Question

Graph between log K and $\frac{1}{T}$ [Where K is rate constant $\left(S^{-1}\right)$ and T is temperature (K)] is a straight line with $OX=5,\theta =ta{n}^{-1}[-\frac{1}{2.303}]$
Hence $E_a$ and log A respectively will be:

• A. $2.303\times 2cal,5$
• B. $\frac{2}{3.303}cal,e^5$
• C. 2 cal, 5
• D. None of these

Answer 1

The correct option is C 2 cal, 5

$K+Ae=\frac{E_a}{RT}$
$logK=logA-\frac{E_a}{2.303R}\times \frac{1}{T}$
$Slope=tan\theta [-\frac{1}{2.303}]=-\frac{E_a}{2.303R}$
So, $E_a=R=2cal/mol$