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H H C H

H H C H. H H C H. Cl. H. OH. H H C H. Cl. H. OH. H H C H. Cl. H. OH. Cl. H. OH. H H C H. H H C H. OH. Cl. H. H H C H. Cl. HO. H. Cl. HO. H. H H C H. HO. H. Cl. KINETICS.

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H H C H

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  1. H H C H H H C H Cl H OH H H C H Cl H OH H H C H Cl H OH Cl H OH H H C H H H C H OH Cl H H H C H Cl HO H Cl HO H H H C H HO H Cl KINETICS REACTION ENERGETICS THERMODYNAMICS CHAPTER 9

  2. HEAT ABSORBED BY SYSTEM = + WORK DONE ON SYSTEM STUDY OF RELATIONSHIP OF CHEMICAL REACTIONS AND CHEMISTRY THERMO DYNAMICS ENERGY 1st LAW OF THERMODYNAMICS 1st LAW OF THERMODYNAMICS 1st LAW OF THERMODYNAMICS DEuniv = DE + DEsur = 0 Efinal - Einit DE = - DEsur q > 0, HEAT IS ABSORBED DE = q + w q < 0, HEAT RELEASED q > 0, ENDOTHERMIC q < 0, EXOTHERMIC w > 0, WORK ON SYSTEM w < 0, WORK BY SYSTEM

  3. MOST REACTIONS OCCUR AT CONSTANT TEMPERATURE AND PRESSURE .....BUT SOME ENERGY MAY BE LOST HEAT OF REACTION HEAT ABSORBED IN A REACTION CARRIED OUT AT CONSTANT PRESSURE ENTHALPY, H CHANGE IN ENTHALPY, DH: >0, REACTION ABSORBS HEAT ENDOTHERMIC DH: < 0, REACTION RELEASES HEAT EXOTHERMIC IS A STATE FUNCTION! PROPORTIONAL TO NUMBER OF MOLES OPPOSITE SIGN FOR REVERSE REACTION

  4. PRODUCT!!! REACTANT!! ENTHALPY H > 0 ENDOTHERMIC

  5. ENTHALPY REACTANTS!!! PRODUCTS!! H< 0 EXOTHERMIC

  6. Hcomb = HEAT ABSORBED WHEN 1 MOLE OF A SUBSTANCE REACTS WITH OXYGEN AT CONSTANT P C6H12O6 (s) + 6O2 (g)  6CO2 (g) + 6H2O (g) Hocomb = -2816 kJ STANDARD STATE MOST STABLE FORM AT 1 atm AND THE SPECIFIED TEMPERATURE FOR DISSOLVED SUBSTANCE, 1 M HOW MUCH HEAT IS RELEASED IF 10 g GLUCOSE IS BURNED? mol glucose = 10 g x 1 mol/180 g = 0.056 mol H = -2816 kJ/mol x 0.056 mol = -157.7 kJ  158 kJ of heat is released

  7. TABLE 9.1 H2C=CH2 + HCl H3C-CH2Cl BOND ENERGY: ENERGY NEEDED TO BREAK 1 MOLE OF BONDS IN THE GASEOUS STATE ALWAYS > 0 BREAKING: DHo : FORMATIONDHo : ALWAYS < 0 ESTIMATE: DHo ~ S BE BROKEN - S BE FORMED 1 C-Cl 1 C=C 1 H-Cl 1 C-H 1 x 612 1 x 431 1 x 413 AND 1 x 234 DHo ~ 1043 - 647 = ~ 396 kJ/mol

  8. q T S = 2nd LAW OF THERMODYNAMICS EVERY PROCESS INCREASES DISORDER IN THE UNIVERSE FOR A SPONTANEOUS PROCESS, SUNIV > 0 J/K AT WHICH HEAT IS ADDED Sgas > Sliquid > Ssolution > Ssolid

  9. DSuniv = DSsur + DS DSuniv > 0 NON-SPONTANEOUS PROCESS SPONTANEOUS PROCESS DSuniv > 0 WHERE THE NUMBER OF MOLES OF GAS INCREASES

  10. H20 (s) H20 (l) H20 (g) N2 (g) + 3H2 (g) 2NH3 (g) CO2 @ 20 oC CO2 @ 0 oC NaCl (s) Na1+ (aq) + Cl1- (aq) Ag (s) + NaCl (s) AgCl (s) + Na (s) CaCO3 (s) + H301+(aq) Ca2+ (aq) + 3H20 (l) + CO2 (g) ESTIMATING ENTROPY CHANGE: COMPARE PRODUCTS TO REACTANTS DS < 0 >0 >0 >0 >0 < 0 ~0

  11. Suniv = Ssur + S -TSuniv G G < 0 AT CONSTANT P: Ssur = -H/T FREE ENERGY CHANGE = H- TS FOR A SPONTANEOUS REACTION: Suniv > 0 THE ENERGY OF THE PROCESS MUST DECREASE AND THE UNIVERSE MUST BECOME MORE RANDOM!!!!

  12. DRIVING FORCES FOR A CHEMICAL REACTION: H -- ENERGY REQUIRED TO CHANGE TO POTENTIAL ENERGY OF REACTANTS TO THAT OF PRODUCTS -TS -- ENERGY TO MAKE THE SYSTEM MORE ORDERED <0 RELATE TO DG = DH - TDS DH DS SPONTANEOUS? - + ALWAYS - AT ANY T + - NEVER - AT ANY T + + AT HIGH T - - AT LOW T

  13. ENERGY DIFFERENCES ONLY! THERMODYNAMICS WHAT IS POSSIBLE WHAT IS NOT POSSIBLE KINETICS CONCERNED WITH PATH WHAT HAPPENS HOW FAST IT HAPPENS

  14. CH3Br + OH 1- CH3OH + Br 1- - TRANSITION STATE STERIC EFFECTS: MUST HAVE PROPER ORIENTATION H-O C O-H C Ea Ea(reverse) DE MINIMUM AMOUNT OF ENERGY FOR COLLISION TO ACHIEVE TRANSITION STATE POT ENTIAL E OK NR NEED COLLISION OF PROPER ENERGY AND ORIENTATION FOR ELECTRONS TO BE SHARED OR TRANSFERRED

  15. M O L L SEC RATE OF DISAPPEARANCE RATE SLOWS WITH TIME RELATED TO NUMBER OF REACTING PARTICLES FOR RATE OFAPPEARANCE Rf = kf[A]X[B]Y

  16. CH3Br + OH 1-<−> CH3OH + Br 1- Rf = kf[A]X[B]Y IN IT’S SIMPLEST FORM: Rf = kf[CH3Br][OH1-] N2 + 3H2<−> 2NH3 Rf = kf[N2][H2]3 2NO2<−> N2O4 Rf = kf[NO2]2 HF (aq) + NH3 (g) <−> NH41+ (aq) + F1- (aq) Rf = kf[HF][NH3] CATALYSTS & INHIBITORS

  17. Rf Rr [PRODUCTS] [REACTANTS] [HI]2 [H2][I2] = = EQUILIBRIUM RATE 1 OR Rf H2 (g) + I2 (g) <−> 2HI (g) RATE 2 OR Rr 2HI (g) <−> H2 (g) + I2 (g) Rf = Rr H2 (g) + I2 (g) <−> 2HI (g) K =

  18. K = K = K = [PRODUCTS] [REACTANTS] F1- (aq) + HNO2 (aq) <−> HF (aq) + NO21- (aq) [X] = MOLAR CONCENTRATIONS 2HCl (g) <−> H2 (g) + Cl2 (g) CAN ALSO USE CONCENTRATIONS CaF2 (s) + 2H3O1+ (aq) <−> Ca2+ (aq) + 2HF (aq) + 2H2O (l)

  19. F1- (aq) + HNO2 (aq) <−> HF (aq) + NO21- (aq) K = HCN (aq) + H2O (l) <−> CN1- (aq) + H3O1+ (aq) K= [Pb2+][Br1-]2 PbBr2 (s) <−> Pb2+ (aq) + 2Br1- (aq) SP

  20. Go = - RTlnK EXTENSIVE K >> 1 LARGE AMOUNT OF PRODUCT EXOTHERMIC PROCESSES NOT EXTENSIVE SMALL AMOUNT OF PRODUCT ENDOTHERMIC (IF NO CHANGE IN MOLE OF GAS INVOVED) K << 1 K VARIES ONLY WITH TEMPERATURE!!!!!!

  21. K = [PRODUCTS] [REACTANTS] + HEAT REACTANTS <−> PRODUCTS LeCHATELIER’S PRINCIPLE: A SYSTEM AT EQUILIBRIUM WILL RESPOND TO A STRESS IN A WAY TO MINIMIZE THE EFFECT OF THE STRESS ADD PRODUCT: FAVOR REACTANTS ADD REACTANT: FAVOR PRODUCTS DRIVE TO LEFT DRIVE TO RIGHT

  22. TO RIGHT OR FAVORS PRODUCTS PbBr2 (s) <−> Pb2+ (aq) + 2Br1- (aq) DHo= 37.2 kJ/MOL b) REMOVING SOME Br1- NO CHANGE!!! c) ADDING PbBr2 (s) d) INCREASING TEMPERATURE e) DOUBLING THE VOLUME f) ADDING Pb2+

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