It has been found that for a chemical reaction with rise in temperature by 9K the rate constant gets doubled. Assuming a reaction to be occurring at 300K, the value of activation energy is found to be _____ ${kJmol}^{- 1}$ . [nearest integer]
(Given $\ln 10 = 2.3, R = 8.3 {JK}^{- 1} {mol}^{- 1}, \log 2 = 0.30$ )

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 59.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\log_{10} \frac{K_{2}}{K_{1}} = \frac{E_{a}}{2.303 R} (\frac{1}{300} - \frac{1}{309})$
$0.3 = \frac{E_{a}}{2.303 \times 8.3} (\frac{9}{300 \times 309})$

$E_{a} = \frac{0.3 \times 2.303 \times 8.3 \times 300 \times 309}{9} = 59065.04 J E_{a} = 59.06 kJ$

Watch 3-min video & get full concept clarity

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Get Expert Academic Guidance – Connect with a Counselor Today!