The equation $k = (6.5 \times 10^{12} \, s^{-1}) e^{-26000 \, K / T}$ is followed for the decomposition of compound $A$. The activation energy for the reaction is $..... \, kJ \, mol^{-1}$. [nearest integer] (Given: $R = 8.314 \, J \, K^{-1} \, mol^{-1}$)

$A$ reaction takes place in three steps with individual rate constant and activation energy,

Step	Rate constant and Activation energy
$Step \ 1$	$k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$
$Step \ 2$	$k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$
$Step \ 3$	$k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$

Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.

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 $T$?

Assertion : According to transition state theory for the formation of an activated complex,one of the vibrational degrees of freedom is converted into a translational degree of freedom.
Reason : Energy of the activated complex is higher than the energy of reactant molecules.

The first order rate constant for the decomposition of $CaCO_3$ at $700 \ K$ is $6.36 \times 10^{-3} \ s^{-1}$ and activation energy is $209 \ kJ \ mol^{-1}$. Its rate constant (in $s^{-1}$) at $500 \ K$ is $x \times 10^{-6}$. The value of $x$ is ..... (Nearest integer)
Given $R=8.31 \ J \ K^{-1} \ mol^{-1} ; \log(6.36 \times 10^{-3})=-2.19 ; [10^{-4.79}=1.62 \times 10^{-5}]$

The velocity of the chemical reaction doubles every $10^\circ C$ rise of temperature. If the temperature is raised by $50^\circ C$,the velocity of the reaction increases to about .......... times.