Studio dei parametri cinetici

Non-isot -> converse

repeating

pieces

Approximation on kinetics

X conversion

Y yield

S selectivity

K.P.I. = Criteria of

Xi

i: is a cause

something that

is detecting,

commoning

is a key reagent

(more expensive)

amount of consumed

amount of react. at iniz

open reactor

fractional conversion 100

perc. conversion

We can pass from RB (Ni, Ni', Ci...) -> RB {xi} but xi is referring to single reaction

NIR = 1

it is v

eeer

RB batch, r = kcA

XA = (C_A⁰ - C_A(t)) / C_A⁰

XA = 1 - xA(t)

C_A⁰ = C_A⁰(1-xA) = C_A(t) - C_A⁰(1-xA)

replacement RB

RBA (xA)

dX_A(t) / dt = k(1-xA)

xA(0) = 0

xA ↔ CA

R = kCA = kCA⁰(1-xA)

exp. A + B -> .

R = k CA CB

= k CA^a(1-xA)(C₀ - a*CA xA)

d xA/dt = k(1-xA)(C₀-a*xA) = x e n(t xA) -> Not simple to solve

Not use it when there are more reactions

xA = No use it when there are more reactions

We can have

xA consumed => instantaneous velocity

xA produced => distance velocity

xA instead/delta t

xA = dCA/C₀

so = rx dt / C₀

remember that i of xi isnt a product

Yi (yield) = amount produced of i / amount consumed of reactant

product at the end of reaction / at end-of(i) at the beginning thus feed

or feed

reactant

Y_k2,cv = N_k2^out - N_k2ⁱⁿ / N_cvⁱⁿ - N_cv^out = 2.7 / 4 so we need |vout/vin| of virtual reactor that converts all to k2

maximum?

0 minimum

1 mole in

sr 3 + W5 1 - 4

react 4

maximum values of k2 that can produce

Delta N_k2 / -Delta N_CH4 = 1/4 = 2.7/4 = 0.6 60%, CH4 forms only 60% of k2 we expect 20%

Y_i,j = closed

Ni(t) - Ni(0) / Nj(0) - Nj(t)

of same reaction connecting i to j

FRA site open =

Ni^out / Niⁱⁿ

Njⁱⁿ - Nj^out

commuted values

if we set t small or dt x t we can identify intermediate y

Y_i,j^int closed | Delta Ni / Delta Nj = -

NI = 1

Ne = Ne R

eangy years

if Y_k2,ca1 = 3.2 / 4 = 3/8 so its not true 1 on 1 as we expected and second are 4

there are at least two reactions that are simultaneously

Sig (selectivity)

Both products

amount of produced

amount of f produced

OPT 1

N_i(t)

N_j(t)

OPT 2

N_i(t)-N_i(0)

N_j(t)

OPT 3

N_i(t)-N_i(0)

N_j(t)-N_j(0)

Always multiply

V₁/V₁

SR =>

v_R2=4

V_C0=1

we obtain a conversion of 1/4

the ratio is called

reactivity ratio

example: copolymerization starting from a monomer r₁ = 0.8

r₂ = 1

1/(1+r₁)

Product

the ratio tells us which path is favored and faster than the others

reactants

i (prod)

j (prod)

If I don’t have the reactor how can I obtain VOC and V1 -> cook catalysts

Approximation

simply R and the n_R

We have RC with NR -> reactions occur through a mechanism composed by each consecutive step who is partially in macromechanism and if it is a reactant or product

not elementary - Σ step

Rexp = k(I)C_A + s

unusual, because of reaction’s be reaction

phase

If R = k(I)C_A^1.5 it doesn’t work

the model doesn’t satisfied the data

Int

reactor

CK₃-CKO

-> CH₃ + CO

prop 1

CK₃ + CO -> CH₃CO

prop 2

2 CK₃ -> C₂H₆

chain reaction

Free radical propagation

In case the radical is another radical

1 -> R(+R), 1R₂

Init. = species producing radicals

Propagation

term => 2R -> RS

There is no direct connection between A, B and D

Prop. 1 creates B and prop. 2 creates C0

Overall reaction

The reaction we observe in a batch, where we add heat increasing the temperature (sparks)

1) = c

2) = pt

3) = p2

4) = t

A -> R1 + R2

R1 + A -> B + R3

R3 -> D + R1

2R1 -> E

A -> B + D

R

UiA

Kp1C2B

Kp2CR3

KtCR1²

R = [Ri Kp1 Rp2 Rt]

C = C(t) = [Ca(t) Cr(t) ... ]

A R1 R2 R3 B D E -> species

We have to owed 7 reaction scheme -> tmax = const (unilevel, from one reaction we obtain 2 molecules)

dC/dt = - x1 = sum nuij Rj = - Ri - Rp1

dC1/dt = x1 = sum nuij Rj + R1Ri - Rp1 + Rp2 - 2Rt

C(0) = [u' 0 0 0 0 0 0]

System to solve simultaneously (7 cbs)

Radicals: when consumption and modification are not happening simultaneously, then something in between is created and then consumed

Approximation to get analytical solution -> manipulat[ing] the equation

PSSA

net rate = rate of production - rate of consumption

radical species are short-lived intermediates, concentration of i does not depend on concentration

partial equilibrium

pseudo-steady state hypothesis

if i is very reactive then i is a radical/ion "active" x'i = xci + xei ~= 0

If the consumption is high then x = 0, the production is immediately consumed

dC_i/dt = 0 (steady state) - the concentration of species is no elementary state condition

how could it be? bc it is an approximation under these conditions

radicals are R1, R2, R3 -> {R1, R3}

dC_R1/dt = k1 - R p1 + R p2 - 2 R t ≅ 0

dC_R3/dt = R p1 - R p2 ≅ 0

from second reaction -> R p1 = R p2

in the first reaction -> R1 = 2 R t

R p1 = R p2 -> K p1 C A C E -> K p2 C E C E (C3) = 0 -> C R3 = g(C A, C R1) = kR1/kP2 C A C R1 -> ?

We have removed two reactive balance of the system, obtaining partial information : now R1, R3 are connected to each others

Kp2 RCOO? = Kp2 C2? = Kp2 C1 C2 = Kp2 C2 A -> K' global linear concurrence

R R = C K1 rate of the second propagation = Rp2

i-pose the increase of the product at a given t in a global z.

cons re expanded is consistency with the theoretical expression

A -> B + D

Resp = kA^1.5

R th based on - species - chain - pass through - radical = radicals very reactive

we obtain Rth = k expf

Exxorpe book.

Bromide

R2 + X2 -> 2KX

R2 -> 2K.

X2 -> 2X.

R2 + X2 -> 2X.

R = f(C)

correex

numerically we obtained that:

A and B have asymmetrical behavior but there are radicals below them are not the (radicals, but very small)

i = initial ratio

scale factor 10^-3, 10^-4

C_e1 = 1/

C_e1 << C A

is not compact but is changed

C R1 parallel to C A over 3-4 different orders

CR

R_e > R_a+

R_e > R_e-

With PFR and a CSTR -> _dCa/_dV = ra, _dCa/_dT = kCa and se equilibrate back / CSTR dCa/dt = rxn

(pe) partial equilibrium -> property of exn j { evaluate the assumption that by tends to zero, Rj -> 0, not other reactions, Rj < {R1 R2 R3} where Psa & vti ->

PSSA -> properties of species i

R = (R reactants) - E (products) and stoechiometric correction f(c) product and react

We can assume that R = 0 at partial equilibrium by so: y' = y - y^*

CO + H2O = CO2 + H2

A B B D

PE

Rnet = k' Cac - k'' Cbco = k''(Cac - 1/Keq CbCO)

0 -> k' Cac = k'' CbCO

k'/k'' = Keq = CbCO/Cac

= exp(- dG/RT)

= unknown

CO(equ) = (Ca(V) Cc(V)) / Cb(V) - Keq

PE allows to replace the rtb of B with analytical form f(ca,ac,co) when unknown

PE is valid "locally" at specific T, ?, r, t, p, s

so not use

RDS Rate Determining Step - sequence of items and one of them is controlling the overall rate of the process

The slowest we determine the overall rate, called RDS

If we want the fastest possible, it is convenient act on the slowest step

We study what happens when a fluid reactant will be catalytic

namely the catalyst reduces the energy barrier and thus facilitates the reaction rate

e.g. Pd/Pt... + solid support (e.g. Al2O3, SiO2)

The catalyst is at a reaction catalytic are used -> they help the system to approach the equilibrium and they help the reaction to follow a regular path obtaining wanted products

In our cases catalyst is a solid -> how the fluid reacts with it? - fluid reacts

There are chemical and physical adsorption: adsorption interaction with a surface; the interaction can be strong - physical or chemical - graphene and surface interactions

L-H mechanism -> 1) A + * ↔ A* adsorption/interaction

Langmuir Mechanism

It requires the adsorption and the reaction

1) Ads + A → A*

It is more active so reacts more easily

for continuous law mean molecules volume → how many molecules of A are found in the volume of feed

C^*_A = mols of active sites / surface of solid

moles / cm²

Nb. C = volumetric concentration and C^* = surface concentration

The adsorption process is reversible, so the product can be desorbed

R1 = k₁ C_A C^*_A - k_-1 C_A*

2) B + * ↔ B*

R2 = k₂ C_B C^*_B - k_-2 C_B*

reply: the steps are elementary

3) A* + B* → D* + *

R3 = k₃ C_A* C_B* - k_-3 C_D* C^*_A

4) D* ↔ D + *

R4 = k₄ C_D* - k_-4 C_D C^*_D

the order is 1) then 3) and finally 4)

independency

we are formulating the rp that Rxn 3 (= surface ex) is slowest the rate of generation (1+6+0) in Rtot = R3

we can also consider and by the reverse ads

the catalytic is represented by the reverse ads

R1 ↔ 0 - k₁ C_A C^*_A - k_-1 C_A* → C_A*^* = k₁ C_AC^*_A / k_-1

equilibrium constant and adsorption can be given or assumed, this gives the BTE

C^*_A = f(k₁, k_-1, C_A^ads, C_B^ads, C_in)

substitution on a final D* can be used on Rxn 1

So, we find that: R_ads = R₃ = k₃ C_A* C_B* - k_-3 (C_D*) C^*_A = k₃ k_A^eq,ads C_A k_B^eq,ads C_B (C_in)² - k_-3 k₀^eq,ads C_D (C_in)²

R = f(C_in in the field)

available

clean surface

where all the sites are free, we achieve the saturationincrease of mass over the time, until a constant value

adsorbed reactm

C_t* (i): C_t^T = all active sites interface

We could have the following conditions:

at t > 0 (reaction start)at t = 0 (before the reaction is started)

We need a RB on site

N_t,tot = (N_A + N_ox + N_OH + N_w) : surface free

C_i* = C_i^H + C_i^Oh + C_i^w + C_i^f

How many sites are there?

C_t* = C_t,tot / [1 + Σ_i K_i^eq c_i]

The amount of sites decreases bc ofthe molecule occupation

R₃ = K₃^s K₁^eq K₃^eq c_A c_B - K_-3^eq c_D/ [1 + Σ_i^nc K_i^eq c_i]² C_t,tot²

Observation

So R₃ ≈ r_p,obs = f([c])— E_land

NOT concentration on surface, it exp the concentration of free species

R₃ = k₃ c_A c_B - k_-3 c_D/ [1 + k_A c_A + k_B c_B + k_D c_D]

k₃^* = mixed reaction

arrheniusk₃ k₁ eq k₂ eqnear linear regimeTD

R₃ max

saturation of the surface

Reason:

we are having a catalytic increase for the holes are full even if we increase the amount of A, the rxn doesn’t go on

pre-equilibrate form k₃= exp(-c_A s)exp(-ΔG_ads,A/RT) exp(-ΔG_ads,B/RT)

eqs + ΔG_ads,1,1 + ΔG_ads,2,1

k₃ exp((D₃)/RT)