Question:medium

The activation energy for a reaction at T (K) was found to be \(2.303RT\ J\ mol^{-1}\). The ratio of rate constant to Arrhenius factor is

Show Hint

Remember: \[ 2.303=\ln 10 \] Therefore \[ e^{-2.303}=10^{-1}=0.1 \] A very common shortcut in Arrhenius problems.
Updated On: Jun 22, 2026
  • 0.01
  • 0.1
  • 0.02
  • 0.002 \bigskip
Show Solution

The Correct Option is B

Solution and Explanation

Was this answer helpful?
0

Top Questions on Chemical Kinetics

Questions Asked in TS EAMCET exam