LHHW Kinetics

Langmuir-Hinshelwood Hougen-Watson (LHHW) Kinetics Example

2 CHCl3 + 2 H2O + O2 → 2 CO2 + 6 HCl

kC A   g   rate = 3  cm s  1 + K P CP + K A C A where:

-1

9

(-21,700/RT)

k (s ) = 0.372 x 10 e 3 7 (+2440/RT) KP (cm  /mol) = 0.597 0.597 x 10 e 3 7 (+5330/RT) KA (cm  /mol) = 0.123 0.123 x 10 e 3 CA (mol/cm ) = concentration of CHCl 3 3 CP (mol/cm ) = concentration of HCl  Aspen LHHW kinetics formulation: The rate expression for catalytic reactions can be modeled using the Langmuir. Isotherm which consists of three fundamental steps. 1. The adsorption of the reactants to the catalyst surface. This term is proportional to the surface available for adsorption. 2. The surface reaction for reactants to form products. This term can be represented by a simple power law expression. 3. The desorption of the products from the catalyst surface to the surrounding environment. This term is proportional to the surface already adsorbed. At steady state, the rate of adsorption of reactants is equal to the rate of desorption of  products. From this information the following generic expression can be developed.

rate = (kinetic factor )

(driving force expression ) (adsorption term)

The kinetic factor represents the surface reaction, and is represented by the expression.  E   1

n

1   

  T    R   T − T    kinetic factor = k   e  To   o

The driving force expression representing step 3 (desorption of product from the catalyst surface, whose rate is directly proportional to the global rate of reaction) is given by.

driving force = K 1

∑C

ai i

− K 2 ∑ C j

bj

The adsorption term represents step 1 ( the adsorption of reactant to the catalyst surface, which is indirectly proportional to the global rate). It is expressed by the following:

adsorption term =

(∑ K ∏ C )

ν j m

i

 j

Both the driving force expression and adsorption term are defined by equilibrium constants. For example the equilibrium constants for the adsorption term help define the chemisorption of reactants to the surface of the catalyst (chemisorption being characterized as an exothermic reaction between surface site and the adsorbate). These equilibrium constants are of the form: ln(K) = A + B/T + C ln(T) + D T

1

Specifics for this example: 1) kinetic factor  n [(E/R)(1/T-1/To)] kinetic factor = k [(T/To) ] e n (E/RT) If To is not specified then kinetic factor = k[T ] e For n=0, kinetic factor= k exp (E/RT) 9

(-21,700[cal/mol]/RT)

For this problem, kinetic factor = 0.372 x 10 e

This kinetic expression needs to be converted to SI units since the pre-exponential factor is always in SI units. The expression also needs to be converted from a weight of catalyst to a mole of reaction basis. Here 5 g of catalyst produces a kmol of reaction. 9

(-21,700[cal/mol]/RT)

3

3

3

kinetic factor = [0.372 x10 e ] / (1000 g/kg) x (100 cm  /m ) / (5g / kmol) 9 (-90854000[J/kmol]/RT) kinetic factor (m3/kg-s) = 0.372 x10 e x 200 10 (9.0854E7/RT) kinetic factor (m3/kg-s) = 7.44x10 e In Aspen Plus, the form of the equation is kinetic factor = k exp (E/RT) Hence, the inputs are as follows: the preexponential factor and the Activation Energy

10

k = 7.44 x10 E = 90854000 J/kmol

2) Driving Force Aspen Plus calls the first product sum term1 and the second term2. driving force expression = term1 - term2 For this problem, driving force expression = (Ca)

In the Aspen Plus form of the driving force expression: driving force expression = K 1(CA) - K2(CA) Term1 = C A where K1 = 1 Term2 = 0 where K1=0 driving force expression

= 1 (C A) - 0 (CA)

Since K is in a log form ln(K) = A + B/T + C ln(T) + D T For term 1: ln 1 = 0 where A = 0 For term2: ln 0 = -infinity Since ∞ cannot be entered, it is approximated as -100. (ln 0 ~ -100 where A = -100) 3) Adsorption term Any number of terms are possible. For this problem there are three terms and the adsorption expression exponent m = 1. m adsorption term = (term1 + term2 + term3)

2

For term1: K ln 1 ln K

=1 =0 =A

For term2:

7

(+2440/RT)

7

(+5330/RT)

KP = 0.597x10 e 7 ln(KP) = ln (0.597 x10 ) + (2440/1.9872/T) = 15.602 + 1228/T ln K = A + B / T For term3:

KA = 0.123 x10 e 7 ln(KA) = ln (0.123 x10 ) + (5330/1.9872/T) = 14.023 + 2682/T ln K = A + B / T where r k T To n E R C m K1, K2, Ki v i, j A,B,C,D

= = = = = = = = = = = = =

rate of reaction pre-exponential factor absolute temperature reference temperature temperature exponent activation energy gas law constant Component concentration Adsorption expression exponent Equilibrium constants Concentration exponent Component Index Constants

Summary of Aspen Plus LHHW input: Kinetic factor: k = 7.44 e10 E = 90854000 J/kmol driving force expression: Term 1 CHCL3 exponent = 1 A = 0 Term 2 CHCL3 exponent = 1 A = -100 adsorption term Adsorption expression exponent = 1 Term 1: =0 Term 2 HCl exponent = 1: A = 15.602 B = 1228 Term 3 CHCl 3 exponent = 1: A = 14.023 B = 2682

3