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JEE Main 2-April-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 2-April-2025 Paper - Shift 24 Marks
MCQ
Reactant A converts to product D through the given mechanism (with the net evolution of heat) :
$\mathrm{A} \rightarrow \mathrm{B}$ slow ; $\Delta \mathrm{H}=+\mathrm{ve}$
$\mathrm{B} \rightarrow \mathrm{C}$ fast; $\Delta \mathrm{H}=-\mathrm{ve}$
$\mathrm{C} \rightarrow \mathrm{D}$ fast; $\Delta \mathrm{H}=-\mathrm{ve}$
Which of the following represents the above reaction mechanism?
  • A
    Image
  • B
    Image
  • C
    Image
  • D
    Image
✓

Answer

A.
Image
Image
\begin{array}{ll}
A \rightarrow B & & Slow\\
\Delta H-+ve & E_{a_1} \rightarrow \text { High } \\
B \rightarrow C & & Fast\\
\Delta H=-ve & E_{a_2} \rightarrow \text { Low } \\
C \rightarrow D & \\
\Delta H=-ve & E_{a_3} \rightarrow \text { Low }
\end{array}

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