1 H H C H OH HCl H H C H OH HCl H H C H OH HCl H H H C H OH H H C H OH HCl H H C H HHO Cl H H C H HHO Cl H H C H HHO CHAPTER 9 REACTION ENERGETICS KINETICS

2 STUDY OF RELATIONSHIP OF CHEMICAL REACTIONS AND THERMO DYNAMICS 1 st LAW OF THERMODYNAMICS  E univ =  E +  E sur = 0 EE= -  E sur E final - E init EE HEAT ABSORBED BY SYSTEM + WORK DONE ON SYSTEM = = q + w q > 0, HEAT IS ABSORBED q < 0, HEAT RELEASED q > 0, ENDOTHERMIC q < 0, EXOTHERMIC w > 0, WORK ON SYSTEM w < 0, WORK BY SYSTEM ENERGY CHEMISTRY

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

4  H > 0 REACTANT!! PRODUCT!!! ENDOTHERMIC ENTHALPYENTHALPY

5  H< 0 PRODUCTS!! REACTANTS!!! EXOTHERMIC ENTHALPYENTHALPY

6  H comb = HEAT ABSORBED WHEN 1 MOLE OF A SUBSTANCE REACTS WITH OXYGEN AT CONSTANT P C 6 H 12 O 6 (s) + 6O 2 (g)  6CO 2 (g) + 6H 2 O (g)  H o comb = 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?  H = kJ/mol x mol = kJ mol glucose = 10 g x 1 mol/180 g = mol  158 kJ of heat is released

7 BOND ENERGY: ENERGY NEEDED TO BREAK 1 MOLE OF BONDS IN THE GASEOUS STATE BREAKING:  H o : ALWAYS > 0 FORMATION  H o : ALWAYS < 0 ESTIMATE:  H o ~  BE BROKEN -  BE FORMED H 2 C=CH 2 + HCl H 3 C-CH 2 Cl 1 x 6121 x C=C 1 H-Cl 1 C-H 1 C-Cl TABLE x 413 AND 1 x 234  H o ~ = ~ 396 kJ/mol

8 FOR A SPONTANEOUS PROCESS,  S UNIV > 0 EVERY PROCESS INCREASES DISORDER IN THE UNIVERSE  S = qTqT AT WHICH HEAT IS ADDED J/K S gas > S liquid > S solution > S solid

9  S univ =  S sur +  S  S univ > 0 SPONTANEOUSPROCESS NON-SPONTANEOUSPROCESS  S univ > 0 WHERE THE NUMBER OF MOLES OF GAS INCREASES

10 ESTIMATING ENTROPY CHANGE: COMPARE PRODUCTS TO REACTANTS SS N 2 (g) + 3H 2 (g) 2NH 3 (g)< 0 NaCl (s) Na 1+ (aq) + Cl 1- (aq)>0 CaCO 3 (s) + H (aq) Ca 2+ (aq) + 3H 2 0 (l) + CO 2 (g)>0 < 0 ~0 H 2 0 (s) H 2 0 (l) H 2 0 (g) 20 o C 0 o C Ag (s) + NaCl (s) AgCl (s) + Na (s)

11  S univ =  S sur +  S AT CONSTANT P:  S sur = -  H/T -T  S univ =  H - T  S GG FOR A SPONTANEOUS REACTION:  S univ > 0  G < 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  H  S SPONTANEOUS? + - NEVER - AT ANY T + + AT HIGH T - - AT LOW T RELATE TO  G =  H - T  S <0 - + ALWAYS - AT ANY T

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

14 CH 3 Br + OH 1- POTENTIALEPOTENTIALE - TRANSITION STATE STERIC EFFECTS: MUST HAVE PROPER ORIENTATION H-O C OK O-H CNR EE EaEa MINIMUM AMOUNT OF ENERGY FOR COLLISION TO ACHIEVE TRANSITION STATE E a(reverse) NEED COLLISION OF PROPER ENERGY AND ORIENTATION FOR ELECTRONS TO BE SHARED OR TRANSFERRED CH3OH + Br 1-

15 MOLLMOLL SEC FOR RATE OFAPPEARANCE RATE SLOWS WITH TIME RELATED TO NUMBER OF REACTING PARTICLES RATE OF DISAPPEARANCE R f = k f [A] X [B] Y

16 IN IT ’ S SIMPLEST FORM: R f = k f [A] X [B] Y CH 3 Br + OH 1- CH3OH + Br 1- R f = k f [CH 3 Br][OH 1- ] N 2 + 3H 2 2NH 3 R f = k f [N 2 ][H 2 ] 3 2NO 2 N 2 O 4 HF (aq) + NH 3 (g) NH 4 1+ (aq) + F 1- (aq) Rf Rf = k f [NO 2 ] 2 R f = k f [HF][NH 3 ] CATALYSTS & INHIBITORS

17 H 2 (g) + I 2 (g) 2HI (g) RATE 1 OR R f 2HI (g) H 2 (g) + I 2 (g) H 2 (g) + I 2 (g) 2HI (g) RATE 2 OR R r R f = R r K = = [HI] 2 [H 2 ][I 2 ] RfRrRfRr [PRODUCTS] [REACTANTS] =

18 K = [PRODUCTS] [REACTANTS] F 1- (aq) + HNO 2 (aq) HF (aq) + NO 2 1- (aq) K = [X] = MOLAR CONCENTRATIONS 2HCl (g) H 2 (g) + Cl 2 (g) K = CAN ALSO USE CONCENTRATIONS CaF 2 (s) + 2H 3 O 1+ (aq) Ca 2+ (aq) + 2HF (aq) + 2H 2 O (l)

19 F 1- (aq) + HNO 2 (aq) HF (aq) + NO 2 1- (aq) K = HCN (aq) + H 2 O (l) CN 1- (aq) + H 3 O 1+ (aq) K = [Pb 2+ ][Br 1- ] 2 PbBr 2 (s) Pb 2+ (aq) + 2Br 1- (aq) SP

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

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

22 b) REMOVING SOME Br 1- TO RIGHT OR FAVORS PRODUCTS c) ADDING PbBr 2 (s) NO CHANGE!!! d) INCREASING TEMPERATURE e) DOUBLING THE VOLUME PbBr 2 (s) Pb 2+ (aq) + 2Br 1- (aq)  H o = 37.2 kJ/MOL f) ADDING Pb 2+