For a first-order gaseous reaction,a plot of $\log \, k$ versus $1/T$ gives a straight line with a slope of $-8000$. Calculate the activation energy $(E_a)$ of the reaction in $cal$.

When a biochemical reaction is carried out in a laboratory without the presence of enzymes,the rate of reaction is $10^{-6}$ times slower. What will be the activation energy $(E_a)$ in the presence of enzymes?

The slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)
$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

The rate of a reaction doubles when its temperature changes from $300 \, K$ to $310 \, K.$ Activation energy of such a reaction will be .......... $kJ \, mol^{-1}$. $(R= 8.314 \, J \, K^{-1} \, mol^{-1}$ and $\log 2=0.301)$

The rate constant of a first order reaction is $3.46 \times 10^{-2} \ s^{-1}$ at $298 \ K$. What is the rate constant of the reaction at $350 \ K$ if its activation energy is $50.1 \ kJ \ mol^{-1}$ (in $s^{-1}$)? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$
$(\log 2 = 0.3010)$

Activation energy of a reaction is