Search any question & find its solution
Question:
Answered & Verified by Expert
Activation energy is given by the formula
Options:
Solution:
2331 Upvotes
Verified Answer
The correct answer is:
$\log \frac{K_1}{K_2}=-\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right]$
\(\log _e K_2=\log _e A-\frac{E_A}{R T_2} \ldots \ldots \ldots \ldots . .(1)\)
\(\log _e K_1=\log _e A-\frac{E_A}{R T_1} \ldots \ldots \ldots \ldots . . . . .(2)\)
Eqn. (1) - (2) we get,
\(\log \frac{K_2}{K_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right]\)
\(\log _e K_1=\log _e A-\frac{E_A}{R T_1} \ldots \ldots \ldots \ldots . . . . .(2)\)
Eqn. (1) - (2) we get,
\(\log \frac{K_2}{K_1}=\frac{E_a}{2.303 R}\left[\frac{T_2-T_1}{T_1 T_2}\right]\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.