20.1 The 2 nd Law of Thermodynamics Spontaneous changes occur without any external influence Examples: Aging, rusting, heat transfer from

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1 hemdynamics: he Diectin f Chemical Reactins hemdynamics has tw maj aspects: Cnsevatin f enegy (1 st law) Diectin f pcesses (2 nd law) 20.1 he 2 nd Law f hemdynamics Spntaneus changes ccu withut any etenal influence Eamples: Aging, usting, heat tansfe fm ht t cld, epansin f gases int vacuum, etc. Nnspntaneus changes equie an etenal fce enegy in de t ccu Eamples: Refigeatin (heat tansfe fm cld t ht), evacuatin f flasks by vacuum pumps, etc. If a pcess is spntaneus in ne diectin, it is nt spntaneus in the the Eample: At -20 C, wate feezes spntaneusly, but ice des nt melt spntaneusly: H 2 O(l) H 2 O(s) spntaneus H 2 O(s) H 2 O(l) nnspntaneus Eample: At 25 C: 2Na(s) 2H 2 O(l) 2NaOH(aq) H 2 (g) spntaneus 2NaOH(aq) H 2 (g) 2Na(s) 2H 2 O(l) nnspntaneus Limitatins f the 1 st Law he 1 st law states that the ttal enegy f the univese is cnstant (E univ = cnst.) E univ can be sepaated int tw pats, enegy f the tem, E, and enegy f its suundings, E su E univ = E E su E univ = E E su Since E univ = cnst., E univ = 0 E E su = 0 E = E su Since E = q w (q w) = (q w) su Enegy (heat and/ wk) eleased by the tem is absbed by its suundings and vice vesa he 1 st law desn t eplain the diectin f spntaneus pcesses since enegy cnsevatin can be achieved in eithe diectin he Sign f and Spntaneity he sign f has been used (especially by ganic chemists) t pedict spntaneus changes nt a eliable indicat In mst cases spntaneus eactins ae ethemic ( < 0) cmbustin, fmatin f salts fm elements, neutalizatin, usting, In sme cases, spntaneus eactins can be endthemic ( > 0) disslutin f many salts, melting and vapizatin at high, A cmmn featue f all spntaneus endthemic eactins is that thei pducts ae less deed than the eactants (disde duing the eactin) Eample: Slid Liquid Gas mst deed least deed he Meaning f Disde and Entpy Mac-state the macscpic state f a tem descibed by its paametes (P, V, n,, E, H, etc.) Spntaneus pcesses pceed fm mac-states that ae less pbable twad nes that ae me pbable Mic-state ne f the pssible micscpic ways thugh which the mac-state can be achieved Refes t ne f the pssible ways the ttal enegy can be distibuted ve the quantized enegy levels f the tem his includes enegy levels elated t all pssible mtins f the mlecules (tanslatinal, tatinal, vibatinal, etc.) Me pbable mac-states ae thse that can be achieved in a lage numbe f diffeent ways (mic-states), and theefe, ae me disdeed In themdynamics, the disde f a mac-state is measued by the statistical pbability, W the numbe f diffeent ways the enegy can be distibuted ve the quantized enegy levels f the tem while still having the same ttal enegy, simply the numbe f mic-states thugh which the mac-state can be achieved Eample: Aangement f 2 paticles ( and ) in 3 pssible psitins hee ae 6 pssible aangements (W = 6) Inceasing the # f pssible psitins inceases W

2 W Disde Entpy (S) a state functin elated t the disde f the tem Bltzmann equatin S = k ln(w) k = J/K (Bltzmann cnstant) W Disde S S is a state functin depends nly n the pesent state f the tem and nt n the way it aived in that state he entpy change, S = S final S initial, is path independent If S > 0, the entpy and the disde incease If S < 0, the entpy and the disde decease he entpy and the disde f a tem can be inceased in tw basic ways: Heating (inceases the themal disde) Epansin, miing phase changes (incease the psitinal disde) Nte: Bth themal and psitinal disde have the same igin (the numbe f pssible mic-states) Eample: Gases epand spntaneusly in vacuum. Duing the epansin the ttal enegy des nt change (n heat is echanged and n wk is dne since P et = 0) he vlume inceases s the numbe f pssible psitins available t the paticles inceases W he pcess is diven by incease in the psitinal disde and thus incease in the entpy Entpy and the 2 nd Law he entpy f the tem alne can nt be used as a citein f spntaneity Spntaneus pcesses can have S > 0 S < 0 he tue citein f spntaneity is the entpy f the univese (S univ ) he 2 nd law f themdynamics f any spntaneus pcess, the entpy f the univese inceases ( S univ > 0) S univ = S S su > 0 hee ae n estictins n the signs f S and S su as lng as the sum f the tw is geate than ze Standad Mla Entpies and the 3 d Law he 3 d law f themdynamics the entpy f a pefect cystal at the abslute ze is ze (As 0K, S 0) At the abslute ze, the paticles f the cystal achieve the lwest pssible enegy and thee is nly ne way they can be aanged (W = 1) S = k ln(1) = k (0) = 0 he 3 d law allws the use f abslute entpies he entpy f a substance at a given is equal t the entpy incease accmpanying the heating f the substance fm 0 K that Standad mla entpy (S ) the entpy f 1 ml f a substance in its standad state and a specified tempeatue (usually 298 K) Standad state 1 atm f gases, 1 M f slutins, pue f liquids and slids Units f S J/ml K S can be affected by changing the themal and psitinal disde in diffeent ways empeatue changes f a given substance, S inceases as the tempeatue is inceased S As, the aveage E k f the mlecules and the # f ways t distibute E k amng the paticles s W and S Phase changes f a given substance, S inceases as the substance is cnveted fm slid t liquid t gas S (slid) <S (liquid) < S (gas) In the liquid and especially in the gas phase, the paticles have me feedm t mve aund an thus highe psitinal disde Within each phase, the entpy inceases gadually with inceasing Since phase changes ccu at cnstant, a shap change in entpy is bseved as the passes thugh the melting biling pints Since the gas phase has much highe S than the liquid and slid phases, S vap >> S fus

3 Standad mla entpy (cntinued) S can be affected by changing the themal and psitinal disde in diffeent ways Disslutin f slids liquids Usually S inceases as slids liquids disslve Miing inceases the psitinal disde, s S Disslving liquids & slids S Smetimes S deceases as substances disslve, especially f inic slids with small highly chaged ins (Al 3, Mg 2 ) Etensive in hydatin deceases the psitinal disde f the wate mlecules which vecmes the incease in the disde due t miing, s S Hydatin S Disslutin f gases S deceases as gases disslve in liquids slids he mlecules f the gas ae me esticted in slutin, s S Disslving gases S S inceases as gases ae mied with each the Miing inceases the psitinal disde, s S Miing f gases S Atmic size mlecula cmpleity F elements and f simila cmpunds in the same phase, S inceases with the mla mass (the # f electns, s S ) Mla mass S F cmpunds in the same phase, S inceases with the chemical cmpleity (S with the numbe f atms in the cmpund) # atms S F ganic cmpunds in the same phase, S inceases with the length f the hydcabn chain Length f chain S he effect f the physical state n S usually dminates the effect f mlecula cmpleity Eamples: Which f the fllwing has highe S a) Ai at 25 C ai at 35 C S inceases with b) CH 3 OH(l) at 25 C CH 3 OH(g) at 25 C S inceases fm liquid t gas c) NaCl(s) at 25 C NaCl(aq) at 25 C S typically inceases with disslutin f slids d) N 2 (g) at 25 C N 2 (aq) at 25 C S deceases with disslutin f gases e) HCl(g) at 25 C HB(g) at 25 C S inceases with inceasing the mla mass f) CO 2 (g) at 25 C CH 3 OH(g) at 25 C S inceases with inceasing the mlecula cmpleity (# atms) 20.2 Calculating the Entpy Change f a Reactin Entpy Changes in the System Standad entpy f eactin ( S ) the diffeence between the standad entpies f the pducts and the eactants S = ΣmSº(pducts) - ΣnSº(eactants) (n, m - stichimetic cefficients f eactants pducts) he equatin is simila t the Hess s law epessin f the standad eactin enthalpy º = Σm f º(pducts) - Σn f º(eactants)

4 Eample: Calculate the standad entpy ( S ) f the eactin N 2 O 4 (g) 2NO 2 (g) S = ΣmSº(pducts) - ΣnSº(eactants) S = 2 Sº(NO 2 (g)) 1 Sº(N 2 O 4 (g)) Fm Appendi B: S = 2 ml J/ml K 1 ml J/ml K S = J/K F eactins invlving gases: S > 0 if (# ml gaseus pducts) > (# ml gaseus eactants) S < 0 if (# ml gaseus pducts) < (# ml gaseus eactants) Entpy Changes in the Suundings he suundings functin as a heat sink f the tem (eactin) q su =-q Ethemic eactins heat is lst by the tem and gained by the suundings which inceases the themal disde in the suundings q < 0 q su > 0 and S su > 0 Endthemic eactins heat is gained by the tem and lst by the suundings which educes the themal disde in the suundings q > 0 q su < 0 and S su < 0 S su is pptinal t the amunt f heat tansfeed S su q su S su - q S su is invesely pptinal t the since the heat tansfe changes the disde f the suundings me at lwe S su 1/ S su = - q / At cnstant pessue (q p = ) S su = - / he equatin allws the calculatin f S su fm the eactin enthalpy and the tempeatue (applies stictly nly at cnstant and P) Accding t the 2 nd law, f a spntaneus eactin S univ = S S su > 0 Substituting S su with - / leads t S univ = S > 0 he equatin allws the calculatin f S univ fm the eactin entpy and enthalpy and the tempeatue (applies stictly at cnstant and P) he equatin pvides a citein f spntaneity in any tems at cnstant and P Psitive S univ (spntaneus pcess) is faved by Psitive entpy f eactin ( S > 0) the disde f the tem inceases Negative enthalpy f eactin ( < 0) the disde f the suundings inceases Eample: Is the cmbustin f glucse spntaneus at 25 C? C 6 H 12 O 6 (s) 6O 2 (g) 6CO 2 (g) 6H 2 O(g) Calculate S and using Appendi B S = [6 Sº(CO 2 (g)) 6 Sº(H 2 O(g))] [1 Sº(C 6 H 12 O 6 (s)) 6 Sº(O 2 (g))] S = [6(214) 6(189)] [1(212) 6(205)] = 976 J/K = [6 f º(CO 2 (g)) 6 f º(H 2 O(g))] [1 f º(C 6 H 12 O 6 (s)) 6 f º(O 2 (g))] = [6(-394) 6(-242)] [1(-1273)] = kj S = kj/k S su = / = (-2543 kj)/(298 K) = 8.53 kj/k S univ = = 9.51 kj/k > 0 spntaneus Entpy Changes and the Equilibium State S univ > 0 the fwad eactin is spntaneus S univ < 0 the fwad eactin is nnspntaneus (the evese eactin is spntaneus) S univ = 0 the eactin is at equilibium At equilibium: S univ = S / = 0 At equilibium: S = / he equatin is useful f calculating the entpies f phase changes duing which the tem is at equilibium at cnstant and P Eample: Calculate S vap f H 2 O at its nmal b.p. vap = 40.7 kj pe 1 ml f H 2 O S vap = vap / = 40.7 kj / 373 K = kj/k

5 Spntaneus E and Endthemic Reactins S univ = S S su > 0 F ethemic eactins S su = - / > 0 If S > 0, the eactin is spntaneus If S < 0, the eactin is spntaneus nly if the incease f S su is geate than the decease f S F endthemic eactins S su = - / < 0 If S > 0, the eactin is spntaneus nly if the incease f S is geate than the decease f S su If S < 0, the eactin is nt spntaneus Eample: 2H 2 (g) O 2 (g) 2H 2 O(g) heat Ethemic S su > 0 Less gaseus pducts S < 0 S su dminates spntaneus 20.3 Entpy, Fee Enegy and Wk Gibbs fee enegy (G) a state functin defined as: G = H S Fee Enegy Change and Spntaneity Fm the 2 nd law at cnstant and P, Suniv = S Ssu Suniv = S Multiply the equatin by (-) - S univ = S S = (H f H i ) (S f S i ) = = (H f S f ) (H i S i ) = (G f G i ) = G G = S = S univ At cnstant and P, the entpy change f the univese is elated t the fee enegy change f the tem - S univ = G If S univ > 0 - S univ < 0 G < 0 G < 0 a citein f spntaneity in any tem at cnstant and P G < 0 the fwad eactin is spntaneus G > 0 the fwad eactin is nn-spntaneus (the evese eactin is spntaneus) G = 0 the eactin is at equilibium F simplicity, the subscipt can be mitted fm all state functins elated t the tem, s at cnst. and P G = S he Effect f empeatue n G G depends stngly n Use the symbl G f tempeatue and S depend vey little n and S can be assumed independent f G = S he sign f G depends n the signs f and S and the magnitude f S High G Lw Yes at all s Spntaneus? N at high s; Yes at lw s Yes at high s; N at lw s N at all s If and S have the same sign, G can be psitive negative depending n the value f At a cetain, the tem eaches equilibium G = 0 S = 0 = S = / S at which equilibium is eached At any the, the tem is nt at equilibium, and the pcess eithe is isn t spntaneus he -ange at which the pcess is spntaneus, can be fund fm G <0 S < 0 < S Slving f gives the desied -ange Nte: Multiplying dividing with a (-) numbe changes (<) t (>) Eample: Is the fllwing eactin spntaneus at high lw tempeatues? 3H 2 (g) N 2 (g) 2NH 3 (g) heat Ethemic eactin < 0 Less gaseus pducts S < 0 G = S = () () = () () < 0 at lw Eample: F the same eactin at cetain cnditins, = kj and S = -197 J/K. What is the -ange at which the eactin is spntaneus? G = S < 0 < S kj < ( kj/k) Multiply by (-1) 91.8 kj > (0.197 kj/k) Nte: (<) flips t (>) < 91.8 kj/0.197 kj/k < 466 K -ange

6 Standad Fee Enegy Changes Standad fee enegy f eactin ( G, )the fee enegy change f a eactin in which all eactants and pducts ae pesent in thei standad states at a specified tempeatue Standad fee enegy f fmatin ( G f, )the standad fee enegy f the eactin f fmatin f 1 ml f a substance fm its elements at a specified tempeatue (usually 298 K) G º, = Σm G f º, (pducts) - Σn G f º, (eactants) G f º,298 values ae given in Appendi B he equatin can be used t calculate G º,298 f a eactin (nly f 298 K!) At any the tempeatue ( 298 K), G, is calculated fm the equatin G, = S Eample: Calculate the standad fee enegy f the eactin 2SO 2 (g) O 2 (g) 2SO 3 (g) at (a) 298 K and (b) 1500 K a) G º,298 = 2 G f º,298 (SO 3 (g)) [2 G f º,298 (SO 2 (g)) 1 G f º,298 (O 2 (g))] = = 2(-371) [2(-300) 1(0)] = -142 kj G º,298 < 0 he eactin is spntaneus at standad cnditins and 298 K b) N data f G f º,1500 use G, = S º = 2 f º(SO 3 (g)) [2 f º(SO 2 (g)) 1 f º(O 2 (g))] = = 2(-396) [2(-297) 1(0)] = -198 kj S º = 2 Sº(SO 3 (g)) [2 Sº(SO 2 (g)) 1 Sº(O 2 (g))] = = 2(257) [2(248) 1(205)] = -187 J/K G,1500 = -198 kj 1500 K ( kj/k) = = -198 kj 281 kj = 83 kj G º,1500 > 0 he eactin is nt spntaneus at standad cnditins and 1500 K G and the Wk a System Can D G is equal t the maimum wk btainable fm a tem in which a spntaneus pcess takes place G = w ma G is als equal t the minimum wk that must be dne in de t evese a spntaneus pcess btain the maimum wk fm a tem, the pcess must be caied ut evesibly Revesible pcesses ae caied ut thugh infinitely small steps (infinitely slw) Can be evesed withut leaving pemanent changes in the tem its suundings he tem is in an almst equilibium state duing such pcesses When wk is dne evesibly, the diving fce f the pcess eceeds the ppsing fce by an infinitely small amunt s the pcess is etemely slw Real pcesses ae nt evesible because they ae caied ut faste in a limited numbe f steps he wk btained fm eal pcesses is less than the maimum wk (less than G) he unhanessed ptin f the fee enegy ( G) is lst t the suundings as heat In eal pcesses, ne must cmpmise between the speed and amunt f wk (fee enegy) gained fm the pcess 20.4 Fee Enegy, Equilibium, and Reactin Diectin he diectin f eactin can be pedicted fm the sign f G by cmpaing Q and K If Q < K Q/K < 1 ln Q/K < 0 and G < 0, the fwad eactin pceeds If Q > K Q/K > 1 ln Q/K > 0 and G > 0, the evese eactin pceeds If Q = K Q/K = 1 ln Q/K = 0 and G = 0, the eactin is at equilibium G and ln Q/K have the same signs and ae elated: Q G = R ln = R lnq R ln K K

7 G is cncentatin dependent (depends n Q) G can be viewed as a diffeence between the fee enegy f the tem at the cuent cncentatins f eactants and pducts, Q, and that at equilibium, K At standad-state cnditins, all cncentatins and pessues ae equal t 1, s Q = 1 and G = G G, = R ln1 R lnk = 0 R lnk G, = R ln K he equatin is used t calculate G fm K and vice vesa If K >1 G < 0 pducts ae faved at equilib. If K < 1 G > 0 eactants ae faved at equilib. If K = 1 G = 0 Eample: Calculate K p at 298 K f the eactin 2NO(g) O 2 (g) 2NO 2 (g) Calculate G,298 = 298 K G,298 f values fm Appendi B can be used G º,298 = 2 G f º,298 (NO 2 (g)) [2 G f º,298 (NO(g)) 1 G f º,298 (O 2 (g))] = = 2(51.3) [2(86.6) 1(0)] = kj/ml G º, = -R lnk ln K = - G º, /R K, G ( 70.6 kj/ml) 3 R kj/ml K 298K 12 = e = e = K p >> 1 the pducts ae highly faved = K p Cmbining G = R lnq R lnk with G, = R lnk leads t: G = G, R lnq he equatin is used t calculate G f any nnstandad state fm G, f the espective standad state bth at tempeatue,, and the eactin qutient f the nnstandad state, Q G, is calculated as G, = S R = kj/ml K is the same f all quantities in the equatin G is assciated with a cetain cmpsitin (Q) and a cetain tempeatue () Eample: Calculate the fee enegy change f the eactin H 2 (g) I 2 (g) 2HI(g) at 500 K if the patial pessues f H 2, I 2, and HI ae 1.5, 0.88 and atm, espectively. Nnstandad state at 500 K ( G 500 =?) Calculate G º,500 as G, = S º = 2 f º(HI(g)) [1 f º(H 2 (g)) 1 f º(I 2 (g))] = = 2(25.9) [1(0) 1(62.4)] = kj S º = 2 Sº(HI(g)) [1 Sº(H 2 (g)) 1 Sº(I 2 (g))] = = 2(206) [1(131) 1(261)] = 20 J/K G,500 = kj 500 K (0.020 kj/k) = kj G 500 Q p = G 2 PHI = P P,500 H 2 I2 kj = ml kj kj = ml ml 500 kj G = 44.5 ml 2 (0.065) = = R 500 K lnq 3 3 kj 500K ln( ml K G 500 < 0 he eactin is spntaneus at this nnstandad state and 500 K 3 ) Deiving the van t Hff Equatin he van t Hff equatin gives the tempeatue dependence f the equilibium cnstant, G = R ln K R ln K = S, G = S S ln K = R R If the equatin is applied f tw diffeent s, and and S ae assumed independent f K ln K 1 = R

8 -ln K S ln K = R R - S /R Slpe = /R 1/ If slpe > 0, > 0, endthemic If slpe < 0, < 0, ethemic A plt f ln K vesus 1/ gives a staight line with a Slpe = /R and Intecept = - S /R Allws the epeimental deteminatin f and S fm measuements f K at diffeent s