1. Entropy and Gibbs free energy

2. 2 Exothermic The products are lower in energy than the reactants Releases energy Often release heat

3. 3 C + O 2  CO 2 Energy ReactantsProducts  C + O 2 CO 2 -395kJ + 395 kJ

4. When will a reaction be exothermic A) When breaking the bonds of the reactants takes more energy than making the bonds of the products. B) When breaking the bonds of the reactants takes less energy than making the bonds of the products C) When you put in energy to break the bonds D) When you get energy by breaking bonds

5. 5 Endothermic The products are higher in energy than the reactants Absorbs energy Absorb heat

6. 6 CaCO 3  CaO + CO 2 Energy ReactantsProducts  CaCO 3 CaO + CO 2 +176 kJ CaCO 3 + 176 kJ  CaO + CO 2

7. 7 Heat of Reaction The heat that is released or absorbed in a chemical reaction Equivalent to ΔH C + O 2 (g)  CO 2 (g) +393.5 kJ C + O 2 (g)  CO 2 (g) ΔH = -393.5 kJ In thermochemical equation it is important to say what state H 2 (g) + ½ O 2 (g)  H 2 O(g) ΔH = -241.8 kJ H 2 (g) + ½ O 2 (g)  H 2 O(l) ΔH = -285.8 kJ

8. 8 Energy ReactantsProducts  Change is down ΔH is <0 + heat

9. 9 Energy ReactantsProducts  Change is up ΔH is > 0 Reactants + heat

10. Choose all that apply... C(s) + 2 S(g)   CS 2 (l) ΔH = 89.3 kJ Which of the following are true? A) This reaction is exothermic B) It could also be written C(s) + 2 S(g) + 89.3 kJ   CS 2 (l) C) The products have higher energy than the reactants D) It would make the water in the calorimeter colder

11. 11 Heat of Combustion The heat from the reaction that completely burns 1 mole of a substance at 25°C and 1 atm C 2 H 4 + 3 O 2  2 CO 2 + 2 H 2 O C 2 H 6 + O 2  CO 2 + H 2 O 2 C 2 H 6 + 7 O 2  4 CO 2 + 6 H 2 O C 2 H 6 + (7/2) O 2  2 CO 2 + 3 H 2 O Always exothermic

12. Heat and phase change Melting and vaporizing are endothermic – Breaking things apart Freezing and condensing are exothermic – Forming connections

13. Heat of Fusion Heat of fusion-ΔH fus - heat to melt one gram q = ΔH fus x m For water 80 cal/g or 334 J/g Same as heat of solidification Book uses molar heat of fusion- heat to melt one mole of solid q = ΔH fus x n

14. Calculating Heat If there is a temperature change – q = m ΔT C If there is a phase change – q = ΔH fus x m or q = ΔH solid x m – q = ΔH vap x m or q = ΔH cond x m If there is both, do them separately and add.

15. Example Ammonia has a heat of fusion of 332 cal/g. How much heat to melt 15 g of ammonia?

16. This formula is for all change ΔH = ΣΔH° f (products) - ΣΔΗ° f (reactants)

17. 17 Example CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) CH 4 (g) = -74.86 kJO 2 (g) = 0 kJCO 2 (g) = -393.5 kJH 2 O(g) = -241.8 kJ u ΔH= [-393.5 kJ + 2(-241.8 kJ)] - [-74.86 kJ +2 (0 kJ )] u ΔH= -802.2 kJ

18. Energy ReactantsProducts  reactants products elements

19. Energy ReactantsProducts  reactants products elements

20. Energy Reaction coordinate Reactants Products

21. Energy Reaction coordinate Reactants Products Activation Energy - Minimum energy to make the reaction happen – how hard

22. Energy Reaction coordinate Reactants Products Activated Complex or Transition State

23. Activation Energy Must be supplied to start the reaction Low activation energy – Lots of collision are hard enough – fast reaction High Activation energy – Few collisions hard enough – Slow reaction

24. Energy Reaction coordinate Reactants Products

25. Energy Reaction coordinate Reactants Products Activation Energy - Minimum energy to make the reaction happen – how hard

26. Energy Reaction coordinate Reactants Products Activated Complex or Transition State

27. Activation Energy Must be supplied to start the reaction Low activation energy – Lots of collision are hard enough – fast reaction High Activation energy – Few collisions hard enough – Slow reaction

28. Activation energy If reaction is endothermic you must keep supplying heat If it is exothermic it releases energy That energy can be used to supply the activation energy to those that follow

29. Energy Reaction coordinate Reactants Products Overall energy change

30. Thermodynamics Will a reaction happen?

31. Things that Affect Rate Catalysts- substances that increase the rate of a reaction without being used up.(enzyme). Not a reactant nor a product. Speeds up reaction by giving the reaction a new path. The new path has a lower activation energy. More molecules have this energy. The reaction goes faster.

32. Energy Reaction coordinate Reactants Products

33. Pt surface HHHH HHHH Hydrogen bonds to surface of metal. Break H-H bonds Catalysts

34. Pt surface HHHH Catalysts C HH C HH

35. Pt surface HHHH Catalysts C HH C HH The double bond breaks and bonds to the catalyst.

36. Pt surface HHHH Catalysts C HH C HH The hydrogen atoms bond with the carbon

37. Pt surface H Catalysts C HH C HH HHH

38. Energy Substances tend react to achieve the lowest energy state. Most chemical reactions are exothermic. Doesn’t work for things like ice melting. An ice cube must absorb heat to melt, but it melts anyway. Why?

39. Entropy The degree of randomness or disorder. Better – number of ways things can be arranged S The First Law of Thermodynamics - The energy of the universe is constant. The Second Law of Thermodynamics -The entropy of the universe increases in any change. Drop a box of marbles. Watch your room for a week.

40. Entropy Entropy of a solid Entropy of a liquid Entropy of a gas A solid has an orderly arrangement. A liquid has the molecules next to each other but isn’t orderly A gas has molecules moving all over the place.

41. Entropy increases when... Reactions of solids produce gases or liquids, or liquids produce gases. A substance is divided into parts -so reactions with more products than reactants have an increase in entropy. The temperature is raised -because the random motion of the molecules is increased. a substance is dissolved.

42. Entropy calculations There are tables of standard entropy (pg 407). Standard entropy is the entropy at 25ºC and 1 atm pressure. Abbreviated Sº, measure in J/K. The change in entropy for a reaction is ΔSº= ΣSº(Products)-ΣSº(Reactants). Calculate ΔSº for this reaction CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g)

43. Calculate ΔSº for this reaction CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O(g) For CH 4 Sº = 186.2 J/K-mol For O 2 Sº = 205.0 J/K-mol For CO 2 Sº= 213.6 J/K-mol For H 2 O(g) Sº = 188.7 J/K-mol

44. Spontaneity Will the reaction happen, and how can we make it?

45. Spontaneous reaction Reactions that will happen. Nonspontaneous reactions don’t. Even if they do happen, we can’t say how fast. Two factors influence. Enthalpy (heat) and entropy(disorder).

46. Two Factors Exothermic reactions tend to be spontaneous. – Negative  Δ H. Reactions where the entropy of the products is greater than reactants tend to be spontaneous. – Positive ΔS. A change with positive ΔS and negative ΔH is always spontaneous. A change with negative ΔS and positive ΔH is never spontaneous.

47. Other Possibilities Temperature affects entropy. Higher temperature, higher entropy. For an exothermic reaction with a decrease in entropy (like rusting). Spontaneous at low temperature. Nonspontaneous at high temperature. Enthalpy driven.

48. Other Possibilities An endothermic reaction with an increase in entropy like melting ice. Spontaneous at high temperature. Nonspontaneous at low temperature. Entropy driven.

49. Gibbs Free Energy The energy free to do work is the change in Gibbs free energy. ΔGº = ΔHº - TΔSº (T must be in Kelvin) All spontaneous reactions release free energy. So ΔG <0 for a spontaneous reaction. ΔG is negative

50. Problems Using the information on page 407 and pg 190 determine if the following changes are spontaneous at 25ºC. 2H 2 S(g) + O 2 (g)  2H 2 O(l) + S(rhombic)

51. 2H 2 S(g) + O 2 (g)  2H 2 O(l) + 2S We find ΔH f ° for each component – H 2 S = -20.1 kJO 2 = 0 kJ – H 2 O = -285.8 kJS = 0 kJ Then Products – Reactants ΔH =2 (-285.8 kJ) + 2(0 kJ) - 2 (-20.1 kJ) - 1(0 kJ) = -531.4 kJ

52. 2H 2 S(g) + O 2 (g)  2H 2 O(l) + 2 S we find S for each component – H 2 S = 205.6 J/K O 2 = 205.0 J/K – H 2 O = 69.94 J/K S = 31.9 J/K Then Products – Reactants ΔS= 2 (69.94 J/K) + 2(31.9 J/K) - 2(205.6 J/K) - 205 J/K = -412.5 J/K

53. 2H 2 S(g) + O 2 (g)  2H 2 O(l) + 2 S ΔG = ΔH – T ΔS G = -531.4 kJ - 298K (-412.5 J/K)  G = -531.4 kJ - -123000 J ΔG = -531.4 kJ - -123 kJ ΔG = -408.4 kJ Spontaneous Exergonic- it releases free energy. At what temperature does it become spontaneous?

54. Spontaneous It becomes spontaneous when ΔG = 0 That’s where it changes from positive to negative. Using 0 = ΔH – T ΔS and solving for T 0 - ΔH = - T ΔS - ΔH = -T ΔS T = ΔH = ΔS = 1290 K -531.4 kJ -412.5 J/K = -531400 J -412.5 J/K

55. There’s Another Way There are tables of standard free energies of formation compounds.(pg 414) ΔGº f is the free energy change in making a compound from its elements at 25º C and 1 atm. for an element ΔGº f = 0 Look them up. ΔGº= ΔGº f (products) - ΔGº f (reactants)

56. 2H 2 S(g) + O 2 (g)  2H 2 O(l) + 2S From we find ΔG f ° for each component – H 2 S = -33.02 kJO 2 = 0 kJ – H 2 O = -237.2 kJS = 0 kJ Then Products – Reactants ΔG =2 (-237.2) + 2(0) - 2 (-33.02) - 1(0) = -408.4 kJ