Chapters 29 and 35 Thermochemistry and Chemical Thermodynamics


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1 Chapters 9 and 35 Thermchemistry and Chemical Thermdynamics 1 Cpyright (c) 011 by Michael A. Janusa, PhD. All rights reserved.
2 Thermchemistry Thermchemistry is the study f the energy effects that accmpany chemical reactins. Why d chemical reactins ccur? What is the driving frce f rxn? Answer: Stability, wants t get t lwer E. Fr a rxn t take place spntaneusly the prducts f reactin must be mre stable (lwer E) than the starting reactants. Nnspntaneus means never happen by self. R release E, spn absrb E, nnspn P E P E R higher E, less stable, mre reactive lwer E, mre stable, less reactive
3 ( 9.1 Reactin Enthalpy In chemical reactins, heat is ften transferred frm the system r reactin t its surrundings, r vice versa. system  the substance r mixture f substances under study in which a change ccurs. The surrundings are everything in the vicinity f the thermdynamic system. system r rxn + int system  ut system surrundings 3
4 eat f Reactin eat flw is defined as the energy that flws int r ut f a system. We fllw heat flw by watching the difference in temperature between the system and its surrundings. Often we fllw the surrundings temp (slvent) and must realize that the ppsite is happening t the system. If system is absrbing heat frm the surrundings than the temp f the surrundings must be decreasing. T system (+) T surr () 4
5 eat f Reactin eat flw r heat f reactin is dented by the symbl q and is the amunt f heat required t return a system t the given temperature at the cmpletin f the reactin. Fr an endthermic rxn the sign f q is psitive; heat is absrbed by the system frm the surrundings. Surrundings absrb heat, nnspn (end) P T system T surr E R +q System q > 0 5
6 eat f Reactin Fr an exthermic rxn, the sign f q is negative; heat is evlved (released) by the system t the surrundings. Surrundings T system R release heat, spn (ex) T surr System q E P q < 0 6
7 Enthalpy and Enthalpy Change The heat absrbed r evlved by a reactin depends n the cnditins under which it ccurs. ex. pressure Usually, a reactin takes place in an pen vessel, and therefre under the cnstant pressure f the atmsphere. heat f this type f reactin is dented q p ; this heat at cnstant pressure is named enthalpy and given symbl. is the heat flw at cnstant pressure. 7
8 Enthalpy and Enthalpy Change Enthalpy, dented, is an extensive prperty f a substance that can be used t btain the heat absrbed r evlved in a chemical reactin at cnstant pressure. an extensive prperty  depends n the quantity f substance. Enthalpy is a state functin, a prperty f a system that depends nly n its present state and is independent f any previus histry f the system. 8
9 Enthalpy and Enthalpy Change The reactin enthalpy fr a reactin at a given temperature and pressure (final) (initial) (prducts) (reactants) 9
10 Enthalpy and Enthalpy Change As we already stated the reactin enthalpy is equal t the heat f reactin at cnstant pressure. This represents the entire change in internal energy ( U) minus any expansin wrk dne by the system; therefre we can define enthalpy and internal wrk by the 1st law f thermdynamics: In any prcess, the ttal change in energy f the system, U, is equal t the sum f the heat absrbed, q, and the wrk, w, dne by the system. U = q p + w = + w 10
11 Changes in E manifest themselves as exchanges f energy between the system and surrundings. These exchanges f energy are f tw kinds; heat and wrk  must accunt fr bth. eat is energy that mves int r ut f a system because f a temperature difference between system and surrundings. Wrk is the energy exchange that results when a frce F mves an bject thrugh a distance d; wrk (w) = F d In chemical systems, wrk is defined as a change in vlume at a given pressure, that is: w P V 11
12 w P V negative sign is t keep sign crrect in terms f system. Fr expansin, V, will be a psitive value but expansin invlves the system ding wrk n the surrundings and a decrease in internal energy  negative keeps it neg. Fr cntractin wrk, V, will be a negative value but cntractin invlves the surrundings ding wrk n the system and an increase in internal energy  negative keeps it psitive ( x  = +). Giving us the 1st law f therm is mre useful frm: U P V realize absrb heat (+) W 44 release r evlved heat () 1
13 9.3 Thermchemical Equatins A thermchemical equatin is the chemical equatin fr a reactin (including phase labels {imprtant}) in which the equatin is given a mlar interpretatin, and the enthalpy f reactin fr these mlar amunts is written directly after the equatin. N (g) 3 (g) N (g); kJ If has a superscript like, means therm standard cnditins  5 C (98K) and 1 atm. 13
14 14 The fllwing are tw imprtant rules fr manipulating thermchemical equatins: 1.) When a thermchemical equatin is multiplied by any factr, the value f fr the new equatin is btained by multiplying the in the riginal equatin by that same factr..) When a chemical equatin is reversed, the value f is reversed in sign. Thermchemical Equatins kj ; ) ( 4 ) ( ) ( kj ; ) ( ) ( ) ( g O g O g g O g O g kj ; ) ( ) ( ) ( kj ; ) ( ) ( ) ( g O g g O g O g O g ex end
15 9.5 ess s Law ess s law f heat summatin states that fr a chemical equatin that can be written as the sum f tw r mre steps, the enthalpy change fr the verall equatin is the sum f the enthalpy changes fr the individual steps. Basically, R & P in individual steps can be added like algebraic quantities in determining verall equatin and enthalpy change. 15
16 simple example : Given: A + D E + C = X kj A + B C = Y kj Questin: D B + E =? 1. Crrect side?. Crrect # mles? A + D E + C = X kj C A + B = Y kj D B + E = X Y kj 16
17 Fr example, suppse yu are given the fllwing GIVEN data: S(s) SO 3 (g) O (g) SO ess s Law SO (g) (g); O (g); 97 kj 198 kj use these data t btain the enthalpy change fr the fllwing reactin? x flip S(s) 3O (g) SO 3 (g);? 17
18 S(s) SO If we multiply the first equatin by and reverse the secnd equatin, they will sum tgether t becme the third. S(s) SO 3 (g) (g) S(s) O O O (g) 3O (g) SO (g) (g) SO (g) SO SO (g); 3 SO (g); O 3 (g); (g); (g); 97 kj 198 kj (97 kj) (198 kj) 79 kj x flip () (1) W 45 18
19 9.6 Standard Enthalpies f Frmatin (mlecular scale) The standard enthalpy f frmatin f a substance, dented f, is the enthalpy change fr the frmatin f ne mle f a substance in its standard state frm its cmpnent elements in their standard state (98K & 1 atm). Ag (s) + ½ Cl (g) AgCl (s) f AgCl Nte that the standard enthalpy f frmatin fr a pure element in its standard state and + is zer. This means elements in their standard state has f = 0: metals  slids, diatmic gases, + in. 19
20 Standard Enthalpies f Frmatin Anther way t determine heat f reactin is the The law f summatin f heats f frmatin which states that the enthalpy f a reactin is equal t the ttal frmatin energy f the prducts minus that f the reactants. n f (prducts) (reactants ) is the mathematical symbl meaning the sum f, and n is the cefficients f the substances in the chemical equatin. n f 0
21 Ex. Generic Law f Summatin aa + bb cc + dd n f (prducts) n f (reactants ) 1
22 f A Prblem t Cnsider 4N (g) 5O (g) 4NO(g) 6 O(g) 3 : 45.9kJ / ml What is the standard reactin enthalpy, rxn, fr this reactin? n f (prducts) n f (reactants )
23 f Using the summatin law: 4N (g) 5O (g) 4NO(g) 6 O(g) 3 : 45.9kJ / ml [4mlN ( n f [4mlNO(90.3kJ kJ [361.kJ kJ (prducts) / mlno) / mln ( 3 ) m 6ml 5mlO kJ)] 183.6kJ O( (0kJ 906kJ f (reactants) 41.8kJ / mlo W kJ [( 183.6kJ) )] / ml 0kJ] O)] Be careful f arithmetic signs as they are a likely surce f mistakes. 3
24 35.1. The Secnd Law f Thermdynamics The secnd law f thermdynamics addresses questins abut spntaneity in terms f a quantity called entrpy. Entrpy, S, is a thermdynamic quantity that is a measure f the randmness r disrder f a system. The SI unit f entrpy is jules per Kelvin (J/K) and, like enthalpy, is a state functin. 4
25 R release E, spn (ex) absrb E, nnspn (end) P E P E R Mst sluble salts disslve in water spntaneusly; hwever, mst sluble salts disslve by an endthermic prcess. N 4 NO 3 (s) N 4 + (aq) + NO 3  (aq) = 8.1 kj There is an increase in mlecular disrder r randmness f the system. Slids: high rder/lw disrder, high energy Liquids: middle rder/lw disrder, medium energy Gases: lw rder/high disrder, lw energy 5
26 entrpy (S)  is a thermdynamic quantity that is a measure f hw dispersed the energy f a system is amng the different pssible ways that system can cntain energy, typically in J/K units. One example f entrpy is the amunt f mlecular disrder r randmness in the system. S increases as disrder increases and energy decreases gases have high disrder, lw energy slids have lw disrder, high energy We typically fllw the change in entrpy in the system s we treat it as a state prperty and measure S = S final  S initial + S = increase in entrpy, i.e. disrder increased;  U  S = decrease in entrpy, ie. disrder decreased ; + U This gets us t the secnd law f therm 6
27 Entrpy and the Secnd Law f Thermdynamics The secnd law f thermdynamics states that the ttal entrpy f a system and its surrundings increases fr a spntaneus prcess. 7
28 The tendency f a system t increase its entrpy (+ S) is the secnd imprtant factr in determining the spntaneity f a chemical r physical change in additin t. recap: spntaneus prcess: (system ges t lwer energy state) favred by  (ex) favred by + S (ie. increase disrder) nnspntaneus prcess: (system ges t higher energy state) favred by + (end) favred by  S (ie. decrease in disrder) D bth need t be true fr spn rxn? N, remember sluble salt disslving example. The larger term will dictate verall prcess. 8
29 35.4 Third Law f Thermdynamics The third law f thermdynamics states that the entrpy f all perfect crystalline substances appraches zer as the temperature appraches abslute zer (Kelvin). As temperature is raised the substance becmes mre disrdered as it absrbs heat and becmes a liquid then a gas, where entrpy > 0; S increases as temp increase. The entrpy f a substance is determined by measuring hw much heat is required t change its temperature per Kelvin degree (J/K). 9
30 35.5 Standard Reactin Entrpy The standard entrpy f a substance r in, als called its abslute entrpy, S, is the entrpy value fr the standard state f the species. Similar t heats f frmatin, f, except n abslute nt relative scale. Standard state implies 5 C (98K), 1 atm pressure, and 1 M fr disslved substances.(therm standard state) 30
31 Standard Entrpies and the Third Law f Thermdynamics This means that elements have nnzer values fr entrpy (abslute scale), unlike standard enthalpies f frmatin, f, which by cnventin, are zer (relative scale). The symbl S, rather than S, is used fr standard entrpies t emphasize that they riginate frm the third law and abslute nt relative values. 31
32 Entrpy Change fr a Reactin Yu can calculate the entrpy change fr a reactin using a summatin law, similar t the way yu btained f. S ns (prducts) ms (reactants ) Even withut knwing the values fr the entrpies f substances, yu can smetimes predict the sign f S fr a reactin. 3
33 Entrpy Change fr a Reactin The entrpy usually increases in the fllwing situatins: 1. A reactin in which a mlecule is brken int tw r mre smaller mlecules. AB A + B + S. A reactin in which there is an increase in the mles f gases. A(g) B(g) + C(g) + S 3. A prcess in which a slid changes t liquid r gas, r a liquid changes t gas. A(s) B(l) r B(g) B(l) C(g) + S + S 33
34 Predict S and spn/nnspn based nly n entrpy fr the fllwing rxns: C 4 (g) + Br (g) BrC C Br (l) gas t liquid; decrease in disrder;  S; nnspn based n S nly C 6 (g) + 7 O (g) 4 CO (g) + 6 O (g) 9 mls gas t 10 mls f gas; increase in disrder; + S; spn based n S nly W 47 C 6 1 O 6 (s) C 5 O (l) + CO (g) slid t liquid/gas (decmpse); increase in disrder; + S; spn based n S nly 34
35 A Prblem T Cnsider Calculate the change in entrpy, S, at 5 C fr the reactin in which urea is frmed frm N 3 and CO. N3(g) CO(g) NCON(aq) O(l) The calculatin is similar t that used t btain frm standard enthalpies f frmatin. Gas t liquid; decrease in disrder; predict  S 35
36 A Prblem T Cnsider N3( g) CO ( g) NCON ( aq) O( l) S : 193 J/ml. K S ns (prducts) ms (reactants ) S [(1mlN [(mln 3 )(193J CON / mlk ) )(174J / mlk ) (1mlCO (1ml O)(70J )(14J / mlk )] / mlk )] 356 J/K decrease in disrder as predicted 36
37 35.6 Gibbs Free Energy The questin arises as t hw d we decide if enthalpy r entrpy dictates the spntaneity f a reactin. What is the relatinship between and S? The American physicist J. Willard Gibbs intrduced the cncept f free energy (smetimes called the Gibbs free energy), G, which is a thermdynamic quantity defined by the equatin G= T S T Kelvin scale This quantity gives a direct criterin fr spntaneity f reactin. 37
38 Free Energy and Spntaneity Changes in an S during a reactin result in a change in free energy, G, given by the equatin G At a given temperature and pressure G = 0, the reactin gives an equilibrium mixture with significant amunts f bth reactants and prducts (Temp transfer pint where reactin switches spn/nnspn) G > 0, the reactin is nnspntaneus as written, and reactants d nt give significant amunts f prduct at equilibrium. G < 0, the reactin is spntaneus as written, and the reactants transfrm almst entirely t prducts when equilibrium is reached. T S 38
39 G T S Lets lk at relatinship amng the signs f, S and G and spntaneity. Nte that temperature will dictate which will rule. Als realize T is in K meaning n negative temp. S G Descriptin (ex) spn +disrder spn spn Spntaneus at all T + (end) nn disrder nn + nn Nnspntaneus at all T (ex) Spn disrder nn + r Spntaneus at lw T (rm); > T S;  G Nnspntaneus at high T (1000K); < T S + G Nnspntaneus at lw T; > T S; + G + (end) +disrder + r Spntaneus at high T; < T S;  G Nn spn enthalpy rules at lw temp but entrpy at very high T 39
40 35.7 Gibbs Energy and Equilibrium The standard free energy change, G, is the free energy change that ccurs when reactants and prducts are in their standard states. The next example illustrates the calculatin f the standard free energy change, G, frm and S. G T S 40
41 A Prblem T Cnsider What is the standard free energy change, G, fr the fllwing reactin at 5 C? N(g) 3(g) N3(g) f : kj/ml S : J/ml K predict, spn S, nnspn G, spn 41
42 n [ mln 3 S [(mln (3ml ns 3 f ( (prducts) )(130.6J 45.9kJ (prducts) )(193J / mlk )] / ml)] / mlk )] m ms [0] f [(1mlN 197 J/K (reactants) 91.8 (reactants ) )(191.5J kj / mlk ) kj/k Nw substitute int ur equatin fr G. Nte that S is cnverted t kj/k and Kelvin fr temp. G N( g) 3( g) N3( g) f : 0 S : kj/ml J/ml K T 91.8kJ kj S (98 K)( 33.1 spn rxn as written kj/k) 4
43 Standard Free Energies f Frmatin The standard free energy f frmatin, G f, f a substance is the free energy change that ccurs when 1 ml f a substance is frmed frm its elements in their stablest states at 1 atm pressure and 5 C. By tabulating G f fr substances, yu can calculate the G fr a reactin by using a summatin law. G n G f (prducts) m G f (reactants) 43
44 A Prblem T Cnsider Calculate G fr the fllwing reactin at 5 C using std. free energies f frmatin. C5O ( l) 3O ( g) CO ( g) 3O( g) G f : kj/ml G n G f (prducts) m G f (reactants) G [(1mlC [(mlco )( O )( kJ 174.8kJ / ml) / ml) (3ml 0] O)( 8.6kJ / ml)] G kj spn rxn 44
45 Relating G t the Equilibrium Cnstant The free energy change ( G) when reactants are in nnstandard states (meaning ther than 98K, 1 atm pressure r 1 M) is related t the standard free energy change, G, by the fllwing equatin. G G RTlnQ ere Q is the thermdynamic frm f the reactin qutient ([prducts]/[reactants] nt necessarily at equil); T in kelvin; R=8.31 J/mlK. 45
46 Relating G t the Equilibrium Cnstant G represents an instantaneus change in free energy at sme pint in the reactin appraching equilibrium G=0. G At equilibrium, G=0 and the reactin qutient Q becmes the equilibrium cnstant K. 0 G G RT ln Q RTlnK 46
47 Relating G t the Equilibrium G Cnstant This result easily rearranges t give the basic equatin relating the standard freeenergy change t the equilibrium cnstant. RTlnK When K > 1 (meaning equil lies t the right), the ln K is psitive and G is negative (spn). When K < 1 (meaning equil lies t the left), the ln K is negative and G is psitive (nnspn). 47
48 A Prblem T Cnsider Find the value fr the equilibrium cnstant, K, at 5 C (98 K) fr the fllwing reactin. The standard freeenergy change, G, at 5 C equals 13.6 kj/ml. N3(g) CO(g) NCON(aq) O(l) Rearrange the equatin G = RTlnK t give lnk G RT 48
49 A Prblem T Cnsider lnk G Substituting numerical values int the equatin, RT ln K J / 8.31J/(ml K) ml 98 K 5.49 K 5.49 e
50 Calculatin f G at Varius Temperatures We typically assume that and S are essentially cnstant with respect t temperature. Yu get the value f G T at any temperature T by substituting values f and S at 5 C int the fllwing equatin. G T T S 50
51 A Prblem T Cnsider Find the G fr the fllwing reactin at 5 C and 1000 C. Relate this t reactin spntaneity. CaCO3(s) CaO(s) CO(g) f : kj/ml S : J/ml K 51
52 [( S n f ns (prducts) m f (reactants) 393.5) ( 106.9)]kJ kj (prducts) ms [( ) (9.9)] J / K (reactants ) kj/k G T S T Nw yu substitute, S (= kj/k), and T (=98K) int the equatin fr G f. G G CaCO3(s) CaO(s) CO(g) f : S : kj/ml J/ml K 178.3kJ (98 K)( kj / K) 5 C 5 C kj S the reactin is nnspntaneus at 5 C. 5
53 A Prblem T Cnsider Find the G fr the fllwing reactin at 1000 C. Nw we ll use 1000 C (173 K) alng with ur previus values fr and S because assume des nt change much. G G 178.3kJ (173 K)( kj / K) 1000 C 1000 C 4.1kJ S the reactin is spntaneus at 1000 C. Yu see that this reactin change frm nnspn t spn smewhere between 5 C t 1000 C. w can we determine at what temp this switch ccurred? G=0 is equil, switch pint 53
54 T T determine the minimal temperature fr spntaneity, we can set Gº=0 and slve fr T. G 0 T S T S kj 111K ( kJ / K T S CaCO3(s) CaO(s) CO(g) nnspn < 848 C; CaCO 3 stable spn > 848 C; CaCO 3 decmpses easily C) W 48 Thus, CaCO 3 shuld be thermally stable until its heated t apprximately 848 C. This is way yu culd calculate the nrmal biling pint f a liquid. At G=0, the liquid phase and l g gas phase will be at equilibrium; temperature at which switch frm liquid t gaseus phase. 54
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