Report Issue

JEE MAIN_2026
_280126_S2
Question
$A \longrightarrow B$ (first reaction) $\mathrm{C} \longrightarrow \mathrm{D}$ (second reaction) Consider the above two first-order reactions. The rate constant for first reaction at 500 K is double of the same at 300 K . At $500 \mathrm{~K}, 50 \%$ of the reaction becomes complete in 2 hour. The activation energy of the second reaction is half of that of first reaction. If the rate constant at 500 K of the second reaction becomes double of the rate constant of first reaction at the same temperature; then rate constant for the second reaction at 300 K is $\_\_\_\_$ $\times 10^{-1}$ hour ${ }^{-1}$ (Nearest integer).
Write Your Answer
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
Solution Image
Question Tags
JEE Main
Chemistry
Hard
Start Preparing for JEE with Competishun