Chapter 17: Thermodynamics: Spontaneous and Nonspontaneous Reactions and Processes

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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 rd Laws: Chap. 17) he three laws f thermdynamics are given in the table frm a chemist s pint f view. he sign cnventins fr heat, q, and wrk, w, are given belw: q is psitive fr an endthermic prcess (absrbs heat). q is negative fr an exthermic prcess (releases heat). w is psitive when wrk is dne n the system by the surrundings. w is negative when wrk is dne by the system n the surrundings. Laws f hermdynamics Law Statement Equatin 1 st Law Energy can be cnverted frm ne frm t anther, but cannt be created r destryed. D E = q+ w Remember the sign cnventins fr heat and wrk. 2 nd Law he entrpy f the universe increases in a spntaneus prcess and remains unchanged in an equilibrium prcess. Fr a spntaneus prcess, D Suniverse =D Ssystem +D Ssurrundings > 0 Fr an equilibrium prcess, D S =D S +D S = 0 universe system surrundings 3 rd Law he entrpy f a perfect crystalline substance is zer at the abslute zer f temperature. Spntaneus Prcesses Water runs dwnhill but never uphill. A lump f sugar disslves in cffee but never reverses the prcess. Heat flws frm a htter bject t a clder ne but never reverses the prcess. Irn expsed t water and xygen frms rust, but rust never spntaneusly reverts t irn. A large number f exthermic reactins are spntaneus. Ex. CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2O(l) DH = -890.4 kj/mle Ex. H + (aq) + OH - (aq) H 2O(l) DH = -56.2 kj/mle Hwever, smetimes the assumptin that spntaneus prcesses always decrease a system s energy (i.e. exthermic reactins are always spntaneus) fails: H 2O(s) H 2O(l) DH = +6.01 kj/mle NH 4NO 3(s) NH + 4 (aq) + NO - 3 (aq) DH = +25 kj/mle Exthermicity favrs the spntaneity f a reactin but des nt guarantee it. We cannt decide whether r nt a chemical reactin will ccur spntaneusly slely n the basis f energy changes in the system. make this kind f predictin, we need anther thermdynamic quantity called entrpy.

Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses, Page 2 17.2: hermdynamic Entrpy Entrpy Entrpy, DS, is a measure f disrder r randmness fr a system. When entrpy increases, D S > 0, and when entrpy decreases, D S < 0. Prcesses (r cmparisns) that increase in entrpy: Slid Liquid Gas Slute Slutin Lwer temperature Higher temperature Fewer particles Mre particles Fewer atms in a mlecule Mre atms in a mlecule Standard entrpy is the abslute entrpy f a substance at 1 atm and 25 C. (Recall that the standard state refers nly t 1 atm. he reasn fr specifying 25 C is that many prcesses are carried ut at rm temperature.) Ex. Predict whether the entrpy change is greater r less than zer fr each f the fllwing prcesses: (a) freezing ethanl (b) evaprating a beaker f liquid brmine at rm temperature (c) disslving glucse in water (d) cling nitrgen gas frm 80 C t 20 C. Secnd Law f hermdynamics he cnnectin between entrpy and the spntaneity f a reactin is expressed by the secnd law f thermdynamics: he entrpy f the universe increases in a spntaneus prcess and remains unchanged in an equilibrium prcess. Mathematically, the secnd law f thermdynamics states that: fr a spntaneus prcess, D S =D S +D S > 0, universe system surrundings and fr an equilibrium prcess, D S =D S +D S = 0. universe system surrundings -DH system Entrpy changes in the surrundings are represented by D Ssurrundings = where is the Kelvin temperature. 17.3: Abslute Entrpy and the hird Law f hermdynamics hird Law f hermdynamics Accrding t the third law f thermdynamics, the entrpy f a perfect crystalline substance is zer at the abslute zer f temperature. he imprtant pint abut the third law f thermdynamics is that it enables us t determine the abslute entrpies f substances. 17.4: Calculating Entrpy Changes Suppse that a system is represented by a A + b B c C + d D, then D S (prducts). Als see the related equatins in the table f rxn =SnS -SmS (reactants) Similar hermdynamic Calculatins fr a System. he standard entrpy, S, values fr a large number f cmpunds appear in Appendix 3. Ex. Frm the standard entrpy values in Appendix 3, calculate the standard entrpy changes fr the fllwing reactins at 25 C: (a) CaCO 3(s) CaO(s) + CO 2(g) (b) N 2(g) + 3 H 2(g) 2 NH 3(g) (c) H 2(g) + Cl 2(g) 2 HCl(g) Similar hermdynamic Calculatins fr a System Quantity Calculatin Cmments

Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses, Page 3 Standard enthalpy f reactin, DH rxn Standard entrpy f reactin, Standard free-energy f reactin, DG rxn D H =SnH (prducts) -SmH rxn f f (reactants) D S =SnS (prducts) -SmS (reactants) rxn D G =SnG (prducts) -SmG (reactants) rxn f f he standard enthalpy f frmatin f any element in its mst stable frm is zer. he standard free-energy f frmatin f any element in its mst stable frm is zer. General rules fr quickly evaluating : is psitive if a reactin prduces mre gas mlecules than it cnsumes. is negative if the ttal number f gas mlecules decreases. can be either a small psitive r a small negative number if there is n net change in the ttal number f gas mlecules. Ex. Predict whether the entrpy change f the system in each f the fllwing reactins is psitive r negative. (a) 2 H 2(g) + O 2(g) 2 H 2O(l) (b) NH 4Cl(s) NH 3(g) + HCl(g) (c) H 2(g) + Br 2(g) 2 HBr(g) -DH system Entrpy changes in the surrundings are represented by D Ssurrundings = where is the Kelvin temperature. 17.5: Free Energy Gibbs Free Energy -DH system Substituting D S int surrundings = D Suniverse =D Ssystem +D Ssurrundings > 0 and multiplying bth sides by -1 gives -D S. Replacing universe =DHsystem -D Ssystem < 0 -DS universe with DG gives D G =DHsystem -D Ssystem < 0 fr a spntaneus reactin. hus, the change in free energy, DG, fr a cnstant-temperature prcess is represented by D G =DH -DS in which DH is the enthalpy f the system, DS is the entrpy f the system, and is the Kelvin temperature. If DG is negative, the reactin is spntaneus (it prceeds in the frward directin). If DG is psitive, the reactin is nnspntaneus (it prceeds in the reverse directin). If DG = 0, the system is at equilibrium. here is n net change. he standard free-energy f reactin, DG f, is the free-energy change fr a reactin when it ccurs under standard-state cnditins, when reactants in their standard states are cnverted t prducts in their standard states. Suppse that a system is represented by a A + b B c C + d D, then D G (prducts). Als see the related equatins in the table f rxn =SnGf -SmGf (reactants) Similar hermdynamic Calculatins fr a System. he standard free-energy f frmatin, DG f, f any element in its mst stable frm is zer.

Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses, Page 4 Ex. Calculate the standard free-energy changes fr the fllwing reactins at 25 C. (a) CH 4(g) + 2 O 2(g) CO 2(g) + 2 H 2O(l) (b) 2 MgO(s) 2 Mg(s) + O 2(g) Use Appendix 3 t find the DG f values. 17.6: emperature and Spntaneity At the temperature at which a phase transitin ccurs, the system is at equilibrium ( DG = 0) and DH D G =DH -DS= 0. Frm this, we can write D S =. Fr example, fr the ice water transitin, DH is the mlar heat f fusin and is the melting 6010 J/mle pint. he entrpy change is D S. Similarly, fr the ice water = = 22.0 J/K-mle 273 K -6010 J/mle water ice transitin, the change is D S. water ice = =-22.0 J/K-mle 273 K Ex. he mlar heats f fusin and vaprizatin f benzene are 10.9 kj/mle and 31.0 kj/mle, respectively. Calculate the entrpy changes fr the slid liquid and liquid vapr transitins fr benzene. At 1 atm pressure, benzene melts at 5.5 C and bils at 80.1 C. Under what cnditins will we have a spntaneus reactin (negative DG )? See the Factrs Affecting the Sign f DG table t see hw the signs f DH and DS affect DG. Nte that the thinking prcess t prduce the cnclusins f the table is as imprtant as the cnclusins themselves. Factrs Affecting the Sign f DG in the Relatinship D G =DH -DS Sign f DH Sign f Sign f DG Is Reactin Spntaneus? DS Negative Psitive Always negative Yes, always Psitive Negative Always psitive N, always Psitive Psitive Negative at high temperature Yes, at high temperature Negative Negative Negative at lw temperature Yes, at lw temperature 17.7: Free Energy and Chemical Equilibrium Free Energy and Chemical Equilibrium During the curse f a chemical reactin, nt all the reactants and prducts will be in their standard states. Under this cnditin, the relatinship between DG and DG, which can be derived frm thermdynamics, is D G=D G + Rln Q where R is the gas cnstant (8.314 J/mle-K, is the abslute temperature f the reactin, and Q is the reactin qutient. At equilibrium, by definitin, DG = 0 and Q= Kwhere K is the equilibrium cnstant. hus, D G=D G + RlnQ= 0 and D G =-Rln K. In the equatin D G =-Rln K, K is used fr gases and is used fr reactins in P KC slutin. Nte that the larger the K is, the mre negative DG is.

Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses, Page 5 D G =-Rln K allws chemists t calculate equilibrium cnstants frm thermdynamic data. Nte that it is the sign f DG and nt that f DG that determines the directin f reactin spntaneity. he sign f DG nly tells us the relative amunts f prducts and reactants when equilibrium is reached, nt the directin f the net reactin. See the table fr additinal infrmatin. Relatin Between DG and K as Predicted by the Equatin D G =-Rln K K ln K DG Cmments >1 Psitive Negative Prducts are favred ver reactants at equilibrium. =1 0 0 Prducts and reactants are equally favred at equilibrium. <1 Negative Psitive Reactants are favred ver prducts at equilibrium. Ex. Using the data listed in Appendix 3, calculate the equilibrium cnstant ( ) fr the fllwing reactin at 25 C: 2 H 2O(l) 2 H 2(g) + O 2(g) Ex. In Chapter 16, we discussed the slubility prduct f slightly sluble substances. Using the slubility prduct f silver chlride at 25 C (1.6 x 10-10 ), calculate DG fr the fllwing prcess: AgCl(s) Ag + (aq) + Cl - (aq) Ex. he equilibrium cnstant ( ) fr the reactin N 2O 4(g) 2 NO 2(g) is 0.113 at 298 K, which crrespnds t a standard free-energy change f 5.40 kj/mle. In a certain experiment, the initial pressures are P NO2 = 0.122 atm and P = 0.453 atm. Calculate DG fr the reactin at NO 2 4 these pressures and predict the directin f the net reactin tward equilibrium. 17.8: Influence f emperature n Equilibrium Cnstants Higher reactin temperatures increase the equilibrium cnstant f an endthermic reactin but decrease the equilibrium cnstant f an exthermic reactin. he slpe f a plt f ln K (y-axis) versus 1/ (x-axis) fr an equilibrium system is used t determine the standard enthalpy fr the equilibrium,, and the y-intercept f the plt is used t determine the standard entrpy change, plt is ln K = ' + ()*, K P -. 0 + 2 ()* + /,. DH rxn. he crrespnding equatin f the line fr this 17.9: Driving the Human Engine: Cupled Reactins Many imprtant bichemical prcesses, including glyclysis and phsphrylatin, are made pssible by cupled spntaneus and nnspntaneus reactins. he free energy released in the spntaneus prcesses ging n in the bdy is used t drive nnspntaneus prcesses. 17.10: Micrstates: A Quantized View f Entrpy In 1868, Bltzmann shwed that the entrpy f a system is given by S = klnw in which k is the Bltzmann cnstant 1.38 x 10-23 J/K and W is the number f micrstates fr the system. A system with fewer micrstates (smaller W ) amng which t spread its energy has a lwer entrpy. K P

Chapter 17: hermdynamics: Spntaneus and Nnspntaneus Reactins and Prcesses, Page 6 A system with mre micrstates (larger W ) amng which t spread its energy has a higher entrpy.