A first order reaction 75% completed in 120 min at 300 K temperature and in 40 min at 320 K. The activation energy of reaction is ______kcal/mol.
(given : ln 3 = 1.1)

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

answer is 11.

(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

At 300 K,    $t_{0.5} = \frac{120}{2} = 60 \min$
At 320 K    $t_{0.875} = 1.5 \times 40 = 60 \min$
ln    $\frac{60}{20} = \frac{E_{a}}{2 \times 10^{- 3}} (\frac{1}{300} - \frac{1}{320})$
$E_{a} = 10.56 c a l / m o l$

Watch 3-min video & get full concept clarity

hear from our champions

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

🔥 Start Your JEE/NEET Prep at Just ₹1999 / month - Limited Offer! Check Now!

Get Expert Academic Guidance – Connect with a Counselor Today!