The activation energy of a certain reaction is $87 \ kJ \ mol^{-1}$. When the temperature is decreased from $37^{\circ} C$ to $15^{\circ} C$,what is the ratio of the rate constant at $37^{\circ} C$ to that at $15^{\circ} C$ for this reaction (in $/ 1$)?

$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$
The curves $M$ and $N$ represent the variation of energy with reaction coordinate for the reaction in absence and presence of catalyst.
Which value represents the activation energy $(E_a)$ for the backward reaction in the presence of catalyst?

For a hypothetical reaction $A \to B$, the activation energies for the forward and backward reactions are $19 \ kJ/mol$ and $9 \ kJ/mol$ respectively. The heat of reaction is $.... \ kJ/mol$.

If the rate of a reaction doubles when the temperature increases from $298 \, K$ to $308 \, K$,the activation energy of the reaction is ........... $kJ \, mol^{-1}$.

The decomposition of a hydrocarbon follows the equation $k = (4.5 \times 10^{11} \ s^{-1}) e^{-28000 \ K / T}$. Calculate the activation energy $E_a$.

At a constant temperature,the activation energy of a reaction is found to be $2.303 \, RT \, J \, mol^{-1}$. The ratio of the rate constant to the Arrhenius constant will be $......$.