The following equation is obtained for a first order reaction at 300 K

$$ \log _{10} \frac{k}{A}=0.00174 $$

What is the activation energy (in $\mathrm{J} \mathrm{mol}^{-1}$ ) of the reaction?

$$ \left(R=8314 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}\right) $$

In a first order reaction, the concentration of the reactant is reduced to $1 / 8$ of the initial concentration in 75 minutes. The $t_{1 / 2}$ of the reaction (in minutes) is $(\log 2=0.30, \log 3=0.47, \log 4=0.60)$

At $T(\mathrm{~K})$ the following equation is obtained for a first order reaction $\log \frac{k}{A}=-\frac{x}{T}$. The activation energy for this reaction is equal to ( $R=$ gas constant)