If we plot a graph between $\log \, K$ and $\frac{1}{T}$ by Arrhenius equation,the slope is

According to the collision theory of reaction rates:

When temperature is increased from $27\,{}^{\circ}C$ to $127\,{}^{\circ}C$,the rate of reaction becomes doubled. The activation energy $E_a$ will be ........ $kCal$.

For a first order decomposition of a certain reaction,rate constant is given by the equation $\log k \left( s^{-1} \right) = 7.14 - \frac{1 \times 10^4 \ K}{T}$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$ (in $.1$)

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 $50^{\circ}C$.

In a laboratory,a biochemical reaction performed outside the human body in the absence of an enzyme proceeds at a rate $10^{-6}$ times slower. If this reaction is performed in the presence of an enzyme,what will be the activation energy of the reaction?