The rate constant is doubled when temperature increases from $27\,^oC$ to $37\,^oC.$ Activation energy in $kJ$ is

The temperature coefficient of the saponification reaction of an ester by $NaOH$ is $1.75$. The activation energy of the reaction is .......... $kcal \ mol^{-1}$. (in $.21$)

For the gaseous reactions,calculate the approximate value of temperature at which $k_1 = k_2$. $[\ln\, 10 = 2.3]$.
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.314\, T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.314\, T}$

For a particular reaction,the rate constant becomes double on increasing temperature from $27^{\circ} C$ to $37^{\circ} C$. Calculate the approximate activation energy (in $kcal \ mol^{-1}$,$R=2 \ cal \ mol^{-1} \ K^{-1}$).

In a reaction,the rate of reaction doubles with a $10^\circ C$ rise in temperature. If the temperature is increased from $10^\circ C$ to $100^\circ C$,by how many times will the rate of reaction increase?

An endothermic reaction with high activation energy for the forward reaction is given by the diagram: