1. Thermodynamics The systematic study of energy is (classically) called thermodynamics. Energy: is the capacity to do work, or supply heat. Energy = Work + Heat

2. Flavors of energy Kinetic Energy: is the energy of motion. – E K = 1 / 2 mv 2 (1 Joule = 1 kg  m 2 /s 2 ) – (1 calorie = 4.184 J) Thermal Energy is the kinetic energy of molecular motion (translational, rotational, and vibrational) and is proportional to the temperature in an absolute temperature scale (Kelvin, for instance). E thermal  T(K) Heat is the amount of thermal energy transferred between two objects at different temperatures

3. Energy – more flavors Potential Energy: is energy of position. Chemical Energy is the potential energy that exists because of the attractions present in chemical bonds.

4. The universe simplified Thermodynamicists are simple folk. The universe, for them, is made of system and surroundings. In an experiment: Reactants and products are the system; everything else is the surroundings. Energy that flows from the system to the surroundings has a negative sign (loss of energy). Energy that flows from the surroundings to the system has a positive sign (gain of energy).

5. Thermodynamics Closed System: Only energy can be lost or gained. Isolated System: No matter or energy is exchanged. Open System: Both energy and matter is exhanged.

6. Units of energy Joule (J) = Generally this unit is too small for chemistry, so kJ is used instead. Calorie (cal) ( = 4.184 J) Nutritional calorie (Cal) ( = 1000 cal) Work has the same units as energy! KE = ½ mv 2

7. Conserve your energy! The law of the conservation of energy: Energy cannot be created or destroyed. The energy of an isolated system must be constant. The energy change in a system equals the work done on the system + the heat added.  E = E final – E initial = E 2 – E 1 = q + w q = heat, w = work

8. Energy Conservation

9. State Functions State Function: A function or property whose value depends only on the present state (condition) of the system. The change in a state function is zero when the system returns to its original condition. For nonstate functions (path functions), the change is not zero if the path returns to the original condition.

10. State vs. Path Functions

11. Chemical Work Surface Expansion (Surface Tension) Electrochemical Extension (Elasticity) Expansion (Gases) 

12. Expansion Work

13. Work/Energy and Force 1.Work has units of Joules: 2.Work is also: 3.We can re-write this as: 4.Distance has units of: 5.So then force must have units of: 6.This is called a Newton (N).

14. -P  V as Energy/Work Pressure is force per area: So P  V is then: More practically, you need the following conversion:

15. Heat Flow We measure heat flow by measuring temperature change (and by knowing the heat capacity of the substance) What is that heat flow we measure? We already know  E = q + w We already know w = -P  V So algebra tells us q =  E + P  V So the heat flow for any process is just the internal energy change minus the work.

16. Constant volume conditions We have q =  E + P  V If we make our measurement under constant volume conditions we know  V = 0 So then we know: q v =  E That is, under constant volume conditions, the heat flow is the entire energy change of the system.

17. Constant pressure conditions We have q =  E + P  V Normal lab measurements are done in open beakers and flasks. We do NOT have constant volume conditions. Instead it is constant pressure conditions. What does the heat flow measurement mean under these conditions? The best we can say is that the heat flow is the energy change minus the work. We make up a name for this: enthalpy. q p =  E + P  V =  H

18. Enthalpy Enthalpy is not a real thing. It is a bookkeeping quantity we use because of how we do our chemistry in the lab. Energy, pressure, and volume are all state functions. So enthalpy is a state function too. Thus we can say:  H = H final - H initial

19. Chemical Rxns and Enthalpy One of the hallmarks of a chemical reaction is the giving off (or taking in) of heat. That is, we can write Ba(OH) 2 ·8 H 2 O(s) + 2 NH 4 Cl(s) +80.3 kJ of heat  BaCl 2 (aq) + 2 NH 3 (aq) + 10 H 2 O(l) But since most reactions are done under constant-pressure conditions (open air) the heat represents an enthalpy change, so we really write Ba(OH) 2 ·8 H 2 O(s) + 2 NH 4 Cl(s)  BaCl 2 (aq) + 2 NH 3 (aq) + 10 H 2 O(l)  H = +80.3 kJ

20. Enthalpy Changes Enthalpies of Chemical Change: Often called heats of reaction (  H reaction ). – Endothermic: Heat flows into the system from the surroundings and  H has a positive sign. – Exothermic: Heat flows out of the system into the surroundings and  H has a negative sign.

21.  H vs.  E We know that enthalpy and energy changes are different. But how different are they? Consider this reaction: C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(g) Under constant volume conditions,  E = -2045 kJ Under constant pressure conditions,  H = -2043 kJ That implies, w = +2 kJ

22. Enthalpy Changes for Physical Processes

23. How temperature plays in There is a change in enthalpy when going from one phase to another. Therefore the  H rxn will change for a reaction if the reactants are products are of a different phase than the reaction as written. So we define a standard state for reactions: – Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and 25°C; 1 M concentration for all substances in solution. – These are indicated by a superscript ° to the symbol of the quantity reported. – Standard enthalpy change is indicated by the symbol  H°.

24. Enthalpy as State Function This has big implications 1 st implication:

25. Enthalpy as State Function 2 nd implication: Reversing a reaction changes the sign of  H for a reaction. – C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l)  H = –2219 kJ – 3 CO 2 (g) + 4 H 2 O(l)  C 3 H 8 (g) + 5 O 2 (g)  H = +2219 kJ Multiplying a reaction by a factor x increases  H by the same factor. – 3 C 3 H 8 (g) + 15 O 2 (g)  9 CO 2 (g) + 12 H 2 O(l)  H = –6657 kJ Thus we usually write the energies as kJ/mole!! The “mole” refers to moles of reaction, not moles of any particular reactant or product

26. Enthalpy as State Function 3 rd implication: Hess’s Law: The overall enthalpy change for a reaction is equal to the sum of the enthalpy changes for the individual steps in the reaction. This is true even if the steps aren’t really right!!! 3 H 2 (g) + N 2 (g)  2 NH 3 (g)  H° = –92.2 kJ

27. Hess’s Law Example Reactants and products in individual steps can be added and subtracted to determine the overall equation. (1) 2 H 2 (g) + N 2 (g) N 2 H 4 (g)  H° 1 = 95.4 kJ (2) N 2 H 4 (g) + H 2 (g) 2 NH 3 (g)  H° 2 = –187.6 kJ (3) 3 H 2 (g) + N 2 (g) 2 NH 3 (g)  H° 3 = ?  H° 3 =  H° 1 +  H° 2 = (95.4 kJ/mole) + (–187.6 kJ/mole) = -92.2 kJ/mole +

28. Enthalpy as Molar Function The enthalpy change for the reaction above is -2219 kJ if you start with 1 mole of propane and 5 moles of oxygen making 3 moles of carbon dioxide and 4 moles of water. If you start with as many moles of reactant as its stoichiometric coefficient, this will make 1 mole of reaction. Thus, Multiplying a reaction increases  H by the same factor. – 3 C 3 H 8 (g) + 15 O 2 (g)  9 CO 2 (g) + 12 H 2 O(l)  H = –6657 kJ C 3 H 8 (g) + 5 O 2 (g)  3 CO 2 (g) + 4 H 2 O(l)  H = –2219 kJ

29. Enthalpy and Stoichiometry We can treat enthalpy change as if it were a product or reactant in the reaction. Therefore we can redo all of our stoichiometry again with energy/enthalpy thrown into the mix. Yay!

30. Calorimetry and Heat Capacity Calorimetry is the science of measuring heat changes (q) for chemical reactions. There are two types of calorimeters: Bomb Calorimetry: A bomb calorimeter measures the heat change at constant volume such that q =  E. Constant Pressure Calorimetry: A constant pressure calorimeter measures the heat change at constant pressure such that q =  H. Coffee cup calorimeter

31. Calorimeters Constant Pressure Bomb

32. From Temperature to Heat Heat capacity (C) is the amount of heat required to raise the temperature of an object or substance a given amount. –Specific Heat: The amount of heat required to raise the temperature of 1.00 g of substance by 1.00°C. –Molar Heat: The amount of heat required to raise the temperature of 1.00 mole of substance by 1.00°C. Extensive property Intensive properties

33. Specific Heats, etc.

34. Standard Heats of Formation Standard Heats of Formation (  H° f ): The enthalpy change for the formation of 1 mole of substance in its standard state from its constituent elements in their standard states. The standard heat of formation for any element in its standard state is defined as being ZERO.  H° f = 0 for an element in its standard state

35. Standard Heats of Formation H 2 (g) + 1 / 2 O 2 (g)  H 2 O(l)  H° f = –286 kJ/mol 3 / 2 H 2 (g) + 1 / 2 N 2 (g)  NH 3 (g)  H° f = –46 kJ/mol 2 C(s) + H 2 (g)  C 2 H 2 (g)  H° f = +227 kJ/mol 2 C(s) + 3 H 2 (g) + 1 / 2 O 2 (g)  C 2 H 5 OH(g)  H° f = –235 kJ/mol

36. Standard Heats of Formation -1131Na 2 CO 3 (s)+49C6H6(l)C6H6(l)-92HCl(g) -127AgCl(s)-235C 2 H 5 OH(g)+95.4N2H4(g)N2H4(g) -167Cl - (aq)-201CH 3 OH(g)-46NH 3 (g) -207NO 3 - (aq)-85C2H6(g)C2H6(g)-286H 2 O(l) -240Na + (aq)+52C2H4(g)C2H4(g)-394CO 2 (g) +106Ag + (aq)+227C2H2(g)C2H2(g)-111CO(g) Some Heats of Formation,  H f ° (kJ/mol) Appendix B in your textbook You can find this number for pretty much every compound you care about.

37. Consider this balanced reaction: 3 CO 2 (g) + 4 H 2 O(l)  C 3 H 8 (g) + 5 O 2 (g) Let’s say the reaction happens in two steps: – Step A: 3 CO 2 (g) + 4 H 2 O(l)  3 C(s) + 4 H 2 (g) + 5 O 2 (g) – Step B: 3 C(s) + 4 H 2 (g) + 5 O 2 (g)  C 3 H 8 (g) + 5 O 2 (g) Hess’ Law says the  H rxn ° is the sum of the two steps. What can we say about the two steps individually?

38. Consider this (cont.): Step B: 3 C(s) + 4 H 2 (g) + 5 O 2 (g)  C 3 H 8 (g) + 5 O 2 (g) We can cancel the oxygen. The rest is the formation of propane from elements in their standard state. That is  H B ° =  H f °( C 3 H 8 (g) ) = -105 kJ/mole

39. Consider this (cont.): Step A: 3 CO 2 (g) + 4 H 2 O(l)  3 C(s) + 4 H 2 (g) + 5 O 2 (g) This step is the formation of 3 moles of CO 2 and 4 moles H 2 O from elements in their standard states, but in reverse! That is,  H A ° = - (  H f °(CO 2 )*3 +  H f °(H 2 O)*4 ) =

40. Consider this (cont.): Recall that the  H rxn ° of this reaction is the sum of the two steps. So then we have  H rxn ° = Step A + Step B = (2323.7 kJ/mole) + (-105 kJ/mole) = 2218.7 kJ/mole

41. How did we solve that problem? We took the sum of the  H f ° of each product multiplied by its coefficient We took the sum of the  H f ° of each reactant multiplied by its coefficient We subtracted the reactant sum from the product sum. That is,

42. Bond Energies & Reaction Enthalpies In Chapters 6 & 7 we spent a long time discussing the strength of bonds. Bond strengths are measured in kJ/mole Reaction enthalpies are measured in kJ/mole. Can they be related?

43. Bond Dissociation Energy Bond Dissociation Energy (D): Can be used to determine an approximate value for  H° f.  H rxn = D (Bonds Broken) – D (Bonds Formed) For the reaction between H 2 and Cl 2 to form HCl:  H = 2 D(H–Cl) – ∑ {D(H–H) + D(Cl-Cl)}

44. Bond Dissociation Energies In truth, these values change depending on the other bonds around it, but to two sig figs the numbers are pretty good.

45. Combustion Reaction There is nothing chemically distinct about combustion, but because it is such an important reaction it is given its own category:

46. Chemical Spontaneity Why does one chemical reaction happen on its own and not the reverse reaction? (Why is one spontaneous?)

47. Chemical Spontaneity A spontaneous reaction is one that, once started, continues on its own without any input of external energy. A spontaneous reaction might not start on its own. Example: C + O 2  CO 2

48. Spontaneity Markers Exothermicity doesn’t work. – false positives and false negatives Rxns that produce gas are usually spontaneous. Rxns that are exothermic often are spontaneous. Rxns that cause something to be dissolved are usually spontaneous. Processes that cause mixing are usually spontaneous.

49. New Accounting: Entropy The link between all spontaneous processes is a sense of increasing molecular-scale randomness. Entropy (S) is an accounting technique designed to capture that randomness. Entropy has units of J/K. Entropy is a measure of the number of equivalent ways to distribute energy in the system.

50. Entropy Any process is spontaneous if it makes the entropy of the universe go up. (both system and surroundings)  S universe > 0 So we need to recognize things with high entropy versus low entropy

51. High Entropy versus Low Entropy S gas >> S liquid > S solid S high temp > S low temp S high volume > S low volume S mixed > S pure S many pieces > S few pieces

52. Predicting Spontaneity Spontaneous system change conditions: – Producing gas – Mixing – Few parts into many – Getting warmer Spontaneous surroundings change conditions: – Getting warmer

53. Surroundings The entropy of the surroundings makes  S universe hard to use—we need to relate entropy of the surroundings to a property of the system. Since the surroundings is simply a thermal sink with no physical or chemical properties, the only change in entropy in the surroundings is due to heat flowing to/from the system. (That is, we simply need to calculate  H rxn.)

54. System So entropy of the surroundings is now related to the changes in the system. And entropy of the system is already linked to changes in the system. How do we combine them to determine spontaneity?

55. Gibbs Energy The two entropy changes are combined in an accounting scheme called Gibbs energy (G). Chemists use  G to determine whether or not any process is spontaneous.

56. Gibbs Energy – 1 st problem Entropy change of surroundings related to  H rxn (Joules) Entropy change of system in units of entropy (Joules per Kelvin) To unify the units we multiply entropy change of system by temperature (T  S) Now both are in units of energy

57. Gibbs Energy – 2 nd problem How to combine system and surroundings We simply add the change in the surroundings to the change in the system and call this the change in Gibbs energy. But we need the opposite sign of the  H rxn. So then we could have  G = -  H rxn + T  S rxn and then  G>0 is the condition for spontaneity.

58. Not quite For various reasons, we instead say  G =  H – T  S and have the spontaneity condition be  G < 0.

59.  G =  H – T  S Spontaneous when  G rxn < 0 T  S sys > 0 (+) (system entropy increasing) T  S sys < 0 (-) (system entropy decreasing)  H rxn < 0 (-) (exothermic) (-) – (+) = (-) Always spontaneous (-) – (-) = ? Spontaneous at low temps  H rxn > 0 (+) (endothermic) (+) – (+) = ? Spontaneous at high temps (+) – (-) = (+) Never spontaneous

60. Gibbs energy calculations Gibbs energy and entropy are tabulated in the same way as enthalpies of formation (Appendix B) So you can calculate gibbs energy and entropy changes in the same way you calculate enthalpy changes—that is, sum of products minus sum of reactants.

61. Entropy notes Entropy is a state function, like enthalpy. Entropy is a measure of molecular-scale disorder. One cannot have negative disorder, thus entropy cannot become negative. Entropy has units of Joules per Kelvin. One cannot have a negative energy—Joules must be a positive number. Therefore the temperature cannot become negative either. That is, there exists a lowest possible temperature—an absolute zero.

62. Entropy notes continued The temperature at which zero entropy can occur is absolute zero. Therefore the entropy of the thermodynamic standard state is NOT zero.

63. Gibbs energy notes Gibbs energy is a state function like enthalpy and entropy. So, like enthalpy, the  G rxn of the reverse reaction is the negative of the forward reaction. The only time one reaction or the other is going forward is when the two sides are in equilibrium. That is, equilibrium is when  G = 0.