This set of Chemical Reaction Engineering Multiple Choice Questions & Answers (MCQs) focuses on “Mechanisms of Catalyst Deactivation”.
1. State true or false.
In parallel deactivation, the reactant produces a side product which deposits on the catalytic surface and deactivates it.
a) True
b) False
View Answer
Explanation: A parallel deactivation is the one in which reactant A produces the product R and the poison P.
A → R + P↓
2. The catalyst deactivation caused by deposition on surface and pores of catalyst is called ____
a) Adsorption
b) Regeneration
c) Fouling
d) Desorption
View Answer
Explanation: Fouling is caused by the blockage of pores on the catalyst. The life and the efficiency of the catalyst is reduced by fouling.
3. Which of the following represents series type deactivation model? (Where R is the product and P is the poison)
a) A → R → P↓
b) A → P↓
c) A → R + P↓
d) A → R, A → P
View Answer
Explanation: The series deactivation is the mechanism in which the product may decompose to produce the poison or further react to produce the poison. The poison deposits on the catalyst and causes deactivation.
4. The activity of a catalyst pellet is defined as the ratio of ___
a) Rate of adsorption of reactants to the catalyst surface to the rate at which the catalytic pellet converts the reactant
b) Rate at which the catalytic pellet converts the reactant to the rate of adsorption of reactants to the catalyst surface
c) Rate at which the catalytic pellet converts the reactant to the rate of reaction with a fresh pellet
d) The rate of reaction with a fresh pellet to the rate at which the catalytic pellet converts the reactant
View Answer
Explanation: Activity is a measure of the catalyst performance. It is defined as the ratio of rate at which the catalytic pellet converts the reactant to the rate of reaction with a fresh catalyst pellet.
5. If kd is the rate constant of deactivation and d is the order of deactivation, then the rate equations representing independent deactivation are ___
a) -r’A = k’CAna
–\(\frac{da}{dt}\) = kdad
b) -r’A = k’a
–\(\frac{da}{dt}\) = kdad
c) -r’A = k’CAna
–\(\frac{da}{dt}\) = kd
d) -r’A = k’CAn
–\(\frac{da}{dt}\) = kdad
View Answer
Explanation: Independent deactivation is the one that is independent of the concentration. Hence, the rate of change of activity is –\(\frac{da}{dt}\) = kdad.
6. Which of the following reactor arrangements causes fast deactivation?
a) Mixed flow for fluid
b) Plug flow for fluid
c) Fluidised bed reactor
d) Batch for fluid and solid
View Answer
Explanation: Fluidisation causes high catalytic deactivation. The catalytic particles are well mixed by the reactant fluid.
7. The rate expression for independent deactivation for batch solids is ____
a) ln\((ln\frac{C_A}{C_{A∞}})\) = ln \(\frac{Wk’}{Vk_d}\) – kdt
b) ln \(\frac{C_A}{C_{A∞}}\) = ln \(\frac{Wk’}{Vk_d}\) – kdt
c) lnln \(\frac{C_A}{C_{A∞}}\) = ln \(\frac{Wk’}{Vk_d}\)
d) lnln \(\frac{C_A}{C_{A∞}}\) = \(\frac{Wk’}{Vk_d}\) – kdt
View Answer
Explanation: For batch solids – fluid, –\(\frac{-dC_A}{dt} = \frac{W}{V}\)k’CAa and –\(\frac{da}{dt}\) = kdad. Integrating and combining the equations, ln\((ln\frac{C_A}{C_{A∞}})\) = ln \(\frac{Wk’}{Vk_d}\) – kdt.
8. The slope of the curve of ln\((ln\frac{C_A}{C_{A∞}})\) and t for batch solid – fluid independent deactivation system is ___
a) ln(lnkd)
b) ln(kd)
c) kd
d) -kd
View Answer
Explanation: ln\((ln\frac{C_A}{C_{A∞}})\) = ln \(\frac{Wk’}{Vk_d}\) – kdt. The slope of the curve is -kd and the intercept is ln\(\frac{Wk’}{Vk_d}. \)
9. If the rate of deactivation is given by –\(\frac{da}{dt}\) = 0.0064 day-1, the expression relating activity and time is ___
a) a = 0.0064t
b) a = 1-0.0064t
c) a = t
d) a = 0.0064-t
View Answer
Explanation: –\(\frac{da}{dt}\) = 0.0064
–\(\int_1^a\)da = 0.0064\(\int_0^t\)dt
Integrating, a = 1-0.0064t
10. The activity of a catalyst at a time t = 0 is ____
a) Negative
b) Zero
c) Unity
d) ∞
View Answer
Explanation: When the reaction has not yet started to occur, the catalyst surface is not poisoned. The entire surface on the catalyst is available to promote the reaction. At a time t=0, the activity is the highest and equal to 1. The value of ‘a’ lies between 0 and 1 as ‘a’ is the ratio of rate of reaction achieved by the catalyst at any given time to the maximum rate that can be achieved by a fresh catalyst.
Sanfoundry Global Education & Learning Series – Chemical Reaction Engineering.
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