The minimum energy a molecule should possess in order to enter into a fruitful collision is known as

The rate constant for the decomposition of a hydrocarbon is given as $K = (4.5 \times 10^{11} \ s^{-1}) e^{-(28000 \ K) / T}$. The activation energy of the reaction (in $J/mol$) is:

Consider $A \xrightarrow{k_1} B$ and $C \xrightarrow{k_2} D$ are two reactions. If the rate constant $(k_1)$ of the $A \rightarrow B$ reaction can be expressed by the following equation $\log_{10} k = 14.34 - \frac{1.5 \times 10^4}{T/K}$ and activation energy of $C \rightarrow D$ reaction $(Ea_2)$ is $\frac{1}{5}$th of the $A \rightarrow B$ reaction $(Ea_1)$, then the value of $(Ea_2)$ is . . . . . . $kJ \ mol^{-1}$. (Nearest Integer)

For $N_2 + 3H_2 \rightarrow 2NH_3$,$\Delta H = -22 \ kcal$,and $E_a = 70 \ kcal$. Hence $E_a$ for $2NH_3 \rightarrow N_2 + 3H_2$ is $.....$ $kcal$.

$A$ catalyst lowers the activation energy for a certain reaction from $83.314 \, kJ \, mol^{-1}$ to $75 \, kJ \, mol^{-1}$ at $500 \, K$. What will be the rate of reaction as compared to the uncatalysed reaction? Assume other things being equal.

Fill in the blanks:
$(a)$ $A$ catalyst ......... the change of equilibrium in reaction.
$(b)$ Boltzmann and Maxwell used ......... to explain the rate of reaction.