Question:

In the Arrhenius equation, \(k = Ae^{- Ea / RT}\), which factor corresponds to the fraction of molecules that have kinetic energy greater than activation energy ?

Show Hint

As temperature increases, \(e^{-E_a/(RT)}\) increases, hence rate constant increases.
Updated On: Apr 24, 2026
  • \(e^{- Ea / RT}\)
  • \(e^{- Ea / R}\)
  • \(e^{- Ea / T}\)
  • \(e^{- Ea}\)
  • \(Ae^{- Ea / R}\)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
Arrhenius equation: \(k = A e^{-E_a/(RT)}\). \(A\) is frequency factor, \(e^{-E_a/(RT)}\) is Boltzmann factor.

Step 2:
Detailed Explanation:
The exponential term \(e^{-E_a/(RT)}\) represents the fraction of molecules with energy \(\geq E_a\). \(A\) accounts for collision frequency and orientation.

Step 3:
Final Answer:
The fraction of molecules with KE$>$Ea is \(e^{-E_a/(RT)}\).
Was this answer helpful?
0
0

Top KEAM Collision Theory of Chemical Reactions Questions

View More Questions