Thermodynamics 1/16/2013. Thermodynamics

Save this PDF as:
 WORD  PNG  TXT  JPG

Size: px
Start display at page:

Download "Thermodynamics 1/16/2013. Thermodynamics"

Transcription

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)

Thermodynamics and Equilibrium

Thermodynamics and Equilibrium Thermdynamics and Equilibrium Thermdynamics Thermdynamics is the study f the relatinship between heat and ther frms f energy in a chemical r physical prcess. We intrduced the thermdynamic prperty f enthalpy,

More information

Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics

Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics Chapters 9 and 35 Thermchemistry and Chemical Thermdynamics 1 Cpyright (c) 011 by Michael A. Janusa, PhD. All rights reserved. Thermchemistry Thermchemistry is the study f the energy effects that accmpany

More information

AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY

AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY AP CHEMISTRY CHAPTER 6 NOTES THERMOCHEMISTRY Energy- the capacity t d wrk r t prduce heat 1 st Law f Thermdynamics: Law f Cnservatin f Energy- energy can be cnverted frm ne frm t anther but it can be neither

More information

Chapter 17 Free Energy and Thermodynamics

Chapter 17 Free Energy and Thermodynamics 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 First Law f Thermdynamics

More information

Chapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes

Chapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses Learning Objectives 17.1: Spntaneus Prcesses Cmparing and Cntrasting the hree Laws f hermdynamics (1 st Law: Chap. 5; 2 nd & 3

More information

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

CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review Review Accrding t the 2 nd law f Thermdynamics, a prcess is spntaneus if S universe = S system + S surrundings > 0 Even thugh S system

More information

Entropy, Free Energy, and Equilibrium

Entropy, Free Energy, and Equilibrium Nv. 26 Chapter 19 Chemical Thermdynamics Entrpy, Free Energy, and Equilibrium Nv. 26 Spntaneus Physical and Chemical Prcesses Thermdynamics: cncerned with the questin: can a reactin ccur? A waterfall runs

More information

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

CHEM Thermodynamics. Change in Gibbs Free Energy, G. Review. Gibbs Free Energy, G. Review Review Accrding t the nd law f Thermdynamics, a prcess is spntaneus if S universe = S system + S surrundings > 0 Even thugh S system

More information

Part One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review)

Part One: Heat Changes and Thermochemistry. This aspect of Thermodynamics was dealt with in Chapter 6. (Review) CHAPTER 18: THERMODYNAMICS AND EQUILIBRIUM Part One: Heat Changes and Thermchemistry This aspect f Thermdynamics was dealt with in Chapter 6. (Review) A. Statement f First Law. (Sectin 18.1) 1. U ttal

More information

Examples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C?

Examples: 1. How much heat is given off by a 50.0 g sample of copper when it cools from 80.0 to 50.0 C? NOTES: Thermchemistry Part 1 - Heat HEAT- TEMPERATURE - Thermchemistry: the study f energy (in the frm f heat) changes that accmpany physical & chemical changes heat flws frm high t lw (ht cl) endthermic

More information

Unit 14 Thermochemistry Notes

Unit 14 Thermochemistry Notes Name KEY Perid CRHS Academic Chemistry Unit 14 Thermchemistry Ntes Quiz Date Exam Date Lab Dates Ntes, Hmewrk, Exam Reviews and Their KEYS lcated n CRHS Academic Chemistry Website: https://cincchem.pbwrks.cm

More information

Thermodynamics Partial Outline of Topics

Thermodynamics Partial Outline of Topics Thermdynamics Partial Outline f Tpics I. The secnd law f thermdynamics addresses the issue f spntaneity and invlves a functin called entrpy (S): If a prcess is spntaneus, then Suniverse > 0 (2 nd Law!)

More information

1/16/2013. Thermodynamics.

1/16/2013. Thermodynamics. Thermodynamics http://www.chem.purdue.edu/gchelp/howtosolveit/howtosolveit.html 1 1 Study of energy changes and flow of energy Answers several fundamental questions: Is it possible for reaction to occur?

More information

Chapter 4 Thermodynamics and Equilibrium

Chapter 4 Thermodynamics and Equilibrium Chapter Thermdynamics and Equilibrium Refer t the fllwing figures fr Exercises 1-6. Each represents the energies f fur mlecules at a given instant, and the dtted lines represent the allwed energies. Assume

More information

GOAL... ability to predict

GOAL... ability to predict THERMODYNAMICS Chapter 18, 11.5 Study f changes in energy and transfers f energy (system < = > surrundings) that accmpany chemical and physical prcesses. GOAL............................. ability t predict

More information

REVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj

REVIEW QUESTIONS Chapter 18. H = H (Products) - H (Reactants) H (Products) = (1 x -125) + (3 x -271) = -938 kj Chemistry 102 ANSWER KEY REVIEW QUESTIONS Chapter 18 1. Calculate the heat reactin ( H ) in kj/ml r the reactin shwn belw, given the H values r each substance: NH (g) + F 2 (g) NF (g) + HF (g) H (kj/ml)

More information

ALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change?

ALE 21. Gibbs Free Energy. At what temperature does the spontaneity of a reaction change? Name Chem 163 Sectin: Team Number: ALE 21. Gibbs Free Energy (Reference: 20.3 Silberberg 5 th editin) At what temperature des the spntaneity f a reactin change? The Mdel: The Definitin f Free Energy S

More information

CHEM 116 Electrochemistry at Non-Standard Conditions, and Intro to Thermodynamics

CHEM 116 Electrochemistry at Non-Standard Conditions, and Intro to Thermodynamics CHEM 116 Electrchemistry at Nn-Standard Cnditins, and Intr t Thermdynamics Imprtant annuncement: If yu brrwed a clicker frm me this semester, return it t me at the end f next lecture r at the final exam

More information

188 CHAPTER 6 THERMOCHEMISTRY

188 CHAPTER 6 THERMOCHEMISTRY 188 CHAPTER 6 THERMOCHEMISTRY 4. a. ΔE = q + w = J + 100. J = 77 J b. w = PΔV = 1.90 atm(.80 L 8.0 L) = 10.5 L atm ΔE = q + w = 50. J + 1060 = 1410 J c. w = PΔV = 1.00 atm(9.1 L11. L) = 17.9 L atm 101.

More information

Spontaneous Processes, Entropy and the Second Law of Thermodynamics

Spontaneous Processes, Entropy and the Second Law of Thermodynamics Chemical Thermdynamics Spntaneus Prcesses, Entrpy and the Secnd Law f Thermdynamics Review Reactin Rates, Energies, and Equilibrium Althugh a reactin may be energetically favrable (i.e. prducts have lwer

More information

Lecture 4. The First Law of Thermodynamics

Lecture 4. The First Law of Thermodynamics Lecture 4. The First Law f Thermdynamics THERMODYNAMICS: Basic Cncepts Thermdynamics: (frm the Greek therme, meaning "heat" and, dynamis, meaning "pwer") is the study f energy cnversin between heat and

More information

CHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25

CHAPTER Read Chapter 17, sections 1,2,3. End of Chapter problems: 25 CHAPTER 17 1. Read Chapter 17, sectins 1,2,3. End f Chapter prblems: 25 2. Suppse yu are playing a game that uses tw dice. If yu cunt the dts n the dice, yu culd have anywhere frm 2 t 12. The ways f prducing

More information

Thermochemistry. Thermochemistry

Thermochemistry. Thermochemistry Thermchemistry Petrucci, Harwd and Herring: Chapter 7 CHEM 1000A 3.0 Thermchemistry 1 Thermchemistry The study energy in chemical reactins A sub-discipline thermdynamics Thermdynamics studies the bulk

More information

General Chemistry II, Unit II: Study Guide (part 1)

General Chemistry II, Unit II: Study Guide (part 1) General Chemistry II, Unit II: Study Guide (part 1) CDS Chapter 21: Reactin Equilibrium in the Gas Phase General Chemistry II Unit II Part 1 1 Intrductin Sme chemical reactins have a significant amunt

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

CHEM 103 Calorimetry and Hess s Law

CHEM 103 Calorimetry and Hess s Law CHEM 103 Calrimetry and Hess s Law Lecture Ntes March 23, 2006 Prf. Sevian Annuncements Exam #2 is next Thursday, March 30 Study guide, practice exam, and practice exam answer key are already psted n the

More information

Chemistry 1A Fall 2000

Chemistry 1A Fall 2000 Chemistry 1A Fall 2000 Midterm Exam III, versin B Nvember 14, 2000 (Clsed bk, 90 minutes, 155 pints) Name: SID: Sectin Number: T.A. Name: Exam infrmatin, extra directins, and useful hints t maximize yur

More information

Lecture 16 Thermodynamics II

Lecture 16 Thermodynamics II Lecture 16 Thermdynamics II Calrimetry Hess s Law Enthalpy r Frmatin Cpyright 2013, 2011, 2009, 2008 AP Chem Slutins. All rights reserved. Fur Methds fr Finding H 1) Calculate it using average bnd enthalpies

More information

SPONTANEITY, ENTROPY, AND FREE ENERGY

SPONTANEITY, ENTROPY, AND FREE ENERGY CHAER 7 SONANEIY, ENROY, AND FREE ENERGY Questins. Living rganisms need an external surce f energy t carry ut these prcesses. Green plants use the energy frm sunlight t prduce glucse frm carbn dixide and

More information

CHEM 1001 Problem Set #3: Entropy and Free Energy

CHEM 1001 Problem Set #3: Entropy and Free Energy CHEM 1001 Prblem Set #3: Entry and Free Energy 19.7 (a) Negative; A liquid (mderate entry) cmbines with a slid t frm anther slid. (b)psitive; One mle f high entry gas frms where n gas was resent befre.

More information

Chem 75 February 16, 2017 Exam 2 Solutions

Chem 75 February 16, 2017 Exam 2 Solutions 1. (6 + 6 pints) Tw quick questins: (a) The Handbk f Chemistry and Physics tells us, crrectly, that CCl 4 bils nrmally at 76.7 C, but its mlar enthalpy f vaprizatin is listed in ne place as 34.6 kj ml

More information

Chem 112, Fall 05 (Weis/Garman) Exam 4A, December 14, 2005 (Print Clearly) +2 points

Chem 112, Fall 05 (Weis/Garman) Exam 4A, December 14, 2005 (Print Clearly) +2 points +2 pints Befre yu begin, make sure that yur exam has all 7 pages. There are 14 required prblems (7 pints each) and tw extra credit prblems (5 pints each). Stay fcused, stay calm. Wrk steadily thrugh yur

More information

Types of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY

Types of Energy COMMON MISCONCEPTIONS CHEMICAL REACTIONS INVOLVE ENERGY CHEMICAL REACTIONS INVOLVE ENERGY The study energy and its transrmatins is knwn as thermdynamics. The discussin thermdynamics invlve the cncepts energy, wrk, and heat. Types Energy Ptential energy is stred

More information

Chapter Outline 4/28/2014. P-V Work. P-V Work. Isolated, Closed and Open Systems. Exothermic and Endothermic Processes. E = q + w

Chapter Outline 4/28/2014. P-V Work. P-V Work. Isolated, Closed and Open Systems. Exothermic and Endothermic Processes. E = q + w Islated, Clsed and Open Systems 9.1 Energy as a Reactant r a Prduct 9.2 Transferring Heat and Ding Wrk 9.5 Heats f Reactin and Calrimetry 9.6 Hess s Law and Standard Heats f Reactin 9.7 Heats f Reactin

More information

General Chemistry II, Unit I: Study Guide (part I)

General Chemistry II, Unit I: Study Guide (part I) 1 General Chemistry II, Unit I: Study Guide (part I) CDS Chapter 14: Physical Prperties f Gases Observatin 1: Pressure- Vlume Measurements n Gases The spring f air is measured as pressure, defined as the

More information

Matter Content from State Frameworks and Other State Documents

Matter Content from State Frameworks and Other State Documents Atms and Mlecules Mlecules are made f smaller entities (atms) which are bnded tgether. Therefre mlecules are divisible. Miscnceptin: Element and atm are synnyms. Prper cnceptin: Elements are atms with

More information

Lecture 12: Chemical reaction equilibria

Lecture 12: Chemical reaction equilibria 3.012 Fundamentals f Materials Science Fall 2005 Lecture 12: 10.19.05 Chemical reactin equilibria Tday: LAST TIME...2 EQUATING CHEMICAL POTENTIALS DURING REACTIONS...3 The extent f reactin...3 The simplest

More information

How can standard heats of formation be used to calculate the heat of a reaction?

How can standard heats of formation be used to calculate the heat of a reaction? Answer Key ALE 28. ess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 4 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk neatly using dimensinal analysis

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

CHM 152 Practice Final

CHM 152 Practice Final CM 152 Practice Final 1. Of the fllwing, the ne that wuld have the greatest entrpy (if cmpared at the same temperature) is, [a] 2 O (s) [b] 2 O (l) [c] 2 O (g) [d] All wuld have the same entrpy at the

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t eep this site up and bring yu even mre cntent cnsider dnating via the lin n ur site. Still having truble understanding the material? Chec ut ur Tutring

More information

Chemical Thermodynamics

Chemical Thermodynamics Chemical Thermdynamics Objectives 1. Be capable f stating the First, Secnd, and Third Laws f Thermdynamics and als be capable f applying them t slve prblems. 2. Understand what the parameter entrpy means.

More information

Contents and Concepts

Contents and Concepts Contents and Concepts 1. First Law of Thermodynamics Spontaneous Processes and Entropy A spontaneous process is one that occurs by itself. As we will see, the entropy of the system increases in a spontaneous

More information

Contents and Concepts

Contents and Concepts Contents and Concepts 1. First Law of Thermodynamics Spontaneous Processes and Entropy A spontaneous process is one that occurs by itself. As we will see, the entropy of the system increases in a spontaneous

More information

Advanced Chemistry Practice Problems

Advanced Chemistry Practice Problems Advanced Chemistry Practice Prblems Thermdynamics: Gibbs Free Energy 1. Questin: Is the reactin spntaneus when ΔG < 0? ΔG > 0? Answer: The reactin is spntaneus when ΔG < 0. 2. Questin: Fr a reactin with

More information

Chem 116 POGIL Worksheet - Week 3 - Solutions Intermolecular Forces, Liquids, Solids, and Solutions

Chem 116 POGIL Worksheet - Week 3 - Solutions Intermolecular Forces, Liquids, Solids, and Solutions Chem 116 POGIL Wrksheet - Week 3 - Slutins Intermlecular Frces, Liquids, Slids, and Slutins Key Questins 1. Is the average kinetic energy f mlecules greater r lesser than the energy f intermlecular frces

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

More Tutorial at

More Tutorial at Answer each questin in the space prvided; use back f page if extra space is needed. Answer questins s the grader can READILY understand yur wrk; nly wrk n the exam sheet will be cnsidered. Write answers,

More information

Unit 11 Solutions- Guided Notes. What are alloys? What is the difference between heterogeneous and homogeneous mixtures?

Unit 11 Solutions- Guided Notes. What are alloys? What is the difference between heterogeneous and homogeneous mixtures? Name: Perid: Unit 11 Slutins- Guided Ntes Mixtures: What is a mixture and give examples? What is a pure substance? What are allys? What is the difference between hetergeneus and hmgeneus mixtures? Slutins:

More information

Chemistry 114 First Hour Exam

Chemistry 114 First Hour Exam Chemistry 114 First Hur Exam Please shw all wrk fr partial credit Name: (4 pints) 1. (12 pints) Espress is made by frcing very ht water under high pressure thrugh finely grund, cmpacted cffee. (Wikipedia)

More information

Edexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes.

Edexcel IGCSE Chemistry. Topic 1: Principles of chemistry. Chemical formulae, equations and calculations. Notes. Edexcel IGCSE Chemistry Tpic 1: Principles f chemistry Chemical frmulae, equatins and calculatins Ntes 1.25 write wrd equatins and balanced chemical equatins (including state symbls): fr reactins studied

More information

" 1 = # $H vap. Chapter 3 Problems

 1 = # $H vap. Chapter 3 Problems Chapter 3 rblems rblem At 1 atmsphere pure Ge melts at 1232 K and bils at 298 K. he triple pint ccurs at =8.4x1-8 atm. Estimate the heat f vaprizatin f Ge. he heat f vaprizatin is estimated frm the Clausius

More information

A Chemical Reaction occurs when the of a substance changes.

A Chemical Reaction occurs when the of a substance changes. Perid: Unit 8 Chemical Reactin- Guided Ntes Chemical Reactins A Chemical Reactin ccurs when the f a substance changes. Chemical Reactin: ne r mre substances are changed int ne r mre new substances by the

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

When a substance heats up (absorbs heat) it is an endothermic reaction with a (+)q

When a substance heats up (absorbs heat) it is an endothermic reaction with a (+)q Chemistry Ntes Lecture 15 [st] 3/6/09 IMPORTANT NOTES: -( We finished using the lecture slides frm lecture 14) -In class the challenge prblem was passed ut, it is due Tuesday at :00 P.M. SHARP, :01 is

More information

Chapter 8 Reduction and oxidation

Chapter 8 Reduction and oxidation Chapter 8 Reductin and xidatin Redx reactins and xidatin states Reductin ptentials and Gibbs energy Nernst equatin Disprprtinatin Ptential diagrams Frst-Ebswrth diagrams Ellingham diagrams Oxidatin refers

More information

CHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK

CHEM 1413 Chapter 6 Homework Questions TEXTBOOK HOMEWORK CHEM 1413 Chapter 6 Hmewrk Questins TEXTBOOK HOMEWORK 6.25 A 27.7-g sample f the radiatr clant ethylene glycl releases 688 J f heat. What was the initial temperature f the sample if the final temperature

More information

CHAPTER PRACTICE PROBLEMS CHEMISTRY

CHAPTER PRACTICE PROBLEMS CHEMISTRY Chemical Kinetics Name: Batch: Date: Rate f reactin. 4NH 3 (g) + 5O (g) à 4NO (g) + 6 H O (g) If the rate f frmatin f NO is 3.6 0 3 ml L s, calculate (i) the rate f disappearance f NH 3 (ii) rate f frmatin

More information

Thermochemistry. The study of energy changes that occur during chemical : at constant volume ΔU = q V. no at constant pressure ΔH = q P

Thermochemistry. The study of energy changes that occur during chemical : at constant volume ΔU = q V. no at constant pressure ΔH = q P Thermchemistry The study energy changes that ccur during chemical : at cnstant vlume ΔU = q V n at cnstant pressure = q P nly wrk Fr practical reasns mst measurements are made at cnstant, s thermchemistry

More information

University Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia:

University Chemistry Quiz /04/21 1. (10%) Consider the oxidation of ammonia: University Chemistry Quiz 3 2015/04/21 1. (10%) Cnsider the xidatin f ammnia: 4NH 3 (g) + 3O 2 (g) 2N 2 (g) + 6H 2 O(l) (a) Calculate the ΔG fr the reactin. (b) If this reactin were used in a fuel cell,

More information

CHAPTER 6 / HARVEY A. CHEMICAL EQUILIBRIUM B. THERMODYNAMICS AND EQUILIBRIUM C. MANUPULATING EQUILIBRIUM CONSTANTS

CHAPTER 6 / HARVEY A. CHEMICAL EQUILIBRIUM B. THERMODYNAMICS AND EQUILIBRIUM C. MANUPULATING EQUILIBRIUM CONSTANTS CHPTER 6 / HRVEY. CHEMICL B. THERMODYNMICS ND C. MNUPULTING CONSTNTS D. CONSTNTS FOR CHEMICL RECTIONS 1. Precipitatin Reactins 2. cid-base Reactins 3. Cmplexatin Reactins 4. Oxidatin-Reductin Reactins

More information

Chemistry 132 NT. Electrochemistry. Review

Chemistry 132 NT. Electrochemistry. Review Chemistry 132 NT If yu g flying back thrugh time, and yu see smebdy else flying frward int the future, it s prbably best t avid eye cntact. Jack Handey 1 Chem 132 NT Electrchemistry Mdule 3 Vltaic Cells

More information

Contents and Concepts

Contents and Concepts Contents and Concepts 1. First Law of Thermodynamics Spontaneous Processes and Entropy A spontaneous process is one that occurs by itself. As we will see, the entropy of the system increases in a spontaneous

More information

State of matter characteristics solid Retains shape and volume

State of matter characteristics solid Retains shape and volume **See attachment fr graphs States f matter The fundamental difference between states f matter is the distance between particles Gas Ttal disrder Much empty space Particles have cmpletely freedm f mtin

More information

_J _J J J J J J J J _. 7 particles in the blue state; 3 particles in the red state: 720 configurations _J J J _J J J J J J J J _

_J _J J J J J J J J _. 7 particles in the blue state; 3 particles in the red state: 720 configurations _J J J _J J J J J J J J _ Dsrder and Suppse I have 10 partcles that can be n ne f tw states ether the blue state r the red state. Hw many dfferent ways can we arrange thse partcles amng the states? All partcles n the blue state:

More information

Tuesday, 5:10PM FORM A March 18,

Tuesday, 5:10PM FORM A March 18, Name Chemistry 153-080 (3150:153-080) EXAM II Multiple-Chice Prtin Instructins: Tuesday, 5:10PM FORM A March 18, 2003 120 1. Each student is respnsible fr fllwing instructins. Read this page carefully.

More information

How can standard heats of formation be used to calculate the heat of a reaction?

How can standard heats of formation be used to calculate the heat of a reaction? Name Chem 161, Sectin: Grup Number: ALE 28. Hess s Law and Standard Enthalpies Frmatin (Reerence: Chapter 6 - Silberberg 5 th editin) Imprtant!! Fr answers that invlve a calculatin yu must shw yur wrk

More information

Semester 2 AP Chemistry Unit 12

Semester 2 AP Chemistry Unit 12 Cmmn In Effect and Buffers PwerPint The cmmn in effect The shift in equilibrium caused by the additin f a cmpund having an in in cmmn with the disslved substance The presence f the excess ins frm the disslved

More information

CHEM-443, Fall 2013, Section 010 Midterm 2 November 4, 2013

CHEM-443, Fall 2013, Section 010 Midterm 2 November 4, 2013 CHEM-443, Fall 2013, Sectin 010 Student Name Midterm 2 Nvember 4, 2013 Directins: Please answer each questin t the best f yur ability. Make sure yur respnse is legible, precise, includes relevant dimensinal

More information

Chem 111 Summer 2013 Key III Whelan

Chem 111 Summer 2013 Key III Whelan Chem 111 Summer 2013 Key III Whelan Questin 1 6 Pints Classify each f the fllwing mlecules as plar r nnplar? a) NO + : c) CH 2 Cl 2 : b) XeF 4 : Questin 2 The hypthetical mlecule PY 3 Z 2 has the general

More information

Lecture 13: Electrochemical Equilibria

Lecture 13: Electrochemical Equilibria 3.012 Fundamentals f Materials Science Fall 2005 Lecture 13: 10.21.05 Electrchemical Equilibria Tday: LAST TIME...2 An example calculatin...3 THE ELECTROCHEMICAL POTENTIAL...4 Electrstatic energy cntributins

More information

AP Chemistry Assessment 2

AP Chemistry Assessment 2 AP Chemistry Assessment 2 DATE OF ADMINISTRATION: January 8 January 12 TOPICS COVERED: Fundatinal Tpics, Reactins, Gases, Thermchemistry, Atmic Structure, Peridicity, and Bnding. MULTIPLE CHOICE KEY AND

More information

Name: Period: Date: BONDING NOTES HONORS CHEMISTRY

Name: Period: Date: BONDING NOTES HONORS CHEMISTRY Name: Perid: Date: BONDING NOTES HONORS CHEMISTRY Directins: This packet will serve as yur ntes fr this chapter. Fllw alng with the PwerPint presentatin and fill in the missing infrmatin. Imprtant terms

More information

Recitation 06. n total = P total V/RT = (0.425 atm * 10.5 L) / ( L atm mol -1 K -1 * 338 K) = mol

Recitation 06. n total = P total V/RT = (0.425 atm * 10.5 L) / ( L atm mol -1 K -1 * 338 K) = mol Recitatin 06 Mixture f Ideal Gases 1. Chapter 5: Exercise: 69 The partial pressure f CH 4 (g) is 0.175 atm and that f O 2 (g) is 0.250 atm in a mixture f the tw gases. a. What is the mle fractin f each

More information

Hess Law - Enthalpy of Formation of Solid NH 4 Cl

Hess Law - Enthalpy of Formation of Solid NH 4 Cl Hess Law - Enthalpy f Frmatin f Slid NH 4 l NAME: OURSE: PERIOD: Prelab 1. Write and balance net inic equatins fr Reactin 2 and Reactin 3. Reactin 2: Reactin 3: 2. Shw that the alebraic sum f the balanced

More information

CHAPTER 13 Temperature and Kinetic Theory. Units

CHAPTER 13 Temperature and Kinetic Theory. Units CHAPTER 13 Temperature and Kinetic Thery Units Atmic Thery f Matter Temperature and Thermmeters Thermal Equilibrium and the Zerth Law f Thermdynamics Thermal Expansin Thermal Stress The Gas Laws and Abslute

More information

Chapter 19. Electrochemistry. Dr. Al Saadi. Electrochemistry

Chapter 19. Electrochemistry. Dr. Al Saadi. Electrochemistry Chapter 19 lectrchemistry Part I Dr. Al Saadi 1 lectrchemistry What is electrchemistry? It is a branch f chemistry that studies chemical reactins called redx reactins which invlve electrn transfer. 19.1

More information

Chem 163 Section: Team Number: ALE 24. Voltaic Cells and Standard Cell Potentials. (Reference: 21.2 and 21.3 Silberberg 5 th edition)

Chem 163 Section: Team Number: ALE 24. Voltaic Cells and Standard Cell Potentials. (Reference: 21.2 and 21.3 Silberberg 5 th edition) Name Chem 163 Sectin: Team Number: ALE 24. Vltaic Cells and Standard Cell Ptentials (Reference: 21.2 and 21.3 Silberberg 5 th editin) What des a vltmeter reading tell us? The Mdel: Standard Reductin and

More information

Chapter 9 Chemical Reactions NOTES

Chapter 9 Chemical Reactions NOTES Chapter 9 Chemical Reactins NOTES Chemical Reactins Chemical reactin: Chemical change 4 Indicatrs f Chemical Change: (1) (2) (3) (4) Cnsist f reactants (starting materials) and prducts (substances frmed)

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

CHEM 1032 FALL 2017 Practice Exam 4 1. Which of the following reactions is spontaneous under normal and standard conditions?

CHEM 1032 FALL 2017 Practice Exam 4 1. Which of the following reactions is spontaneous under normal and standard conditions? 1 CHEM 1032 FALL 2017 Practice Exam 4 1. Which f the fllwing reactins is spntaneus under nrmal and standard cnditins? A. 2 NaCl(aq) 2 Na(s) + Cl2(g) B. CaBr2(aq) + 2 H2O(aq) Ca(OH)2(aq) + 2 HBr(aq) C.

More information

Solutions to the Extra Problems for Chapter 14

Solutions to the Extra Problems for Chapter 14 Slutins t the Extra Prblems r Chapter 1 1. The H -670. T use bnd energies, we have t igure ut what bnds are being brken and what bnds are being made, s we need t make Lewis structures r everything: + +

More information

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site.

Find this material useful? You can help our team to keep this site up and bring you even more content consider donating via the link on our site. Find this material useful? Yu can help ur team t keep this site up and bring yu even mre cntent cnsider dnating via the link n ur site. Still having truble understanding the material? Check ut ur Tutring

More information

CHEMICAL EQUILIBRIUM

CHEMICAL EQUILIBRIUM 14 CHAPTER CHEMICAL EQUILIBRIUM 14.1 The Nature f Chemical Equilibrium 14. The Empirical Law f Mass Actin 14.3 Thermdynamic Descriptin f the Equilibrium State 14.4 The Law f Mass Actin fr Related and Simultaneus

More information

Heat Effects of Chemical Reactions

Heat Effects of Chemical Reactions eat Effects f hemical Reactins Enthalpy change fr reactins invlving cmpunds Enthalpy f frmatin f a cmpund at standard cnditins is btained frm the literature as standard enthalpy f frmatin Δ (O (g = -9690

More information

Process Engineering Thermodynamics E (4 sp) Exam

Process Engineering Thermodynamics E (4 sp) Exam Prcess Engineering Thermdynamics 42434 E (4 sp) Exam 9-3-29 ll supprt material is allwed except fr telecmmunicatin devices. 4 questins give max. 3 pints = 7½ + 7½ + 7½ + 7½ pints Belw 6 questins are given,

More information

BIT Chapters = =

BIT Chapters = = BIT Chapters 17-0 1. K w = [H + ][OH ] = 9.5 10 14 [H + ] = [OH ] =.1 10 7 ph = 6.51 The slutin is neither acidic nr basic because the cncentratin f the hydrnium in equals the cncentratin f the hydride

More information

NUPOC STUDY GUIDE ANSWER KEY. Navy Recruiting Command

NUPOC STUDY GUIDE ANSWER KEY. Navy Recruiting Command NUPOC SUDY GUIDE ANSWER KEY Navy Recruiting Cmmand CHEMISRY. ph represents the cncentratin f H ins in a slutin, [H ]. ph is a lg scale base and equal t lg[h ]. A ph f 7 is a neutral slutin. PH < 7 is acidic

More information

CHE 105 EXAMINATION III November 11, 2010

CHE 105 EXAMINATION III November 11, 2010 CHE 105 EXAMINATION III Nvember 11, 2010 University f Kentucky Department f Chemistry READ THESE DIRECTIONS CAREFULLY BEFORE STARTING THE EXAMINATION! It is extremely imprtant that yu fill in the answer

More information

lecture 5: Nucleophilic Substitution Reactions

lecture 5: Nucleophilic Substitution Reactions lecture 5: Nuclephilic Substitutin Reactins Substitutin unimlecular (SN1): substitutin nuclephilic, unimlecular. It is first rder. The rate is dependent upn ne mlecule, that is the substrate, t frm the

More information

Chem 115 POGIL Worksheet - Week 8 Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra

Chem 115 POGIL Worksheet - Week 8 Thermochemistry (Continued), Electromagnetic Radiation, and Line Spectra Chem 115 POGIL Wrksheet - Week 8 Thermchemistry (Cntinued), Electrmagnetic Radiatin, and Line Spectra Why? As we saw last week, enthalpy and internal energy are state functins, which means that the sum

More information

Chem 116 POGIL Worksheet - Week 4 Properties of Solutions

Chem 116 POGIL Worksheet - Week 4 Properties of Solutions Chem 116 POGIL Wrksheet - Week 4 Prperties f Slutins Key Questins 1. Identify the principal type f slute-slvent interactin that is respnsible fr frming the fllwing slutins: (a) KNO 3 in water; (b) Br 2

More information

GASES. PV = nrt N 2 CH 4 CO 2 O 2 HCN N 2 O NO 2. Pressure & Boyle s Law Temperature & Charles s Law Avogadro s Law IDEAL GAS LAW

GASES. PV = nrt N 2 CH 4 CO 2 O 2 HCN N 2 O NO 2. Pressure & Boyle s Law Temperature & Charles s Law Avogadro s Law IDEAL GAS LAW GASES Pressure & Byle s Law Temperature & Charles s Law Avgadr s Law IDEAL GAS LAW PV = nrt N 2 CH 4 CO 2 O 2 HCN N 2 O NO 2 Earth s atmsphere: 78% N 2 21% O 2 sme Ar, CO 2 Sme Cmmn Gasses Frmula Name

More information

A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1

A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1 A.P. CHEMISTRY. SOLUTIONS AND ACID BASE CHEMISTRY. p 1 (Nte: questins 1 t 14 are meant t be dne WITHOUT calculatrs!) 1.Which f the fllwing is prbably true fr a slid slute with a highly endthermic heat

More information

Lecture 23: Lattice Models of Materials; Modeling Polymer Solutions

Lecture 23: Lattice Models of Materials; Modeling Polymer Solutions Lecture 23: 12.05.05 Lattice Mdels f Materials; Mdeling Plymer Slutins Tday: LAST TIME...2 The Bltzmann Factr and Partitin Functin: systems at cnstant temperature...2 A better mdel: The Debye slid...3

More information

Electrochemistry. Half-Reactions 1. Balancing Oxidation Reduction Reactions in Acidic and Basic Solutions

Electrochemistry. Half-Reactions 1. Balancing Oxidation Reduction Reactions in Acidic and Basic Solutions Electrchemistry Half-Reactins 1. Balancing Oxidatin Reductin Reactins in Acidic and Basic Slutins Vltaic Cells 2. Cnstructin f Vltaic Cells 3. Ntatin fr Vltaic Cells 4. Cell Ptential 5. Standard Cell Ptentials

More information

KEY POINTS: NOTE: OCR A

KEY POINTS: NOTE: OCR A KEY: A.J.F.S DEFINITION: KEY POINTS: NOTE: OCR A Chemistry Mdule 3- Energy 2.3 (1) Enthalpy What is chemical energy? - Chemical energy is a special frm f ptential energy that lies within chemical bnds

More information

Experiment #3. Graphing with Excel

Experiment #3. Graphing with Excel Experiment #3. Graphing with Excel Study the "Graphing with Excel" instructins that have been prvided. Additinal help with learning t use Excel can be fund n several web sites, including http://www.ncsu.edu/labwrite/res/gt/gt-

More information