Post on 01-Jan-2016
Outline
Part I: Reaction + Catalyst
1. Thermodynamics of the target reaction
2. Thermodynamics of catalyst: bulk (see classes on solids
and defects) and surface
3. Thermodynamics of interaction between reactant and
catalyst (see class on adsorption)
Part II: Practical Matters
1. Vapor pressure
Target Reaction - Motivation
Why look at TD? …can’t change it anyway by catalysis
Reactants
Products
E
EA
without catalyst
Reaction coordinate
Reactants
Products
E
EA
with catalyst
Reaction coordinateMust look at T
D because we can’t change it!
Target Reaction – Quantities to Look at
Enthalpy of reaction ΔrH
exothermic / endothermicΔrH of side reactions
Free Enthalpy (Gibbs Energy) ΔrG
exergonic / endergonic
Equilibrium Constant K: Equilibrium Limitations
Change of Temperature and Pressure (variables)
Enthalpy of Reaction
Determines reactor setup (see classes on catalyst
testing and reaction engineering)
catalyst formulation / dilution
“hot spots” / heating power
isothermal operation in the lab
Enthalpy of side reactions
parallel / secondary reactions
Enthalpy of Reaction, ΔrH
Reaction enthalpy needs a reaction equation!!!
Calculate from enthalpies of formation of products and reactants
L
iifir HH
1
ΔrH°: standard enthalpy of reactionΔfH°: standard enthalpies of formationvi: stoichiometric factors, positive for products, negative for reactants
DCBA DCBA
Things to Watch in Calculations…..
Stoichiometric factors
Standard conditions
State of the matter (solid, liquid, gaseous)
Which data are available (sometimes only enthalpy of
combustion, ΔcH° )
Standard Conditions (IUPAC)
International Union of Pure and Applied Chemistry
(IUPAC)
Größen, Einheiten und Symbole in der Physikalischen
Chemie
VCH , Weinheim 1996
FHI library 50 E 49 (English version: 50 E 48)
Standard state indicated by superscript ,°
www.iupac.org
Standard Conditions (IUPAC)
„Standard state pressure“(IUPAC 1982)
p° = 105 Pa
„Standard atmosphere“ (before 1982)
p° = 101 325 Pa = 1 atm
„Standard concentration“ c° = 1 mol dm-3
„Standard molality“ m° = 1 mol kg-1
„Standard temperature“ ??
Standard Conditions (Textbooks)
AtkinsSTP „Standard temperature and pressure““p = 101 325 Pa = 1 atm, T° = 273,15 K SATP „Standard ambient temperature and pressure“p° = 105 Pa = 1 bar, T° = 298,15 K
Wedler „Standarddruck“p = 1.013 bar = 1 atm = 101.325 kPa„Standardtemperatur“T° = 298,15 K
Standard Conditions (Other)
Catalysis Literature
NTP „Normal temperature and pressure““
20°C and 760 torr
70 degrees F and 14.7 psia (1 atmosphere)
ALWAYS CHECK / SPECIFY THE CONDITIONS !!
ALWAYS CHECK / SPECIFY THE CONDITIONS !!
Sources for Thermodynamic Data
CRC Handbook of Thermophysical and Thermochemical
Data
Eds. David R. Lide, Henry V. Kehiaian
CRC Press Boca Raton New York 1994
FHI library 50 E 55
D'Ans Lax
Taschenbuch für Chemiker und Physiker
Ed. C. Synowietz
Springer Verlag 1983
FHI library 50 E 54
Some Examples: Combustion
Combustion of hydrogen (Knallgasreaktion)
ΔcH° = -286 kJ mol-1
)()()(2
1222 gOHgHgO
Combustion of carbon
ΔcH° = -394 kJ mol-1
)()()( 22 gCOgOsC
Reactions with CO2, H2O or other very stable molecules as products are usually strongly exothermic, however….
State of the Matter
Formation of benzene at 298.15 K
)()(3)(6 662 gHCgHsC
ΔfH° = 49.0 kJ mol-1
)()(3)(6 662 lHCgHsC
Enthalpy of evaporation of benzene?
ΔvapH° = 30.8 kJ mol-1 at 80°C
ΔfH° = 82.93 kJ mol-1
Partial Oxidation of Propene
Oxidation of propene to acrolein
ΔrH° = ??? kJ mol-1
OHOHCOHC 243263 2
1
Partial Oxidation
Only enthalpy of combustion, ΔcH°, of acrolein is given
ΔcH° = -1633 kJ mol-1
)(2)(3)(5.3)( 22243 gOHgCOgOgOHC
Enthalpies of combustion are easily determined quantities (e.g. from quantitative combustion in a bomb calorimeter)
Use Hess’s Law
)(2)(3)(4)(2)(3 2222 gOHgCOgOgHsC ΔcH° = -1754 kJ mol-1
ΔcH° = -1633 kJ mol-1)(2)(3)(5.3)( 22243 gOHgCOgOlOHC
)()(5.0)(2)(3 4322 lOHCgOgHsC ΔfH° = -121 kJ mol-1
Enthalpy is a State Function
Partial vs. Total Oxidation
Oxidation of propene to acrolein
ΔrH° = -427 kJ mol-1
OHOHCOHC 243263 2
1
Oxidation of acrolein to CO2
ΔcH° = -1633 kJ mol-1
)(2)(3)(5.3)( 22243 gOHgCOgOgOHC
Reactants
Partial Oxidation Product
EEA
Reaction coordinate
Total Oxidation Products
EA
Dehydrogenation vs. Oxidative Dehydrogenation
)()()( 284104 gHgHCigHCi
Oxidative dehydrogenation of isobutane to isobutene
ΔrH° = -124 kJ mol-1
)()()(5.0)( 2842104 gOHgHCigOgHCi
Dehydrogenation of isobutane to isobutene
ΔrH° = 117 kJ mol-1
Combustion of isobutene
ΔcH° = - 2525 kJ mol-1
Oxidative Dehydrogenation:Thermodynamic Traps
)(4)(4)(6)( 22284 gOHgCOgOgHCi
Nevertheless, the oxidative dehydrogenation of isobutene is in commercial operation (CrO3/Al2O3 or supported Pt catalyst)
Dehydrogenation of ethylbenzene to styrene
ΔrH° = 117 kJ mol-1
Dehydrogenation
)()()( 288108 gHlHClHC
Change of ΔrH with Temperature
Most of the time, we are not interested in room temperature
Reactants, T1
Products, T1
Enthalpy
Reaction coordinate
Products, T2
Reactants, T2 ΔrH1
ΔrH2
How to Calculate ΔrH as Function of T
Each enthalpy in the reaction equation changes according to
Kirchhoff’s law
TCdTCHd p
T
T
p
E
A
E
A
T
T
pdTCHHdHH 112
And, if Cp = constant over the temperature range of interest
2
1
12
T
T
pTrTr dTCHH
How to Calculate ΔrH as Function of T
If there is a phase transition within the temperature range, it must be
accounted for
E
U
U
A
T
T
pU
T
T
p dTCHdTCHd 21
...2
21
K
Tb
K
TaCC p
Cp as a function of temperature is usually a polynomial expression such as
Consistency check....
Isomerization of butane
ΔrH° = - 7 kJ mol-1
ΔrS° = -15 J mol-1
ΔrG°= - 2.3 kJ mol-1
Isomerization
)()( 104104 gHCigHCn
STHG
Free Enthalpy ΔrG, and Equilibrium Constant K
Relation between ΔrG° and K in equilibrium, ΔrG=0
thr KRTG ln
i
ithiaK
(dimensionless)
L
iirr
iaRTGG1
ln
Thermodynamic equilibrium constant
Composition dependence of ΔrG
correlation between Kth and Kp
Different Equilibrium Constants K
i
ipipK
For low pressures (a few bars and less), the fugacity coefficients are about 1All pressures, including po should be in the same units.
[Pai]
L
ii
L
iioth
iii fppK
poth KpK i
Kp
With and
Isomerization of butane
ΔrG°= - 2.3 kJ mol-1
Isomerization Equilibrium
53.2RT
G
th eK
xv
p KpK i
at 298 K
)()( 104104 gHCigHCn
28 % 72 %
poth KpK i
Equilibrium Constant Temperature Dependence
2
ln
RT
H
T
K
p
van’t Hoff’s Equation
.ln constRT
HK p
Indefinite integration
211,
2, 11
TTR
H
K
K
pTp
Tp Definite integration
Equilibrium Temperature Dependence
Start your research by calculating the thermodynamics of your reaction!
H = const.
0102030405060708090
100
200 250 300 350 400 450 500 550 600 650 700
Temperature / K
Frac
tion
%
Isobutane
n -Butane
H= f(T); Cp = const.
0102030405060708090
100
200 250 300 350 400 450 500 550 600 650 700
Temperature / K
Frac
tion
%
Isobutane
n -Butane
Part II: Practical Matters
Vapor pressure and saturators
Saturator, 100 ml Methanol79.17 g, is 2.47 mol
Gas outGas in
Heat Consumed by Evaporation
Assumption: saturator is adiabatic, evaporate 20 ml of
methanol, all energy for evaporation taken from remaining
80 ml methanol
20 ml is about 0.5 mol, need about 17.7 kJ for evaporation
80 ml is about 2 mol, Cp of liquid MeOH is 81.6 J mol-1 K-1
The temperature of the methanol would theoretically drop
by 108 K
The Clausius-Clapeyron Equation
VT
H
V
S
T
p
coex
.
General differential form of the Clausius-Clapeyron Equation
2. RT
Hp
T
p
coex
For sublimation and evaporation
assumes ideal behavior of the gas phase
21
11ln
1
2
TTR
H
p
p
T
T August’s vapor pressure formulaassumes enthalpy is constant within given temperature range
Vapor Pressure and Temperature
At 64.4°C, the vapor pressure of methanol is 755 torr
and the enthalpy of evaporation is 35.4 kJ mol-1
T1 = 337.6 K, p = 100.66 kPa
21
12
11
TTR
H
TT epp
The carrier gas will dissolve in the liquid and the vapor pressure will be lowered
Methanol Vapor Pressure
H assumed constant
0
50
100
150
200
250
300
350
280
290
300
310
320
330
340
350
360
Temperature / K
Va
po
r P
res
su
re /
kP
a
H assumed constant
0
5
10
15
20
25
30
285
287
289
291
293
295
297
299
301
303
Temperature / K
Va
po
r P
res
su
re /
kP
a
Small temperature changes can cause significant changes in vapor pressure