Question

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

• A. 0.01
• B. 0.1
• C. 0.02
• D. 0.002 \bigskip

Hint:

Remember: \[ 2.303=\ln 10 \] Therefore \[ e^{-2.303}=10^{-1}=0.1 \] A very common shortcut in Arrhenius problems.

Answer 1

The Correct Option is B

Concept: Arrhenius equation: \[ k=Ae^{-E_a/RT} \] where
• \(k\) = rate constant
• \(A\) = Arrhenius factor
• \(E_a\) = activation energy Therefore \[ \frac{k}{A} = e^{-E_a/RT} \]

Step 1: Substitute the given activation energy.
\[ E_a=2.303RT \] Hence \[ \frac{k}{A} = e^{-2.303} \]

Step 2: Convert exponential form into logarithmic form.
Since \[ 2.303=\ln 10 \] therefore \[ e^{-2.303} = e^{-\ln 10} \] \[ = \frac{1}{10} \] \[ = 0.1 \]

Step 3: Write the final answer.
\[ \boxed{\frac{k}{A}=0.1} \] Hence Option (B) is correct.