rudolf Žitný, Ústav procesní a zpracovatelské techniky Čvut fs 2010 this course is...
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Rudolf Žitný, Ústav procesní a zpracovatelské techniky ČVUT FS 2010
This course is approximately at this level
CHEMISTRY E182019 CH8
Reaction equilibriaRate of chemical reactions
CHEMICAL REACTIONSCH8
Typical reactions
Addition C+O2 → CO2
Decomposition CaCO3→CaO+CO2 (calcium carbonate)
Neutralization H2SO4+2NaOH→Na2SO4+2H2O (sodium sulfate)
Reversible C2H4+H2O↔C2H5OH (ethylalcohol)
REACTION PROGRESSCH8
0 0 0 0
0 0 0 0[ ]
[ ] i-th component in the R-th reaction
kA B C D
A A B B C C D D
A B C D
A A B B C C D D
A B C D
iR iR R
A B C D
n n n n n n n n
c c c c c c c c
V
dc d
During chemical reaction the number of moles of participating reactants and products are changing according to stoichiometry of reaction. These changes and also corresponding molar concentrations are expressed by only one scalar value
REACTION PROGRESS
Changes of concentrations depend only upon the changes of reaction progress and are independent of initial concetrations
[A] means molar
concentration of A therefore it is the same as
cA
nA-nA0 corresponds only to one reaction. In the case of more
reactions with the component A it is necessary to sum up of all reactions.
REACTION RATECH8
1 [ ]
1 1i i
i i
d dr
V dt dtdn dc
rV dt dt
Reaction progress is a positive parameter (concentration of reactants decreases during reaction) and its value increases from zero up to a limiting value at equilibrium. This increase is a monotonous function and therefore the reaction rate is also positive and monotonically decreasing towards zero
for i= A,B,C,D
RATE EQUATIONCH8
Reaction rate of forward reaction depends upon concentration of reactants, temperature (reaction rate always increases with temperature) and to a lesser extent also upon pressure. It is independent of concentrations of products!
[ ]A Bm m
A B
dr kc c
dt
Rate coefficient (dependent upon temperature)
Reaction order with respect A (it need not be an integer or even positive number for complicated reactions)
Overall reaction order m = mA + mB
ARRHENIUS LAWCH8
Reaction constant depends upon temperature according to Arrhenius law
aE
RTk Ae
Activation energy of chem.reaction
J/mol
Preexpontial factor (the same unit as
k)
Relative amount of molecules having kinetic energy greater than Ea is given by Maxwell distribution of energies /E RTe
The greater is temperature the more molecules have energy greater than E
Collision theory: only the molecules having energy greater than Ea react at mutual collision
ARRHENIUS LAWCH8
Reaction rate can be increased either by the temperature increase or by decreasing activation energy
aE
RTk Ae
Preexponential factor A is not a constant and slightly depends upon temperature.
Activation energy of chem.reaction can be decreased by catalyst
(changing reaction mechanism with intermediate transition complex)
Ea and CATALYSISCH8
Activation energy Ea is a barrier which must be overcome so that the colliding molecules can react. The activation energy Ea depends on the bonding energy of the so called activated complex, a temporary molecule having only partially formed bonds. The smaller the energy of the activated complex the higher the probability that a collision will result in a chemical change. There are usually many ways to decompose a summary chemical reaction into intermediate elementary reactions (e.g., decomposition of reactants to free elements - radicals and subsequent formation of products). Every elementary reaction has its own activation energy and the actual reaction mechanism (sequences of elementary reactions) leads through the valley of the lowest activation energies. Sometimes species not explicitly enumerated in lists either of reactants or products take part in the intermediate reactions. These species, which are not consumed in the overall chemical reaction, are called catalysts. Catalysts open other possible reaction paths, characterised by lower activation energies, and thus their presence increases the overall reaction rate.
Monomolecular reactionCH8
AA
A
A
AB
C
AB+C
dc
dtkcA
A lnc
cktA
A0
Reaction of the first order, exponentially decreasing concentration of reactant. I do not know why the decomposition of A could not be initiated by collisions with products B and C.
Bimolecular reactionCH8
AA B
dckc c
dt
Reaction is of the second order.
Precious information of bimolecular reaction data are available at NIST
AA
A
A
A
A
A
A
A A
A
A
A
A
A
B
B
B
B
B
BB
C
B
C
C
A
ABB
A
B B
A
A
BC
A + B CNothing happens at A-A or B-B collision
Nothing happens at low energy collision
Only if A-B energy of impact is > Ea the components react
Reversible reactionsCH8
1 2
1 2
/ /1 2
1
2
E RT E RTA B C D
E EC D RT
A B
Ae c c A e c c
c c AK e
c c A
AA
A
A
A
A
A
A
A A
A
A
A
A
A
B
B
B
B
B
BB
C
B
C
C
A + B C + D/
11
dc E RTC A e c cA Bdt
/2
2
dc E RTC A e c cDCdt
A
BC
AB
C
At equilibrium the rate of forward reaction is the same as the rate of backward reaction
K-equilibrium constant (will be discussed later)
Forward reaction
Backward reaction
Calculation of concentrationsCH8
A B C DA B C D Problem: Given initial concentrations cA0, cB0, cC0, cD0 calculate evolution of concentrations at time for specified temperature and pressure. During reaction T,p is assumed constant and only number of moles of participating reactants and products are changing according to stoichiometry of reaction.
Solution: Concentrations in rate equation must be expressed in terms of reaction progress
0
0
A A A
B B B
c c c
c c c
0 0( ) ( )a
A B
Em mRT
A A B B
dcAe c c c c
dt
Ordinary differential equation for unknown c with initial condition c=0 must be solved
numerically for real reaction orders
Calculation of concentrationsCH8
Analytical solution exists for the case of bimolecular reactions when mA=mB=1. Let us assume unit stoichiometric coefficients for simplicity
0
0
A A
B B
c c c
c c c
0 0
0 0
0 0 0 0
0 00 0
0 0
0 00 0
0 0
( )( )
( )( )
1 1( )
( )
ln ln ( )
ln( ) ( )
A B
c
A Bo
c
A B B Ao
A BB A
A B
A BB A
B A
dck c c c c
dt
dckt
c c c c
dckt
c c c c c c
c c c cc c kt
c c
c c cc c kt
c c c
( )
( )0
( 1)( )
Bo Ao
Bo Ao
k c c tAo Bo
k c c tAo B
c c ec t
c e c
Final result (time course of concentrations)
Calculation of concentrationsCH8
What to do if reactants A,B are in stoichiometric ratio?
cAo=cB0
1Ao
A BAo
cc c
c kt
( )
( )0
0
( 1)( )
( )
( ) ( ) 1
Bo Ao
Bo Ao
k c c tAo Bo
k c c tAo B
Ao Bo Bo Ao Ao Bo
Ao Bo Ao B Ao Ao
c c ec t
c e c
c c k c c t c c kt
c k c c t c c c kt
Use Taylor’s expansion (only linear term)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 10 20 30 40
t [s]
c
Reaction progress
Stoichiometric ratio
cB
cA
Calculation of concentrationsCH8
What are equilibrium concentrations at t ?
If cB0>cA0
cA=0
cB=cB0-cAo
cC=cC0+cAo
cD=cD0+cAo
else
cA=cAo-cBo
cB=0
cC=cC0+cB0
cD=cD0+cBo
C D
A B
c cK
c c
Equilibrium constant is infinitely large, because it is not reversible reaction
Tutorial NO2 reduction (1/2)CH8
2 2NO CO NO CO
21 2
[NO ][NO ] [CO]
dk
dt
Rate equation for bimolecular equations
holds only at high temperatures above 225 oC and at lower temperatures different rate equation should be applied
222 2
[NO ]2 [NO ]
dk
dt
Tutorial NO2 reduction (2/2)CH8
2 2
2 2
3
3
NO NO NO
CO
NO
CO + ON NO
Overall reaction can be decomposed to two reactions
Slow reaction
222 2
[NO ]2 [NO ]
dk
dt
Fast reaction. Resulting reaction rate determines the
slowest reaction in the reaction chain
Tutorial NO-removal (1/4)CH8
2 2 22H +2NO N +2H OThis overall reaction assumes simultaneous collision of 4 molecules. This is improbable and therefore overall reaction is substituted by several simpler reaction steps
1
2
3
4
212 2 1 1
22 2 2 2 2 2
32 2 2 2 2 3 3 2 2 2
2 2 2 2
[ ]2 [ ]
[ ]2 [ ]
[ ] [ ][ ]
k
k
k
k
dNO N O r k NO
dtd
N O NO r k N Odtd
N O H N O H O r k N O Hdt
N O H N H O r
44 4 2 2
[ ][ ][ ]
dk N O H
dt
Tutorial NO-removal (2/4)CH8
1
2
3
4
212 2 1 1
22 2 2 2 2 2
32 2 2 2 2 3 3 2 2 2
2 2 2 2
[ ]2 [ ]
[ ]2 [ ]
[ ] [ ][ ]
k
k
k
k
dNO N O r k NO
dtd
N O NO r k N Odtd
N O H N O H O r k N O Hdt
N O H N H O r
44 4 2 2
[ ][ ][ ]
dk N O H
dt
Production of intermediate compounds N2O2 and N2O are determined from previous rate equations. Assuming that the production rate of these intermediates is fast and close to equilibrium, these concentrations can be calculated without necessity to solve differential equations
222 2 1
2 2 2 1 3 2 2 2 2 22 3 2
21 32
3 2 2 2 4 2 2 24 2 3 2
[ ] [ ][ ] [ ] [ ][ ] 0 [ ]
[ ]
[ ][ ][ ][ ] [ ][ ] 0 [ ]
( [ ])
d N O k NOk N O k NO k N O H N O
dt k k H
k k NOd N Ok N O H k N O H N O
dt k k k H
Tutorial NO-removal (3/4)CH8
N2 is produced only from the last reaction
4 42 2 2 2 4 4 2 2
[ ] [ ][ ]k d
N O H N H O r k N O Hdt
The rate equation for the overall reaction
21 3
24 2 3 2
[ ][ ]
( [ ])
k k NON O
k k k H
21 3 22
4 2 22 3 2
[ ] [ ][ ][ ][ ][ ]
[ ]
k k NO Hd Ndr k N O H
dt dt k k H
2 2 22 2 2H NO N H O
Tutorial NO-removal (4/4)CH8
Assuming that the first two reversible reactions are at equilibrium
And therefore rate equation for the overall reaction
22 14 2 2 3 2 2 2 3 2
2
[ ][ ][ ][ ] [ ][ ] [ ] [ ]
d N kdr k N O H k N O H k NO H
dt dt k
2 2 22 2 2H NO N H O
1
2
22 2 1
2 2 2 2 2
21 2 2 2
[ ]2 2 [ ]
[ ]2 2 [ ]
[ ] [ ]
k
k
d NONO N O k NO
dtd NO
N O NO k N Odt
k NO k N O
Conclusion: Reaction is of the second order with respect NO but only of the first order with respect H2.
Tutorial HCl (1/4)CH8
Production of hydrochlorid acid
Cl2 H2
Water sprayGaseous HCl
Hydrochlorid acid
Combustor (reactor)
Tutorial HCl (2/4)CH8
2 2H +Cl 2HClThis overall reaction assumes simultaneous collision of 2 molecules and looks like a standard bimolecular reaction. However, actual reaction mechanisms is more complicated
1
2
3
4
12 1 1 2
222 2 2
32 3 3 2
42 4
[ ]2 [ ]
[ ]2 Cl [ ]
[ ] [ ][ ]
[ ]
k
k
k
k
dCl Cl r k Cl
dtd
Cl r k Cldt
dCl H HCl H r k Cl H
dtd
H Cl HCl Cl r
4 2[ ][ ]k H Cldt
Tutorial HCl (3/4)CH8
Intermediate radicals Cl and H react so fast that their equilibrium values can be assumed
2 11 2 2 3 2 4 2 2
2
3 2 13 2 4 2 2
4 2 2
[ ]2 [ ] 2 [ ] [ ][ ] [ ][ ] 0 [ ] [ ]
[ ][ ][ ][ ] [ ][ ] 0 [ ] [ ]
[ ]
kd Clk Cl k Cl k Cl H k H Cl Cl Cl
dt k
k H kd Hk Cl H k H Cl H Cl
dt k Cl k
1
2
3
4
12 1 1 2
222 2 2
32 3 3 2
42 4
[ ]2 [ ]
[ ]2 Cl [ ]
[ ] [ ][ ]
[ ]
k
k
k
k
dCl Cl r k Cl
dtd
Cl r k Cldt
dCl H HCl H r k Cl H
dtd
H Cl HCl Cl r
4 2[ ][ ]k H Cldt
subtract this term
Tutorial HCl (4/4)CH8
HCl is produced in the third and fourth reactions
3 2 4 2
[ ][ ][ ] [ ][ ]
d HClk Cl H k H Cl
dt
The rate equation for the overall reaction
13 2 2
2
[ ] [ ]2 [ ] [ ]
kd d HClr k H Cl
dt dt k
2 2 2H Cl HCl
First order reaction with respect hydrogen, but only 0.5 order reaction with respect chlorine.
REACTION EquilibriumCH8
A B C DA B C D
1 2
1
2
E EC D RT
A B
c c AK e
c c A
It was demonstrated that for reversible reactions there exists an equilibrium constant, determining concentrations of all participating components
This relationship follows from equality of reaction rates of forward and backward reactions expressed in terms of activation energies E and preexponential factors.
Equilibrium constant K can be derived in a different way, without resorting to reaction rates, just from a general requirement of equilibria at constant pressure:
0reactionG Gibbs reaction change at given pressure p and temperature T
REACTION EquilibriumCH8
, , ,
0reaction i ii A B C D
G g
Molar Gibbs energy of component i at pressure p and temperature T
0 0
, , , , , ,
( )i i i i ii A B C D i A B C D
g g g
0 0,
, , , , , ,
( ) ( )i i i i f i Ti A B C D i A B C D
T s s g
Gibbs energies of formation at standard pressureg=h-Ts
How to calculate entropy changes, corresponding to
partial pressure of i-th component?
Molar Gibbs energy of component i at standard pressure p and temperature T
REACTION EquilibriumCH8
0
0 0
ln lni ii i
v ps s RT RT
v p
Derived previously for ideal gas
0,
, , , , , ,0
ln ( )ii i f i T
i A B C D i A B C D
pRT g
p
0
, , , 0
ln ( ) iireaction
i A B C D
pRT G
p
Partial pressure of component at equilibrium
REACTION EquilibriumCH8
0
, , , 0
ln ( ) iireaction
i A B C D
pRT G
p
Kp - equilibrium constant expressed in terms of partial pressures of participating componentsSpecial case
1 1A B C D
C D C Dp
A B A B
p p c cK
p p c c
01 2
1
2
reactionG E E
RT RTp
AK e e
A
Gibbs energy change for FORWARD reaction at
standard pressure 100kPa
REACTION EquilibriumCH8
01 2
1
2
reactionG E E
RT RTp
AK e e
A
Consequencies:
Negative value of G increases equilibrium constant and shifts equlibrium composition towards the forward reaction products.
Equilibrium composition is independent of applied catalyser (catalyser decreases activation energies Ea, but has no effect upon Gibbs energy). Catalyser only shortens the time necessary for equilibrium achievement.
REACTION EquilibriumCH8
Le Chatelier's principle:
The chemical equilibrium shifts in a way that tends to undo the external stress.
Mother Nature does not like sudden changes, and promotes that reaction (forward or reversal) which helps to restore the previous state. Examples:
A temperature increase shifts the equilibrium towards the endothermic reaction, which consumes the superfluous heat.
If the volume of the products is less than the volume of the reactants (i<0), the total pressure decreases. Therefore the increase of pressure increases the equilibrium constant.
Tutorial Equilibrium steam reformingCH8
2 2 2CO+H O CO +HCalculate equilibrium constants if the molar fraction of carbon dioxide (CO2) in the equilibrium mixture is 27%.
Initial composition Final composition
2 2 2 2
2 2
2
21.378
(1 )CO H CO H
CO H O CO H O
c c n nK
c c n n
2
2 2
1 , 1
0, 0
CO H O
CO H
n mol n mol
n n
2
2 2
1 , 1
,
CO H O
CO H
n n
n n
2
2
2 2 22
COCO
CO H O H CO
ny
n n n n
22 0.54COy
Tutorial Equilibrium steam reformingCH8
2 2 2CO+H O CO +H
Theoretical calculation of equilibrium constant
0
2 2 2 2
2 2
2
2(1 )
GCO H CO H RT
CO H O CO H O
c c p pK e
c c p p
2 2 2
0 0 0 0 0 0298 , 298 , 298 , 298 , 298 , 298( ) ( ) ( ) ( ) ( ) ( )
393.5 0 110.5 + 241.8 41.2 kJ/mol CO
i f i f CO f H f CO f H Oi
H h h h h h
2 2 2
0 0 0 0 0 0298 , 298 , 298 , 298 , 298 , 298( ) ( ) ( ) ( ) ( ) ( )
394.4 137.1 228.8 28.7 kJ/mol CO
i f i f CO f H f CO f H Oi
G g g g g g
2 2 2
0 0 0 0 0 0298 298 298 298 298 298
-1 1
( ) ( ) ( ) ( ) ( ) ( )
0.21369 0.13057 0.19754 0.18872 0.0421 kJ.(mol CO) .
i i CO H CO H Oi
S s s s s s
K
Tutorial Equilibrium steam reformingCH8
2 2 2CO+H O CO +H0 0
298 298( ) ( ) 41.2 0.0421
0.008314
H T S T
TRTpK e e
JANAF tables (NIST) NSRD-NBS-
37
Example Kp=1.378
Tutorial Equilibrium steam reformingCH8
2 2 2CO+H O CO +H0
298( ) 28.7
0.008314
G
TRTpK e e
0.1
1
10
100
1000
10000
100000
200 300 400 500 600 700 800 900 1000 1100 1200
T [K]
Kp
this only very rough approximation. It is better to
use entropy and enthalpy