Chemical Reaction Engineering Questions and Answers Part-14

1. The overall reaction order for A + B → C + D is two. It implies that ______
a) reaction may be non-elementary and molecularity is two
b) reaction may be elementary and molecularity is two
c) reaction is non-elementary and molecularity is two
d) reaction is elementary and molecularity is two

Answer: d
Explanation: Order is equal to molecularity when reaction is elementary.

2. Reaction corresponding to stoichiometric equation is called __________
a) Non-elementary
b) Elementary
c) Heterogeneous
d) Homogeneous

Answer: b
Explanation: Elementary reactions are the ones with single reaction step and the reaction rate corresponds to stoichiometric equation.

3. The rate constant for a first order reaction depends on _____________
a) Pressure
b) Temperature
c) Concentration
d) Time

Answer: b
Explanation: The reaction rate constant depends on temperature only.

4. The mlecularity of the following single step reaction is ___________
aA + bB → pP
a) a+b-p
b) a+b+p
c) a+b
d) a-p

Answer: c
Explanation: The above single step reaction is an elementary reaction. Molecularity is the number of molecules participating in the reaction.

5. Which of the following reactions follows elementary rate law?
a) Formation of hydrogen bromide
b) Vapor phase decomposition of ethanal
c) Cis-trans isomerization
d) Reversible catalytic decomposition of isopropylbenzene

Answer: c
Explanation: Cis-trans isomerization follows elementary rate laws. Hydrogen bromide formation reaction as well as the reversible catalytic decomposition of isopropylbenzene are non-elementary in nature. Rate of vapor phase decomposition of ethanal is proportional to Cethanal3/2.

6. Langmuir Hinshelwood kinetics is followed by the reaction:
a) Formation of hydrogen bromide
b) Vapor phase decomposition of ethanal
c) Cis-trans isomerization
d) Reversible catalytic decomposition of isopropylbenzene

Answer: d
Explanation: C6H5CH(CH3)2 ←→ C6H6 + C3H6
Let’s represent this reaction symbolically as C ←→ B + P
the reaction follows the rate law -r’C = [k(PC – PBPP/KP)] / [1 + KCPC + KBPB] which is the Langmuir Hinshelwood model.

7. Which of the following is true for fluidized catalytic beds?
a) They cannot be used for multi-phase chemical reactions
b) The bulk density is a function of the flow rate through the bed
c) They come under the category of batch reactors
d) There is no pressure drop

Answer: b
Explanation: Fluidized catalytic beds are mostly used for multi-phase reactions, for example, cumene decomposition. They come under flow reactors. The inclusion of fluidized bed offers more resistance to flow and hence results in a pressure drop.

8. The reaction H2 + Br2 → 2HBr proceeds via which mechanism?
a) Free-radical
b) Ionic substitution
c) Elimination
d) Pericyclic

Answer: a
Explanation: Bromine forms a free-radical and reacts with H2. Thus, free-radical mechanism is followed. There is no elimination and no cyclic structure is formed as product or intermediate state.

9. Which of the following does not hold true for gas-solid catalyzed reactions?
a) Rate law is preferably written in terms of partial pressures
b) Rate law is preferably written in terms of concentrations
c) Rate law can be written only in terms of partial pressures
d) Rate law can be written only in terms of concentrations

Answer: a
Explanation: In most cases, for gas-solid catalyzed reactions we write the rate law in terms of partial pressures as it makes the analysis simpler. Also, it is preferable because measuring pressure is easier than measuring concentration. Partial pressure and concentration are directly related and the rate law can very easily be represented in either form.

10. For the reaction A + 2B ←→ C, what is the value of KC if the rate law is given by
–rA = k(CA2/3/CB – KCCC)? Given that initial concentration of A is equal to that of B = 5M and the equilibrium conversion of B is found to be 0.3. k = 100.
a) Kc = 0
b) Kc = 0.5
c) Kc = 1
d) Kc = 1.5

Answer: c
Explanation: XA = CBoaXB/CAob = 0.15
CA = CAo(1 – XA)
CB = CAo(1 – 2XA)
CC = CAoXA
At equilibrium, take –rA equal to 0 and solve for KC.