1. THERMODYNAMICS REVIEW

2. ENERGY ABILITY TO DO WORK UNITS– JOULES (J), WE WILL USE “KJ” CAN BE CONVERTED TO DIFFERENT TYPES ENERGY CHANGE RESULTS FROM FORMING AND BREAKING CHEMICAL BONDS IN REACTIONS

3. SYSTEM VS. SURROUNDINGS

4. HEAT (Q) ENERGY TRANSFER BETWEEN A SYSTEM AND THE SURROUNDINGS DUE TO A TEMPERATURE CHANGE TRANSFER IS INSTANTANEOUS FROM HIGH----LOW TEMPERATURE UNTIL THERMAL EQUILIBRIUM TEMPERATURE— MEASURE OF HEAT, “HOT/COLD”

5. HEAT (Q) CONTINUED KINETIC THEORY OF HEAT HEAT INCREASE RESULTING IN TEMPERATURE CHANGE CAUSES AN INCREASE IN THE AVERAGE MOTION OF PARTICLES WITHIN THE SYSTEM. INCREASE IN HEAT RESULTS IN ENERGY TRANSFER INCREASE IN BOTH POTENTIAL AND KINETIC ENERGIES

6. THERMODYNAMICS 101 FIRST LAW OF THERMODYNAMICS ENERGY IS CONSERVED IN A REACTION (IT CANNOT BE CREATED OR DESTROYED)---SOUND FAMILIAR??? MATH REPRESENTATION: Δ E TOTAL = Δ E SYS + Δ E SURR = 0 Δ = “CHANGE IN” ΔΕ = POSITIVE (+), ENERGY GAINED BY SYSTEM ΔΕ = NEGATIVE (-), ENERGY LOST BY SYSTEM TOTAL ENERGY = SUM OF THE ENERGY OF EACH PART IN A CHEMICAL REACTION

7. EXOTHERMIC TEMPERATURE INCREASE (--ISOLATED SYSTEM) HEAT IS RELEASED TO SURROUNDINGS (--OPEN/CLOSED SYSTEM) Q = - VALUE CHEMICAL  THERMAL ENERGY

8. ENDOTHERMIC TEMPERATURE DECREASE (--ISOLATED SYSTEM) ALL ENERGY GOING INTO REACTION, NOT INTO SURROUNDINGS HEAT ABSORBED BY SYSTEM, SURROUNDINGS HAVE TO PUT ENERGY INTO REACTION Q = + VALUE THERMAL  CHEMICAL ENERGY

9. ENTHALPY (H) MEASURES 2 THINGS IN A CHEMICAL REACTION: 1)ENERGY CHANGE 2)AMOUNT OF WORK DONE TO OR BY CHEMICAL REACTION 2 TYPES OF CHEMICAL REACTIONS: 1)EXOTHERMIC—HEAT RELEASED TO THE SURROUNDINGS, GETTING RID OF HEAT, - ΔΗ 2)ENDOTHERMIC—HEAT ABSORBED FROM SURROUNDINGS, BRINGING HEAT IN, + ΔΗ **ENTHALPY OF REACTION—HEAT FROM A CHEMICAL REACTION WHICH IS GIVEN OFF OR ABSORBED, UNITS = KJ/MOL ENTHALPY OF REACTION HEAT FROM A CHEMICAL REACTION WHICH IS GIVEN OFF OR ABSORBED AT CONSTANT PRESSURE UNITS = KJ/MOL

11. EXAMPLE 2: CALCULATE THE Δ H FOR THE FOLLOWING REACTION WHEN 12.8 GRAMS OF HYDROGEN GAS COMBINE WITH EXCESS CHLORINE GAS TO PRODUCE HYDROCHLORIC ACID. H 2 + CL 2  2HCL Δ H = -184.6 KJ

12. 1 ST LAW OF THERMODYNAMICS (CONSERVATION OF ENERGY) ENERGY CANNOT BE CREATED OR DESTROYED. WITH PHYSICAL AND CHEMICAL CHANGES, ENERGY CAN BE TRANSFERRED OR CONVERTED. TOTAL ENERGY = Σ ENERGY OF ITS COMPONENTS Δ U = Q + W, Δ E TOTAL = Δ E SYS + Δ E SURR = 0

13. METHODS FOR DETERMINING Δ H 1)CALORIMETRY 2)APPLICATION OF HESS’ LAW

14. CALORIMETRY HOW DO WE FIND THE CHANGE IN ENERGY/HEAT TRANSFER THAT OCCURS IN CHEMICAL REACTIONS???

15. CALORIMETRY EXPERIMENTALLY “MEASURING” HEAT TRANSFER FOR A CHEMICAL REACTION OR CHEMICAL COMPOUND CALORIMETER INSTRUMENT USED TO DETERMINE THE HEAT TRANSFER OF A CHEMICAL REACTION DETERMINES HOW MUCH ENERGY IS IN FOOD OBSERVING TEMPERATURE CHANGE WITHIN WATER AROUND A REACTION CONTAINER ** ASSUME A CLOSED SYSTEM, ISOLATED CONTAINER NO MATTER, NO HEAT/ENERGY LOST CONSTANT VOLUME

16. SPECIFIC HEAT CAPACITY AMOUNT OF HEAT REQUIRED TO INCREASE THE TEMPERATURE OF 1G OF A CHEMICAL SUBSTANCE BY 1°C UNITS--- J/G  °K UNIQUE TO EACH CHEMICAL SUBSTANCE AL (S) = 0.901J/G  °K H 2 O (L) = 4.18 J/G  °K

17. Q = SM Δ T

18. “COFFEE CUP” CALORIMETER (CONT.) Q CHEMICAL = -Q WATER

19. EXAMPLE 2: USING THE FOLLOWING DATA, DETERMINE THE METAL’S SPECIFIC HEAT. METAL MASS = 25.0G WATER MASS = 20.0G TEMPERATURE OF LARGE WATER SAMPLE = 95°C INITIAL TEMPERATURE IN CALORIMETER = 24.5°C FINAL TEMPERATURE IN CALORIMETER = 47.2°C SPECIFIC HEAT OF WATER = 1.00 CAL/G  °C OR 4.184 J/G  °K (KNOW!!!!)

20. ENTHALPY

21. Δ Q RXN HEAT GAINED/LOST IN EXPERIMENT WITH CALORIMETER Δ H RXN HEAT GAINED/LOST IN TERMS OF THE BALANCED CHEMICAL EQUATION

22. CALCULATING Δ H RXN. BY BOND ENTHALPIES (4 TH METHOD) LEAST ACCURATE METHOD Δ H = Σ BE (BONDS BROKEN) - Σ BE (BONDS FORMED)

23. HESS’ LAW ENTHALPY CHANGE FOR A CHEMICAL REACTION IS THE SAME WHETHER IT OCCURS IN MULTIPLE STEPS OR ONE STEP Δ H RXN = ΣΔ H A+B+C (SUM OF Δ H FOR EACH STEP) ALLOWS US TO BREAK A CHEMICAL REACTION DOWN INTO MULTIPLE STEPS TO CALCULATE Δ H ADD THE ENTHALPIES OF THE STEPS FOR THE ENTHALPY FOR THE OVERALL CHEMICAL REACTION

24. GUIDELINES FOR USING HESS’ LAW MUST USE DATA AND COMBINE EACH STEP IN A WAY THAT GIVES THE CHEMICAL REACTION WITH THE UNKNOWN Δ H SET UP STEPS SO CHEMICAL COMPOUNDS NOT IN THE FINAL REACTION ARE CANCELLED REVERSE A REACTION IF NECESSARY AND CHANGE THE SIGN ON Δ H CHECK FOR CORRECT MOLE RATIOS

25. EXAMPLE 1: H 2 O (L)  H 2 O (G) Δ H° = ? BASED ON THE FOLLOWING: H 2 + ½ O 2  H 2 O (L) Δ H° = -285.83 KJ/MOL H 2 + ½ O 2  H 2 O (G) Δ H° = -241.82 KJ/MOL

26. ENTHALPY OF FORMATION ( Δ H F °) ENTHALPY FOR THE REACTION FORMING 1 MOLE OF A CHEMICAL COMPOUND FROM ITS ELEMENTS IN A THERMODYNAMICALLY STABLE STATE. ELEMENTS PRESENT IN “MOST THERMODYNAMICALLY STABLE STATE” 25°C°, 1ATM

27. APPLY HESS’ LAW---- REALLY Δ H F (PRODUCTS) - Δ H F (REACTANTS) CALCULATE Δ H RXN BASED ON ENTHALPY OF FORMATION ( Δ H F ) AA + BB  CC + DD Δ H° =[C ( Δ H F °) C + D( Δ H F °) D ] - [A ( Δ H F °) A + B ( Δ H F °) B ]

28. BOND ENTHALPIES HOW DOES A CHEMICAL REACTION HAVE ENERGY?

29. BOND ENERGY ENERGY REQUIRED TO MAKE/BREAK A CHEMICAL BOND ENDOTHERMIC REACTIONS PRODUCTS HAVE MORE ENERGY THAN REACTANTS MORE ENERGY TO BREAK BONDS EXOTHERMIC REACTIONS REACTANTS HAVE MORE ENERGY THAN PRODUCTS MORE ENERGY TO FORM BONDS

30. BOND ENTHALPY FOCUSES ON THE ENERGY/HEAT BETWEEN PRODUCTS AND REACTANTS AS IT RELATES TO CHEMICAL BONDING AMOUNT OF ENERGY ABSORBED TO BREAK A CHEMICAL BOND--- AMOUNT OF ENERGY RELEASED TO FORM A BOND. MULTIPLE CHEMICAL BONDS TAKE MORE ENERGY TO BREAK AND RELEASE MORE ENERGY AT FORMATION AMOUNT OF ENERGY ABSORBED = AMOUNT OF ENERGY RELEASED TO BREAK CHEMICAL BOND TO FORM A CHEMICAL BOND

31. EXAMPLE 1: USING AVERAGE BOND ENTHALPY DATA, CALCAULATE Δ H FOR THE FOLLOWING REACTION. CH 4 + 2O 2  CO 2 + 2H 2 O Δ H = ? BondAverage Bond Enthalpy C-H413 kJ/mol O=O495 kJ/mol C-O358 kJ/mol C=O799 kJ/mol O-H467 kJ/mol

32. ENTROPY

33. SPONTANEOUS VS. NONSPONTANEOUS 1)SPONTANEOUS PROCESS OCCURS WITHOUT HELP OUTSIDE OF THE SYSTEM, NATURAL MANY ARE EXOTHERMIC—FAVORS ENERGY RELEASE TO CREATE AN ENERGY REDUCTION AFTER A CHEMICAL REACTION EX. RUSTING IRON WITH O 2 AND H 2 O, COLD COFFEE IN A MUG SOME ARE ENDOTHERMIC EX. EVAPORATION OF WATER/BOILING, NACL DISSOLVING IN WATER

34. SPONTANEOUS VS. NONSPONTANEOUS 2) NONSPONTANEOUS PROCESS REQUIRES HELP OUTSIDE SYSTEM TO PERFORM CHEMICAL REACTION, GETS AID FROM ENVIRONMENT EX. WATER CANNOT FREEZE AT STANDARD CONDITIONS (25°C, 1ATM), CANNOT BOIL AT 25°C **CHEMICAL PROCESSES THAT ARE SPONTANEOUS HAVE A NONSPONTANEOUS PROCESS IN REVERSE **

35. ENTROPY (S) MEASURE OF A SYSTEM’S DISORDER DISORDER IS MORE FAVORABLE THAN ORDER Δ S = S (PRODUCTS) - S (REACTANTS) Δ S IS (+) WITH INCREASED DISORDER STATE FUNCTION ONLY DEPENDENT ON INITIAL AND FINAL STATES OF A REACTION EX. EVAPORATION, DISSOLVING, DIRTY HOUSE

36. THERMODYNAMIC LAWS 1 ST LAW OF THERMODYNAMICS ENERGY CANNOT BE CREATED OR DESTROYED 2 ND LAW OF THERMODYNAMICS THE ENTROPY OF THE UNIVERSE IS ALWAYS INCREASING. NATURALLY FAVORS A DISORDERED STATE

37. WHEN DOES A SYSTEM BECOME MORE DISORDERED FROM A CHEMICAL REACTION? ( Δ S > 0) 1)MELTING 2)VAPORIZATION 3)MORE PARTICLES PRESENT IN THE PRODUCTS THAN THE REACTANTS 4C 3 H 5 N 3 O 9 (L)  6N 2 (G) + 12CO 2 (G) + 10H 2 O (G) + O 2 (G) 4)SOLUTION FORMATION WITH LIQUIDS AND SOLIDS 5)ADDITION OF HEAT

38. 3 RD LAW OF THERMODYNAMICS THE ENTROPY ( Δ S) OF A PERFECT CRYSTAL IS 0 AT A TEMPERATURE OF ABSOLUTE ZERO (0°K). NO PARTICLE MOTION AT ALL IN CRYSTAL STRUCTURE ALL MOTION STOPS

39. HOW DO WE DETERMINE IF A CHEMICAL REACTION IS SPONTANEOUS? 1)CHANGE IN ENTROPY ( Δ S) 2)GIBBS FREE ENERGY ( Δ G)

40. CHANGE IN ENTROPY ( Δ S) FOR A CHEMICAL REACTION TO BE SPONTANEOUS ( Δ S T > 0), THERE MUST BE AN INCREASE IN SYSTEM’S ENTROPY ( Δ S SYS > 0) AND THE REACTION MUST BE EXOTHERMIC ( Δ S SURR > 0). EXOTHERMIC REACTIONS ARE FAVORED, NOT ENDOTHERMIC REACTIONS. EXOTHERMIC ( Δ H 0) ENDOTHERMIC ( Δ H > 0, Δ S < 0) Δ S T = Δ S SYS + Δ S SURR IF Δ S T > 0, THEN THE CHEMICAL REACTION IS SPONTANEOUS

41. EXAMPLE 1: WILL ENTROPY INCREASE OR DECREASE FOR THE FOLLOWING? a)N 2 (G) + 3H 2 (G)  2NH 3 (G) b)2KCLO 3 (S)  2KCL (S) + 3O 2 (G) c)CO (G) + H 2 O (G)   CO 2 (G) + H 2 (G) d)C 12 H 22 O 11 (S)  C 12 H 22 O 11

42. HOW DO WE CALCULATE THE ENTROPY CHANGE ( Δ S) IN A CHEMICAL REACTION? SAME METHOD AS USING THE ENTHALPIES OF FORMATION TO CALCULATE Δ H AND USE THE SAME TABLE. AA + BB  CC + DD Δ S° =[C ( Δ S° C ) + D( Δ S° D )] - [A ( Δ S° A ) + B ( Δ S° B )]

43. EXAMPLE 2: CALCULATE Δ S° FOR THE FOLLOWING REACTION AT 25°C…. 4HCL (G) + O 2 (G)  2CL 2 (G) + 2H 2 O (G)

44. GIBBS FREE ENERGY

45. CHANGE IN GIBBS FREE ENERGY ( Δ G) Δ G = Δ H – T Δ S RELATES ENTHALPY AND ENTROPY TO DETERMINE WHICH HAS MORE IMPORTANCE IN DETERMINING WHETHER A REACTION IS SPONTANEOUS COMBINES ENERGY TRANSFER AS HEAT ( Δ H) AND ENERGY RELEASED TO CONTRIBUTE TO DISORDER ( Δ S)

46. CHANGE IN GIBBS FREE ENERGY ( Δ G) Δ G = Δ H – T Δ S Δ G < 0, SPONTANEOUS REACTION ENERGY AVAILABLE TO DO WORK Δ G > 0, NONSPONTANEOUS REACTION ENERGY DEFICIENCY, NO LEFTOVER ENERGY AND NOT ENOUGH ENERGY FOR REACTION

47. HOW CAN WE APPLY THE GIBBS EQUATION TO DETERMINE SPONTANEITY OF REACTION? ΔHΔH ΔSΔS ΔGΔG Result - + - Spontaneous (all temperatures) + - + Nonspontaneous (all temperatures) - - - Spontaneous (low temperatures) + + + Nonspontaneous (low temperatures) Δ G = Δ H – T Δ S

48. TWO PATHS TO CALCULATING Δ G 1) Δ G = Δ H – T Δ S DETERMINE Δ H. WHAT METHODS CAN WE USE? DETERMINE Δ S. THEN CALCULATE Δ G

49. TWO PATHS TO CALCULATING Δ G 2) USE STANDARD FREE ENERGY OF FORMATION ( Δ G F °) VALUES TO DETERMINE Δ G STANDARD FREE ENERGY OF FORMATION ( Δ G F °) --- Δ G° FOR THE FORMATION OF 1 MOLE OF A CHEMICAL COMPOUND IN ITS STANDARD STATE. Δ G F ° FOR ELEMENT FORMATION IN THEIR MOST STABLE STATE = 0. AA + BB  CC + DD Δ G° =[C ( Δ G F °) C + D( Δ G F °) D ] - [A ( Δ G F °) A + B ( Δ G F °) B ]

50. EXAMPLE 2: A)FIND Δ G FOR A CHEMICAL REACTION GIVEN Δ H = -218 KJ AND Δ S = -765 J/K AT 32°C. B) AT WHAT TEMPERATURE DOES THIS REACTION BECOME SPONTANEOUS? ASSUME ONLY TEMPERATURE CHANGES.

51. EXAMPLE 3: CALCULATE Δ G° RXN UNDER STANDARD CONDITIONS FOR THE FOLLOWING REACTION USING Δ G F ° VALUES. FE 2 O 3 (S) + 2AL (S)  2FE (S) + AL 2 O 3 (S)

52. A SPONTANEOUS REACTION IS NOT NECESSARILY FAST!!!! REACTION RATE INVOLVES KINETICS ! !

53. FREE ENERGY AND EQUILIBRIUM

54. Δ G°  Δ G CONCENTRATION AND PARTIAL PRESSURES CAN CHANGE FOR REACTANTS AND PRODUCTS. Δ G = Δ G° + RTLNQ WHAT HAPPENS WHEN WE LEAVE THE STANDARD STATE CONDITONS?

55. AT EQUILIBRIUM, Δ G = 0, SO REACTION QUOTIENT (Q) = EQUILIBRIUM CONSTANT (K) AT EQUILIBRIUM Δ G° = - RTLNK ENABLES THE REACTION’S EQUILIBRIUM CONSTANT (K) TO BE CALCULATED FROM THE CHANGE IN FREE ENERGY ( Δ G°) Δ G = Δ G° + RTLNQ

56. RATIO OF THE REACTANTS’ CONCENTRATIONS AND THE PRODUCTS’ CONCENTRATIONS AT EQUILIBRIUM. MORE ON THIS VALUE LATER….. EQUILIBRIUM CONSTANT (K C )

57. THE MAGNITUDE OF Δ G° INDICATES HOW FAR THE CHEMICAL REACTION IN ITS STANDARD STATE IS FROM EQUILIBRIUM. Δ G° = 0, EQUILIBRIUM Δ G°= LARGE VALUE, FAR FROM EQUILIBRIUM Δ G° = SMALL VALUE, CLOSE TO EQUILIBRIUM THE SIGN (+, - ) INDICATES WHICH DIRECTION THE REACTION NEEDS TO SHIFT TO ACHIEVE EQUILIBRIUM POSITIVE (+) -------- SHIFT TO LEFT, NO REACTION NEGATIVE (-) -------- SHIFT TO RIGHT, REACTION GOES TO COMPLETION WHAT IS THE RELATIONSHIP BETWEEN FREE ENERGY( Δ G) AND K?

58. APPLY HESS’ LAW---- REALLY Δ H F (PRODUCTS) - Δ H F (REACTANTS) CALCULATE Δ H RXN BASED ON ENTHALPY OF FORMATION ( Δ H F ) AA + BB  CC + DD Δ H° =[C ( Δ H F °) C + D( Δ H F °) D ] - [A ( Δ H F °) A + B ( Δ H F °) B ]

59. EXAMPLE 5 ISOPROPYL ALCOHOL (RUBBING ALCOHOL) UNDERGOES A COMBUSTION REACTION 2(CH 3 ) 2 CHOH + 9O 2  6CO 2 + 8H 2 O Δ H° = -4011 KJ/MOL CALCULATE THE STANDARD ENTHALPY OF FORMATION FOR ISOPROPYL ALCOHOL.