Question

For the Lindemann–Hinshelwood mechanism of gas phase unimolecular reactions, the true statement(s) is(are)

• A. Only molecules with three or more atoms can follow the Lindemann–Hinshelwood mechanism
• B. Lindemann–Hinshelwood mechanism involves bimolecular elementary steps
• C. The overall reaction is of second order at low pressure
• D. The overall reaction is of second order at high pressure

Hint:

L–H = bimolecular activation (A + M) followed by unimolecular breakdown (A* → P). Low \(p\): rate \(\propto [A][M]\) (2nd order); high \(p\): rate \(\propto [A]\) (1st order). Polyatomic molecules (\(\geq\) 3 atoms) efficiently redistribute vibrational energy needed for unimolecular decay.

Answer 1

The Correct Option is A

Step 1: Nature of species that show unimolecular behavior.
The Lindemann–Hinshelwood (L–H) scheme explains unimolecular reactions such as isomerizations and decompositions in the gas phase. Energy redistribution inside an activated molecule requires many internal degrees of freedom. This is generally possible only in polyatomic molecules (≥ 3 atoms). Hence, statement (A) is true.

Step 2: Elementary steps in the L–H mechanism.
The mechanism involves two stages:
\[ \mathrm{A + M \xrightleftharpoons[k_{-1}]{k_1} A^{*} + M} \quad \text{(bimolecular activation/deactivation)} \]
\[ \mathrm{A^{*} \xrightarrow{k_2} \text{Products}} \quad \text{(unimolecular decomposition)} \]
Thus, the scheme does include bimolecular steps (A + M collisions), making statement (B) true.

Step 3: Pressure dependence of the overall rate law.
At low pressure, collisions between A and M are rare, so the formation of A* is slow. The rate is proportional to \([A][M]\), giving an overall second-order rate law. Hence, statement (C) is true.

At high pressure, activation to A* is rapid and essentially in equilibrium. The slow step is the unimolecular decomposition of A*, so the overall reaction becomes first order in [A]. Therefore, statement (D) is false.

Final Answer: The correct true statements are A, B, and C.