Reactant A converts to product D through the given mechanism (with the net evolution of heat) : \(\mathrm{A} \rightarrow \mathrm{B} \quad\) slow; \(\Delta \mathrm{H}=+\mathrm{ve}\) \(\mathrm{B} \rightarrow \mathrm{C} \quad\) fast; \(\Delta \mathrm{H}=-\mathrm{ve}\) \(\mathrm{C} \rightarrow \mathrm{D} \quad\) fast; \(\Delta \mathrm{H}=-\mathrm{ve}\) Which of the following represents the above reaction mechanism?

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Chemical Kinetics

Reactant A converts to product D through the given mechanism (with the net evolution of heat) :

A \to B

slow;

Δ H = + ve

B \to C

fast;

Δ H = - ve

C \to D

fast;

Δ H = - ve

Which of the following represents the above reaction mechanism?

(A)(see structure)
✓ Correct
(B)(see structure)
(C)(see structure)
(D)(see structure)

Answer: (A)

Solution

The first step

A B

is slow, so it crosses the highest activation barrier, and it is endothermic, so the first intermediate B lies above the reactant A. The next two steps

B C

and

C D

are fast, each crossing only a small barrier, and both are exothermic, so the energy drops at each stage. Because the overall change evolves heat, the product D lies below A. The correct energy profile rises steeply over a large first barrier to an intermediate above A, then descends through two small barriers to a product lying below the starting level.

JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Chemical Kinetics

Solution

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