1. Chapter 18 Thermodynamics and Equilibrium

2. The Three (4) Laws of Thermodynamics First Law: Matter cannot be created nor destroyed, it can only change form. Second Law: Disorder is always increasing. Third Law: All processes cease and entropy becomes a minimum as temperatures approach absolute zero. Zeroth Law: Is a generalization about thermal equilibrium.

3. The First Law of Thermodynamics The Law of Conservation of Energy applies to thermochemistry and thermodynamic systems. http://library.thinkquest.org/2745/data/lawce1.htm

4. Internal Energy As we said in Chapter 6, the internal energy, U, is the sum of the kinetic and potential energies of the particles making up the system. ◦ KE is the energy of motion of electrons, nuclei, and molecules. ◦ PE is the result of chemical bonding of atoms and from the attractions between molecules.

5. Internal Energy Internal energy is a state function—a property of a system that depends only on its present state; the state is determined by variables such as pressure and temperature. ◦ U changes when the state changes and can be calculated:   U = U f - U i

6. Internal Energy Normally, all we want to know is the changes in internal energy which are measured by noting changes between the system and its surroundings—either measured as heat or work.

7. Heat Heat will move in or out of a system due to a temperature difference.

8. Work Work is the energy exchange that results when a force, F, moves an object through a distance, d. ◦ W = Fd If you increase the energy in the cylinder, the piston moves up. By decreasing the energy of the surroundings by 165J, the LCE states that the internal energy of the system must have increased by this much (q = +165J)

9. q When q is positive, heat is absorbed by the system. When q is negative, heat is given off by the system. If we increase the temperature of the gas in the vessel, we increase the pressure when the volume remains fixed. Once you allow the piston to move, the gas will expand moving the piston up and hence doing work.

10. Internal Energy and Work The energy gained, E, is equal to the force applied times the distance traveled. E = Fd ◦ Again, work done on the system is positive; work done by the system is negative. If the work done is 92J, then W = -92J, because this is the work done by the system to the surroundings.

11. Internal Energy and Work In general, the net change of internal energy equals heat plus work. ◦ U = q + w The First Law of Thermodynamics states that the change in the internal energy of a system,  U = q + w

12. Internal Energy and Work Consider the following chemical system:  Zn(s) + 2HCl(aq)  ZnCl(aq) + H 2 (g) http://i.ytimg.com/vi/oQz5YEsx7Fo/0.jpg

13. Internal Energy and Work Consider the following chemical system:  Zn(s) + 2HCl(aq)  ZnCl(aq) + H 2 (g) This reaction is exothermic, giving off 152.4kJ of heat per mole of Zn. q p = -152.4kJ; “p” is the process that occurs at constant pressure. http://i.ytimg.com/vi/oQz5YEsx7Fo/0.jpg

14. Internal Energy and Work The H 2 gas given off increases the volume of the system. As the H 2 is evolved, work must be done by the system to push back on the atmosphere. http://i.ytimg.com/vi/oQz5YEsx7Fo/0.jpg

15. Internal Energy and Work To calculate the work: ◦ Imagine the atmosphere is replaced by a piston and weights whose downward force, F, from gravity creates a pressure on the gas equivalent to that of the atmosphere.

16. Internal Energy and Work The pressure, P, equals (F/area of the piston). ◦ The volume increases due to the H 2 production =  V Volume of the cyliner = h Area  V = h A h =  V/A W = -Fh = -F  V/A = -F/A  V ◦ It’s negative because work is done by the system and represents energy lost by the system to the surroundings. F/A = P W = -P  V

17. Internal Energy and Work For example: when 1.00 mol Zn reacts with excess HCl, 1.00 mol of H 2 is produced. At 25°C and 1.00 atm, this amount of H 2 is equal to 24.5 L. The work done is: ◦ W = -P  V ◦ = -(1.01 x 10 5 Pa) (24.5 x 10 -3 m 3 ) ◦ = -2.47 x 10 3 J

18. Internal Energy and Work Applying the First Law, the change in internal energy,  U, ◦ U = q p + w = q p – P  V For the Zn + HCl rection: q p = -152.4kJ, w = -P  V = -2.47kJ So, ◦ U = -152.4kJ – 2.47kJ = -154.9kJ

19. Internal Energy and Work Here’s a summary of what happens: ◦ When 1.00mol of Zn is reacted with excess HCl, the internal energy changes as the KE and PE change as the reactants go to products. ◦ The energy change,  U, is -154.9kJ. ◦ Most of this energy is heat, q p = -152.4kJ, but some of it is given off as expansion work— w = -2.47kJ

20. Enthalpy and Enthalpy Change We can now look at enthalpy as the quantity U + PV ◦ H = U + PV Enthalpy is a state function because U, P and V are all state functions. (For a given T and P, a given amount of substance has a definite enthalpy.) There is a relationship between enthalpy,  H, and heat of reaction, q p.

21. Enthalpy and Enthalpy Change  H = H f – H i ◦ H = (U f – PV f ) – (U i + PV i ) ◦ H = (U f – U i ) + P(V f – V i ) =  U + P  V From earlier:  U = q p – P  V ◦ H = (q p – P  V) + P  V = q p And from chapter 6,  H = q p

22. Standard Heats of Reaction  H° =  n  H° f(products) –  m  H° f(reactants)  H° f = standard heats of reaction, found in appendix C.

23. Standard Heats of Reaction Consider the following: ◦ How much heat is absorbed or evolved when NH 3 and CO 2 produce urea and H 2 O? ◦ At 25°C,  H° f  NH 3 (g) = -45.9 kJ/mol  CO 2 (g) = -393.5 kJ/mol  NH 2 CONH 2 (aq) = -319.2 kJ/mol  H 2 O(l) = -285.8 kJ/mol

24. Standard Heats of Reaction  H° = [(-319.2-285.8)-((-2x45.9)-393.5)]kJ = -119.7kJ From this, heat is evolved and the reaction is exothermic.

25. Entropy We often think of something that is spontaneous as “fast.” This may or may not be the case in chemistry but, by definition, a spontaneous process is a physical or chemical change that occurs by itself—nothing from the outside is required to make it happen. ◦ For instance, iron quickly rusts in air and a diamond slowly breaks down into graphite. Both of these processes are spontaneous.

26. Spontaneous Reactions There are many processes which are spontaneous: ◦ Rock rolling down a hill ◦ Heat flowing from hot to cold ◦ Iron rusting in moist air http://thebsreport.files.wordpress.com/2009/11/rusty-nail-tetanus-1.jpg http://www.g9toengineering.com/resources/heatflow.gif

27. Nonspontaneous Reactions Nonspontaneous reactions require an outside force (work). ◦ Getting iron from rust (iron ore) ◦ Moving a rock to the top of a hill ◦ Moving heat from cold to hot http://www.revisionworld.co.uk/files/iron.jpg http://www.explainthatstuff.com/how-heat-exchangers-work.gif

28. Entropy The First Law of Thermodynamics cannot be used to predict if a chemical change is spontaneous or not. The Second Law of Thermodynamics, which deals with entropy, S, can be used to predict the spontaniety of a reaction. Entropy, S, is a measure of the disorder of a system.

29. Entropy In a spontaneous reaction, entropy (S) always increases. ◦ Cooling coffee—the energy flows from the coffee to the table and the surrouding air. The energy gets dispersed and entropy increases. ◦ When the gas moves from the container into the vacuum, randomness or entropy increases. http://mechgrads.com/Documents/HTTTP.1.gif

30. Entropy The unit for entropy is J/K. Entropy is a state function. It depends on the variables that determine the state of the substance (such as temperature and pressure).

31. Entropy Ice is more ordered than liquid water and has less entropy. The entropy change,  S, can be calculated. ◦ S = S f – S i ◦ H 2 O(s)  H 2 O(l) ◦ S = (63 - 41)J/K = 22J/K

32. Entropy When 1 mole of ice melts @ 0°C, the water increases in entropy by 22J/K ◦ The Second Law of Thermodynamics states that the total entropy of a system and its surroundings always increases for a spontaneous process. ◦ Entropy can be produced (or created) during a spontaneous process, but energy of course, cannot. http://dclips.fundraw.com/zobo500dir/ice_cube_jarno_vasamaa_.jpg

33. Entropy When we look at chemical reactions, we can restate the Second Law so as to only include the system. ◦ Heat flow is a flow of entropy because it is a dispersal of energy. In general, the  S associated with heat flow, q, at absolute temperature, T = q/T, that is,  S = q/T  S = entropy created + q/T

34. Entropy Entropy created during a spontaneous process is a positive quantity. So, we can rewrite the equation: ◦ S > q/T We can also say the Second Law of Thermodynamics the following way: ◦ For a spontaneous process at a given temperature, the  S of the system is greater than q/T.

35. Entropy At equilibrium:  S = q/T =  H fus /T ◦ q is heat flow ◦ T is temperature We can apply  S = q/T to get the entropy changes for a phase change. ◦ H fus for 1.00 mol of ice is 6.0 kJ ◦ S = 6.0kJ/273K = 6.0x10 3 J/273K = 22J/K

36. Entropy We can now apply thermodynamics to determine whether or not a reaction is spontaneous. 2NH 3 (g) + CO 2 (g)  NH 2 CONH 2 (aq) + H 2 O(l) Is this reaction spontaneous? ◦ Recall  S>q/T. Thus if we know  S and  H, we can answer the question. ◦ Intuitively, we can also answer the question.

37. Spontaneous Reactions The heat of reaction at constant pressure: ◦ q p =  H. The second law for a spontaneous reaction at constant T and P becomes: ◦ S > q p /T =  H/T where T is the average absolute temperature. ◦ S >  H/T from a spontaneous reaction ◦ 0 >  H/T –  S for a spontaneous reaction at constant T and P. ◦ H – T  S for a spontaneous reaction at constant T and P.

38. Spontaneous Reactions If  H – T  S is negative, then the reaction is spontaneous as written from left to right. ◦ H – T  S = - # If it is positive, it is nonspontaneous as written from left to right. ◦ H – T  S = + # When it is zero, the reaction is in equilibrium. ◦ H – T  S = 0

39. The Third Law of Thermodynamics The Third Law of Thermodynamics says that a substance that is perfectly crystalline at 0K has an entropy of zero. By breaking the temperature changes up into small, incremental changes, we can calculate the entropy change for when heat is absorbed at a temperature, T.

40. Standard Entropies and the Third Law of Thermodynamics Entropy increases gradually as temperature changes, but sharply at a phase change. The standard entropy, S°, is the entropy value for the standard state of a species (pure substance at 1 atm). For a species in solution, the standard state is a 1 M solution.

41. Entropy In general, entropy generally increases when: ◦ 1. A reaction occurs in which a molecule is broken into 2 or more smaller molecules. ◦ 2. A reaction in which there is an increase in the number of moles of a gas (could, in this case be related to rule #1) ◦ 3. A process where a solid changes to a liquid or a liquid changes to a gas. Predictions are good for qualitative work.

42. Entropy For quantitative work, we need to find  S°. ◦ To find  S°, we need to subtract the standard entropies of reactants from the standard entropies of products (similar to getting  H°).  S° =  nS° (products) –  mS° (reactants)

43. Free Energy Free energy, G, is a thermodynamic quantity defined as G = H – TS, and gives a direct criterion for spontonaiety. As a reaction proceeds at a given temperature and pressure, reactants form products and the enthalpy (H) and entropy (S) change. ◦ These changes result in a change in free energy:  G =  H - T  S

44. Free Energy If  G is negative, the reaction is spontaneous. We need to use standard states when we calculate this. They are: ◦ 1 atm for solids and liquids ◦ 1 atm partial pressure for gases ◦ 1M for solutions ◦ T is usually 25°C, or 298K

45. Free Energy Thus,  G° =  H° - T  S°  G° f is the standard free energy of formation. It is the free energy change that occurs when 1 mol of a substance is formed from its elements in their stable states at 1 atm and 298K. We can obtain the values from a table of free energies of a formation and plug them into the following equation. ◦ G° =  n  G° f(products) –  m  G° p(reactants)

47. Free Energy The following rules are useful for judging the spontonaiety of a reaction: ◦ 1. When  G° is large and negative (more than -10kJ), the reaction is spontaneous as written, and the reactants transform almost entirely to products when equilibrium is reached.

48. Free Energy The following rules are useful for judging the spontonaiety of a reaction: ◦ 2. When  G° is large and positive (larger than +10kJ), the reaction is nonspontaneous as written, and reactants do not give significant amounts of products.

49. Free Energy The following rules are useful for judging the spontonaiety of a reaction: ◦ 3. When  G° has small negative or positive values (less than ±10kJ), the reaction gives an equilibrium mixture with significant amounts of both reactants and products.

50. Free Energy In theory, spontaneous reactions can be used to obtain useful work—move objects like cars, weights, or to generate electricity. In principle, if a reaction is carried out to obtain maximum useful work, no entropy is produced. W max =  G

51. Free Energy The amount of free energy available to do useful work is where the term “free energy” comes from. ◦ The concept is an idealization. In any real reaction, some entropy is created and less than maximum work is expended.

52. Free Energy Think of burning gas on a stove and in a car. ◦ In both cases, there is entropy being created, but in the car, some of the decrease in free energy shows up as work done to move the car. http://www.publicdomainpictures.net/pictures/4000/nahled/1-1246450967k9xk.jpg

53. Equilibrium Constants The thermodynamic equilibrium constant, K, is a constant in which the concentrations of gases are expressed in parital pressures in atmospheres, whereas the concentrations of solutes in liquid are expressed in molarities. ◦ K c is for reactions involving solutes in liquid solutions. ◦ K p is for reactions involving gases.

54. Free Energy If we want the free energy change,  G°, of a reaction under standard conditions, we use the information from data tables.

55. Free Energy To calculate the free energy change when reactants under nonstandard conditions are converted to products under nonstandard conditions use: ◦ G =  G° + RT ln Q ◦ Q is the thermodynamic form of the reaction quotient. It has the same general appearance as K (the thermodynamic equilibrium constant) but concentrations and partial pressures are those for a mixture at some instant. ◦ Thus, we can get  G from  G° by adding RT ln Q

56. Free Energy  G is the free energy that occurs when the reaction mixture has the composition expressed by Q. ◦ At equilibrium,  G = 0, and Q = K So, 0 =  G° + RT ln K Rearranging gives:  G° = -RT ln K ◦ Looking at it this way, when the equilibrium constant is greater than 1, ln K is positive and  G° is negative.

57. Free Energy Also, when the equilibrium constant is less than 1, ln K is negative and  G° is positive. To find  G° or K at a temperature other than standard temperature, we have to do involved calculations, so we will use approximate methods instead.

58. Free Energy Change with Temperature Here, we assume  H° and  S° are constant with respect to temperature. We get  G T ° at any temperature, T, by substituting values of  H° and  S° at 25°C into: ◦ G T ° =  H° - T  S° Consider the reactions in Table 18.3

59. Free Energy Change with Temperature In the first reaction  H° is negative and  S° is positive. We are breaking the molecules into 2 or more products, more stable bonds are forming and heat is released,  S° increases (becomes more positive),  H° is negative.

60. Free Energy Change with Temperature Consider the following: ◦ CaCO 3 is quite stable at room temperature. ◦ CaCO 3 (s)  CaO(s) + CO 2 (g) ◦ At 25°C,  G = +130.9kJ, the equation of partial pressure of CO 2 is 1.1x10 -23 atm.  The small partial pressure and the large  G (much greater than +10kJ) both indicate stability.

61. Free Energy Change with Temperature We can find the tempertaure at which the decomposition of CaCO 3 goes from nonspontaneous to spontaneous at 1atm. ◦ At this temperature,  G° = 0. ◦ G° = 0 =  H° - T  S°  Solve for T: T  S° =  H° ◦ T =  H°/  S° For the decomposition of CaCO 3, ◦ T = 178.3kJ/0.1590kJ/K = 1121K (848°C)