Why a particular process occurs? Is it due to decrease of energy? Model study 1: An adiabatic ( 絕熱 ) system insulation

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1 17 Spontaneity, Entropy ( 熵 ) and Free Energy ( 自由能 ) Question: Why a particular process occurs? Is it due to decrease of energy? Model study 1: An adiabatic ( 絕熱 ) system insulation ideal gas vacuum q = 0 w = P V = 0 (note: expansion against zero P) E = q + w = 0 This is a spontaneous process but no change of E!! Model study 2 KI(s) K + (aq) + I (aq) H = +21 kj Endothermic yet spontaneous!! Model study 3 H 2 O(s) H 2 O(l) H > 0 Spontaneous above 0 o C What's in common? Increase randomness or disorder The measurement of randomness: entropy ( 亂度 or 熵 ) 1

2 How to measure randomness? Probability A model of positional probability A B Arrange two balls in two boxes: 2 2 = 4 possibilities Possible arrangements Probability A,B -- ¼ -- A,B ¼ A B ¼ B A ¼ ½ (more probability) Four balls: A, B, C, D (2 4 = 16 possibilities) Possible arrangements Probability ABCD -- 1/16 ABC D 4/16 = 1/4 (4 3 2/3! = 4) AB CD 6/16 = 3/8 (4 3/2! = 6) More probability Boltzmann defines: S = k B lnω (k B N A = R; Ω: the number of microstates) Ω = 1 S = 0 Boltzmann constant Avogadro s number S = S 2 S 1 = k B lnω 2 k B lnω 1 = k B ln(ω 2 /Ω 1 ) 2

3 Ex. 1 mol N 2 gas at 1 atm, 25 o C 1 mol N 2 gas at atm, 25 o C Larger V Higher S Entropy and the second law of thermodynamics The second law: In any spontaneous process there is always an increase in the entropy of the universe S univ = S sys + S surr > 0 for spontaneous process S univ < 0 spontaneous for the reverse process S univ = 0 at equilibrium Ex. In a living cell, large molecules are assembled from simple ones. Against the 2 nd law? S sys is negative Need to consider S surr also 3

4 Reversible process Model study: An ideal gas, isothermal (constant T) P P i ideal gas (P, V, T) P ext P f V i V f V For ideal gas E = 3 / 2 kt E = 3 / 2 k T T = 0 for isothermal process E = 0 = q + w q = w P P i ideal gas (P, V, T) P ext P f Case I: P ext changes to P f suddenly V i expands to V f V i V f V w = P f (V f V i ) q = P f (V f V i ) Impossible to go back through the same path irreversible 4

5 P ideal gas (P, V, T) P ext P i P f V i V f V P ext decreases infinitesimal amount along the path until V = V f Moves along the PV line A reversible process Note: Conclusion: S = q rev T (for a reversible process) For a reversible process at constant pressure: q rev = q P = H qrev H S = T T Ex. H 2 O(l) H 2 O(g) H vaporization This is an equilibrium process at 100 o C and reversible H = q rev Hvaporization S = T 5

6 The effect of q and T System H sys Surroundings When H sys is negative: Increases randomness of atoms in the surroundings Qualitatively consider the surroundings as a huge system With constant P, V and T S surr is primarily determined by the heat flow Heat surroundings Heat surroundings S surr increases S surr decreases S surr also depends on T For the same amount of heat: Lower T more randomness Higher T less randomness System Surroundings Exothermic S surr = + Endothermic S surr = We can define heat (J) T (K) heat (J) T (K) S surr = H sys T (Note: H surr = H sys ) 6

7 Ex. Sb 2 S 3 (s) + 3Fe(s) 2Sb(s) + 3FeS(s) S surr? At 25 o C, 1 atm H = 125 kj S surr = 125 kj 298 K = 419 J/K The unit of entropy A quick analysis S sys S surr S univ spontaneous non-spontaneous +? +? depends on the relative size Free energy S univ = S sys + S surr > 0 for spontaneous process H S univ = S sys sys > 0 (at constant P, q P = H) T T S univ = T S sys H sys > 0 T S sys H sys = T(S f S i ) (H f H i ) = TS f H f TS i + H i Now, define Gibbs free energy: G = H TS T S sys H sys State function = G f + G i = G sys at constant T and P For a spontaneous process: G > 0 or G < 0 at constant T and P 7

8 Ex. H 2 O(s) H 2 O(l) H o = Jmol 1, S o = 22.1 JK 1 mol 1 T ( o C) H o (J/mol) S o (JK 1 mol 1 ) T S o (J/mol) G o = H o T S o (J/mol) (Assume H o and S o do not change at different T) Qualitatively: H T S = G + spontaneous (exothermic) (more random) + + non-spontaneous (endothermic) (less random) spontaneous at low T non-spontaneous at high T + + spontaneous at high T non-spontaneous at low T In general: S is more important at higher T 8

9 Ex. Br 2 (l) Br 2 (g) H o = 31.0 kjmol 1, S o = 93.0 JK 1 mol 1 At what T is the reaction spontaneous at 1 atm? Soln. Find the equilibrium point first G o = 0 = H o T S o H o = T S o T = H o S o = When T > 333 K T S o > H o G o < 0 Spontaneous = 333 K Equilibrium T = bp (bp: boiling point) Entropy changes in chemical reactions N 2 (g) + 3H 2 (g) 2NH 3 (g) volume decreases S: 4NH 3 (g) + 5O 2 (g) 4NO(g) + 6H 2 O(g) volume increases S: + CaCO 3 (s) CaO(s) + CO 2 (g) S: + 2SO 2 (g) + O 2 (g) 2SO 3 (g) volume decreases S: 9

10 In general: for homogeneous reaction A + B C S: A B + C S: + A + B C + D S: difficult to estimate Absolute entropies E and H can not be determined But S can be determined The third law of thermodynamics: Entropy of a perfect crystal at 0 K is zero 0 K T S = 0 S S T From experiment Can be determined Standard entropy: S o (at 298 K, 1 atm) S o rxn = n p So products n r So reactants reaction coefficient 10

11 Ex. Al 2 O 3 (s) + 3H 2 (g) 2Al(s) + 3H 2 O(g) S o Al2O3(s) = 51 JK 1 mol 1 S o H2(g) = 131 S o Al(s) = 28 S o H2O(g) = 189 Soln. S o rxn = 2S o Al + 3S o H2O 3S o H2 So Al2O3 = 2(28) + 3(189) 3(131) 51 = 179 JK 1 mol 1 Note: O H H In general: has more rotational and vibrational freedom than H 2 自由度 the more complex the molecule the higher the entropy Free energy and chemical reactions G is measured indirectly Method 1 Ex. G o = H o T S o 2SO 2 (g) + O 2 (g) 2SO 3 (g) H fo (SO 2 ) = 297 kj/mol S o (SO 2 ) = 248 JK 1 mol 1 H fo (SO 3 ) = 396 S o (SO 3 ) = 257 S o (O 2 ) = 205 Soln: H o rxn = 2( 396) 2( 297) = 198 kj S o rxn = 2(257) 2(248) 1(205) = 187 J/K = kj/k G o rxn = 198 (298)( 187 x 10 3 ) = 142 kj 11

12 Method 2 Free energy is a state function The change is independent of the pathway Derive from known reactions Ex. 2CO(g) + 4H 2 O(g) 2CH 4 (g) + 3O 2 (g) 2CH 4 (g) + 4O 2 (g) 2CO 2 (g) + 4H 2 O(g) G o = 1088 kj G o = 1602 kj 2CO(g) + O 2 (g) 2CO 2 (g) G o = = 514 kj Method 3 From standard free energy of formation (same as enthalpy treatment) Ex. G o rxn = n p Go products n r Go reactants 2CH 3 OH(g) + 3O 2 (g) 2CO 2 (g) + 4H 2 O(g) G o rxn =? Known G o f for CH 3 OH = 163 kj/mol CO 2 = 394 kj/mol H 2 O = 229 kj/mol Soln. G o rxn = 2( 394) + 4( 229) 2( 163) = 1378 kj 12

13 The dependence of G on P From definition: G = H TS H = E + PV G = E + PV TS One can show that dg = RTd(lnP) (1 mol and constant T) Set a reference state with G = G o and P = 1 atm Integrate from this state to any state of interest G G o = RTln(P/1) = RTlnP G = G o + RTlnP free energy of interest When P = 1 atm G = G o (free energy of the gas at 1 atm) a function of T Ex. N 2 (g) + 3H 2 (g) 2NH 3 (g) G rxn = n p G products n r G reactants = 2G G NH 3 N2 3G H2 From the relationship of G and P: G = NH 3 Go NH3 + RTlnP NH3 G N 2 G H 2 = Go N2 + RTlnP = N2 Go H2 + RTlnP H2 G rxn = 2G o NH3 + 2RTlnP NH3 Go N2 RTlnP N2 3G o H2 3RTlnP H2 = (2G o NH3 Go N2 3Go H2 ) + RT(2lnP NH3 lnp N2 3lnP H2 ) = G o rxn + RTln P 2 NH3 P N 2 P3 H2 Quotient of the reaction = Q 13

14 G = G o + RTlnQ Reaction quotient Standard free energy change (P = 1 atm for all gases) Ex. CO(g) + 2H 2 (g) CH 3 OH(l) G? At 25 o C start from P CO = 5.0 atm, P H = 3.0 atm 2 Known: G o f (CH 3 OH) = 166 kj G o f (H 2 ) = 0 kj G o f (CO) = 137 kj Soln. G o = ( 166) ( 137) 2(0) = 29 kj G = G o + RTlnQ = (8.31)(298)ln (5.0)(3.0) 2 = J per mol of CO [More negative than G o (at 1 atm): agrees with Le Chatelier s principle] Note about significant figures ln(1/45) = 2.303log (1/45) = 2.303log45 = 2.303log (4.5 10) = 2.303[( log 4.5) + 1] exact number 0.65 two significant figures = 2.303(1.65) =

15 Free energy and equilibrium H = X A H A + X B H B G rxn = n p G products n r G reactants H A H H = H rxn T S rxn A B S S B H B Randomness due to mixing A and B S A S A reactant 1 0 mole of A (or n A ) 0 1 mole of B (or n B ) B product A G = H TS B G A The equilibrium point: The point at which there is no driving force to go towards either direction A B G G B G = G o + RTlnQ At equilibrium: G = 0 Q = K (equilibrium constant) 0 = G o + RTlnK G o = RTlnK The standard free energy change with all reactants and products at 1 atm partial pressure 15

16 G = G o + RTlnQ = RTlnK + RTlnQ = RTln(Q/K) When Q = K, G = RTln1 = 0 When Q > K, G > 0 When Q < K, G < 0 at equilibrium backward forward Agrees with Le Châtelier s principle!! Ex. 4Fe(s) + 3O 2 (g) 2Fe 2 O 3 (s) K? H o f (kj/mol) S o (JK 1 mol 1 ) Fe 2 O 3 (s) Fe(s) 0 27 O 2 (g) Soln. H o = 2( 826) = 1652 kj S o = 2(90) 4(27) 3(205) = 543 JK 1 = kjk 1 G o = H o T S o = 1652 (298)( 0.543) = kj = J = RTlnK = (8.31)(298)lnK lnk = logK = 601 logk = 261 K = a highly favorable process 16

17 The temperature dependence of K G o = RTlnK = H o T S o H o S o lnk = + RT R H = o 1 S o ( ) + R T R Assume H o, S o are T independent (not quite so) A linear relationship of lnk and 1/T with slope = ( H o /R) intercept = S o /R Experimentally this is very useful When H o > 0 (endothermic) slope: ( ) lnk T 1/T lnk 1/T When H o < 0 (exothermic) lnk slope: ( ) T 1/T lnk 1/T 17

18 Consider T 1 T 2 K 1 K 2 lnk 2 = Ho RT 2 S o + lnk R 1 = Ho RT 1 + S o R lnk 2 lnk 1 = Ho R 1 1 T 2 T 1 The van t Hoff equation: ln(k 2 /K 1 ) = Ho R 1 1 T 2 T 1 Ex. N 2 (g) + 3H 2 (g) 2NH 3 (g) K = at 900 K H o = 92 kj K at 550 K? Soln. K 550 ln ( 92000) x 10-6 = lnk 550 = log K 550 = 4.8 log K 550 = 2.1 = K 550 =

19 Free energy and work Qualitatively G: to know whether a reaction will occur or not Quantitatively G = w max (maximum possible work obtainable) For a spontaneous reaction G is the energy that is free to do work For a non-spontaneous reaction G is the minimum amount of work that must be expended Ex. Battery Motor Useful work is wasted as heat higher current more heat lower current less heat zero current zero heat Maximum amount of work can be obtained (no waste) Only hypothetical 19

20 Ex. w 1 Battery charged Battery discharged w 2 > w 1 unless the current 0 (a reversible process) w 2 All real processes are irreversible Work is always wasted as heat Released into the surroundings S goes up Biological relevance 自由能 (G) 的變化是反應的驅動力自由能越低越好 : G 愈負, 驅動力愈強 生命體系最重要的能量轉換系統 : ATP + H 2 O ADP + HPO H + G o = 30.5 kj/mol NH 2 NH 2 HO O P O O O P O O O P O O O N N N N HO O P O O O P O O O N N N N OH ATP OH OH ADP OH O + H 2 O + HO P O O * G : at ph 7 and based on water concentration of 55.5 M 20

21 例子 CHO H OH HO H H OH H OH CH 2 OH + HOPO 3 2 CHO H OH HO H + H 2 O G' o = 13.8 kj/mol H OH H OH 2 CH 2 OPO 3 Glucose Glucose 6-phosphate 正值 : 無法進行 解決之道 : Glucose 2 + HOPO 3 Glucose 6-phosphate + H 2 O G' o = 13.8 kj/mol ATP + H 2 O ADP + HOPO H + G' o = 30.5 kj/mol Glucose + ATP Glucose 6-phosphate + ADP + H + G' o = 16.7 kj/mol 透過 ATP 的幫助, 葡萄糖的磷酯化現在可行了! 21