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JEE Main 20252 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
2 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Chemical Kinetics

JEE Main 2025 · 2 April · Evening Shift · Chemistry · Chemical Kinetics

Reactant A converts to product D through the given mechanism (with...

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

Options

  1. AAn organic compound undergoes first order decomposition. The time taken for decomposition to ((1)/(8))^ th and
  2. BAn organic compound undergoes first order decomposition. The time taken for decomposition to ((1)/(8))^ th and
  3. CAn organic compound undergoes first order decomposition. The time taken for decomposition to ((1)/(8))^ th and
  4. DAn organic compound undergoes first order decomposition. The time taken for decomposition to ((1)/(8))^ th and

Correct answer

A. An organic compound undergoes first order decomposition. The time taken for decomposition to ((1)/(8))^ th and

Step-by-step solution

A→BSlowΔH=+veEa1→ High B→CfastΔH=−veEa2→ Low C→DΔD=−veEa3→ Low \begin{array}{ll}\mathrm{A} \rightarrow \mathrm{B} Slow & \\ \Delta \mathrm{H}=+\mathrm{ve} & \mathrm{E}_{\mathrm{a}_1} \rightarrow \text{ High } \\ \mathrm{B} \rightarrow \mathrm{C} fast& \\ \Delta \mathrm{H}=-\mathrm{ve} & \mathrm{E}_{\mathrm{a}_2} \rightarrow \text{ Low } \\ \mathrm{C} \rightarrow \mathrm{D} & \\ \Delta \mathrm{D}=-\mathrm{ve} & \mathrm{E}_{\mathrm{a}_3} \rightarrow \text{ Low }\end{array}
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