Catalyst deactivation

Mechanisms of catalyst deactivation, activity and deactivation kinetics, conversion decay in packed beds and temperature compensation, with worked kd and time-on-stream calculations.

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Why it matters

Every industrial catalyst loses activity — in seconds for fluid catalytic cracking, in years for ammonia synthesis. Deactivation sets the cycle length, the regeneration scheme, the choice between fixed, moving and fluidised beds, and the temperature schedule operators follow to hold production. Designing a reactor without a deactivation model means designing for the first day only.

Key ideas

Activity. a(t) = rate on the used catalyst / rate on fresh catalyst at the same conditions, so a = 1 fresh and falls toward 0.

Main mechanisms.

  • Poisoning: strong chemisorption of a feed impurity on active sites (sulphur on nickel or platinum, lead and phosphorus on exhaust catalysts, water on some acid catalysts). Can be reversible or permanent. Remedy: guard beds, feed purification.
  • Fouling / coking: carbonaceous or heavy deposits physically cover sites and plug pores; common in hydrocarbon processing. Usually reversible by controlled burn-off with air.
  • Sintering (thermal degradation): at high temperature metal crystallites migrate and merge, or the support loses surface area or changes phase. Usually irreversible. Remedy: stay below the catalyst's maximum temperature, stabilise with promoters and supports.
  • Loss of active phase: volatilisation or leaching; attrition: mechanical breakage, important in fluidised and moving beds.

Deactivation kinetics. Written as an empirical rate for the activity: −da/dt = kd·a^d·(concentration terms). Common cases:

  • Independent (size-independent) deactivation, d = 1, rate not depending on fluid composition: a = exp(−kd·t). Typical for many sintering and slow structural processes when approximated.
  • Second order in a, often used for sintering: a = 1/(1 + kd·t).
  • Parallel deactivation: the reactant itself forms the poison or coke (rate ∝ CA·a); series: the product does (rate ∝ CR·a); side-by-side: an impurity P does (rate ∝ CP·a). Parallel coking hits the reactor inlet hardest; series coking the exit. kd follows an Arrhenius law; deactivation energies are often larger than reaction activation energies, so hotter operation shortens life.

Using activity in design. Replace the fresh rate with a(t) × fresh rate. For first-order reaction (rate constant k' per catalyst mass) and first-order independent deactivation, with batch of solids and steady fluid flow:

  • Mixed flow of fluid: CA0/CA − 1 = k'·τ'·a.
  • Plug flow of fluid: ln(CA0/CA) = k'·τ'·a. Here τ' = W·CA0/FA0 is the weight time (kg·s/m³). Plotting ln[ln(CA0/CA)] (plug flow) or ln(CA0/CA − 1) (mixed flow) against t gives a straight line of slope −kd — the standard way to get kd from plant data.

Operating strategies. Raise temperature with time so that k·a stays constant (constant-conversion policy, until the maximum temperature is reached); cut feed rate; regenerate in place (swing reactors); or move the catalyst continuously between reactor and regenerator (moving bed, FCC riser + regenerator) for very fast deactivation.

Formulas

a(t) = (−r'A)(t)/(−r'A)fresh

−da/dt = kd·a^d (general, composition-independent)

  • kd: deactivation rate constant (h⁻¹ for d = 1); d: order of deactivation.

a = exp(−kd·t) (d = 1); a = 1/(1 + kd·t) (d = 2)

kd = kd0·exp(−Ed/(R·T))

  • Ed: deactivation energy (J/mol).

ln(CA0/CA) = k'·τ'·a (plug flow of fluid, first-order reaction)

CA0/CA − 1 = k'·τ'·a (mixed flow of fluid, first-order reaction)

τ' = W·CA0/FA0

  • W: catalyst mass (kg); FA0: molar feed (mol/s); k': rate constant per mass (m³/kg·s).

ln[ln(CA0/CA)] = ln(k'·τ') − kd·t (plug flow, a = e^(−kd·t); plot to find kd)

k(T2)·a(t) = k(T1) (temperature compensation to hold conversion)

Worked examples

Example 1 (standard, deactivation constant). A catalyst's activity falls to 0.6 after 50 h at constant conditions, following first-order (size-independent) deactivation. Find kd and the time to reach a = 0.3.

  1. a = exp(−kd·t) → kd = ln(1/a)/t = ln(1/0.6)/50 = 0.5108/50 = 0.0102 h⁻¹.
  2. t for a = 0.3: t = ln(1/0.3)/kd = 1.204/0.01022 = 117.8 h.

kd = 0.0102 h⁻¹; a = 0.3 after about 118 h.

Example 2 (GATE level, conversion decay and temperature compensation). A packed bed (plug flow of gas, first-order reaction) gives XA = 0.90 on fresh catalyst at 600 K. Deactivation is first order with kd = 0.01 h⁻¹. (a) Find XA after 100 h at 600 K. (b) To what temperature must the bed be raised at 100 h to restore 90 % conversion, if E = 100 kJ/mol (neglect the effect of T on kd over this step)?

(a)

  1. Fresh: k'τ' = ln[1/(1 − 0.9)] = ln 10 = 2.303.
  2. a(100 h) = exp(−0.01 × 100) = e⁻¹ = 0.368.
  3. k'τ'·a = 2.303 × 0.368 = 0.847 → XA = 1 − e^(−0.847) = 0.571.

(b) 4. Need k(T2)·a = k(600) → k(T2)/k(600) = 1/a = e = 2.718. 5. ln 2.718 = 1 = (E/R)·(1/600 − 1/T2) → 1/T2 = 1/600 − 8.314/100 000 = 1.5835 × 10⁻³ K⁻¹. 6. T2 = 631.5 K.

XA falls to 0.571 after 100 h; raising the bed to about 632 K restores 90 %.

Common mistakes

  • Treating "5 % activity loss per hour" as linear when the model is exponential (100 → 59.9 after 10 h, not 50).
  • Applying the activity to the conversion instead of to the rate constant.
  • Forgetting that raising temperature also speeds deactivation (Ed is often large).
  • Confusing poisoning (chemical, site-specific) with fouling (physical deposit).
  • Using the mixed-flow plot for a plug-flow reactor (or vice versa) when finding kd.

For GATE CH

Expect conceptual MCQs on deactivation mechanisms and remedies, NAT problems finding kd from activity or conversion–time data, conversion after a given time on stream, and the temperature rise needed to compensate. Practise both the plug-flow and mixed-flow fluid forms with exponential activity decay.

Quick check

  1. Which deactivation mechanism is usually irreversible: coking or sintering?
  2. For a = exp(−kd·t) with kd = 0.02 h⁻¹, what is a after 50 h?
  3. What plot gives kd for a plug-flow bed with first-order reaction and first-order deactivation?
  4. Why do FCC units use a separate regenerator?

Answers: 1. sintering; 2. e⁻¹ = 0.368; 3. ln[ln(CA0/CA)] against time (slope −kd); 4. because coke deactivates the catalyst within seconds, so it must be burned off continuously.

See it move

All Chemical animations
Catalyst Deactivation Visualization

Adjust the sliders to see how the concentration of active sites decreases over time due to deactivation. Observe how different deactivation rate constants affect the rate of deactivation.

Equations used
  • r_d = -k_d * C_A — r_d: Rate of deactivation (mol/s), k_d: Deactivation rate constant (s⁻¹), C_A: Concentration of active sites (mol/m³)

Interview questions

All Chemical Reaction Engineering interview questions

Try answering each one aloud before you open it.

  1. 1.What is catalyst deactivation?Concept

    Catalyst deactivation refers to the loss of catalytic activity and/or selectivity over time. This can occur due to various mechanisms such as poisoning, fouling, thermal degradation, or sintering. Deactivation reduces the efficiency of the catalyst in facilitating chemical reactions.

  2. 2.Explain the mechanism of catalyst poisoning.Concept

    Catalyst poisoning occurs when a foreign substance, often an impurity in the feed, binds to the active sites of the catalyst. This prevents the reactants from accessing these sites, thereby reducing the catalyst's activity. Common poisons include sulfur, lead, and phosphorus compounds.

  3. 3.How does thermal degradation lead to catalyst deactivation?Concept

    Thermal degradation involves the breakdown of the catalyst structure due to high temperatures. This can lead to the loss of surface area, changes in pore structure, or phase transformations, all of which reduce the catalyst's effectiveness. It is particularly a concern in processes that operate at elevated temperatures.

  4. 4.Why is catalyst regeneration important in industrial processes?Application

    Catalyst regeneration is important because it restores the activity and selectivity of a deactivated catalyst, extending its useful life. This is crucial for maintaining process efficiency and reducing costs associated with catalyst replacement. Regeneration can involve processes like calcination, washing, or chemical treatment.

  5. 5.What happens if a catalyst is not regenerated in a timely manner?Application

    If a catalyst is not regenerated in a timely manner, the process efficiency will decline due to reduced reaction rates. This can lead to lower product yields, increased energy consumption, and higher operational costs. In some cases, it may also result in the formation of undesirable by-products.

  6. 6.Describe how sintering affects catalyst performance.Concept

    Sintering involves the agglomeration of catalyst particles at high temperatures, leading to a reduction in surface area and active sites. This decreases the catalyst's ability to facilitate reactions, thus reducing its activity and selectivity. Sintering is often irreversible and can significantly impact catalyst performance.

  7. 7.What role does catalyst support play in preventing deactivation?Application

    Catalyst support provides a stable structure that helps disperse the active catalytic material, increasing its surface area and resistance to sintering. It can also enhance thermal stability and reduce the impact of poisoning by providing additional sites for adsorption of impurities.

  8. 8.A catalyst starts at 100 units of activity and loses 5% of its current activity every hour. What activity remains after 10 hours, and how does this differ from a linear loss of 5 units per hour?Numerical

    Losing 5% of the current activity each hour is first-order (exponential) decay: a = 100 × 0.95^10 = 59.9 units, equivalent to a = exp(−kd·t) with kd = −ln 0.95 = 0.0513 h⁻¹. A linear loss of 5 units per hour would leave 100 − 50 = 50 units. The exponential model is the usual one for size-independent deactivation, and confusing the two is a common exam error.

  9. 9.A catalyst has a surface area of 200 m²/g. After sintering, the surface area reduces to 150 m²/g. Calculate the percentage decrease in surface area.Numerical

    Initial surface area = 200 m²/g. Final surface area = 150 m²/g. Decrease in surface area = 200 - 150 = 50 m²/g. Percentage decrease = (50/200) × 100% = 25%.

  10. 10.How can fouling lead to catalyst deactivation, and what are common methods to mitigate it?Application

    Fouling occurs when deposits form on the catalyst surface, blocking active sites and reducing accessibility for reactants. This can be caused by coke formation or deposition of heavy hydrocarbons. Common mitigation methods include periodic cleaning, using feedstock with fewer impurities, and optimizing reaction conditions to minimize deposit formation.

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