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: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$

Activation energy of any reaction depends on

For a first-order reaction $A \to P$,the rate constant equation is given by $\log K = -2000 \, (1/T) + 6.0$. The pre-exponential factor $A$ and the activation energy $E_a$ are,respectively:

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)

The rate constant of a chemical reaction at a very high temperature will approach:

The rate of a reaction quadruples when the temperature changes from $300 \, K$ to $310 \, K$. The activation energy of this reaction is ........... $kJ \, mol^{-1}$ (Assume activation energy and pre-exponential factor are independent of temperature; $\ln 2 = 0.693$; $R = 8.314 \, J \, mol^{-1} \, K^{-1}$)