Thermodynamics Partial Outline of Topics

Transcription

1 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!) (1) This means that if a prcess is spntaneus, Suniv increases. This des NOT mean that Ssys always increases!!! UNIV = SYS + SURR Suniverse Ssys + Ssurr (2) Fr sme hypthetical prcesses: -If entrpy increases in the system and the surrundings, then the prcess is bviusly spntaneus -If entrpy decreases in the system and the surrundings, then the prcess is bviusly nt spntaneus (and reverse prcess wuld be spntaneus) -If entrpy increases in the system and decreases in the surrundings, then the prcess is spntaneus if the increase in the system is bigger than the decrease in the surrundings! (i.e., if Ssys > Ssurr ; because that will mean that Ssys + Ssurr > 0) -If entrpy decreases in the system and increases in the surrundings, then the prcess is spntaneus if the decrease in the system is smaller than the increase in the surrundings! (i.e., if Ssys < Ssurr ; because that will mean that Ssys + Ssurr > 0) This is why we care s much abut entrpy(ies)! It is change in ttal entrpy (fr the universe) which ultimately (always!) dictates whether r nt a prcess is spntaneus. II. Sme patterns regarding entrpy f substances: 1. Fr a given amunt f a given substance, entrpy increases a. if energy flws int it: i) If T ges up ii) If it turns frm: s l; l g; r s g (Nte: gases tend t have a much greater entrpy (per mle) than slids r liquids; see Figure 17.5 and Table 17.2 in Tr) b. if vlume increases (applies nly t gases and disslved slutes) 2. The amunt f entrpy in a sample f a given substance depends n the amunt f that substance. Tables list the amunt f entrpy per mle f a given substance (at a given T). J Units are [r JK -1 ml -1 ] (read Jules per Kelvin per mle [f substance]) K ml 3. Different substances (r species) have a different amunt f entrpy (per mle). In general, fr a given physical state f a substance (see 1b abve), substances have a greater entrpy (per mle) if they are: a. Mre cmplex (mre atms/bnds/ins per frmula unit) b. Mre massive (per frmula unit) III. Predicting the sign f Ssys fr prcesses Given the abve pattens in II abve, yu shuld be able t predict whether entrpy increases r decreases (in the system), when certain prcesses ccur (hypthetically). See Examples Ssys Ssurr (NOTE: There likely will be ther tpics and bardwrk befre the next sectin f utline is cvered.) 1

2 IV. Secnd Law, Reprise-Cnnectin t Gibbs Free Energy Functin A. Cnceptual Cnnectins-I Fr any prcess Suniverse > 0 prcess is spntaneus (in frward directin) Suniverse < 0 prcess is nnspntaneus (reverse is spnt.) Suniverse = 0 prcess is nt spntaneus in either directin Fr a prpsed chemical (r physical) change (represented by a chemical equatin), & ccurring under any specific cnditins (T, P cnstant; cncentratins "fixed") spntaneus means "frward rxn is spntaneus" nnspntaneus means "reverse rxn is spntaneus" nt spntaneus in either directin means at equilibrium! Q < K Q > K Q = K Because f ideas intrduced in ur equilibrium unit! S (fr a chemical rxn at fixed T, P, and reactant and prduct cncentratins): Suniverse > 0 Q < K frward rxn is spntaneus Suniverse < 0 Q > K reverse rxn is spntaneus Suniverse = 0 Q = K at equilibrium B. Switching t all system variables (Hw is Ssurr related t system variables?) Recall: If n chemical change is ccurring in the surrundings, if energy flws int the surrundings, its entrpy will increase (Ssurr > 0). Furthermre, if energy flws int the surrundings, it had t have cme frm...the system!! -- Assertin: Ssurr = T (Nte: Hsys < 0 Ssurr > 0) Suniverse = Ssys + Ssurr > 0 spnt. Ssys + T Nw, d sme algebra (fr specific purpse): (Fr a prcess at cnstant T and P) (3) Start with 2 nd Law [Eq. (1)] and Eq.(2), cmbined > 0 spnt. Substitute Eq. (3) int the abve! TSsys + (Hsys) > 0 spnt. (multiplied thrugh by T) TSsys + (Hsys) < 0 spnt. (multiplied thrugh by -1) Hsys TSsys < 0 spnt. Mdified frm f 2nd Law (system variables nly): (swapped first tw terms) Hsys TSsys < 0 spnt. (if T, P cnstant) (1 ) **If we can calculate Hsys and Ssys, we can determine spntaneity. **The Hsys term "matters" because it ultimately is related t the entrpy change f the surrundings. **Nte: TSsys has units f energy C. Define a new functin: G, Gibbs Free Energy such that G = HTS (4) 2

3 Then, (1 ) becmes simply: Gsys < 0 spnt. (if T, P cnstant) Nte: By substituting Eq. (2) and Eq. (3) int Eq. (1), yu can shw that: S universe G T sys (1 ), using G (5) D. Cnceptual Cnnectins-II Thus fr a chemical r physical change at cnstant T & P, but any (i.e. nn-standard) cncentratins: Gsys < 0 spntaneus Q < K ( Suniverse > 0) Gsys > 0 reverse rxn spntaneus Q > K ( Suniverse < 0) Gsys 0 system is at equilibrium Q K ( Suniverse = 0) The relatins abve apply at any fixed set f cncentratins (r partial pressures). (i.e., any "cnditins", nt necessarily "standard state" cnditins). Wherever a "Q" is well defined, a Gsys is als defined (fr thse same set f cncentratins). Nte: As an "actual" reactin ccurs and cncentratins change, Q changes, and s des the Gsys. It is imprtant t realize that if [ ]'s change, the entrpies f species change, the Suniv changes, the Q changes, and the G changes! RECALL: "Standard State Cnditins" means a very specific set f cnditins: It means that all species (R's and P's) are in their standard state, which means: 1) All pure substances are in their mst stable physical state (s, l r g) at the T specified. 2) All species in slutin have a cncentratin f 1 M. 3) All gases have a partial pressure f 1 bar (~1 atm) (relates t [ ]) If a reactin is cnsidered t ccur at "standard state cnditins" THEN THE (very general) EXPRESSIONS ABOVE BECOME THE MUCH MORE SPECIFIC ONES BELOW: < 0 Spntaneus at st. state 1 < K ( P favred!) > 0 Reverse rxn is Spnt at st. state 1 > K ( R favred!) 0 System is at equilibrium at st. state 1 K ( neither R nr P favred) Please nte the fllwing: c d [ C] [ D] Since "standard state" means "all cncentratins are 1 M", Q a b [ A] [ B] 1 at standard state cnditins (i.e., when all species are in their standard state). "Prduct favred" means K > 1, and that means that IF all cncentratins were initially 1 M, frward reactin wuld ccur t reach equilibrium. "Prduct favred" is nw als assciated with the standard free energy change being negative (i.e., < 0). G and are NOT the same thing! The first is MUCH mre general; the secnd is quite specific: G < 0 means a reactin is spntaneus at whatever cnditins (cncentratins) are specified. It des NOT mean "prduct favred" (at equilibrium). K culd be large r small! (Q < K) 3 < 0 means a reactin is spntaneus at standard state cnditins (cncentratins), which means that the prcess is "prduct favred" (at equilibrium). K must be larger than 1! (K > 1)

4 Recall the distinctin that was made (during the discussin f equilibrium) between "prduct favred" (K > 1) and "prduct deficient" (Q < K; t few prducts t be at equilibrium; "net frward reactin ccurs t reach equilibrium [frm this pint]). This is the SAME distinctin I'm trying t get yu t see nw: If G < 0, then frward reactin will ccur t reach equilibrium [frm the specified "pint", where Q is bviusly < K]. If < 0, then frward reactin wuld ccur if the system were t start at standard state cnditins. E. The explicit mathematical cnnectin between Q and G: G = G + RT ln Q The free energy change fr a chemical reactin under any cnditins is equal t what it wuld be if all reactants and prducts were (and remained) in their standard states plus sme crrectin factr t accunt fr the fact that the system is nt at standard state cnditins. Sme imprtant "upshts" f all this: We can tabulate standard Gf's fr all substances, and use them t calculate given prcess. (r we can calculate frm T ) 's fr any This (alne) can tell us whether a prcess is prduct favred r reactant favred (i.e., spntaneus under standard state cnditins) It tells us in which directin there is a driving frce fr reactin t happen if yu started with all R's and P's at 1 M We can even get the specific value f K! G = -RT ln K (Where des this cme frm? Simple! If a system is at equilibrium, we nw knw that TWO things must be true: 1) Q K (frm befre); and 2) G 0. If yu substitute thse tw "values" int G G + RT ln Q, what d yu get? 0 G + RT ln K, which rearranges t the equatin at the left!) K e G RT If the cnditins are NOT standard (generally, Q 1), then we can get the "driving frce" at these (nn-standard) cnditins using G = G + RT ln Q F. Revisit Gsys HsysTSsys at standard state cnditins: T-dependence f It turns ut that Hsys and Ssys are nt very T-dependent. We generally assume that they are cmpletely temperature independent s that we can determine the T-dependence f fr a particular chemical reactin equatin: 4

5 = T = (a cnstant) T (anther cnstant) 1) Think abut the extremes (this is nw dne effectively in the PwerPint presentatin as well): a) If T ges t abslute zer, then T ges t zer, and s. Thus, at lw temperature: endthermic prcesses are nn-spntaneus at standard state cnditins ( exthermic prcesses are spntaneus at standard state cnditins ( < 0) > 0), and This can be ratinalized in entrpy terms as fllws: At very lw T, the entrpy change fr the surrundings will dminate S univ. If energy flws int the system (endthermic), then S fr the surrundings ges dwn by a huge amunt, making S univ < 0 (nnspntaneus). If energy flws ut f the system (exthermic), then S fr the surrundings ges up by a huge amunt, making S univ > 0 (spntaneus). b) If T gets very large (e.g., tward infinity), then T infinity), and s ~ T. Thus, at high temperature: gets very large in magnitude (e.g., tward prcesses that increase entrpy in the system are spntaneus at standard state cnditins ( < 0), and prcesses that decrease entrpy in the system are nnspntaneus at standard state cnditins ( > 0) This can be ratinalized in entrpy terms as fllws: At very high T, the entrpy change fr the surrundings will be negligible, s S univ will essentially equal! 2) Fr prcesses in which the signs f and are the SAME (which means that the signs f and Ssurr are ppsite): there will be sme T between 0 and infinity at which there is a "breakeven" pint where 0 and the system is at equilibrium at standard state cnditins (K = 1). Find this T by putting in 0 fr (Remember, and, and then slving the equatin [0 T ] fr T! are cnsidered t be cnstant fr a given reactin equatin!) 5