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, B, C

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 randomization within an \emph{activated} molecule requires multiple internal degrees of freedom; therefore it is typically valid for polyatomic (}\(\ge 3\)\textbf{-atom) molecules. Hence statement (A) is true. Step 2: Elementary steps in the L–H mechanism.
The mechanism proceeds via:
\[ \mathrm{A + M \xrightleftharpoons[k_{-1}]{k_1} A^{*} + M} \quad \text{(bimolecular activation/deactivation)} \] \[ \mathrm{A^{*} \xrightarrow{k_2} \text{Products}} \quad \text{(unimolecular decomposition)} \] Thus, it \emph{does} involve bimolecular elementary steps (A + M), so (B) is true. Step 3: Pressure dependence of the overall rate law.
At low pressure, activation collisions are infrequent, and the rate is proportional to \([A][M]\), i.e., second order. Hence (C) is true.
At high pressure, rapid activation makes decomposition of \(A^{*}\) rate-determining, giving a first-order overall rate. Therefore (D) is false. \[ \boxed{\text{True statements: A, B, and C.}} \]