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19_FreeEnergy.pdf

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Chemistry
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CHEM 1E03
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Weisner

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Entropy, the 2 ndLaw of Thermodynamics Gibbs Free Energy & Spontaneity & the connection to Equilibrium → Which rxns go & which do not? ST → 1 Law just keeps track of energy There is a natural direction of physical & chemical processes not explained by the 1 Law. Eg. 1 A gas expands spontaneously into a vacuum ♣ The reverse of this process is never seen though it is allowed by the 1stlaw. Eg. 2 Heat flows from a hot body (on the right) to a cold body (on the left). ⇒ T i1creases & T de2reases until the two systems achieve the same temperature – thermal equilibrium T1 T 2 T1< T 2 q →Heat never flows the other way, though the 1 Lawst would allow it. 115 Egs. 3. Spontaneous mixing of gases, miscible liquids, (natural) dissolution of soluble salts ♣ Methanol & water are miscible - they spontaneously form a solution - a more disordered state when added together. NaCl(s) dissolves spontaneously when added to water - the resulting solution is more disordered than solid salt + water. Though these examples include energy considerations, they illustrate that increasing disorder is a driving force causing some processes to proceed, while prohibiting the reverse Spontaneous mixing of two ideal gases is the best illustration. In this case, interactions between molecules can be neglected (i.e. no energy considerations), yet the Random motions of individual molecules leads to irreversible mixing of gases. 116 • an ordered state → a disordered state • disorder increases In Eg. 2, kinetic energy spreads from the hot body to the cold body. • energy is more spread out • a more disordered state In Eq. 3, dissolution of salt in water ... • the ions have greater freedom in solution - due to increased volume (in comparison to solid) • final state has more freedom more disorder In a natural (i.e. spontaneous) process there is a spontaneous increase in disorder. → Entropy, S, measures this disorder → S increases 123 true for S of entire universe How do we measure "disorder"? or How is S defined? 117 Consider a deck of cards - continuously reshuffled. → If the card shuffler doesn’t cheat, then any ordering of the card is possible ⇒ the cards are disordered. → However, if the card shuffler cheats - eg. producing an ordered deck of cards with each shuffle (eg. always ensuring that the dealer gets a royal flush) ⇒ the cards are more ordered. supplementary - you are not responsible for the formulae from here to the next dividing line ... Boltzmann proposed the following formula for entropy in terms of the "number of arrangements of the system which produce the same state" S = k ln W number of arrangements ... entropy Boltzmann constant k = R/N A = 1.3807 × 10-23J/K This formula is engraved on Boltzmann's headstone. In the case of the shuffled cards, the "number of arrangements ...", W, is just the number deck states sampled in the continuous reshuffling. 118 Unshuffled deck: W = 1 → completely ordered case ⇒ S = O Properly shuffled deck: W = n! → disordered deck of n cards ⇒ S = k ln n! → proper shuffling produces much larger S Always-royal-flush-to-dealer shuffled deck: W' = 4 (n-5)! → There are 4 royal flushes - the remaining n - 5 cards are arranged randomly ⇒ S' ≅ k [ln 4 + ln (n-5)!] ∆S = S' - S = k [ln 4 + ln 47! − ln 52!← case of 52 cards = -2.51 × 10-2J/K → negative ∆S ⇒ the system (deck of cards in this case) has become more ordered Of course the dealer might get a royal flush without cheating. However, the probability is W'/W = exp[∆S /k] -8 = 4/[5251 50•49•48] = 1.3 × 10 i.e. 1 chance in about 100 million - very unlikely 119 |∆S| for processes we consider in chemistry are much larger - the corresponding probabilities of these events are mind- bogglingly small. eg. What is the probability of all the air, at STP, inside a (10 m) × (10 m) × (10 m) room, finding itself inside an open 1 L pop bottle (soda pop having been removed)? Solution: 3 6 The room has a volume of 1000 m = 1.0 × 10 L. ⇒ There is 1 chance in 1.0 × 10 of finding one specific air 6 -6 molecule in the bottle - prob. = 1/(1.0 × 10 ) = 1.0 × 10 However, there are pV/RT = 6 -1 -1 = (1.0 × 10 L atm)/[(0.0821 L atm K mol )(298 K)] = 4.1 × 10 moles of air 28 = 2.5 × 10 molecules of air Probability of all air molecules inside bottle = 29 -6 2.5 × 1028 -1.5 × 10 (1.0 × 10 ) = 1.0 × 10 → this is such a ridiculously small number, it is hard to take it seriously → Gases simply do not voluntarily constrain themselves in this way - evidenced by the extraordinary unlikelihood of such an event ... 2ndLaw of Thermodynamics: Entropy increases for all spontaneous processes - i.e. those actually occuring in nature - natural processes 120 But wait a minute ... Don't we ever see processes with negative ∆S ? Yes, but the entropy of one system decreases only at the expense of an even greater entropy increase in another system (the surroundings). Change in the total entropy of the universe, ∆ S = ∆ S + ∆ S > O total system surroundings ⇒ ∆ S system> ∆ S surroundings ♣ A. A process for which ∆ system 0 A is always accompanied by q < 0. Heat leaving the system enters the surroundings & heat increases entropy (hotter ⇒ more disorder). In fact, the entropy increase of the B surroundings always exceeds the decrease in entropy of the system. B. If the system increases its entropy, then the surroundings can actual decrease in entropy, but not by more than the system increased its entropy. C C. There are restrictions, if both system & surroundings increase in entropy. 121 Properties of entropy: → S is proportional to amount sofstance (like E & H - also n & V) eg. SSS total2 S → S depends on temperature & on phase - increases with T, including discrete (positive) jumps when solid → liquid & liquid → gas ♣ 5 4 3 2 1 0 K 122 1. 3 Law of Thermodynamics: S = O for a pure crystalline substance at T = 0 K. → a perfectly ordered state 2. Heating increases disorder of solid. → atoms vibrate about mean positions 3. When the melting point is reached further heating produces a liquid wherein molecules are much freer to move about (diffusive motion). 4. Entropy of liquid (like the solid) increases with T until the b.p. is reached. 5. Vaporization increases S significantly - molecules are free to explore the entire containment vessel. ∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗∗ Standard molar entropy , S ° ° → entropy of 1 mol of substance at 1 atm & 25 C → units of S = J K −1 mol −1 note - Joules not kJ egs. S °[H2O(l)] = 70.0 J K −1 mol −1 S °[H2O(g)] = 188.7 J K −1 mol−1 123 ° ° ° NOTE ∆S vap = S [H O2g)] − S [H O2l)] = 118.7 J K −1 > 0 more egs. ° −1 −1 S [H2(g)] = 130.6 J K mol S ° [O2(g)] = 205.0 J K −1 mol−1 → both gases - however, S [O 2g)] > S [H 2g)] → heavier molecules have more entropy → more complex molecules also have more entropy eg. ♣ Molecules with more bonds have more ways of spreading out vibrational energy. Thermal energy takes the form of translational, rotational & vibrational energy - at the microscopic level. 124 Standard reaction entropy ° ° ° ∆S = Σ nprodS prod − Σ nreacS reac prod reac eg. H 2(g) + ½ O2(g) → H 2(l) ∆S = 1 × S [H O(2)] − (1 × S [H2(g)] + ½ × S [O 2g)]) = 70.0 -(130.6 + 205.0/2) J K−1 −1 = - 163.1 J K → S is a state function like E & H → ∆S add for sequential processes - i.e. Hess's Law → Gases have the most entropy ⇒ a rxn increasing moles gas generally has a positive ∆S while decreasing moles of gas means negative ∆S eg. H 2(g) + ½ O2(g) → H 2(l) 123 123 1½ moles of gas → 0 moles of gas ∆n gas= -1½ ° −1 Recall ∆S = −163.1 J K < 0 125 another eg. 2 NH 4O (3) → 2 N (g2 + O (g2 + 4 H O(2) ° ∆n gas= +7 & ∆S > 0 Let's return to the 2w of Thermodynamics. It's utilization requires knowledge surroundings Entropy change in surroundings is due to heat transferred to/from system - hence from/to surroundings It can be shown that ∆Ssurr= q surr = −q /sys = − ∆H /T (under constant pressure conditions) sys → exothermic ⇒ ∆S surr> 0 → system heats surroundings → endothermic ⇒ ∆S surr< 0 → system cools surroundings (system entropy must increase in this case) NOTE: effect of transferred heat is greater at low T 126 Consider H 2g) + ½ O (2) → H O(l)2 ∆H ° = −285.8 kJ ( heat of formation of H O(l) ) sys 2 ⇒ ∆S = + 285.8 × 1000 J K −1 surr 298.15 = 958.6 J K −1 ° ° ° ∆S tot ∆S sys ∆S surr = −163.1 + 958.6 J K −1 −1 = 795.5 J K > 0 ⇒ a spontaneous process Consider 4 Fe(s) + 3 O 2g) → 2Fe O (2) 3 ° ∆H = −1649 kJ ← −ve ⇒ exothermic ° −1 ∆S = −549 JK ← −ve ( note - ∆n gas = -3 ) ° ° ° −1 ∆S tot= ∆S − ∆H = 4980 JK > 0 at 298 K T ⇒ spontaneous → energetics (i.e. ∆H) drives this rxn 127 Gibbs free energy Criterion for spontaneity i∆S tot> 0 or ∆S −
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