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 Spntaneus and Nnspntaneus Prcesses (18.1, 18.2) One knws frm experience that there is a natural directin fr all kinds f changes in the physical universe. This natural directin is implicitly assuming a directin fr time. What factrs determine the natural directin f a prcess? While these factrs may appear t be vague, thermdynamics ffers a precise definitin. Furthermre, thermdynamics defines a functin f any chemical system which plays the same rle fr chemical prcesses that ptential energy des fr rdinary mechanics f macrscpic bjects. We seek the factrs determining the spntaneity f a reactin and the functin, which will nw be dented by F. This is the functin we have been seeking. F F reactants prducts reactants prducts spntaneus nnspntaneus ΔF = 0 => equilibrium spntaneus directin examples f reversible (equilibrium) and irreversible (spntaneus) prcesses; nnspntaneus prcesses d nt ccur (withut utside interventin)! Reversible (equilibrium) prcess Expansin f a gas when P P ext. A slight increase in P ext will cause the gas t cntract. Ice in water at 0 C. A slight heat flw int the icewater mixture will melt sme ice; a slight withdrawal f heat frm the mixture will freeze sme water. A chemical reactin at equilibrium, e.g., CH 3 COOH + H 2 O < = > CH 3 COO - + H 3 O + A slight increase in the cncentratin f CH 3 COOH will prduce mre ins; a slight increase in the cncentratin f either in will prduce mre CH 3 COOH (exactly in accrd with LeChâtelier s principle). Irreversible (spntaneus) prcess Expansin f a gas int a vacuum. N slight change in pressure can stp the expansin. Ice in water at 20 C. The ice will melt even if the water is cled several degrees (t any temperature abve C). A chemical reactin far frm equilibrium, e.g., a piece f zinc drpped int acid Zn + 2H + Zn 2+ + H 2 N small (r even large) increase in the cncentratin f Zn 2+ r in the pressure f H 2 can reverse the directin f the reactin.
-2- predictin => state functin => is it ΔH? 1) NH 4 NO 3 (s) NH + 4 (aq) + NO - 3 (aq) ΔH = 25.7 kj ml -1 > 0 (but spntaneus) (cld packs) 2) H 2 O(l, T > 0 C) H 2 O(s) ΔH = - ΔH fus < 0 (but nnspntaneus) predictin => state functin => is it ΔE? 1) unmixing f gases Why des this nt ccur? 2) heat flw frm cld t ht Why des this nt ccur? Entrpy and the Secnd Law f Thermdynamics (18.3, 18.4) Entrpy FIG I - Expansin f an Ideal Gas int a Vacuum Niclas Lénard Sadi Carnt spntaneus directin (fr a change in an islated system) - the ne which leads t the greatest number f micrstates (has the highest prbability fr the energy t be distributed r dispersed as randmly, chatically, disrderly as pssible) - due t randm behavir f the large numbers f mlecules in macrscpic systems
-3- Why reactin may nt be spntaneus: 1. 2. statistical definitin f entrpy (micrscpic) rdered (lw S) disrdered (high S) state functin S R statistical definitin due t Bltzmann: S = k ln W = ln W N W is the number f energetically equivalent ways t arrange the cmpnents f the system (atms and mlecules) amng their available micrstates. EX 1. What is S fr ne mle f crystalline N 2 O at 0 K if each mlecule can exist in ne f tw pssible rientatins? Arrangements with same energy: NNO and ONN thermdynamic definitin f entrpy (macrscpic) entrpy is a state functin its value is determined by the heat transferred n a reversible path whether the actual prcess is reversible r nt fr a prcess whse temperature is cnstant VERY USEFUL Rudlf Julius Emanuel Clausius The energy f the universe is cnstant. The entrpy f the universe tends t a maximum. secnd law f thermdynamics (ur spntaneity criterin!) fr any spntaneus prcess the entrpy f the universe increases: ΔS univ > 0 The Entrpy Assciated with a Change f State GAS LIQUID The entrpy f liquid brmine is 152 and that f the gas is 245 J K -1 ml -1
-4- What Effects S? vlume FIG II - Entrpy n, the Mlecular Level: Translatinal, Vibratinal, and Rtatinal Mtin pressure temperature physical state phase change FIG III - Entrpy: Effect f Temperature and Change f State EX 2. The nrmal biling pint f ethanl, C 2 H 5 OH, is 78.3 C and its mlar enthalpy f vaprizatin is 38.56 kj/ml. What is the change in entrpy if 25.0 g f C 2 H 5 OH(g) at 1 atm pressure cndenses t a liquid at the nrmal biling pint? atmic, mlecular size cmplexity FIG IV - Disslutin disslutin
-5- SPONTANEITY CRITERION.................... ΔS univ 2nd law f thermdynamics (18.3): fr any spntaneus prcess the entrpy f the universe (ΔS univ = ΔS sys + ΔS surr ) increases SPONTANEITY CRITERION.................... ΔG sys Gibbs Free Energy (18.5, 18.6): ΔG = ΔH TΔS maximum useful wrk which can be extracted frm a clsed system J. Willard Gibbs
-6- Which d we use, ΔS univ r ΔG? EX 3. Predict the spntaneity f the melting f ice. H 2 O(s) H 2 O(l) at -10 C, 0 C, and 10 C given that ΔH fus = 6030 J ml -1 and ΔS fus = 22.1 J K -1 ml -1. Use bth criteria f ΔS univ and ΔG. Assume that ΔH fus and ΔS fus are cnstant in the range frm -10 C t 10 C. T ( C) T (K) ΔH fus (J ml -1 ) ΔS fus (J K -1 ml -1 ) ΔH fus ΔS surr = T (J K -1 ml -1 ) ΔS univ = ΔS fus + ΔS surr (J K -1 ml -1 ) TΔS fus (J ml -1 ) ΔG fus = ΔH fus TΔS fus (J ml -1 ) -10 263 6030 22.1-22.9-0.8 5810 +220 0 273 6030 22.1-22.1 0 6030 0 10 283 6030 22.1-21.3 +0.8 6250-220
-7- Entrpy Changes in Chemical Reactins (18.7) Standard Mlar Entrpies (Abslute Entrpies) and the Third Law f Thermdynamics FIG V - Entrpy f a Crystal 3rd Law f Thermdynamics The entrpy f a perfect crystal, in equilibrium, at abslute zer (0 K) is zer. S = 0 at T = 0 perfect crystal at 0 K W = 1 S = 0 T determine the value f S, it is necessary t measure the energy transferred as heat under reversible cnditins fr the cnversin frm 0 K t the desired final temperature. As this cnversin is dne at cnstant pressure, in the thermdynamic definitin f entrpy, ΔS = q rev /T, we can recgnize that, since q rev = c P ΔT we can just divide the heat capacity by T and perfrm the integratin frm 0 t T final. This gives the abslute entrpy. Since it is necessary t add energy as heat t raise the temperature, all substances have psitive entrpy values at temperatures abve 0 K. FIG VI Determining S fr Platinum
-8- Standard Mlar Entrpies, S FIG VII - Standard Mlar Entrpies, S f the Elements at 298 K Table 18.2 Selected Standard Mlar Entrpies, J K -1 ml -1 at 298 K Gases S Liquids S Slids S H 2 (g) 130.57 H 2 O(l) 69.91 Li(s) 29.12 N 2 (g) 191.50 CH 3 OH(l) 126.8 Na(s) 51.21 O 2 (g) 205.03 Br 2 (l) 152.23 K(s) 64.18 Br 2 (g) 245.35 I 2 (s) 116.14 H 2 O(g) 188.72 NaCl(s) 72.11 CH 3 OH(g) 239.70 MgO(s) 26.85 CH 4 (g) 186.15 CaCO 3 (s) 91.7 C 2 H 6 (g) 229.49 C(s,gr) 5.74 C 3 H 8 (g) 269.91 C(s,d) 2.377 EX 4. Determine the entrpy change fr 2H D H H + D D 143.80 130.57 144.96
Gibbs Free Energy Changes in Chemical Reactins (18.8, 18.9 calculating ΔG) -9-1) standard free energy fr a reactin, ΔG = ΔH TΔS calculate frm values f ΔH and ΔS fr the reactin using the standard temperature (usually 298 K, 25 C) 2) ΔG is a state functin use "Hess's law" fr cmbining Gibbs free energies f reactin as is dne fr ΔH EX 5. Frm the fllwing Gibbs energy changes at 1 atm and 25 C 1) C(s, diamnd) + O 2 (g) CO 2 (g) ΔG 1 = -397 kj 2) C(s, graphite) + O 2 (g) CO 2 (g) ΔG 2 = -394 kj Find: C(s, diamnd) C(s, graphite) ΔG = 3) standard free energy calculated frm standard free energies f frmatin, GG f by the same manner that standard enthalpies f frmatin are calculated standard mlar Gibbs energy f frmatin, GG f, f a cmpund is the Gibbs energy change fr the reactin frming ne mle f the cmpund frm its elements in their standard states. GG f (element in standard state) = 0 4) ΔG(T, P=1 atm) = ΔG (T) ΔH TΔS (calculatin fr nnstandard temperatures) fr a pressure f 1 atm and a temperature ther than the standard 25 C assume that ΔH and ΔS d nt change much with temperature and use them with the target temperature t calculate ΔG. (VERY USEFUL) FIG VIII - Effect f Temperature Fr example, the entrpy change f the reactin 3 NO(g) N 2 O(g) + NO 2 (g) varies less than 5% between 0 C and 300 C; the enthalpy change varies even less. The Gibbs energy change in the reactin shifts greatly ver the temperature range as the magnitude f TΔS increases. This gives credence t the cmmn apprximatin that ΔH and ΔS are independent f temperature ver the temperature range being cnsidered. 5) ΔG(T,P) = ΔH (T,P) TΔS(T,P) use values f ΔH and ΔS at the T and P f interest
-10- ΔG and Temperature EX 6. Calculate, fr the fllwing reactin 2 CuCl 2 (s) 2 CuCl(s) + Cl 2 (g) H f (kj ml -1 ) -220.1-137.2 0 S (J K -1 ml -1 ) 108.07 86.2 222.96 GG f (kj ml -1 ) -175.7-119.88 0 a) ΔH b) S c) ΔG at 590 K, assuming H r and SS r are independent f temperature. d) ΔG at 590 K, using the measured mlar values f ΔH(590) = 158.36 kj and ΔS(590) = 177.74 J K -1 Effect f the Sign f ΔH and ΔS n the Spntaneity f a Prcess
-11- EX 7. Predict at which temperature, if any, the fllwing reactin is spntaneus N 2 O 4 (g) 2 NO 2 (g) HH f (kj ml -1 ) 9.16 33.18 S (J K -1 ml -1 ) 304.18 239.95 GG f (kj ml -1 ) 97.82 51.29 EX 8. Predict at which temperature, if any, the fllwing reactin is spntaneus H 2 O(l) H 2 O(g) HH f (kj ml -1 ) -285.8-241.8 S (J K -1 ml -1 ) 69.91 188.7 Free Energy Changes fr Nnstandard Pressure, Cncentratin (18.9, 18.10) review equilibrium cnstant and prduct qutient, Chapter 15.2-15.7 ΔG(T, P) = ΔG (T) + RT ln Q(T, P) prduct qutient, Q(T, P) Gibbs Free Energy and Equilibrium van t Hff Equatin Jacbus Henricus van't Hff
-12- EX 9. Fr the reactin at 298 K 2 NO 2 (g) N 2 O 4 (g) GG f (kj ml -1 ) 51.29 97.82 a) What is the value f the equilibrium cnstant at 298 K? b) If the partial pressure f NO 2 is 1 atm, fr which partial pressures is the reactin spntaneus? c) If the partial pressure f NO 2 is 1 atm, fr which partial pressures is the reactin nnspntaneus? FIG IX - Hw ΔG Changes During the Curse f a Reactin The Gibbs energy f a reactin is pltted against its prgress frm pure reactants (left) t pure prducts (right). ΔG r can be assciated with the slpe f this line. Equilibrium cmes when the Gibbs energy is minimized, where ΔG = 0. On the reactant side f equilibrium, the slpe is negative, ΔG r < 0 and Q < K. On the prduct side, the slpe is psitive, ΔG r > 0 and Q > K. ΔG = ΔG + RT ln Q = RT ln (Q/K)
-13- Temperature Dependence f the Equilibrium Cnstant (18.10) Chapter 11.5- Clausius-Clapeyrn Equatin- van't Hff equatin applied t vaprizatin Nte: applicatin f linear regressin Benît Paul Émile Clapeyrn EX 10. The value f K is 3.7 x 10-6 at 900 K fr the ammnia synthesis reactin. Assuming the value f ΔH fr this reactin is -92 kj, calculate the value f K at 550 K. EX 11. Fr the vaprizatin at 25 C CH 3 CH 2 OH(l) CH 3 CH 2 OH(g) HH f (kj ml -1 ) -277.69-235.10 GG f (kj ml -1 ) -174.89-168.57 a) What is the vapr pressure? b) What is the equilibrium cnstant at 400 K?
-14- EX 12. The nrmal biling pint f ethanl, CH 3 CH 2 OH, is 351.7 K. Estimate the vapr pressure at 298.15 K. The enthalpy f vaprizatin is 42.59 kj/ml. The Gibbs Free Energy and Wrk (Why Free Energy is Free ) (pp. 868-869)