Thank you for reporting, we will resolve it shortly
Q.
The rate constant for a first order decomposition reaction is given by :$\log K =10-\frac{1000}{ T }$. Then, what will be
activation energy in $kcal / mol$ ?
Solution:
$\log \,K =\log A -\frac{ E _{ a }}{2.303 RT }\ldots$(1)
$\log \,k =10-\frac{1000}{ T } $ (given) $\ldots(2)$
Compairing equ. (1) & ( 2$)$
$\frac{ E _{ a }}{2.303 R }=1000$
$E _{ a }=4.60\, k\, cal\, mol ^{-1}$