Rate constant of a reaction is given as $\log_{10} K = -\frac{2000}{T} + 6.0$. Activation energy is ....... $kcal$.

Activation energy is defined as:

For a first order reaction $A \rightarrow P$,the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation $\log k = -(2000) \frac{1}{T} + 6.0$. The pre-exponential factor $A$ and the activation energy $E_{a}$,respectively,are

$A$ sample of milk splits after $60 \; min.$ at $300 \; K$ and after $40 \; min.$ at $400 \; K$ when the population of $Lactobacillus \; acidophilus$ in it doubles. The activation energy (in $kJ / mol$) for this process is closest to ............. $kJ / mol$.
(Given: $R = 8.3 \; J \; mol^{-1} \; K^{-1}$,$\ln(1.5) = 0.405$)

Activation energy for the hydrolysis of sucrose by acid is $X \ kJ \ mol^{-1}$ whereas activation energy for the hydrolysis of sucrose by sucrase is $Y \ kJ \ mol^{-1}$. $X$ and $Y$ respectively are

For a certain reaction,a large fraction of molecules has energy more than the threshold energy,yet the rate of reaction is very slow. Why?