What is the slope of the straight line for the graph drawn between $\ln k$ and $\frac{1}{T}$,where $k$ is the rate constant of a reaction at temperature $T$?

The rate constant for the decomposition of $N_{2}O_{5}$ at various temperatures is given below:

$T / ^{\circ}C$	$0$	$20$	$40$	$60$	$80$
$10^{5} \times k / s^{-1}$	$0.0787$	$1.70$	$25.7$	$178$	$2140$

Draw a graph between $\ln k$ and $1 / T$ and calculate the values of $A$ and $E_{a}.$ Predict the rate constant at $30^{\circ}C$ and $50^{\circ}C$.

$A$ reaction rate constant is given by $k = 1.2 \times 10^{14} \, e^{-25000/RT} \, sec^{-1}$. It means

$A$ reaction has an activation energy of $209 \, kJ \, mol^{-1}$. The rate increases $10$-fold when the temperature is increased from $27^{\circ} C$ to $X^{\circ} C$. The temperature $X$ is closest to
[Gas constant,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$ ]

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

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