250+ TOP MCQs on Non – Elementary Rate Laws and Answers

Chemical Reaction Engineering Multiple Choice Questions & Answers on “Non – Elementary Rate Laws”.

1. 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.

2. 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.

3. 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.

4. 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.

5. 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.

6. 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.

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