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