Download Kinetics & Equilibrium

H
H
C HOH
H
KINETICS
Cl
H
Cl
H
OH
H C
H
Cl H
H C OH
H
H
OH Cl
H C
H
H
H
H C OH Cl H H
H C H Cl
H
Cl
HH C
H
CHAPTER 9
H
H C
H
HO H
HO H
Cl
HO H
1
STUDY OF RELATIONSHIP OF
CHEMISTRY
THERMO DYNAMICS
CHEMICAL REACTIONS AND
ENERGY
st LAW
1st LAW
OFTHERMODYNAMICS
THERMODYNAMICS
OF
THERMODYNAMICS
1st1LAW
OF
DEuniv = DE + DEsur = 0
Efinal - E
DE
init = - DEsur
HEAT ABSORBED BY SYSTEM
q > 0, HEAT IS ABSORBED
=
=
q
+
w
DE + WORK DONE ON SYSTEM
q < 0, HEAT RELEASED
w > 0, WORK ON SYSTEM
w < 0, WORK BY SYSTEM
q > 0, ENDOTHERMIC
q < 0, EXOTHERMIC
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MOST REACTIONS OCCUR AT CONSTANT TEMPERATURE
AND PRESSURE .....BUT SOME ENERGY MAY BE LOST
HEAT OF REACTION
HEAT ABSORBED IN A
ENTHALPY,
H
CHANGE
IN ENTHALPY,
DH: REACTION CARRIED OUT
AT CONSTANT PRESSURE
>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
3
E
N
T
H
A
L
P
Y
H>0
PRODUCT!!!
ENDOTHERMIC
REACTANT!!
4
E
N
T
H
A
L
P
Y
H< 0
REACTANTS!!!
EXOTHERMIC
PRODUCTS!!
5
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
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BOND ENERGY: ENERGY NEEDED TO BREAK 1 MOLE
OF BONDS IN THE GASEOUS STATE
BREAKING: DHo : ALWAYS > 0
FORMATION DHo : ALWAYS < 0
ESTIMATE: DHo ~ S BE BROKEN - S BE FORMED
H2C=CH2 + HCl
1 C=C
1 H-Cl
1 x 612 1 x 431
H3C-CH2Cl
1 C-H
TABLE 9.1
1 C-Cl
1 x 413 AND 1 x 234
DHo ~ 1043 - 647 = ~ 396 kJ/mol
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EVERY PROCESS INCREASES
DISORDER IN THE UNIVERSE
FOR A SPONTANEOUS PROCESS, SUNIV > 0
q
S=
J/K
T AT WHICH HEAT IS ADDED
Sgas > Sliquid > Ssolution > Ssolid
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DSuniv = DSsur + DS
DSuniv > 0
SPONTANEOUS
PROCESS
NON-SPONTANEOUS
PROCESS
DSuniv > 0 WHERE THE NUMBER OF
MOLES OF GAS INCREASES
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ESTIMATING ENTROPY CHANGE:
COMPARE PRODUCTS TO REACTANTS
N2 (g) + 3H2 (g)
NaCl (s)
2NH3 (g)
Na1+ (aq) + Cl1- (aq)
CaCO3 (s) + H301+(aq)
H20 (s)
H20 (l)
CO2 @ 20 oC
Ag (s) + NaCl (s)
DS
<0
>0
Ca2+ (aq) + 3H20 (l) + CO2 (g) >0
H20 (g)
>0
>0
CO2 @ 0 oC
<0
AgCl (s) + Na (s)
~0
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DSuniv
= DSsur + DS
AT CONSTANT P: DSsur = -DH/T
-TDS
DGuniv = DH - TDS
FOR A SPONTANEOUS REACTION:
DSuniv > 0
DG < 0
THE ENERGY OF THE PROCESS MUST DECREASE AND
THE UNIVERSE MUST BECOME MORE RANDOM!!!!
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DRIVING FORCES FOR A CHEMICAL REACTION:
DH -- ENERGY REQUIRED TO CHANGE TO POTENTIAL
ENERGY OF REACTANTS TO THAT OF PRODUCTS
-TDS -- ENERGY TO MAKE THE SYSTEM
MORE ORDERED
RELATE TO DG = DH - TDS
DH
+
+
-
DS
+
+
-
<0
SPONTANEOUS?
ALWAYS - AT ANY T
NEVER - AT ANY T
AT HIGH T
AT LOW T
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ENERGY DIFFERENCES ONLY!
WHAT IS POSSIBLE
WHAT IS NOT POSSIBLE
CONCERNED WITH PATH
WHAT HAPPENS
HOW FAST IT HAPPENS
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CH3Br + OH
1-
TRANSITION
STATE
P
O
T
E
N
T
I
A
L
Ea
DE
H H
C H
Br
OH
-
CH3OH + Br 1-
STERIC EFFECTS: MUST
HAVE PROPER ORIENTATION
MINIMUM
OK OF
H-O
CAMOUNT
ENERGY FOR COLLISION
O-H
C
NR
TO ACHIEVE TRANSITION
STATE
Ea(reverse) NEED COLLISION OF
PROPER ENERGY AND
ORIENTATION FOR ELECTRONS
TO BE SHARED OR TRANSFERRED
E
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RATE OF DISAPPEARANCE
M
O
L
L
RATE SLOWS WITH TIME
RELATED TO NUMBER OF
REACTING PARTICLES
FOR RATE OFAPPEARANCE
Rf = kf[A]X[B]Y
SEC
15
X[B]Y
R
=
k
[A]
f
IN IT’S SIMPLEST FORM: f
CH3Br + OH 1- <−> CH3OH + Br 1-
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
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H2 (g) + I2 (g) <−> 2HI (g)
RATE 1 OR Rf
2HI (g) <−> H2 (g) + I2 (g)
RATE 2 OR Rr
H2 (g) + I2 (g) <−> 2HI (g)
Rf = Rr
Rf
[HI]2
K = R = [H ][I ]
r
2
2
= [PRODUCTS]
[REACTANTS]
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K=
F1-
[PRODUCTS]
[REACTANTS]
(aq) + HNO2 (aq) <−> HF (aq) + NO2
1-
1
[
HF
][
NO
(aq)
2 ]
K = 1
[F ][ HNO 2 ]
[X] = MOLAR CONCENTRATIONS
2HCl (g) <−> H2 (g) + Cl2 (g)
K=
PH 2 PCl 2
2
(PHCl
)
CAN ALSO USE CONCENTRATIONS
CaF2 (s) + 2H3O1+ (aq) <−> Ca2+ (aq) + 2HF (aq) + 2H2O (l)
[Ca 2 ][ HF]2
K
[H 3O1 ]2
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F1- (aq) + HNO2 (aq) <−> HF (aq) + NO21- (aq)
[HF][ NO12 ]
K = 1
[F ][ HNO 2 ]
HCN (aq) + H2O (l) <−> CN1- (aq) + H3O1+ (aq)
[CN1 ][ H3O1 ]
K
[HCN]
PbBr2 (s) <−> Pb2+ (aq) + 2Br1- (aq)
K SP= [Pb2+][Br1-]2
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DGo = - RTlnK
K >> 1
K << 1
EXTENSIVE
LARGE AMOUNT OF PRODUCT
EXOTHERMIC PROCESSES
NOT EXTENSIVE
SMALL AMOUNT OF PRODUCT
ENDOTHERMIC (IF NO CHANGE IN MOLE OF
GAS INVOVED)
K VARIES ONLY WITH TEMPERATURE!!!!!!
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REACTANTS <−> PRODUCTS + HEAT
K=
[PRODUCTS]
[REACTANTS]
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
DRIVE TO LEFT
ADD REACTANT: FAVOR PRODUCTS
DRIVE TO RIGHT
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PbBr2 (s) <−> Pb2+ (aq) + 2Br1- (aq)
DHo= 37.2 kJ/MOL
b) REMOVING SOME Br1-
c) ADDING PbBr2 (s)
TO RIGHT OR
FAVORS PRODUCTS
NO CHANGE!!!
d) INCREASING TEMPERATURE
e) DOUBLING THE VOLUME
f) ADDING Pb2+
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