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Zumdahl’s Chapter 16 Spontaneity, Entropy, Free Energy, and Why All Things Happen … “The Universe Becomes Less Predictable”

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Presentation on theme: "Zumdahl’s Chapter 16 Spontaneity, Entropy, Free Energy, and Why All Things Happen … “The Universe Becomes Less Predictable”"— Presentation transcript:

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2 Zumdahl’s Chapter 16 Spontaneity, Entropy, Free Energy, and Why All Things Happen … “The Universe Becomes Less Predictable”

3 Chapter Contents  Spontaneous Process and Entropy, S  2 nd Law of Thermo- dynamics,  S univ  0  Entropy’s Change with Temperature  Change in S During Chemical Reactions  “Free Energy”, G, & Chemical Reactions  G’s Dependence on Pressure  Pointing the Way to Equilibrium  G’s Relation to K  Non-PV Work & G

4 Spontaneity “Sponte” is Latin for “voluntarily.”  We’re willing to concede that highly exothermic reactions are spontaneous. –While the First Law assures that the enthalpy released could be used to resurrect reactants, we know from experience that hot things cool off, and disperse q to the environment, so that it is unavailable to reverse the reaction. –But why do some endothermic reactions go?

5 Punctuality  For that matter, why do some highly exothermic reactions hesitate, requiring a kick start, to do their spontaneous thing? –Or proceed lethargically once started?  While last slide’s question is one Thermo can address, the questions above lie in the realm of later chemical topics, viz., Kinetics and Dynamics.

6 Norse Mythology  Valhalla is the abode of the Norse gods. –But, contrary to many other mythologies, Norse gods are not immortal. –Valhalla is held up by a giant tree, the roots of which are being gnawed by a serpent. –The serpent will succeed, and when it does, Valhalla and the Universe will fall.  The serpent’s name is haos

7 Universal Chaos, S univ  The Norsemen were right! –There is Chaos growing in the Universe all the time at the expense of Order. It is now a fundamental principle of Science. –It’s called “entropy,” S, and is a state function that must always increase for the Universe as a whole, but some System’s S may decrease. –It is a (logarithmic) measure of the combinations of wave functions available to the Universe!

8 S = k log e W (Boltzmann’s Headstone!)  S = k ln W in modern symbolism. –W is an actual count of how many different ways the Universe could be arranged without being detectably different macroscopically. And it is usually enormous! For example, how many different poker hands might be in some player’s possession? W  (52)(51)(50)(49)(48) / 5! or 2,598,960. For 4 players, that’s ~1.48  different games. Over twice Avogadro’s Number!

9 Poker Microstates  One microstate in poker might be a flush; all cards of the same suit. –W flush = 4(13)(12)(11)(10)(9) / 5! = 5148 as the number of ways to get a flush on the deal. –But W flush / W total gives ~505:1 odds against. –So flushes-on-the-deal are fairly ignorable.  In k ln W, the most likely microstate is used to calculate W *. It overwhelms others.

10 Chemical Microstates  Positional –In a solid, molecules are frozen in position. –But a liquid can swap molecular positions without macroscopic consequence: S liq > S solid –A gas is far more chaotic: S gas >> S liquid ! –Therefore, it’s a safe bet that if  n gas > 0 for a reaction, so is  S. –And, of course,  n gas < 0 makes  S negative.

11 Structure and Microstates  Since the more modes of motion in a molecule, the more places it can hide energy (higher heat capacity), larger molecules have higher S than smaller ones.  Still, decomposition reactions have  S > 0! –Although the products have to be smaller molecules, there are more of them, so Nature can fool you as to where the atoms are!

12 2 nd Law of Thermodynamics  “In any spontaneous process, the entropy of the Universe increases.” –We must include consideration of a system’s environment to apply this law. For example, condensing a gas implies a large decrease in the system’s entropy!  S sys << 0 Fortunately, the (latent) heat of vaporization gets released to force the surroundings to occupy higher energy levels, so  S surr >> 0 and  S univ > 0!  S univ =  S sys +  S surr  0

13 Entropy Rules Everywhere  Photosynthesis makes few large molecules (CH 2 O) n from smaller ones (CO 2 & H 2 O). –So definitely  S sys < 0 –But the absorption of light releases heat into the environment. More importantly … –It then casts many long IR photons into the universe having absorbed fewer short VIS. –So even growth of Life makes  S univ > 0

14 Perhaps even where it shouldn’t  Over a century ago, Darwin published The Origin of Species and coined “the survival of the fittest.” (…condemning us to Reality TV) –Social Darwinism used that to excuse all the excesses of predatory Capitalism.  Economists are turning to Ilya Prigogine. –His notion that processes win that make S grow most quickly is ripe for similar abuse.

15 Entropy and Temperature –Increased heat, q, should correlate with S since it makes available high energy states. –But the chaos of q makes  S more impressive if initial states are more ordered ( lower T ).  And  S = q / T codifies both notions. (units?)  At constant P,  S =  H / T if only q happens. –So  S surr = –  H sys / T since exothermicity flows into the surroundings.

16 0 th Law of Thermodynamics  “If two system are in equilibrium with a third, they are in equilibrium with one another.” –Take T as a measure; we presume 2 or more systems in contact come to the same T equil. If T 2 > T 1, then q = q 1 = – q 2 > 0  S 1 = q / T 1 > 0 by more than  S 2 = – q / T 2 < 0 And  S univ =  S 1 +  S 2 > 0 until T 2 = T 1. Whereupon  S univ = 0 and q stops flowing.

17 Le Châtlier Confirmed!  Suppose a reaction has an exothermicity of  H. Then a q surr = –  H > 0  And  S surr = q surr / T > 0 aids spontaneity.  Le Châtlier claims that higher T makes such a reaction less spontaneous!  Assuming q varies insignificantly with T (true), then higher T makes  S surr a smaller value! Le Châtlier Confirmed!

18 S, an Extensive State Function   S rxn  =  n p S products  –  n r S reactants  where  ’s seem to be missing on the right side! –This version of Hess’s Law is correct for S.  3 rd Law: S for perfect crystal at 0 K is 0. –W = 1 since all atoms frozen in fixed places! –  S   0 since we can warm solids up from 0 to 298 K via dS = q / T = (C P / T ) dT Even elements have non-zero S . Enthalpy may be relative, but Entropy is Absolute.

19 Imperfect Crystals  Imagine the molecule NH 2 D where an H has been replaced by deuterium, i.e., 2 H.  The deuteroammonia has the same crystal structure as regular NH 3, but each D can be in one of three possible places at random.  S(0 K) = k ln W = k ln(3) = k –That’s per molecule. Per mole: W Nav instead. –ln(3 Nav ) = N Av ln 3, so S(0 K) = R

20 Perfect Solutions –Assuming no molecular interactions differ between pure solutions, they mix perfectly.  The Entropy of Mixing quantifies Nature’s need to scramble stuff to confuse you:   S mix = – R  X i ln X i (mole fractions) –which is entirely consistent with R ln W –E.g., NH 2 D at 0 K has  S mix = – R ln(1/3) –Since X i = 1/3 for all 3 “kinds” of NH 2 D

21 Hiding the Surroundings  Since  S surr = –  H sys / T, and   S univ =  S sys +  S surr  0, and therefore  T  S univ = T  S sys + T  S surr  0, then  T  S sys –  H sys  0 is also the 2 nd Law.   H sys – T  S sys  0 is too.   G sys   H sys – T  S sys  0 is our choice!  Gibb’s Free Energy, G  H – TS

22 Spontaneity and Equilibrium   G 0.   G > 0 means that the reverse process is the spontaneous one!  But  G = 0 means neither the process nor its reverse is spontaneous. So   G = 0 means EQUILIBRIUM.

23 Freezing Point of Mercury  Hg(solid)  Hg(liquid) –  H fusion  ~ 2.16 kJ / mol –  S fusion  ~ 9.3 J / mol K –  G fusion  =  H fusion  – T   S fusion  = – 6.11 kJ –OK, that’s spontaneous; Hg should be liquid at 298 K. –T fusion   H fusion /  S fusion since  G fusion = 0 –T fusion ~  H fusion  /  S fusion  = 232 K = – 41ºC –The actual T fusion = – 39ºC so H and S are T-dependent.

24 Hydrogenation of Ethene  C 2 H 4 (g) + H 2 (g)  C 2 H 6 (g) –We’re not sanguine about this since  n gas < 0. –Indeed  S  = S  ( ethane ) – S  ( ethene ) – S  ( H 2 )  S  = (270) – (219) – (131) = – 120 J/mol K but… –  H  =  H f  ( ethane ) –  H f  ( ethene ) –  H f  ( H 2 )  H  = ( – 84.7) – (52) – (0) = – 137 kJ/mol and  G  = ( – 32.9) – (68) – (0) = – 101 kJ/mol < 0 –So reaction is spontaneous at std. conditions.

25 Improving Le Châtlier’s Odds  Since  H  < 0, we don’t want to heat the reaction, or we’d reduce spontaneity. –We would expect  G  to be increased.  But since  n gas < 0, we do want to apply additional pressure to drive it to products. –We’d expect  G  to become more negative.  So what was that again about G’s pressure dependence?

26 G’s Pressure Dependence  dE = q + w = TdS – PdV But H = E + PV so dH = dE + PdV + VdP  dH = TdS + VdP (used before with fixed P, so dP=0) But G = H – TS so dG = dH – TdS – SdT  dG = VdP – SdT or, at fixed T, dG = VdP  G – G  =  dG =  V ideal dP = R T  P –1 dP  G – G  = R T ln(P / P  ) = R T ln P dG = RT lnP

27 G and K (equilibrium constant)   G –  G° =  n G products –  m G reactants   G –  G° = R T [  n ln P p –  m ln P r ]  (  G –  G°) / R T =  ln P p n –  ln P r m ]  (  G –  G°) / R T = ln  P p n – ln  P r m  (  G –  G°) / R T = ln (  P p n /  P r m ) = ln Q –But Q  K when  G  0 so  +  G° = – R T ln K Mass Action Quotient

28 G and Reaction Progress,   0 (pure reactants) 1 (pure products) G equilibrium G°G° G minimizes at equilibrium.  G=0 for any small variation there.

29 Equilibrium Constant  K = e –  G° / R T is that relation’s inverse.  For the hydrogenation,  G° = – 101 kJ/mol  K = e +101,000 J / J/K (298 K) = 5.1  –well and truly spontaneous!  Remember, while K is clearly dependent upon T, it is independent of P total. It’s the partial Ps that adjust to render  G = 0.

30 K’s Temperature Dependence  ln K = –  G° /R T = –  H° /R T +  S° /R  ln K = – (  H° /R) T –1 + (  S° /R) –We expect a plot of ln K vs. 1 / T to be ~ linear. That’s if H and S are weak functions of T themselves. True if we don’t change T much.  d(lnK) = + (  H° /R) T –2 dT (van’t Hoff) It says that ln K increases with T when the reaction is endothermic; decreases otherwise. – Le Châtlier! But the increase becomes less impressive at high T.

31 Maximizing Work   G =  VdP –  SdT + w non-PV We’ve been ignoring the non-PV work all this time, but it’s really been there in E, H, and G. –Here it means that at fixed P & T, the first two terms vanish, and  G = w non-PV, the maximum (non-PV) work of which the system is capable. If you want maximum total w, the physicists need to tell you about  A. (A = E – TS, the “work function.”) In either case, we must be so gentle as to be at equilibrium all the time; “reversible work!”


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