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## Chemical Reaction Engineering ### Catalyst Deactivation * Catalyst activity decreases with time due to: Fouling, Poisoning, Sintering * This leads to: Decreased conversion, changed product distribution, increased operating temperatures, excessive pressure drop across the catalyst bed. * D...

## Chemical Reaction Engineering ### Catalyst Deactivation * Catalyst activity decreases with time due to: Fouling, Poisoning, Sintering * This leads to: Decreased conversion, changed product distribution, increased operating temperatures, excessive pressure drop across the catalyst bed. * Deactivation can be reduced by: proper catalyst selection, pretreatment of feedstreams, modifying operating conditions, catalyst regeneration or replacement. ### Types of Catalyst Deactivation 1. Fouling: * Physical deposition of species from the fluid phase onto the catalyst surface. * This blocks active sites and/or pores, preventing reactants from accessing the interior of the catalyst pellet Examples: * Coke formation in cracking $C_nH_m \rightarrow nC + \frac{m}{2}H_2$ Coke * Metal deposition $Ni(CO)_4 \rightarrow Ni + 4CO$ (Metal carbonyl formation) * Steam condensation 2. Poisoning: * Strong chemisorption of species onto active sites. * These species are called poisons, and they block active sites from reactant adsorption. Examples: * $H_2S, CO$ on Pt * Basic amines on acidic sites * Lead on catalytic converters * Arsenic on reforming catalysts 3. Sintering: * Loss of active surface area due to: * Crystal growth * Support collapse * Pore closure * This is more prevalent at high temperatures Examples: * Alumina support collapses at $T>800^oC$ * Small metal particles migrate and coalesce to form larger particles. 4. Attrition: * Mechanical destruction of the catalyst. * This can lead to pressure drop problems in fixed beds, and loss of catalyst in fluidized beds. ### Kinetics of Catalyst Deactivation * Deactivation is a function of time on stream and concentration of the deactivating species: $r_d = f(t, C_i)$ * The rate of reaction on a catalyst that is deactivating is given by: $r' = a(t) \cdot r'(t=0)$ Where: $r'$ is the rate of reaction at time $t$ $a(t)$ is the activity of the catalyst $r'(t=0)$ is the initial rate of reaction (i.e. rate on a fresh catalyst) * $a(t)$ can range from 0 (completely deactivated) to 1 (fresh catalyst) * The form of $a(t)$ must be determined experimentally. * The rate law for the deactivation can be either 1) independent of the main reaction, or 2) dependent on the main reaction. * We will consider the case where deactivation occurs independently of the main reaction. ### Independent Deactivation * Here we assume that the deactivation rate law is not dependent on the concentrations of the reactants or products in the main reaction. * Most commonly, the rate of deactivation is proportional to the activity: $-\frac{da}{dt} = k_d a^{d}$ Where: $k_d$ is the deactivation rate constant $d$ is the order of the deactivation reaction * Integrating this equation from $a=1$ at $t=0$ gives: $a = [1 + (d-1)k_dt]^{1/(1-d)}$ for $d \neq 1$ $a = exp(-k_dt)$ for $d = 1$ * The rate law for the main reaction can be substituted into the mole balance. And "a" can be substituted as a function of "t". * There are two ideal reactor schemes for running reactions with catalyst deactivation: 1. Moving-bed reactors 2. Fluidized-bed reactors * In these reactors, the catalyst is constantly mixed, so the catalyst activity is uniform throughout the reactor. ### Example: A fluidized CSTR is used to carry out the following catalytic reaction: $A \rightarrow B$ The rate law for this reaction is: $-r_A' = akC_A$ Where the rate constant $k = 0.72 \frac{dm^3}{kg \cdot s}$ The catalyst deactivation is first order with $k_d = 0.15 s^{-1}$ The feed stream is 500 $dm^3 / s$ with $C_{A0} = 2 \frac{mol}{dm^3}$ The reactor contains 50 kg of catalyst. What is the conversion as a function of time? ### Solution: #### Mole Balance $V_0C_{A0} - V_0C_A + r_A'W = 0$ Solving for $C_A$: $C_A = \frac{V_0C_{A0}}{V_0 + akW}$ #### Conversion $X = \frac{C_{A0} - C_A}{C_{A0}} = \frac{akW}{V_0 + akW}$ Substituting in $a = exp(-k_dt)$: $X = \frac{kWexp(-k_dt)}{V_0 + kWexp(-k_dt)}$ $X = \frac{0.72 \frac{dm^3}{kg \cdot s} \cdot 50kg \cdot exp(-0.15 s^{-1} \cdot t)}{500 \frac{dm^3}{s} + 0.72 \frac{dm^3}{kg \cdot s} \cdot 50kg \cdot exp(-0.15 s^{-1} \cdot t)}$ $X = \frac{0.072 \cdot exp(-0.15t)}{1 + 0.072 \cdot exp(-0.15t)}$