Thermodynamics vs. Kinetics

Transcription

1 Thermodynamics

2

3 Thermodynamics vs. Kinetics Kinetics is concerned with reaction rates which depend on the activation energy. Thermodynamics is concerned with the difference in energy between reactants and products Thermodynamics is concerned with the position of equilibrium, related to Keq

4 Laws of Thermodynamics First Law: the law of conservation of energy

5 Spontaneous Processes and Entropy

6 In chemistry we are interested in whether a particular reaction will go usually means a favorable equilibrium constant and a conveniently rapid rate Or will not go either an unfavorable equilibrium constant or a rate too slow to be useful

7 Spontaneous Processes We describe a process as spontaneous or nonspontaneous. spontaneous means you get more products than reactants at equilibrium. Keq > 1 Whether a reaction is spontaneous or not has nothing to do with its rate. A spontaneous reaction can be very fast or can be so slow that it appears not to take place at all

8 Examples of spontaneous processes Waterfalls run downhill spontaneously, but not uphill. A gas expands into a vacuum spontaneously, but does not flow out of its container to form a vacuum spontaneously. Water freezes below 0ºC spontaneously; ice melts above 0 ºC spontaneously. 2Na(s) + 2H 2 O (l) spontaneous H 2 (g) + 2NaOH (aq) H 2 (g) + 2NaOH (aq) nonspontaneous 2Na(s) + 2H 2 O (l)

9 We have emphasized enthalpy in our earlier discussions of thermodynamics Our expectation is that a reaction that leads to a decrease in the total energy of the system should be spontaneous (ΔH is negative; reaction is exothermic). But observation tells us that enthalpy alone is an insufficient indicator of spontaneity. some endothermic reactions are spontaneous spontaneity depends on temperature; some reactions are spontaneous at one temperature, but nonspontaneous at another

10 Two examples of spontaneous endothermic processes ice melts spontaneously at temperatures above 0 ºC but not below 0 ºC. H 2 O(s) H 2 O(l) ΔH º = kj ammonium nitrate dissolves in water NH 4 NO 3 (s) NH 4+ (aq) + NO 3 (aq) ΔH º = + 25 kj

11 Four possibilities; examples of all four are known exothermic spontaneous exothermic nonspontaneous endothermic spontaneous endothermic nonspontaneous

12 Entropy A process is spontaneous if it leads to an increase in the entropy of the universe. Entropy is a measure of the randomness or disorder of a system. Entropy is related to probability.

13 Probability a probable event is one that can happen in many ways an improbable event is one that can happen in only one way

14 Entropy and Probability Expansion of ideal gas into a vacuum is spontaneous, but migration of gas molecules into one region of a container is nonspontaneous.

15 expansion of ideal gas into a vacuum is spontaneous

16 expansion of ideal gas into a vacuum is spontaneous but migration of gas molecules into one region of a container is nonspontaneous

17 Why is the spontaneous process the one that gives equal numbers of gas molecules in both flasks? It is the most probable state -- the one that has the most ways of being achieved.

18 How many ways may 1 gas molecule be arranged in a two-bulb container? probability that the gas molecule will be in left bulb =.5 Left Right n where n = number of molecules

19 What is the probability that two gas molecules will be in the same bulb of a two-bulb container? Left Right both 0 green blue blue green 0 both probability that both gas molecules will be in left bulb = n = where n = number of molecules

20 What is the probability that 3 gas molecules will be in the same bulb of a two-bulb container? Left Right ways ways 0 3 probability that all gas molecules will be in left bulb = n = where n = number of molecules

21 What is the probability that one mole of gas molecules will be in the same bulb of a two-bulb container? where n = Avogadro s number of molecules probability that all gas molecules will be in left bulb is very small 1 2 n = 1 2 N

22 What is the probability that 4 gas molecules will be equally distributed in a two-bulb container? 6 16 Left Right No. of ways ways 3 1 4ways 16 possibilities ways 1 3 4ways ways Equal distribution is the most probable out come

23 An ordered state has a low probability of occurring and a small entropy, while a disordered state has a high probability of occurring and a high entropy

24 How are the entropy of different phases related? solid liquid gas S solid < S S liquid < gas

25 Entropy v.s Enthalpy it is possible to determine absolute entropy (S) as opposed to enthalpy (H) S = J/K ΔH = kj and per mole: S º = J/Kmol ΔH º = kj/mol

26 Third (and last) Law of Thermodynamics - Absolute Entropy the entropy of a perfect crystalline substance is zero at the absolute zero of temperature : 0 K ( -273º C ) important in the determination of absolute entropy values

27 Some standard entropy values ( S º) substance S º (J/K mol ) H 2 O(l ) H 2 O(g ) Br 2 (l ) Br 2 ( g ) I 2 ( s ) I 2 ( g ) C (diamond) 2.44 C ( graphite) 5.69

28 Second Law of Thermodynamics

29 First Law: the total energy of the universe is a constant Second Law: The entropy of the universe increases in a spontaneous process, and remains unchanged in a process at equilibrium

30 Columbia University Chemistry Ph.D., author of many works of science fiction including I, Robot, and the short story The Last Question

31 Entropy and the second law A process is spontaneous if ΔS univ is positive ΔS univ = ΔS system + ΔS surr > 0 A process is at equilibrium if: ΔS univ = ΔS system + ΔS surr = 0 Therefore we need to consider how the entropy of the system and the surroundings change during a process

32 Entropy changes in the System

33 Entropy changes in the System ΔS system is + if disorder increases ΔS system is - if products are more ordered than reactants Can be calculated from tables of thermodynamic values ΔS rxn = ΣnS (products) Σm S (reactants) We can often make good guesses as to the sign of ΔS system

34 Entropy changes in the System Calculate the standard entropy change for: 2CO(g ) + O 2 (g ) 2CO 2 (g ) qualitative prediction: 2 moles of gas on the right, 3 on the left ; the products are more ordered than reactants; the sign of ΔS is -

35 Entropy changes in the System If a reaction produces excess gas ΔS system is + If a reaction produces no net change in gas molecules ΔS system may be ( +) or ( - ) but the change will have a small value General Rule: a reaction that increases the total number of molecules or ions will increase ΔS system

36 Example: Calculate the standard entropy change for: 2CO(g ) + O 2 (g ) 2CO 2 (g ) S (reactants) 2 mol(197.9 J/K mol) + 1mol(205 J/K mol) = J/K S (products) 2 mol(213.6 J/K mol) = J/K ΔSº rxn = J/K J/K ΔSº rxn = J/K

37 Entropy changes in the surroundings

38 Entropy changes in the surroundings How are surroundings affected by heating and cooling? exothermic reactions increases entropy of surroundings endothermic reactions decrease entropy of surroundings ΔH system ΔS surr = q T ΔS surr = T

39 Surroundings Surroundings System Heat System

40 Surroundings Surroundings System Heat Entropy System Heat

41 Surroundings Surroundings System Heat Entropy System Heat Entropy

42 What are the possibilities? ΔS sys + = ΔS surr ΔS univ +?? spontaneous? yes no

43 Spontaneity and temperature a reaction may be spontaneous at one temperature and nonspontaneous at a different temperature ΔS univ = ΔS system + ΔS surr ΔH system T

44 Gibbs Free Energy ΔS universe = ΔS surroundings + ΔS system we study the system; therefore reference surroundings in terms the system ΔS universe = ΔH system / T + ΔS system TΔS universe = ΔH system + TΔS system TΔS universe = ΔH system TΔS system ΔG = ΔH system TΔS system

45 Gibbs Free Energy Free Energy (ΔG) is a measure energy available to do work. a release of free energy during a chemical reaction is spontaneous. takes into account both enthalpy (heat released or absorbed) and entropy (disorder).

46 Criterion for spontaneity is the Gibbs free energy change The meaning of its +/- signs are opposite DSuniverse. At constant temperature and pressure, if ΔG is system negative,the reaction is spontaneous positive, the reaction is not spontaneous zero, the system is at equilibrium

47 Standard Free-Energy Changes

48 Standard Free Energies of Formation the change in free energy that accompanies the formation of 1 mole of a substance from its constituent elements at standard conditions ΔGf = kj ΔGº = kj/mol f

49 From Standard Free Energies of Formation ΔGº rxn = ΣnΔGº f (products) Σm ΔGº f (reactants) the standard free energies of formation of any element in its stable form equals zero

50 Example: Calculate the standard free-energy changes for the following reaction at 25ºC : 2C 2 H 6 (g ) + 7O 2 (g ) 2mol( 32.9 kj/mol) + 7mol(0 kj/mol) 4CO 2 (g ) + 6H 2 O(l ) 4mol( kj/mol) + 6mol( kj/mol) 66 kj 3001 kj ΔGº = 3001 ( 66) = 2935 kj

51 Applications of ΔG = ΔH TΔS

52 ΔG = ΔH TΔS ΔH ΔS ΔG spontaneous rxn? - (exo) + (endo) + (endo) - (exo) + (disorder ) - (disorder ) + (disorder ) - (disorder ) - yes, at all temps + no, at all temps?? depends on temp. T favors spontaneity of forward rxn depends on temp. T favors spontaneity of forward rxn

53 Endothermic dissolution: at 25ºC ( 298 K) NH Cl(s ) 4 NH + (aq ) 4 + Cl - (aq ) ΔHº f kj/mol kj/mol kj/mol ΔH rxn Sº 94.6 J/mol K J/mol K 56.5 J/mol K = [ kj/mol + ( kj/mol) ] - ( kj/mol) = 15.4 kj/mol ΔS = [112.8 J/mol K J/mol K ] - (94.6 J/mol K) = 74.7 J/mol K =.0747 kj/mol K

54 Endothermic dissolution: at 25 º C ( 298 K) NH Cl(s ) 4 NH + (aq ) 4 + Cl - (aq ) ΔHº rxn = 15.4 kj/mol ΔSº =.0747 kj/mol K Τ = 298 K ΔG = ΔH TΔS ΔG = 15.4 kj/mol 298 K (.0747 kj/mol K) ΔG = kj/mol

55 Endothermic dissolution: at 25 º C ( 298 K) NH Cl(s ) 4 NH + (aq ) 4 + Cl - (aq ) ΔHº f kj/mol kj/mol kj/mol Sº 94.6 J/mol K J/mol K 56.5 J/mol K ΔGº f kj/mol kj/mol kj/mol ΔGº rxn = -6.8 kj/mol

56 Free Energy and Equilibrium

57 kinetic definition of equilibrium position: the point at which rates of forward and reverse reactions are equal thermodynamic definition : the point of minimum free energy; ΔG = 0

58 The relationship between ΔG and ΔG ΔGº = ΔHº TΔSº reactants in their standard state completely changing into products in their standard state as soon as the reaction starts the standard state condition no longer exists ΔG = ΔH TΔS the absolute change in free energy change

59 The relationship between ΔG and ΔG ΔG = ΔGº + RT ln(q ) ΔG = 0 at equilibrium ΔG = ΔGº + RT ln [products] [reactants]

60 The relationship between ΔG and ΔG ΔG = ΔG + RT ln(q) ΔG = 0 at equilibrium a large negative value ΔG 0 = ΔG + RT ln a large relative value for products needed to make (RTlnQ) equivalent to ΔG [products] [reactants]

61 The relationship between ΔG and ΔG a large positive value ΔG ΔG = ΔG + RT ln(q ) ΔG = 0 at equilibrium a small relative value for products needed to make (RTlnQ) equivalent to ΔG 0 = ΔG + RT ln [products] [reactants]

62 The relationship between ΔG and ΔG ΔG = ΔG + RT ln(q ) at equilibrium: ΔG = 0 and Q = K 0 = ΔG + RT ln K ΔG = RT ln K

63 Example ΔG = kj for the reaction at 25ºC: N 2 (g ) + 3H 2 (g ) 2NH 3 (g ) Is the reaction going forward in the direction written under these conditions? P(NH 3 ) = 12.9 atm, P(N 2 ) = 0.87 atm, and P(H 2 ) = 0.25 atm?

64 Example N 2 (g ) + 3H 2 (g ) 2NH 3 (g ) 0.87 atm 0.25 atm 12.9 atm ΔG = ΔG + RT ln(q ) ΔG= kj + (8.314 J/mol K) (298K) ln P2 (NH 3 ) P(N 2 ) P 3 (H 2 )

65 Example N 2 (g ) + 3H 2 (g ) 2NH 3 (g ) 0.87 atm 0.25 atm 12.9 atm ΔG = ΔG + RT ln(q ) ΔG= kj + (8.314 J/mol K) (298K) ln (12.9) 2 (0.87) (0.25) 3 ΔG = J J = J Reaction is not at equilibrium; the direction is to the right

66 Applications calculate the equilibrium constants from tables of standard free energies of formation ΔG = RT ln K K K a K sp

67 Example Calculate the equilibrium constant ( Kp ) for the reaction shown at 25 º C. 2O 3 (g ) 3O 2 (g ) ΔG f = 2mol (163.4 kj/mol) ΔG = kj ΔG = RT ln K 3mol (0 kj/mol) J/mol = -(8.314 J/mol K)(298K)ln K ln K = K = e 132 = 1.93 x 10 57

68 What is the K for the reaction of ammonia with hydrochloric acid in aqueous solution? NH (aq ) + H + (aq ) NH + (aq ) 3 4 ΔG f = kj/mol 0 kj/mol kj/mol ΔG = -53 kj ΔG = RT ln K J/mol = (2478 J/mol)ln K ln K = 21.4 K = e 21.4 = 1.9 x 10 9

69 What is the K for the reaction of ammonia with hydrochloric acid in aqueous solution? NH (aq ) + H + (aq ) NH + (aq ) 3 4 K = e 21.4 = 1.9 x 10 9 Note: the reverse reaction defines K a for NH K a = 1.9 x 10 9 = 5 x 10-10

70 Free Energy and Equilibrium Summary

71 The relationship between ΔG and ΔG and equilibrium ΔG rxn < 0 ΔG rxn > 0 ΔG rxn = 0 spontaneous nonspontaneous K > 1 K < 1 K = 1