1

H

H C

H

OH

HCl

H

H C

H

OH HCl

H

H C

H

OH HCl

HCl

H

H C

H

OH

H

H C

H

OH HCl

Cl

H

H C

H

HHO

Cl

H

H C

H

HHO

Cl

H

H C

H

HHO

CHAPTER 9

REACTION

ENERGETICS

THER

MODYN

AMICS

THER

MODYN

AMICS

KINETICS

2

STUDY OF RELATIONSHIP OFCHEMICAL REACTIONS AND THERMO DYNAMICS

1st LAW OF THERMODYNAMICS1st LAW OF THERMODYNAMICS1st LAW OF THERMODYNAMICSEuniv = E + Esur =

0E = - EsurEfinal - Einit

E HEAT ABSORBED BY SYSTEM+ WORK DONE ON SYSTEM

= = q + w q > 0, HEAT IS ABSORBEDq < 0, HEAT RELEASED

q > 0, ENDOTHERMIC q < 0, EXOTHERMIC

w > 0, WORK ON SYSTEMw < 0, WORK BY SYSTEM

ENERGY

CHEMISTRY

3

MOST REACTIONS OCCUR AT CONSTANT TEMPERATUREAND PRESSURE .....BUT SOME ENERGY MAY BE LOST

HEAT OF REACTION

ENTHALPY, HCHANGE IN ENTHALPY, H:HEAT ABSORBED IN A REACTION CARRIED OUT AT CONSTANT PRESSURE

H:>0, REACTION ABSORBS HEAT ENDOTHERMIC

< 0, REACTION RELEASES HEAT EXOTHERMIC

IS A STATE FUNCTION!

PROPORTIONAL TO NUMBER OF MOLES

OPPOSITE SIGN FOR REVERSE REACTION

4

H > 0

REACTANT!!

PRODUCT!!!

ENDOTHERMIC

ENTHALPY

5

H< 0

PRODUCTS!!

REACTANTS!!!

EXOTHERMIC

ENTHALPY

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 STATEMOST STABLE FORM AT 1 atm AND THE SPECIFIED TEMPERATUREFOR DISSOLVED SUBSTANCE, 1 M

HOW MUCH HEAT IS RELEASED IF 10 g GLUCOSE IS BURNED?

H = -2816 kJ/mol x 0.056 mol = -157.7 kJ

mol glucose = 10 g x 1 mol/180 g = 0.056 mol

158 kJ of heat is released

7

BOND ENERGY: ENERGY NEEDED TO BREAK 1 MOLE OF BONDS IN THE GASEOUS STATE

BREAKING: Ho : ALWAYS > 0

FORMATION Ho : ALWAYS < 0

ESTIMATE: Ho ~ BE BROKEN - BE FORMED

H2C=CH2 + HCl H3C-CH2Cl

1 x 612 1 x 431

1 C=C 1 H-Cl 1 C-H 1 C-Cl

TABLE 9.1

1 x 413 AND 1 x 234

Ho ~ 1043 - 647 = ~ 396 kJ/mol

8

FOR A SPONTANEOUS PROCESS, SUNIV > 0

EVERY PROCESS INCREASES DISORDER IN THE UNIVERSE

S = qT AT WHICH HEAT IS ADDED

J/K

Sgas > Sliquid > Ssolution > Ssolid

9

Suniv = Ssur + S

Suniv > 0

SPONTANEOUS SPONTANEOUS PROCESSPROCESS

NON-SPONTANEOUS NON-SPONTANEOUS PROCESSPROCESS

Suniv > 0 WHERE THE NUMBER OF

MOLES OF GAS INCREASES

10

ESTIMATING ENTROPY CHANGE:

COMPARE PRODUCTS TO REACTANTS S

N2 (g) + 3H2 (g) 2NH3 (g) < 0

NaCl (s) Na1+ (aq) + Cl1- (aq) >0

CaCO3 (s) + H301+(aq) Ca2+ (aq) + 3H20 (l) + CO2 (g) >0

>0>0

< 0

~0

H20 (s) H20 (l) H20 (g)

CO2 @ 20 oC CO2 @ 0 oC

Ag (s) + NaCl (s) AgCl (s) + Na (s)

11

Suniv = Ssur + S

AT CONSTANT P: Ssur = -H/T

-TSuniv = H - TSG

FOR A SPONTANEOUS REACTION:

Suniv > 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 POTENTIALENERGY OF REACTANTS TO THAT OF PRODUCTS

-TS -- ENERGY TO MAKE THE SYSTEM MORE ORDERED

H S SPONTANEOUS?

+ - NEVER - AT ANY T

+ + AT HIGH T

- - AT LOW T

RELATE TO G =H - TS <0

- + ALWAYS - AT ANY T

13

WHAT IS POSSIBLE WHAT IS NOT POSSIBLE

WHAT HAPPENSHOW FAST IT HAPPENS

ENERGY DIFFERENCES ONLY!

CONCERNED WITH PATH

14

CH3Br + OH 1-

POTENTIAL

E

CBr

H HH

OH-

TRANSITIONSTATE STERIC EFFECTS: MUST

HAVE PROPER ORIENTATION

H-O C OK

O-H C NR

E

Ea

MINIMUM AMOUNT OFENERGY FOR COLLISIONTO ACHIEVE TRANSITION

STATE

Ea(reverse)NEED COLLISION OF

PROPER ENERGY AND ORIENTATION FOR ELECTRONS

TO BE SHARED OR TRANSFERRED

CH3OH + Br 1-

15

MOLL

SEC

FOR RATE OFAPPEARANCE

RATE SLOWS WITH TIMERELATED TO NUMBER OF REACTING PARTICLES

RATE OF DISAPPEARANCE

Rf = kf[A]X[B]Y

16

IN IT’S SIMPLEST FORM: Rf = kf[A]X[B]Y

CH3Br + OH 1- <−> CH3OH + Br 1-

Rf = kf[CH3Br][OH1-]

N2 + 3H2 <−> 2NH3 Rf = kf[N2][H2]3

2NO2 <−> N2O4

HF (aq) + NH3 (g) <−> NH41+ (aq) + F1- (aq)

Rf = kf[NO2]2

Rf = kf[HF][NH3]

CATALYSTS & INHIBITORS

17

H2 (g) + I2 (g) <−> 2HI (g) RATE 1 OR Rf

2HI (g) <−> H2 (g) + I2 (g)

H2 (g) + I2 (g) <−> 2HI (g)

RATE 2 OR Rr

Rf = Rr

K = = [HI]2

[H2][I2]Rf

Rr

[PRODUCTS] [REACTANTS]=

18

K = [PRODUCTS] [REACTANTS]

F1- (aq) + HNO2 (aq) <−> HF (aq) + NO21- (aq)

K = ]HNO][F[

]NO][HF[

21

12

[X] = MOLAR CONCENTRATIONS

2HCl (g) <−> H2 (g) + Cl2 (g)K =

)P(

PP2HCl

ClH 22

CAN ALSO USE CONCENTRATIONS

CaF2 (s) + 2H3O1+ (aq) <−> Ca2+ (aq) + 2HF (aq) + 2H2O (l)

213

22

]OH[

]HF][Ca[K

19

F1- (aq) + HNO2 (aq) <−> HF (aq) + NO21- (aq)

K = ]HNO][F[

]NO][HF[

21

12

HCN (aq) + H2O (l) <−> CN1- (aq) + H3O1+ (aq)

]HCN[

]OH][CN[K

13

1

K = [Pb2+][Br1-]2PbBr2 (s) <−> Pb2+ (aq) + 2Br1- (aq)SP

20

K >> 1

Go = - RTlnK

EXTENSIVELARGE AMOUNT OF PRODUCT

EXOTHERMIC PROCESSES

K << 1NOT EXTENSIVESMALL AMOUNT OF PRODUCTENDOTHERMIC (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 STRESSIN 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 Br1- TO RIGHT OR

FAVORS PRODUCTS

c) ADDING PbBr2 (s) NO CHANGE!!!

d) INCREASING TEMPERATURE

e) DOUBLING THE VOLUME

PbBr2 (s) <−> Pb2+ (aq) + 2Br1- (aq) Ho= 37.2 kJ/MOL

f) ADDING Pb2+