DASAR-DASAR KATALIS & KATALISIS - Spada UNS

DASAR-DASAR

KATALIS & KATALISIS

DASAR-DASAR KATALIS & KATALISIS

Pendefinisian katalis:

• Katalis merupakan suatu zat atau substansi yang dapat mempercepat reaksi, tanpa terkonsumsi oleh reaksi, namun bukannya tanpa bereaksi.

• Katalis bersifat mempengaruhi kecepatan reaksi, tanpa mengalami perubahan secara kimiawi pada akhir reaksi. Peristiwa / fenomena / proses yang dilakukan oleh katalis ini disebut katalisis.

• Istilah negative catalyst (atau inhibitor) merujuk

kepada zat yang berperan menghambat atau

memperlambat berlangsungnya reaksi.

TENTANG KATALIS

1.Katalis berperan mempercepat reaksi

2.Katalis tidak muncul di dalam persamaan stoikiometri reaksi. Katalis muncul di dalam mekanisme reaksi, serta muncul dalam persamaan kecepatan reaksi.

3.Kuantitas atau banyaknya katalis tidak mengalami perubahan selama reaksi berlangsung.

4.Komposisi kimiawi suatu katalis tidak berubah pada akhir reaksi.

5.Katalis dibutuhkan oleh suatu reaksi dalam kuantitas yang sangat sedikit. Contoh: 1 gram katalis logam Pt dibutuhkan untuk reaksi penguraian 108 liter H

₂

O

₂

.

6.Jika lebih dari 1 (satu) reaksi berlangsung secara

simultan pada saat yang bersamaan, maka pada

umumnya katalis mempengaruhi arah atau selektivitas

TENTANG KATALIS

7.Katalis tidak mengubah atau menggeser kesetimbangan reaksi, termasuk semua sifat termodinamikanya, seperti kecenderungan keberlangsungan reaksi, besarnya panas reaksi (∆H), harga tetapan kesetimbangan reaksi (K), dan konversi maksimum reaksi (Xe) yang dapat dicapai pada kondisi tertentu. Katalis hanya berpengaruh terhadap sifat kinetika reaksi.

8.Katalis tidak memulai berlangsungnya suatu reaksi, tetapi mempengaruhi kecepatan reaksinya. Katalis hanya mempromosikan reaksi-reaksi yang perubahan energi bebas Gibbs (∆G)-nya berharga negatif.

9.Katalis hanya mempercepat reaksi untuk mencapai

kesetimbangan

Karena tetapan kesetimbangan reaksi (K) yang merupakan perbandingan antara tetapan kecepatan reaksi ke kanan terhadap tetapan kecepatan reaksi ke kiri tidak mengalami perubahan, maka katalis bersifat mempercepat reaksi dalam kedua arah. Artinya, katalis yang mempercepat reaksi ke kanan juga akan mempercepat

5

TENTANG KATALIS

10 Katalis mempunyai suhu operasi optimum

11. Katalis dapat teracuni oleh suatu zat dalam jumlah yang sangat sedikit yang disebut racun katalis.

Contoh:

12. Keaktifan katalis dapat diperbesar oleh suatu zat yang disebut pemercepat katalis (promotor).

Contoh: Efisiensi katalis CuO-ZnO yang digunakan untuk mengkatalisis reaksi shift conversion

(CO (g) + H O (g) ↔ CO (g) + H (g)

6

13. Pada reaksi-reaksi tertentu, terdapat salah satu produk reaksi yang dapat berfungsi sebagai katalis untuk reaksi yang bersangkutan. Zat atau produk reaksi ini disebut autokatalis, sedangkan reaksinya biasa disebut reaksi autokatalitik.

14.Katalis yang dapat menghambat atau memperlambat kecepatan reaksi disebut katalis negatif (atau inhibitor).

Contoh:

Katalis mempunyai tiga fungsi katalitik

1. Aktivitas, (berkaitan dengan kemampuannya mempercepat reaksi),

2. Selektivitas atau spesifisitas, (berkaitan dengan kemampuannya mengarahkan suatu reaksi), dan

3. Stabilitas atau lifetime, (berkaitan dengan

kemampuannya menahan hal-hal yang dapat

mengarahkan terjadinya deaktivasi katalis).

BAGAIMANA KATALIS DAPAT MEMPERCEPAT REAKSI?

Katalis dapat mempercepat reaksi dengan cara menurunkan energi aktivasi reaksi. Energi aktivasi reaksi merupakan banyaknya energi minimum yang dibutuhkan oleh reaksi agar reaksi dapat berlangsung.

Keterangan gambar:

• Ea1 = energi aktivasi reaksi tanpa katalis

• Ea2 = energi aktivasi reaksi dengan katalis

• ∆Hr = panas reaksi

= H

_reaktan

- H

produk reaksi

PENGGOLONGAN KATALIS

Penggolongan katalis berdasarkan fasenya di dalam sistem reaksi:

(1) Katalis homogen

– Yakni jika fase katalis sama dengan fase reaktan dan fase produk reaksi Yang paling umum berupa fase cair, dengan katalis dan reaktan berada dalam larutan.

– Sifat-sifat katalis homogen:

Keunggulan: aktivitas dan selektivitasnya tinggi, tidak mudah teracuni oleh keberadaan pengotor, mudah dioperasikan, mudah dimodifikasi, mudah untuk dipelajari.

Kekurangan: sulit dipisahkan dari campuran reaksi, kurang stabil pada suhu tinggi.

Contoh katalis homogen:

Reaksi berkatalis homogen, fase gas

CO (g) + O

₂

(g) → CO

₂

(g) katalis: NO (g)

CH

₃

CHO (g) → CH

₄

(g) + CO (g) katalis: uap I

₂ ¹⁰

PENGGOLONGAN KATALIS

Reaksi berkatalis homogen, fase cair

C

₁₂

H

₂₂

O

₁₁

+ H

₂

O → C

₆

H

₁₂

O

₆

+ C

₆

H

₁₂

O

₆

katalis: asam

CH

₃

COOC

₂

H

₅

+ H

₂

O → CH

₃

COOH + C

₂

H

₅

OH katalis: asam

• Proses katalitik pada reaksi berkatalis homogen

berlangsung melalui pembentukan senyawa kompleks

dan penyusunan ulang antara molekul-molekul reaktan

dengan ligan katalis.

(2) Katalis heterogen

• Fase katalis tidak sama dengan fase reaktan dan/atau fase produk reaksi : fase katalis → padatan ; fase reaksi → gas

• Sifat-sifat katalis heterogen: Mudah dipisahkan dari campuran reaksi, tahan dan stabil terhadap suhu relatif tinggi, mudah disiapkan dalam bentuk pellet katalis padat, konstruksinya sederhana.

• Contoh:

Katalis padat Fe untuk Proses Haber pada pembuatan amonia:

N

₂

(g) + 3 H

₂

(g) ↔ 2 NH

₃

(g)

Katalis padat Fe

₂

O

₃

-BiO

₂

untuk oksidasi amonia pada pembuatan asam nitrat:

4 NH

₃

(g) + 5 O

₂

(g) ↔ 4 NO (g) + 6 H

₂

O (g) Katalis padat Ni pada hidrogenasi hidrokarbon:

R

₁

CH=CHR

₂

(l) + H2 (g) → R

₁

CH

₂

CH

₂

R

₂

(l) (minyak tak jenuh) (lemak jenuh)

Katalis arang (C) pada pembuatan asam khlorida:

H

₂

(g) + Cl

₂

(g) → 2 HCl (g)

¹²

Catalysis and Catalytic

Reactors

Catalysts & Catalysis

• ~1/3 of the GNP of materials produced in the US involve a catalytic process

• A Catalyst is a substance that speeds up the rate of reaction but is not changed by the reaction

• A catalyst lowers the energy barrier by promoting a different molecular pathway (mechanism) for the reaction

Many catalyst are porous (high surface area)

• Homogeneous catalysis: catalyst is in solution with at least 1 reactant

• Heterogeneous catalysis: more than 1 phase, usually solid and fluid or solid and gas is present. Reaction occurs at solid/liquid or gas interface.

catalyst

1. Mass transfer of A to surface

2. Diffusion of A from pore mouth to internal catalytic surface

3. Adsorption of A onto catalytic surface

4. Reaction on surface

5. Desorption of product B from surface

6. Diffusion of B from pellet interior to pore mouth 7. Diffusion of B from external

surface to the bulk fluid (external diffusion)

Steps in a Heterogeneous Catalytic

Reaction

Adsorption Step

A(g) + S ⇌ A—S

S: open (vacant) surface site A—S: A bound to a surface site The adsorption of A (gas phase) on an active site S is represented by:

A I -S-S-S-

Rate of adsorption = rate of attachment – rate of detachment

AD A A v A A S

r =k P C −k₋ C _⋅ partial pressure of A

• Rate is proportional to # of collisions with surface, which is a function of P_A

• Rate is proportional to # of vacant (active) sites, C_v, on the surface

• Active site: site on surface that can form a strong bond with adsorbed species Molar conc of vacant sites on surface

A -S-S-S-

AD A A v A A S

r =k P C −k₋ C _⋅

In terms of the adsorption equilibrium constant K_A where _A ^A

A

K k

k₋

=

AD A A v A A S

A

r k P C k C

k

− ⋅

 

→ =  − 

 

AD A A v A S

A

r k P C C

K

 ⋅ 

→ =  − 

  Equation I

Conc of sites occupied by A

Site Balance

C_t: Total number of active sites per unit mass of catalyst divided by Avogadro’s # (mol/g cat)

C_v: Number of vacant sites per unit mass of catalyst divided by Avogadro’s #

C_t = C_v + C_A—S + C_B—S Surface

Vacant active site

A

Active site occupied by A

B

Active site occupied by B

Site balance:

In the absence of catalyst deactivation, assume the total number of active sites remains constant:

C_v is not measurable, but the total number of sites C_t can be measured

Langmuir Isotherm Adsorption

Adsorption of carbon monoxide onto a surface: CO + S ⇌ CO—S

AD A CO v A CO S

r =k P C −k₋ C _⋅ AD A CO v ^{CO S}

A

r k P C C

K

 ⋅ 

→ =  − 

 

A A

A

K k

k₋

=

Put C_v in terms of C_t using the site balance; only CO can absorb on the surface:

t v CO S

C = C +C _⋅

At equilibrium, r_AD = 0: _{AD A CO v} ^{CO S}

A

r 0 k P C C

K

 ⋅ 

= =  − 

 

Determine the concentration of CO adsorbed on the surface at equilibrium

Rearrange &

solve for C_CO—S

CO S CO v A

C P C

K

→ ⋅ =

t CO S v

C C _⋅ C

→ − = Insert into eq for C_CO—S from rxn rate

CO S A CO v

C _⋅ =K P C ^{→ C}_{CO S⋅} ^{=K P}_{A CO}

(

^C_t ^−C_{CO S⋅}

)

Solve for C_CO—S

CO S A CO t A CO CO S

C _⋅ K P C K P C _⋅

→ = − →C_{CO S⋅} +K P_{A CO CO S}C _⋅ =K P_{A CO}C_t

A CO t CO S

A CO

K P C

C ^⋅ 1 K P

→ =

+ Concentration of CO adsorbed on surface vs P_CO→ Langmuir Isotherm

CO S A CO v

C _⋅ K P C

→ =

Surface Reaction Step

After the molecule is adsorbed onto the surface, it can react by a few different mechanisms

1. Singe site mechanism: Only the site to which the reactant is absorbed is involved in the reaction

A I

-S- ⇌ B

I -S-

A—S ⇌ B—S _{S S A S} ^{B S}

S

r k C C

K

⋅ ⋅

 

=  − 

 

S S

S

where K k

k₋

=

2. Dual site mechanism: Adsorbed reactant interacts with another vacant site to form the product

A I

-S-S-S ⇌

B I -S-S-S-

A—S + S ⇌ S+ B—S

B S v S S A S v

S

C C

r k C C

K

⋅ ⋅

 

=  − 

 

Equation IIa

Equation IIb 3. Eley-Rideal mechanism: reaction between adsorbed reactant and a

molecule in the gas phase

⇌

C I

A—S + B(g) ⇌ C—S

S S A S B C S

r k C _⋅ P C ^⋅ 

=  −  Equation IIc

A I B

Desorption Step

Products are desorbed into the gas phase

C I -S-S-S-

⇌ C

-S-S-S-

C—S ⇌ C + S

D,C D C S C v

D,C

r k C P C

⋅ K

 

=  − 

 

D,C D

D

where K k

k₋

=

Equation III

Note that the desorption of C is the reverse of the adsorption of C

D,C AD,C

r = −r

Also the desorption equilibrium constant K_D,C is the reciprocal of the adsorption equilibrium constant K_C

D,C

C

K 1

= K

[ ]

D,C D C S C C v

r = k C _⋅ −K P C

Substituting 1/K_C for K_D,C in the rate equation for product desorption gives:

Derive a Rate Law for Catalytic Rxn

• Postulate catalytic mechanism, and then derive the rate law for that mechanism

• Assume pseudo-steady state hypothesis (rate of adsorption = rate of surface reaction = rate of desorption)

• No accumulation of species on the surface or near interface

• Each species adsorbed on the surface is a reactive intermediate

• Net rate of formation of species i adsorbed on the surface is 0, r_i—S=0

• One step is usually rate-limiting

• If the rate-limiting step could be sped up, the entire rxn would be faster

• Although reactions involve all 7 steps, (for chapter 10) only adsorption, surface reaction, or desorption will be rate limiting

• The surface reaction step is rate limiting ~70% of the time!

• Steps to derive the rate law

• Select among types of adsorption, surface reaction, and desorption

• Write rate laws for each individual step, assuming all are reversible

• Postulate which step is rate limiting

• Use non-rate-limiting steps to eliminate the surface concentration

Consider A ⇌ B and assume the following mechanism is correct:

AD A A v A S

A

r k P C C

K

 ⋅ 

=  − 

 

1. Adsorption:

2. Surface reaction:

A(g) + S ⇌ A—S

A—S + S ⇌ S+ B—S _{S S A S v} ^{B S v}

S

C C

r k C C

K

⋅ ⋅

 

=  − 

 

3. Desorption: B—S ⇌ B + S _{D D B S} ^{B v}

D

r k C P C

⋅ K

 

=  − 

 

We need to select one of these 3 reactions as the rate limiting step, then derive the corresponding rate equation, and see if this rate eq matches experimental data. Which step is the most logical to start with?

a) Adsorption

b) Surface reaction c) Desorption

d) None of the above

e) Any of these would be “logical” - they all have equal probability of being the rate limiting step

The surface reaction step as is rate limiting ~70% of the time

Consider A ⇌ B and assume the following mechanism is correct:

AD A A v A S

A

r k P C C

K

 ⋅ 

=  − 

 

1. Adsorption:

2. Surface reaction:

A(g) + S ⇌ A—S

A—S + S ⇌ S+ B—S _{S S A S v} ^{B S v}

S

C C

r k C C

K

⋅ ⋅

 

=  − 

 

3. Desorption: B—S ⇌ B + S _{D D B S} ^{B v}

D

r k C P C

⋅ K

 

=  − 

 

Derive the rate equation for when the surface reaction is rate limiting (true

~70% of the time)

• For surface reaction-limited mechanisms, k_S is small and k_A and k_D are relatively large

• Therefore r_AD/k_A and r_D/k_D are very small, and can be approximated as equal to zero

B S v

A S S A S v

S

C C

r ' r k C C

K

⋅ ⋅

 

− = =  − 

 

1. C_A—S, C_v, and C_B—S need to be expressed in terms of measurable quantities

2. Use this relationship to eliminate C and C from their respective rate

Consider A ⇌ B and assume the following mechanism is correct:

AD A A v A S

A

r k P C C

K

 ⋅ 

=  − 

 

1. Adsorption 2. Surface reaction

B S v S S A S v

S

C C

r k C C

K

⋅ ⋅

 

=  − 

 

3. Desorption

B v

D D B S

D

r k C P C

⋅ K

 

=  − 

 

Derive the rate equation for when the surface reaction is rate limiting

B S v

A S S A S v

S

C C

r ' r k C C

K

⋅ ⋅

 

− = =  − 

 

Use r_AD/k_A =0 and r_D/k_D =0 to eliminate C_A—S and C_B—S from their respective rate equations and the site balance to eliminate C_V

AD A A v A S

D

r k P C C

K

 ⋅ 

=  − 

 

A S

AD A v

A A

r C

0 P C

k K

→ = = − ⋅ ^{A S A v}

A

C P C

K

→ ⋅ = →C_{A S⋅} =K P C_{A A v}

B v

D D B S

D

r k C P C

⋅ K

 

=  − 

 

D B v

B S

D D

r P C

0 C

k ^⋅ K

→ = = − ^{B v} _{B S}

D

P C C

K ^⋅

→ =

Use r_AD/k_A =0 & r_AD

equation to solve for C_A—S:

Use r_D/k_D =0 & r_D equation to solve for C_B—S:

Consider A ⇌ B and assume the following mechanism is correct:

AD A A v A S

A

r k P C C

K

 ⋅ 

=  − 

 

1. Adsorption 2. Surface reaction

B S v S S A S v

S

C C

r k C C

K

⋅ ⋅

 

=  − 

 

3. Desorption

B v

D D B S

D

r k C P C

⋅ K

 

=  − 

 

Derive the rate equation for when the surface reaction is rate limiting

B S v

A S S A S v

S

C C

r ' r k C C

K

⋅ ⋅

 

− = =  − 

 

A S A A v

C _⋅ =K P C

r_AD/k_A=0 & r_D/k_D =0 ^{B v} B S

D

P C C

K = ^⋅

Use site balance to solve for C_V: C_t = C_v +C_{A S⋅} +C_{B S⋅}

v t A S B S

C C C _⋅ C _⋅

→ = − − Make substitutions for C_A—S & C_B—S, solve for C_v

B v

v t A A v

D

C C K P C P C

→ = − − K _{v A A v} ^{B v} _t

D

C K P C P C C

→ + + K =

v A A B t

C 1 K P P C

K

 

→  + +  =

 

v t

C C

→ = P

Consider A ⇌ B and assume the following mechanism is correct:

AD A A v A S

A

r k P C C

K

 ⋅ 

=  − 

 

1. Adsorption 2. Surface reaction

B S v S S A S v

S

C C

r k C C

K

⋅ ⋅

 

=  − 

 

3. Desorption

B v

D D B S

D

r k C P C

⋅ K

 

=  − 

 

Derive the rate equation for when the surface reaction is rate limiting

B S v

A S S A S v

S

C C

r ' r k C C

K

⋅ ⋅

 

− = =  − 

 

A S A A v

C _⋅ =K P C ^{B v} B S D

P C C

K = ^⋅ _v ^t

A A B D

C C

1 K P P K

= + +

Substitute in C_A—S, C_B—S, &C_v

2 2

t B t

A S S A A

A A B D S D A A B D

C P C

r ' r k K P

1 K P P K K K 1 K P P K

     

 

→ − = =   + +  −  + +  

2

t B

A S S A A

A A B D S D

C P

r ' r k K P

1 K P P K K K

 

 

→ − = =  + +    − 

This is the rate equation in terms of measurable species and rate constants for the mechanism given in the problem statement

Evaluating a Catalytic Reaction Mechanism

• Collect experimental data from test reactor

• See if rate law is consistent with data

• If not, then try other surface mechanism (i.e., dual-site

adsorption or Eley-Rideal) or choose a different rate-limiting

step (adsorption or desorption)

Consider A ⇌ B and assume the following mechanism is correct:

AD A A v A S

A

r k P C C

K

 ⋅ 

=  − 

 

1. Adsorption 2. Surface reaction

B S v S S A S v

S

C C

r k C C

K

⋅ ⋅

 

=  − 

 

3. Desorption

B v

D D B S

D

r k C P C

⋅ K

 

=  − 

 

Now derive the rate equation for when adsorption is rate limiting:

 ⋅ 

− = =  − 

 ^v 

A AD A A A

A

r ' r k P S

K C C

Conc of vacant and occupied sites must be eliminated from the rate equation

⋅ ⋅

→ = = _{A S} − ^B

S S

S v

S v C

r C C

k 0 C K

⋅ ⋅

→ ^{B S v} = _v

S A S

K C

C C

C → ^⋅ = _⋅

S A

B S S

K C

C

→ = = _B⋅ − ^B

D

v D

D S

r P

0 C

k C K → ^{B v} = _{B S⋅}

D

P C C K

If adsorption is rate limiting, k_S>>k_AD, so r_S/k_S can be approximated as 0. Then:

⋅ ⋅

 

=  − 

 

B v

S S v S

S A S

r k C C

C CK

If adsorption is rate limiting, k_D>>k_AD, so r_D/k_D can be approximated as 0. Then:

⋅

 

=  − 

 

B S B v

D D

D

r k C P C

K

Need to put C_B—S in measureable terms

Consider A ⇌ B and assume the following mechanism is correct:

AD A A v A S

A

r k P C C

K

 ⋅ 

=  − 

 

1. Adsorption 2. Surface reaction

B S v S S A S v

S

C C

r k C C

K

⋅ ⋅

 

=  − 

 

3. Desorption

B v

D D B S

D

r k C P C

⋅ K

 

=  − 

 

Now derive the rate equation for when adsorption is rate limiting:

Conc of vacant and occupied sites must be eliminated from the rate eq

⋅ = ⋅

S A

B S S

K

C C ^B = _{B S⋅}

D

P v

K C C

Make substitutions for C_A—S & C_B—S

→ = _v + ^{B S}⋅ + ^B

D

t v

S

C C P C

K K

C

→ = _v + ^B + ^B

D

t v v

D

KSK C

P P C

C K

C  

→ =  + + 

 

B B

t

S D D

v

P P

C 1

K K K

C → =

+ +

t v

B B

S D D

C C

P P

1 K K K

Solve for C_v using the site balance equation

⋅ ⋅

= _v + _{A S} + _{B S}

t C C C

C

Substitute C_v into the

expression ^⋅  

+

=

 + 

t

B B

B S B

C

P P

1 C P

K

Substitute

C_B—S into ^⋅ =  

+ +

 

B t

B B

A S

P C

P P

1 K K C

 ⋅ 

− = =  − 

 ^v 

A AD A A A

A

r ' r k P S

K C C

Consider A ⇌ B and assume the following mechanism is correct:

AD A A v A S

A

r k P C C

K

 ⋅ 

=  − 

 

1. Adsorption 2. Surface reaction

B S v S S A S v

S

C C

r k C C

K

⋅ ⋅

 

=  − 

 

3. Desorption

B v

D D B S

D

r k C P C

⋅ K

 

=  − 

 

Now derive the rate equation for when adsorption is rate limiting:

⋅  

+ +

 

 

= ^{B t}

A S

B B

S D

S D D

C P C

P P

K K 1

K K K

⋅  

+ +

 

 

= ^{B t}

B S

B B

D

S D D

C P C

P P

K 1

K K K + +

t =

B B v

S D D

C C

P P

1 K K K

 

 

 

→ = −

 

+ +

 

 +



 + 

 



AD A A

A

B t

B B

S D

S D D

t

B B

S D D

P C

P P

K K C

P P

1 K K 1

K K

K K K

r k P Use these eqs

to replace C_A—S

& C_v in r_AD:

 

 

 

→ = −

 + +  + + 

  

 

 

A B

AD B B B B

A S D A t

S D D S D D

P P

r 1 P P K K K 1 P P

K K K K

C

K k

K k

 ⋅ 

− = =  − 

 ^v 

A AD A A A

A

r ' r k P S

K C C

Factor out C_t:

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