The rate of reaction becomes double when the temperature of the reaction is increased from $27\,^{\circ}C$ to $57\,^{\circ}C$. The activation energy of the reaction will be ....... $k\,cal$.

The activation energy of a reaction is zero. Its rate constant at $280 \ K$ is $1.6 \times 10^{-6} \ s^{-1}$,the rate constant at $300 \ K$ is

What will be the effect on the fraction of molecules and the rate constant on increasing temperature?

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

Consider $A \xrightarrow{k_1} B$ and $C \xrightarrow{k_2} D$ are two reactions. If the rate constant $(k_1)$ of the $A \rightarrow B$ reaction can be expressed by the following equation $\log_{10} k = 14.34 - \frac{1.5 \times 10^4}{T/K}$ and activation energy of $C \rightarrow D$ reaction $(Ea_2)$ is $\frac{1}{5}$th of the $A \rightarrow B$ reaction $(Ea_1)$, then the value of $(Ea_2)$ is . . . . . . $kJ \ mol^{-1}$. (Nearest Integer)

Decomposition of a hydrocarbon follows the equation $k = (5.5 \times 10^{11} \text{ s}^{-1}) e^{\frac{-28000 \text{ K}}{T}}$. The activation energy of the reaction is . . . . . . $\text{kJ mol}^{-1}$. (Nearest Integer) Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$