1. Chapter 19 Chemical Thermodynamics

2. Spontaneity of Physical & Chemical Changes Thermodynamics is concerned with the question: can a reaction occur? First Law of Thermodynamics: energy is conserved. Any process that occurs without outside intervention is spontaneous. When two eggs are dropped they spontaneously break. The reverse reaction (two eggs leaping into your hand with their shells back intact) is not spontaneous. We can conclude that a spontaneous process has a direction.

3. Spontaneity of Physical & Chemical Changes Spontaneous changes happen without any continuing outside influences. A spontaneous change has a natural direction. rusting of iron - occurs spontaneously Have you ever seen rust turn into iron metal without man made interference? melting of ice at room temperature - occurs spontaneously Will water spontaneously freeze at room temperature?

4. The Two Parts of Spontaneity Exothermicity does not ensure spontaneity Freezing of water exothermic spontaneous only below 0 o C

5. Spontaneous Processes A process that is spontaneous in one direction is not spontaneous in the opposite direction. The direction of a spontaneous process can depend on temperature: Ice turning to water is spontaneous at T > 0C, Water turning to ice is spontaneous at T 0C, Water turning to ice is spontaneous at T < 0C.

6. The Second Law of Thermodynamics In spontaneous changes the universe tends towards a state of greater disorder. Spontaneous processes require: free energy change of system must be negative entropy of universe must increase – system must be capable of doing useful work on surroundings

7. Entropy, S Entropy is a measure of the disorder or randomness of a system. As with H, entropies have been measured and tabulated in Appendix C as S o 298. When: S is positive disorder increases (favors spontaneity) S is negative disorder decreases (disfavors spontaneity)

8. Entropy, S From Second Law of Thermodynamics, for a spontaneous process

9. Entropy, S For a reversible process: S univ = 0. For a spontaneous process (i.e. irreversible): S univ > 0 Note: the second law states that the entropy of the universe must increase in a spontaneous process. It is possible for the entropy of a system to decrease as long as the entropy of the surroundings increases.

10. The Molecular Interpretation of Entropy A gas is less ordered than a liquid that is less ordered than a solid. In general: S gas > S liquid > S solid Any process that increases the number of gas molecules leads to an increase in entropy.

11. The Molecular Interpretation of Entropy NO (g) and O 2 (g) react to form NO 2 (g)

12. The Molecular Interpretation of Entropy There are three atomic modes of motion: translation (the moving of a molecule from one point in space to another), vibration (the shortening and lengthening of bonds, including the change in bond angles), rotation (the spinning of a molecule about some axis).

13. The Molecular Interpretation of Entropy Energy is required to get a molecule to translate, vibrate or rotate. The more energy stored in translation, vibration and rotation, the greater the degrees of freedom and the higher the entropy. In a perfect crystal at 0 K there is no translation, rotation or vibration of molecules. Therefore, this is a state of perfect order. Therefore, this is a state of perfect order. Third Law of Thermodynamics: the entropy of a perfect crystal at 0 K is zero.

14. The Molecular Interpretation of Entropy

15. As we heat a substance from absolute zero, the entropy must increase. Entropy will increase when liquids or solutions are formed from solids,liquids or solutions are formed from solids, gases are formed from solids or liquids,gases are formed from solids or liquids, the number of gas molecules increase,the number of gas molecules increase, the temperature is increased.the temperature is increased.

16. Entropy, S Third Law of Thermodynamics states that the entropy of a pure, perfect, crystalline solid at 0 K is zero. – allows us to measure absolute values of entropy for substances – cool them down to 0 K, or as close as possible, then measure entropy increase as substance warms up Entropy changes for reactions can be determined similarly to H for reactions.

17. Entropy, S Third Law of Thermodynamics states that the entropy of a pure, perfect, crystalline solid at 0 K is zero. allows us to measure absolute values of entropy for substances cool them down to 0 K, or as close as possible, then measure entropy increase as substance warms up Entropy changes for reactions can be determined similarly to H for reactions. Units: J/mol-K. Note units of H: kJ/mol. Standard molar entropies of elements are not zero.

18. Entropy, S Calculate the entropy change for the following reaction at 25 o C. Use appendix.

19. Entropy, S  negative sign indicates system is more ordered  reverse the reaction and sign changes  S o 298 = +0.1758 kJ/K where the + sign indicates system is more disordered

20. Entropy, S  Calculate S o 298 for the reaction below. Use appendix.

21. Entropy, S Changes in S are usually quite small compared to H.

22. Free Energy Change,  G, and Spontaneity J. Willard Gibbs determined the relationship of enthalpy and entropy that best describes the maximum useful energy obtainable in the form of work from a process at constant T & P. The relationship also describes the spontaneity of a system. The relationship is a new state function, G, the Gibbs Free Energy.

23. Free Energy Change,  G, and Spontaneity The relationship is a new state function, G, the Gibbs Free Energy.

24. Free Energy Change,  G, and Spontaneity The change in the Gibbs Free Energy is a reliable indicator of spontaneity of a physical process or chemical reaction. does not tell us the speed of the process That is kinetics When: G is > 0 reaction is nonspontaneous G is = 0 system is at equilibrium G is < 0 reaction is spontaneous

25. Free Energy Change,  G, and Spontaneity Changes in free energy obey the same type of relationship we have described for enthalpy and entropy changes.

26. Free Energy Change,  G, and Spontaneity Calculate G o 298 for the reaction:

27. Free Energy Change,  G, and Spontaneity G o 298 is negative, so the reaction is spontaneous at standard state conditions. If we reverse the reaction: G o 298 is positive, so the reaction is nonspontaneous at standard state conditions.

28. The Temperature Dependence of Spontaneity The general relationship of G, H, and S is

29. The Temperature Dependence of Spontaneity The general relationship of G, H, and S is This gives us 4 possibilities among the signs

30. The Temperature Dependence of Spontaneity H SGTherefore - +- forward rxn spontaneous at all T’s - +- forward rxn spontaneous at all T’s - -? forward rxn spontaneous at low T’s - -? forward rxn spontaneous at low T’s + +? forward rxn spontaneous at high T’s + +? forward rxn spontaneous at high T’s + -+ forward rxn nonspontaneous at all T’s + -+ forward rxn nonspontaneous at all T’s

31. The Temperature Dependence of Spontaneity

33. Example: Calculate S o 298 for the following reaction. We found that H o 298 = -2219.9 kJ, and we found that G o 298 = -2108.5 kJ.

34. The Temperature Dependence of Spontaneity Example: Calculate S o 298 for the following reaction. We found that H o 298 = -2219.9 kJ, we found that G o 298 = -2108.5 kJ.

35. The Temperature Dependence of Spontaneity

36.  S o 298 = -374 J/K which indicates that the disorder of the system decreases. For the reverse reaction, 3 CO 2(g) + 4 H 2 O (l) C 3 H 8(g) + 5 O 2(g)  S o 298 = +374 J/K which indicates that the disorder of the system increases.

37. Free Energy and the Equilibrium Constant For a spontaneous process, the free energy of the system decreases until it reaches a minimum value. When this minimum value is reached the system is in a state of Equilibrium. For a spontaneous process, the free energy of the system decreases until it reaches a minimum value. When this minimum value is reached the system is in a state of Equilibrium.

38. Free Energy and the Equilibrium Constant G and K (equilibrium constant) apply to standard conditions. G and K (equilibrium constant) apply to standard conditions. G and Q (equilibrium quotient) apply to any conditions. G and Q (equilibrium quotient) apply to any conditions. It is useful to determine whether substances under any conditions will react: R = Ideal gas law constant 8.314 J/molK

39. Free Energy and the Equilibrium Constant At equilibrium, Q = K c and G = 0, so From the above we can conclude: If G 1. The more negative G is the larger the equilibrium constant. If G = 0, then K c = 1. If G > 0, then K c 0, then K c < 1.

40. Synthesis Question When it rains an inch of rain, that means that if we built a one inch high wall around a piece of ground that the rain would completely fill this enclosed space to the top of the wall. Rain is water that has been evaporated from a lake, ocean, or river and then precipitated back onto the land. How much heat must the sun provide to evaporate enough water to rain 1.0 inch onto 1.0 acre of land?  1 acre = 43,460 ft 2

41. Synthesis Question

43. Group Question When Ernest Rutherford, introduced in Chapter 5, gave his first lecture to the Royal Society one of the attendees was Lord Kelvin. Rutherford announced at the meeting that he had determined that the earth was at least 1 billion years old, 1000 times older than Kelvin had previously determined for the earth’s age. Then Rutherford looked at Kelvin and told him that his method of determining the earth’s age based upon how long it would take the earth to cool from molten rock to its present cool, solid form

44. Group Question was essentially correct. But there was a new, previously unknown source of heat that Kelvin had not included in his calculation and therein lay his error. Kelvin apparently grinned at Rutherford for the remainder of his lecture. What was this “new” source of heat that Rutherford knew about that had thrown Kelvin’s calculation so far off? was essentially correct. But there was a new, previously unknown source of heat that Kelvin had not included in his calculation and therein lay his error. Kelvin apparently grinned at Rutherford for the remainder of his lecture. What was this “new” source of heat that Rutherford knew about that had thrown Kelvin’s calculation so far off?