 16

17 Entrpy If Energy = mney Entrpy (S) describes number f different ways f cunting it Criterin fr Spntaneity Clear rder t things Things get rusty spntaneusly Dn't get shiny again Sugar disslves in cffee Stir mre it desn't undisslve Ice liquid water at RT Oppsite des NOT ccur Fire burns wd, smke ges up chimney Can't regenerate wd Cmmn factr in all f these: Increase in randmness and disrder f system Smething that brings abut randmness mre likely t ccur than smething that brings rder Entrpy, S Measure f randmness and disrder Measure f chas State functin Independent f path S = Change in Entrpy Als state functin S = S final S initial Fr chemical reactin S = Sprducts S reactants 51 17

18 Entrpy S prducts > S reactants means S+ Entrpy 's Prbability f state 's Randmness 's Favrs spntaneity S prducts < S reactants means S Entrpy 's Prbability f state 's Randmness 's Des nt favr spntaneity Any reactin that ccurs with in entrpy tends t ccur spntaneusly 52 Effect f Vlume n Entrpy Fr gases, Entrpy as Vlume A. Gas separated frm vacuum by partitin B. Partitin remved C. Gas expands t achieve mre prbable particle distributin Mre randm, higher prbability, mre psitive S 53 Effect f Temperature n Entrpy As T, entrpy A. T = 0 K, particles ( ) in equilibrium lattice psitins and S relatively lw B. T > 0 K, mlecules vibrate, S C. T further, mre vilent vibratins ccur and S higher than in B 54 18

19 Effect f Physical State n Entrpy Crystalline slid very lw S Liquid higher S, mlecules can mve freely Mre ways t distribute KE amng them Gas highest S, particles randmly distributed thrughut cntainer Many, many ways t distribute KE 55 Entrpy Affected by Number f Particles Adding particles t system number f ways energy can be distributed in system S all ther things being equal Reactin that prduces mre particles will have psitive S 56 Summary Larger V, greater S Expansin f gas S + Higher T, greater S Higher T, means mre KE in particles, mve mre, s randm distributins favred S slid < S liquid << S gas Slids mre rdered than liquids, which are much mre rdered than gases Reactins invlving gases Simply calculate change in number f mle gas, n gas If n gas +, S + If n gas, S 57 19

20 Entrpy Change fr a Reactin S may be psitive fr reactins with the fllwing: 1. The reactin is ne in which a mlecule is brken int tw r mre smaller mlecules. 2. The reactin is ne in which there is an increase in mles f gas. 3. The prcess is ne in which a slid changes t a liquid r a liquid changes t a gas. Did Yu Get It! Which represents an increase in entrpy? A. water vapr cndensing t liquid B. carbn dixide subliming C. liquefying helium gas D. prteins frming frm amin acids 59 Entrpy Changes in Chemical Reactins Ex. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) n reactant = 4 n prduct = 2 n = 2 4 = 2 Predict S rxn < 0 Higher psitinal prbability Lwer psitinal prbability 20

21 Entrpy Changes in Chemical Reactins Reactins withut gases Simply calculate number f mle mlecules n = n prducts n reactants If n +, S + If n, S Mre mlecules, means mre disrder Usually the side with mre mlecules, has less cmplex mlecules Smaller, fewer atms per mlecule Ex. 2 Predict Sign f S fr Fllwing Reactins CaCO 3 (s) + 2H + (aq) Ca 2+ (aq) + H 2 O(l) + n gas = 1 ml 0 ml = 1 ml n gas +, S + 2 N 2 O 5 (g) 4 NO 2 (g) + O 2 (g) n gas = 4 ml + 1 ml 2 ml = 3 ml n gas +, S + OH (aq) + H + (aq) H 2 O (l) n gas = 0 ml n = 1 ml 2 ml = 1 ml n, S CO 2 (g) Predict Sign f S in Fllwing: Dry ice carbn dixide gas CO 2 (s) CO 2 (g) psitive Misture cndenses n a cl windw H 2 O(g) H 2 O(l) negative AB A + B psitive A drp f fd clring added t a glass f water disperses psitive 2Al(s) + 3Br 2 (l) 2AlBr 3 (s) negative 21

22 Did Yu Get It? Which f the fllwing has the mst entrpy at standard cnditins? A. H 2 O(l) B. NaCl(aq) C. AlCl 3 (s) D. Can t tell frm the infrmatin Which reactin wuld have a negative entrpy? A. Ag + (aq) + Cl - (aq) AgCl(s) B. N 2 O 4 (g) 2NO 2 (g) C. C 8 H 18 (l ) + 25/2 O 2 (g) 8CO 2 (g) + 9H 2 O(g) D. CaCO 3 (s) CaO(s) + CO 2 (g) 64 Bth Entrpy and Enthalpy Can Affect Reactin Spntaneity Smetimes they wrk tgether Building cllapses PE H Stnes disrdered S + Smetimes wrk against each ther Ice melting (ice/water mix) Endthermic H + nnspntaneus Disrder f mlecules S + spntaneus Which Prevails? Hard t tell depends n temperature! At 25 C, ice melts At 25 C, water freezes S three factrs affect spntaneity: H S T 66 22

23 Secnd Law f Thermdynamics When a spntaneus event ccurs, ttal entrpy f universe s ( S ttal > 0) In a spntaneus prcess, S system can decrease as lng as ttal entrpy f universe increases S ttal = S system + S surrundings It can be shwn that S surrundin gs = q surrundings T 2 nd Law: The ttal entrpy f a system and its surrundings increases fr a spntaneus prcess. Prcess ccurs naturally as a result f energy dispersal In the system. S = entrpy created + q/t S > q/t Fr a spntaneus prcess at a given temperature, the change in entrpy f the system is greater than the heat divided by the abslute temperature. 68 Des This Make Sense? Yes? As heat added Disrder r entrpy, S q But hw much T, S, depends n T at which it ccurs Lw T, larger S High T, smaller S S 1/T 23

24 What d yu think? When ice melts in yur hand (assume yur hand is 30 C), A. the entrpy change f the system is less then the entrpy change f the surrundings. B.the entrpy change f the surrundings is less than the entrpy change f the system. C.the entrpy change f the system equals the entrpy change f the surrundings. S sys = H/273K S surr = H/303K S sys > S surr Law f Cnservatin f Energy Says q lst by system must be gained by surrundings q surrundings = q system If system at cnstant P, then q system = H S q surrundings = H system and S surrundings q = surrundings T H = T system Thus Entrpy fr Entire Universe is S ttal = S system H T system Multiplying bth sides by Twe get T S ttal = T S system H system r T S ttal = ( H system T S system ) Fr reactin t be spntaneus T S ttal > 0 (+) S, ( H system T S system ) < 0 ( ) fr reactin t be spntaneus 24

25 Standard Entrpy (Abslute Entrpy): Entrpy value fr the standard state f a species. See Table B-13 p. B-17 and Table B-16 p. B-28 Entrpy values f substances must be psitive. S must be >0 but H can be plus r minus (Why?) Hw abut inic species? S fr H 3 O + is set at zer. 73 Predict The Entrpy sign fr the fllwing reactins: a. C 6 H 12 O 11 (s) 2CO 2 (g)+ C 2 H 5 OH (l) b. 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) c. CO (g) + H 2 O (g) CO 2 (g) + H 2 (g) d. Stretching a rubber band Exercise 13.2 p S Calculating S fr a reactin = Σn S (prducts) - Σm S (reactants) Calculate the entrpy change fr the fllwing reactin At 25 C. 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) S 2 x S = Σn S (prducts) - Σm S (reactants) S = [( ) - (2x )] J/K= -356 J/K See exercise 13.3 p 584 and prblems 8-12 and

26 Third Law f Thermdynamics At abslute zer (0 K), Entrpy f perfectly rdered, pure crystalline substance is zer S= 0 at T= 0 K Since S= 0 at T= 0 K Define abslute entrpy f substance at higher temperatures Standard entrpy, S Entrpy f 1 mle f substance at 298 K (25 C) and 1 atm pressure S = S fr warming substance frm 0 K t 298 K (25 C) 76 Cnsequences f Third Law 1. All substances have psitive entrpies as they are mre disrdered than at 0 K Heating randmness S is biggest fr gases mst disrdered 2. Fr elements in their standard states S 0 (but H f = 0) Units f S J/(ml K) Standard Entrpy Change T calculate S fr reactin, d Hess's Law type calculatin Use S rather than entrpies f frmatin S = n S (prducts) m S (reactants) 77 Learning Check Calculate S 0 fr the fllwing: CO 2 (s) CO 2 (g) S 0 (J/ml K) S 0 = ( ) J/ml K S 0 = 26.1 J/ml K CaCO 3 (s) CO 2 (g) + CaO(s) S 0 (J/ml K) S 0 = ( ) J/ml K S 0 = 161 J/ml K 78 26

27 Ex. 3.Calculate S fr reductin f aluminum xide by hydrgen gas Al 2 O 3 (s) + 3 H 2 (g) 2 Al (s) + 3 H 2 O (g) Substance S (J/ K ml) Al (s) 28.3 Al 2 O 3 (s) H 2 (g) H 2 O (g) S = nps prducts S = [2S Al ( s ) [ S Al O n + 3S H O ( 2 3 ( s ) 2 + 3S H r S reactants g ) ] 2 ( g ) ] 79 Ex. 3 S 28.3 J J = 2 ml * + 3 ml * ml K ml K J 130.6J 1 ml * + 3 ml * ml K ml K S = 56.5 J/K J/K J/K J/K S = J/K 80 Gibbs Free Energy Wuld like ne quantity that includes all three factrs that affect spntaneity f a reactin Define new state functin Gibbs Free Energy Maximum energy in reactin that is "free" r available t d useful wrk G H TS At cnstant P and T, changes in free energy G = H T S 81 27

28 G = H T S Gibbs Free Energy G < 0 Spntaneus prcess G = 0 At equilibrium G > 0 Nnspntaneus G state functin Made up f T, H and S = state functins Has units f energy Extensive prperty G = G final G initial 82 Criteria fr Spntaneity? At cnstant P and T, prcess spntaneus nly if it is accmpanied by in free energy f system H S Spntaneus? + G = ( ) [T(+)] = Always,regardless f T + G = (+) [T( )] = + Never,regardless f T + + G = (+) [T(+)] =? Depends; spntaneus at hight, G G = ( ) [T( )] =? Depends; spntaneus at lwt, G 83 Summary When Hand Shave same sign, Tdetermines whether spntaneus r nn-spntaneus Temperature-cntrlled reactins are spntaneus at ne temperature and nt at anther 84 28

29 G = H - T S 87 1/16/2013 Free Energy Cncept G = H T S H T S Can Serve as a criteria fr Spntaneity 2 NH 3 (g) + CO 2 (g) NH 2 CONH 2 (aq) + H 2 O (l) H = kj S = -365 J/K = kj/k H T S = ( kj) (298 K) x (-0.365kJ/K = kj H T S is a negativity quantity, frm which we can cnclude that the reactin is spntaneus under standard cnditins. 85 Free Energy and Spntaneity Free Energy: Thermdynamic quantity defined by the equatin G = H - TS G = H - T S If yu can shw that G fr a reactin at a given temperature and pressure is negative, yu can predict that the reactin will be spntaneus 86 Standard Free Energy Changes Standard Cnditins: 1 atm pressure 1 atm partial pressure 1 M cncentratin Temperature f 25 C r 298 K Standard free energy is free-energy change that takes place when reactants in their standard states are cnverted t prducts in their standard states. 29

30 Standard Free Energy Changes Standard Free Energy Change, G Gmeasured at 25 C (298 K) and 1 atm Tw ways t calculate, depending n what data is available Methd 1. G = H ( K) S Ex. 4. Calculate G fr reductin f aluminum xide by hydrgen gas Al 2 O 3 (s) + 3 H 2 (g) 2 Al(s) + 3H 2 O(g) 88 Ex. 4. Methd 1 Al 2 O 3 (s) + 3 H 2 (g) 2 Al (s) + 3 H 2 O (g) Step 1: Calculate H fr reactin using H f Substance H f (kj/ml) H = Al (s) 0.0 Al 2 O 3 (s) H 2 (g) 0.0 H 2 O (g) np Hf prducts H = [2 H [ H f Al( s ) f Al O ( s ) H f H O( H nr Hf reactants g ] ) f H ( 2 g ] ) 89 Ex. 4. Methd 1 Step 1( H ) kj kj H = 2 ml* + 3 ml* ml ml kj kj 1 ml* + 3 ml* ml ml H = 0.0 kj kj 0.00 kj ( kj) H = kj 90 30

31 Ex. 4 Methd 1 Step 2: Calculate S see Ex. 4 S = J/K Step 3: Calculate G = H ( K) S G = kj (298 K)(179.9 J/K)(1 kj/1000 J) G = kj 53.6 kj = kj G = + nt spntaneus 91 Methd 2 Use Standard Free Energies f Frmatin G f Energy t frm 1 mle f substance frm its elements in their standard states at 1 atm and 25 C G = np Gf prducts nr Gf reactants 92 Ex. 4. Methd 2 Calculate G fr reductin f aluminum xide by hydrgen gas. Al 2 O 3 (s)+ 3 H 2 (g) 2 Al (s)+ 3 H 2 O (g) Substance Al (s) 0.0 G f (kj/ml) Al 2 O 3 (s) H 2 (g) 0.0 H 2 O (g) G = [2 G [ G f Al( s ) f Al O ( s ) G + 3 G f H O( 2 f H ( 2 g ] ) g )] 93 31

32 Ex. 4. Methd kj kj G = 2 ml* + 3 ml* ml ml kj kj 1 ml* + 3 ml* ml ml G = 0.0 kj kj 0.00 kj ( kj) G = kj Bth methds same within experimental errr 94 G as a Criterin fr Spntaneity G = H - T S 1. When G is a large negative number (mre negative than abut 10 kj), the reactin is spntaneus as written, and reactants transfrm almst entirely int prducts when equilibrium is reached. 2. When G is a large psitive number (mre psitive than abut + 10 kj), the reactin is nt spntaneus as written, and reactants transfrm almst entirely int prducts when equilibrium is reached. 3. When G has a small psitive r negative value(less than abut 10 kj), the reactin mixture gives an equilibrium mixture with significant amunts f bth reactants and prducts.. 95 Interpreting the Sign f G Calculate H and G fr the fllwing reactin 2 KClO 3 (s) 2 KCl (s) + 3 O 2 (g) Interpret the signs f H and G H f are as fllws: KClO 3 (s) = kj/ml KCl (s) = kj/ml O 2 (g) = 0 G f are as fllws: KClO 3 (s) = kj/ml KCl (s) = kj/ml O 2 (g) =

33 2 KClO 3 (s) 2 KCl (s) + 3 O 2 (g) H f 2 x ( ) 2 x (-436.7) 0 kj G f 2 x (-296.3) 2 x (-408.8) 0 kj Then: H = [2 x (-436.7) 2 x (-397.7)] kj = -78 kj G = [2 x (-408.8) 2 x (-296.3)] kj = -225 kj The reactin is exthermic, liberating 78 kj f heat. The large negative value f G indicates that the equilibrium is mstly KCl and O Spntaneus Reactins Prduce Useful Wrk Fuels burned in engines t pwer cars r heavy machinery Chemical reactins in batteries Start cars Run cellular phnes, laptp cmputers, mp3 players Energy nt harnessed if reactin run in an pen dish All energy lst as heat t surrundings Engineers seek t capture energy t d wrk Maximize efficiency with which chemical energy is cnverted t wrk Minimize amunt f energy transfrmed t unprductive heat 98 Thermdynamically Reversible Prcess that can bereversed and is always very clse t equilibrium Change in quantities is infinitesimally small Example - expansin f gas Dne reversibly, it des mst wrk n surrundings 99 33

34 G = Maximum Pssible Wrk Gis maximum amunt f energy prduced during a reactin that can theretically be harnessed as wrk Amunt f wrk if reactin dne under reversible cnditins Energy that need nt be lst t surrundings as heat Energy that is free r available t d wrk 100 Ex. 5 CalculateΔG fr reactin belw at 1 atm and 25 C, given ΔH = 246.1kJ/ml, ΔS = /(ml K). H 2 C 2 O 4 (s) + ½ O 2 (g) 2CO 2 (g) + H 2 O(l) G 25 = H T S G = kj ml ΔG =( )kJ/ml ΔG = 358.5kJ/ml J 1J (298K ) K ml 1000kJ 101 Yu Try! Calculate G fr the fllwing reactin, H 2 O 2 (l ) H 2 O(l ) + O 2 (g) given H = kj ml -1 and S = J K -1 ml -1. A kj ml -1 B kj ml -1 C kj ml -1 D. 3.7 x 10 5 kj ml -1 G = kj ml K ( kj K -1 ml -1 ) G = kj ml

35 System at Equilibrium Neither spntaneus nr nnspntaneus In state f dynamic equilibrium G prducts = G reactants G= 0 Cnsider freezing f water at 0 C H 2 O(l) H 2 O(s) System remains at equilibrium as lng as n heat added r remved Bth phases can exist tgether indefinitely Belw 0 C, G < 0freezing spntaneus Abve0 C, G> 0 freezing nnspntaneus Define equilibrium 103 Ex. 6 CalculateT bp fr reactin belw at 1 atm and 25 C, given ΔH = 31.0kJ/ml, ΔS = 92.9 J/(ml K) Br 2 (l) Br 2 (g) H 31.0kJ / ml T bp = 334K S kJ /( ml K ) Fr T > 334 K, ΔG < 0 and reactin is spntaneus (ΔS dminates) Fr T < 334 K, ΔG > 0 and reactin is nnspntaneus (ΔH dminates) Fr T = 334 K, ΔG = 0 and T = nrmal biling pint 104 Free energy change during reactin

36 Free energy change during reactin 106 Free Energy and Equilibrium Cnstant Very imprtant relatin is the relatin between free energy and the equilibrium cnstant. Thermdynamic Equilibrium Cnstant- the equilibrium cnstant in which the cncentratin f gases are expressed in partial pressures in atmspheres, whereas the cncentratin f slutes in liquid are expressed in mlarities. K = Kc fr reactins invlving nly liquid slutins K = Kp fr reactins invlving nly gases 107 Equilibrium Expressin aa + bb cc + dd Kc = [C] c [D] d [A] a [B] b Slids and liquids cnsidered unity (1)

37 Write Kp and Kc Cnsider again N 2 (g) + 3H 2 (g) 2 NH 3 (g) K c = [NH 3 ] 2 [N 2 ][H 2 ] In terms f partial pressures, N 2 (g) + 3H 2 (g) 2 NH 3 (g) K p = [P NH3 ] 2 [P N2 ][P H2 ] K c and K p are related. PV = nrt P = [n/v]rt K p =K c (RT) n n = (number f mles f prduct gas) (number f mles f reactant gas) Fr this reactin, n = -2. N 2 (g) + 3H 2 (g) 2 NH 3 (g)

38 N Wrk Dne at Equilibrium G = 0 N free energy available t d wrk Cnsider fully charged battery Initially All reactants, n prducts G large and negative Lts f energy available t d wrk As battery discharges Reactants cnverted t prducts G less negative Less energy available t d wrk At Equilibrium G = G prducts G reactants = 0 N further wrk can be dne Dead battery 112 Phase Change = Equilibrium H 2 O(l) H 2 O(g) G= 0 = H T S Only ne temperature pssible fr phase change at equilibrium Slid-liquid equilibrium Melting/freezing temperature (pint) Liquid-vapr equilibrium Biling temperature (pint) H Thus H = T S and S = T r H T = S

39 G and Psitin f Equilibrium When G > 0 (psitive) Psitin f equilibrium lies clse t reactants Little reactin ccurs by the time equilibrium is reached Reactin appears nnspntaneus When G < 0 (negative) Psitin f equilibrium lies clse t prducts Mainly prducts exist by the time equilibrium is reached Reactin appears spntaneus 115 G and Psitin f Equilibrium When G = 0 Psitin f equilibrium lies ~ halfway between prducts and reactants Significant amunt f bth reactants and prducts present at time equilibrium is reached Reactin appears spntaneus, whether start with reactants r prducts Can Use G t Determine Reactin Outcme G large and psitive N bservable reactin ccurs G large and negative Reactin ges t cmpletin 116 Learning Check Ex. 7 Given that ΔH = 97.6 kj/ml, ΔS = 122 J/(ml K), at 1atm and 298K, will the fllwing reactin ccur spntaneusly? MgO(s)+ 2HCl(g) H 2 O(l)+ MgCl 2 (s) G = H T S = 97.6kJ/ml 298K( kj/ml K) G = 97.6kJ/ml +36.4kJ/ml = 61.2 kj/ml

40 Effect f Temperature n G Reactins ften run at T s ther that 298 K Psitin f equilibrium can change as G depends n T G = H T S Fr T s near 298 K, expect nly very small changes in H and S Fr reactin at T,we can write: T T T G = H T S G T H 298 T S Ex. 8 Determining Effect f T n Spntaneity Calculate G at 25 C and 500 C fr the Haber prcess N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Assume that H and S d nt change with T Slving Strategy Step 1.Using data frm Tables, calculate H and S fr the reactin at 25 C H = kj S = J/K 119 Ex. 8 Determining Effect f T n Spntaneity Step 2.Calculate G fr the reactin at 25 C using H and S N 2 (g) + 3 H 2 (g) 2 NH 3 (g) H = kj 1kJ S = J/K 1000J G = H T S G = kj (298 K)( J/K) G = kj kj = 33.3 kj S the reactin is spntaneus at 25 C

41 Ex. 8 Determining Effect f T n Spntaneity Step 3.Calculate G fr the reactin at 500 C using H and S. T = 500 C = 773 K 1kJ H = kj 1000J S = J/K G = H T S G = kj (773 K)( J/K) G = kj+ 153 kj = 61 kj S the reactin is NOT spntaneus at 500 C 121 Ex. 8 Des this answer make sense? G = H T S H = kj S = J/K Since bth H and S are negative At lw T G will be negative and spntaneus At high T T S will becme a bigger psitive number and G will becme mre psitive and thus eventually, at high enugh T, will becme nnspntaneus 122 Effect f Change in Pressure r Cncentratin n G Gat nnstandard cnditins is related t G at standard cnditins by an expressin that includes reactin qutient Q G = G + RT lnq This imprtant expressin allws fr any cncentratin r pressure Recall: y [ prducts] Q = x [ reactants]

42 Ex. 9 Calculating G at Nnstandard cnditins Calculate Gat 298 K fr the Haber prcess N 2 (g) + 3 H 2 (g) 2 NH 3 (g) G = 33.3 kj Fr a reactin mixture that cnsists f 1.0 atm N 2, 3.0 atm H 2 and 0.5 atm NH 3 Step 1 Calculate Q Q = P 2 PNH 3 N PH (0.50) = (1.0)(3.0) 3 = Ex. 9 Calculating G at Nnstandard cnditins Step 2Calculate G = G + RT lnq G = 33.3 kj/ml + (8.314J/K ml)*(1kj/1000j)* (298K)*ln( ) = 33.3 kj/ml + (2.479 kj/ml)*ln( ) = 33.3 kj + ( 11.6 kj/ml) = 44.9 kj/ml At standard cnditins all gases (N 2, H 2 and NH 3 ) are at 1 atm f pressure Gbecmes mre negative when we g t 1.0 atm N 2, 3.0 atm H 2 and 0.5 atm NH 3 Indicates larger driving frce t frm NH 3 P reactants > P prducts 125 Hw K is related t G Use relatin G = G + RT lnq t derive relatinship between K and G At Equilibrium G = 0 and Q = K S 0= G + RT lnk G = RTlnK Taking antilg (e x ) f bth sides gives K= e G /RT

43 At Equilibrium G = RTlnK and K= e G /RT Prvides cnnectin between G and K Can estimate K s at varius T s if knw G Can get G in knw K s Relatinship between K and G K eq G Reactin > 1 Spntaneus Favred Energy released < 1 + nn-spntaneus Unfavrable Energy needed = 1 0 At Equilibrium 127 Ex. 10 Calculating G frm K K sp fr AgCl(s) at 25 C is Determine G fr the prcess Ag + (aq) + Cl (aq) AgCl (s) Reverse f K sp equatin, s K = 1 K sp 1 = = G = RT lnk = (8.3145J/K ml)(298k)* ln( )*(1kJ/1000J) G = 56 kj/ml Negative G indicates precipitatin will ccur Ex. 11 Calculating K frm G Calculate K at 25 C fr the Haber prcess N 2 (g) + 3 H 2 (g) 2 NH 3 (g) G = 33.3 kj/ml = 33,300 J/ml G / RT K = e Step 1 Slve fr expnent ( 33,300J / ml ) = = 13.4 RT (8.3145J / K ml )(298K ) Step 2Take e x t btain K G / RT K = e = e = 7 10 Large Kindicates NH 3 favred at RT G

44 Yu Try! Calculate the equilibrium cnstant fr the decmpsitin f hydrgen perxide at 298 K given G = kj ml. A. 8.5 x B. 1.0 x C. 3.4 x D x G ( 234,300 J / ml ) = = RT ( J / K ml )(298 K ) K = e = 1.17 x Ex. 12 Calculating K frm G, First Calculate G Calculate the equilibrium cnstant at 25 C fr the decarbxylatin f liquid pyruvic acid t frm gaseus acetaldehyde and CO 2 O O H 3 C C C OH H 3 C C H + CO 2 O 131 First Calculate G frm G f Cmpund G f, kj/ml CH 3 COH CH 3 COCOOH CO G = G G f ( 3 CH 3COH ) + G f ( CO2 ) G f ( CH COCOOH ) = ( ) ( ) G = kJ

45 Next Calculate Equilibrium Cnstant K = e G RT G 64.28kJ 1000J = = RT (8.314J / K)(298K) kj ( ) K = e = e K = Temperature Dependence f K G = RT lnk = H T S Rearranging gives H 1 S lnk = + R T R Equatin fr line Slpe = H /RT Intercept = S /R Als way t determine Kif yu knw H and S 134 Ex. 12 Calculate K given H and S Calculate Kat 500 C fr Haber prcess N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Given H = kj and S = J/K Assume that H and S d nt change with T H S lnk = + RT R 92,380J 198.4J / K lnk = + (8.314J / K )(773K ) (8.314J / K ) lnk = = 9.49 K = e 9.49 =

46 Calculatin f G at Varius Temperatures Cnsider the fllwing reactin: CaCO 3 (s) CaO (s) + CO 2 (g) At 25 C G = and K p = 1.1 x atm What d these values tell yu abut CaCO 3? What happens when the reactin is carried ut at a higher temperature? 136 Calculating G and K at Varius Temperatures a. What is G at 1000 C fr the calcium carbnate reactin? CaCO 3 (s) CaO (s) + CO 2 (g) Is this reactin spntaneus at 1000 C and 1 atm? b. What is the value f K p at 1000 C fr this reactin? What is the partial pressure f CO 2? 137 Strategy fr slutin a. Calculate H and S at 25 C using standard enthalpies f frmatin and standard entrpies. Then substitute int the equatin fr G f. b. Use the G f value t find K (=K p )

47 a. Frm Data Table yu have the fllwing: CaCO 3 (s) CaO (s) + CO 2 (g) H f : kj S : J/K H = [( ) ( )] = kj S = [( ) (92.9)] = J/K G T = H T S = 178.3kJ (1273 K)(0.1590kJ/K) = kj G is negative reactin is spntaneus 139 b. Substitute the values f G at 1273 K, which equals x 10 3 J, int the equatin relating ln K and G. ln K = G -RT = x x 1273 = K = K p = e = 9.76 K p = P CO2 = 9.76 atm 140 Where des the reactin change frm spntaneus t nn-spntaneus? Slve fr T; G = 0 = H - T S T = H S = kj kj/k T = 1121 K = 848 C

48 Bnd Energy Amunt f energy needed t break chemical bnd int electrically neutral fragments Useful t knw Within reactin Bnds f reactants brken New bnds frmed as prducts appear Bnd breaking 1 st step in mst reactins One f the factrs that determines reactin rate Ex. N 2 very unreactive due t strng N N bnd 142 Bnd Energies Can be determined spectrscpically fr simple diatmic mlecules H 2, O 2, Cl 2 Mre cmplex mlecules, calculate using thermchemical data and Hess s Law Use H frmatin enthalpy f frmatin Need t define new term Enthalpy f atmizatin r atmizatin energy, H atm Energy required t rupture chemical bnds f 1 mle f gaseus mlecules t give gaseus atms 143 Determining Bnd Energies Ex. CH 4 (g) C(g) + 4 H(g) H atm = energy needed t break all bnds in mlecule H atm /4 = average bnd C H dissciatin energy in methane D = bnd dissciatin energy Average bnd energy t required t break all bnds in mlecule Hw d we calculate this? Use H f fr frming gaseus atms frm elements in their standard states Hess s Law

49 Determining Bnd Energies Path 1: bttm Frmatin f CH 4 frm its elements = H f Path 2: tp 3 step path Step 1 break H H bnds Step 2 break C C bnds Step 3: frm 4 C H bnds 1.2H 2 (g) 4H(g) H 1 = 4 H f (H,g) 2.C(s) C(g) H 2 = H f (C,g) 3.4H(g) + C(g) CH 4 (g) H 3 = H atm 2H 2 (g) + C(s) CH 4 (g) H = H f (CH 4,g) 145 Calculating H atm and Bnd Energy H f (CH 4,g) = 4 H f (H,g) + H f (C,g) H atm Rearranging gives H atm = 4 H f (H,g) + H f (C,g) H f (CH 4,g) Lk these up in Table 18.3, 6.2 r appendix C H atm = 4(217.9kJ/ml) kJ/ml ( 74.8kJ/ml) H atm = kj/ml f CH kj/ml bnd energy = 4 = kj/ml f C H bnds 146 Table 19.4 Sme Bnd Energies

50 Using Bnd Energies t Estimate H f Calculate H f fr CH 3 OH(g) (bttm reactin) Use 4 step path Step 1 break 1C C bnds Step 2 break 2H H bnds Step 3: break 1O O bnd Step 4: frm 3 C H, 1 O H, & 1O C bnds 148 Using Bnd Energies H f (CH 3 OH,g) = H f (C,g) + 4 H f (H,g) + H f (O,g) H atm (CH 3 OH,g) H f (C,g) + 4 H f (H,g) + H f (O,g) = { (4*217.9) }kJ = kj H atm (CH 3 OH,g) = 3D C H + D C O + D O H = (3*412) = 2059 kj H f (CH 3 OH,g) = kj 2059 kj = 222 kj Experimentally find H f (CH 3 OH,g) = 201 kj/ml S bnd energies give estimate within 10% f actual 149 Chapter 13/19 Chemical Thermdynamics 1. Spntaneus Chemical and Physical Prcesses 2. Entrpy and Disrder 3. Entrpy and the Secnd Law f Thermdynamics 4. Standard-State Entrpies f Reactin 5. The Third Law f Thermdynamics 6. Calculating Entrpy Changes fr Chemical Reactins 7. Gibbs Free Energy 8. The Effect f Temperature n the Free Energy f a Reactin 9. Beware f Oversimplificatin 10. Stand-State Free Energies f Reactin 11. Equilibria Expressed in Partial Pressures 12. Interpreting Stand-State Free Energy f Reactin Data 13. Relatinship Between Free Energy and Equilibrium Cnstants 14. Temperature Dependence f Equilibrium Cnstants 15. Gibbs Free Energies f Frmatin and Abslute Entrpies 16. Calculate H with bnd energies

51 Entrpy Change fr a Phase Transitin S > q/t (at equilibrium) What prcesses can ccur under phase change at equilibrium? Slid t liquid Liquid t gas Slid t gas 151 Slutin: S = H vap /T = (39.4 x 10 3 J/ml)/ 298 K = 132 j/(ml K) Entrpy f Vapr = ( ) J/(ml K) = 348 J/(ml K)

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