1. Rate Laws and stoichiometry

2. Rate Laws: basic definitions

3. Rate laws and stoichiometry
What we will see:
Order of reaction and molecularity
Elementary and nonelementary rate laws
Reversible and irreversible reactions
Reaction rate constant – Arrhenius factor and activation energy
Stoichiometric tables for batch and flow systems

4. Relative Rates of Reaction
aA + bB  cC + dD
Which is the relationship among (-rA) and (-rB), (rC) and (rD) ?

5. The rate law expression
The algebraic equation that relates -rA to the species concentrations is called the kinetic expression or rate law
(-rA) = [k Af (T)] * [f´(CA, CB, ..)]
Reaction rate constant

6. Dependence of the reaction rate on the concentrations of the species
Reaction: aA + bBcC + dD
-rA = k CAa CBb Power Law model
a = reaction order with respect to reactant A
b = reaction order with respect to reactant B
n = a + b = overall order of the reaction
If a = a e b = b, the reaction is «elementary»
Elementary
H2 +I2  2HI -rHI = kCH2 CI2
Non-elementary
CO + Cl2  COCl2 -rCO = k CCO CCl2 3/2

7. UNITS OF k (rA) = k * [fn(conc)]
Units of k = units of (rA )/ units of [fn(conc)]
Zero-order reaction
(rA) = k k in mol/m3 ·s
First-order reaction
(rA) = kCA k in s-1
Second-order reaction
(rA) = kCA k in m3/mol ·s

8. Elementary reactions and molecularity
We say that a reaction follows an ELEMENTARY RATE LAW when the reaction orders are identical with the stoichiometric coefficients of the reacting species for the reaction as written
The MOLECULARITY of a reaction is the number of atoms, ions, or molecules involved (colliding) in a reaction step
The terms unimolecular, bimolecular, and termolecular refer to reactions involving one, two, or three atoms (or molecules) interacting or colliding in any one reaction step

9. Nonelementary Rate Laws
A large number of both homogeneous and heterogeneous reactions do not follow simple rate laws [ = k(T) * f(conc.) ]:
Reactions involving radicals
Catalytic reactions
Enzymatic reactions
Biomass reactions

10. Apparent reaction order
Sometimes reactions have complex rate expressions that cannot be separated into solely temperature-dependent and concentration-dependent portions
We can only speak of reaction order under certain limiting conditions:
at very low concentrations of oxygen, the reaction would be "APPARENT" first order with respect to N2O
if the concentration of O2 were large enough, the apparent reaction order would be -I with respect to O2 and first order with respect to N2O giving and overall apparent zero order

11. Reversible reactions All reactions are …. reversible!!!
Reaction aA + bB  cC + dD
At equilibrium, the rate of reaction is identically zero for all species (i.e., - rA= 0 )
All rate laws for reversible reactions must reduce to the thermodynamic relationship relating the reacting species concentrations at equilibrium

12. Rate laws for rev elementary reactions
Reaction: aA + bB cC + dD
krev
kfor
Rate of consumption of A for the forward reaction (-rA_forward)
= kfor CAa CBb
Rate of formation of A for the forward reaction (rA_forward)
= – kfor CAa CBb
Rate of formation of A for the reverse reaction (rA_reverse)
= krev CCc CDd
Net rate of formation of A = (rA_reverse) + (rA_forward)

13. Rate laws for elementary reactions
Net rate of formation of A (rA)net = (rA_reverse) + (rA_forward)
= krev CCc CDd – kfor CAa CBb
At equilibrium (rA)net MUST be =0
krev CCc CDd – kfor CAa CBb = 0

14. Temperature dependence of Kc
From the integration of the van’t Hoff equation

15. The reaction rate constant (k)
(-rA) = kA * [fn(con)]
where
kA = f(T)
ln(k)
1/T
Rxn1: High Ea
Arrhenius equation
k = A exp(-Ea/RT)
A = pre-exponential factor or frequency factor
Ea = activation energy, J/mol or cal/mol
R = gas constant = J/mol · K = cal/mol · K
T = absolute temperature, K
Rxn 2: Low Ea
ln( k) = ln A - (Ea/R) x 1/T

16. Why is there an activation energy?
Molecules need energy
to distort or stretch their bonds so that they break and form new bonds
to overcome the steric and electron repulsive forces as they come close together
One way to view the barrier to a reaction is through the use of the reaction coordinates
potential energy of the system as a function of the progress along the reaction path (see next slide)
Energy of formation of the reactants  literature
Energy of formation of the transition state  QM calculation (Gaussian)

17. Activation energy (Ea)
Activation energy: can be thought of as a barrier to energy transfer between reacting molecules
A+B
C+D Ea
HR
Reaction coordinate
Poetential energy

18. Example - Determination of the Activation Energy
Calculate the activation energy for the decomposition of benzene diazonium chloride to give chlorobenzene and nitrogen
Using the following information for a first-order reaction:

20. T dependence of the reaction rate
The larger the Ea, the more temperature-sensitive is the rate of reaction
Reference values (order of magnitudine)
A: s-1
E: 300 kJ/mole
Correlations exist to relate the activation energy to the heat of reaction
Activation energies can be estimated from first principles (molecular modelling)

21. Factors affecting the reaction rate
Temperature (T)
Heat transfer
Pressure/Temperature
Catalyst
Concentration (C)
Mass transfer
rA = f (T, C)

22. Effect of P & T on concentration
In gas phase
EOS for real gas: PV = z n RT con C = n/V = P/zRT
It is clear that in gas phase concentration is sensitive to P and T
In liquid phase
Concentration = moli per volume unit = N/V
The density of a liquid DOESN’T change much with T
Liquids possess low compressibility – effect of P on density is small
Concentrations in liquid hase are «generally» not affected by changes of P and T

23. WHERE WE ARE….

24. Design equations for isothermal reactors
REACTOR DIFF. EQN ALGEBR. EQN INTEGRAL FORM
BATCH
CSTR
PFR

25. Design equations for isothermal reactors
From the previous table we see that if we know the relationship –rA = g(XA) then it is possible to size batch, CSTRs, PFRs (and PBRs) operating at the same conditions under which -rA = g(X) was obtained
In general the rate law is known -rA = k [f(Ca, Cb,…)]
Thus, if we know Cj =hj(X)
then in fact we have -rA as a function of X and this is all that is needed to evaluate the isothermal design equations

26. stoichiometric tables
Batch systems
Constant volume reactions
Flow systems
Variable volume reactions

27. Stoichiometric tables
A stoichiometric table presents the stoichiometric relationships between reacting molecules for a single reaction.
t = t NA NB NC ND NI
t = 0
NA0
NB0
NC0
ND0
NI0
Batch reactor

28. Stoichiometric tables for Batch Reactors
Reaction:
Concentration in batch reactors:

29. Stoichiometric tables for Batch Reactors
From stoichiometric table we have
Ni = f(X)
The goal is to find the value of V needed to obtain a given concentration
For constant volume reactors:
V = VR = VO (gas phase reaction)
V = VO (liquid phase reaction)
For constant volume reactors (gas phase reaction):
V has to be expressed as a function of P, T
The increment in the number of moles has to be considered

30. Stoichiometric tables for Batch Reactors VARIABLE VOLUME
Ideal gas
PV = ZNT RT
POVO = ZO NTO RTO
Def. “d” – change in the total number of moles per mole of A reacted
NT = NTO + d NAO X
Def. “e”
Change in total number of moles for complete conversion
Total moles fed

31. Stoichiometric tables for Batch Reactors
Reaction:
Concentration in batch reactor:

34. Stoichiometric tables for FLOW Reactors
Reaction
Concentration in flow reactor:
The form of the stoichiometric table for a continuous-flow system is virtually identical to that for a batch system except that we replace Nj0 by Fj0 and Nj by Fj

35. Concentrations for flow reactors
From the stoichiometric table, we have
Fi = f(X)
which value of the volumetric flow rate v, should we use?
Liquid phase reactions
v = vO
Gas phase reactions
For isothermal and isobaric reactors, without change in the number of moles
v = vo

36. Concentrations for flow reactors – volume change
Per gas ideale
PV = ZNT RT
POVO = ZO NTO RTO
CT = FT /v = P/(ZRT)
Def. “d” – change in the total number of moles per mole of A reacted
FT = FTO + d FAO X
Def. “e”
Change in total number of moles for complete conversion
Total moles fed

37. Concentrations for flow reactors – volume change
To derive the concentrations in terns of conversion

38. Concentrations for flow reactors – gas pahse, volume change

39. Example
The following reaction (liquid phase) is irreversible and follows an elementary cinetic law: A+B  C The reaction is conducted in a PFR of volume V1 dm3 at T1 K and in a CSTR of volume V2 dm3 working at T2 K. The feed is obtained mixing two streams, one containing only A, the other only B. The concentration of each reactant in the stream is CO mol/L and the total volumetric flow entering the reactor is vo dm3 /min. Find the relationship needed to calculate which X can be achieved in each reactor.
Further information: at T1 K, k = 0.07 dm3/mol-min Ea = J/mol-K
ATTENZIONE, mancano alkcuni dati numerici che sono forniti durante l’esercizio

40. Steps Draw a scheme containing all important information
Write the equations:
Rate law as a function of concentration
Velocity constant as a function of T and Ea
Express concentrations as a function of X
Design equation for your reactor
Solve for X

41. Solution A A PFR V1 ; T1 B CAO = 0.5 CO CSTR CBO = 0.5 CO B V2 ; T2
0.5 vO CO A
CAO = 0.5 CO
CBO = 0.5 CO vO
0.5 vO CO A B
PFR V1 ; T1
CSTR
V2 ; T2
CAO = 0.5 CO
CBO = 0.5 CO vO

42. Solution A+B  C 1. Rate law (Elementary)
(-rA) = k CA CB = (-rB) = (rC)
We need k at T1 e T2 ; CA and CB expressed as function of X
2. Velocity constant (k) k = A exp(-Ea/RT)
k a T1 = k1 = A exp (-Ea/RT1)
= 9.1 L/mol.min
3. Concentration as a function of X (Stoichiometric tables)
CA = CAO (1-X)
CB = CAO (B –b/aX) = CAO (1-X) [ B = 1]

43. Solution 4.Reaction rate as a function of X
dFA=FA0 dXA= v0 CA0
4.Reaction rate as a function of X
(-rA) = k CAO2 (1-X)2 = (-rB) = (rC)
5. Design equation from the mole balance
5.6
182
XPFR = 0.84
XCSTR = 0.928

44. Expressing concentration as a function of X

45. Expressing concentration as a function of X

47. …