In a first order reaction at $27\,^oC$ and $47\,^oC$,$50\%$ of the reaction is complete in $30\,\min$ and $10\,\min$ respectively. Calculate the energy of activation $(E_a)$.

For a reaction $A \rightarrow B$,the enthalpy of reaction is $-4.2 \ kJ \ mol^{-1}$ and the enthalpy of activation is $9.6 \ kJ \ mol^{-1}$. The correct potential energy profile for the reaction is shown in which option?

Define the following terms:
$1.$ Collision
$2.$ Frequency of collision

The reaction $X \to Y$ is an exothermic reaction. The activation energy of the forward reaction $X \to Y$ is $150\,kJ\,mol^{-1}$. The enthalpy of the reaction is $-135\,kJ\,mol^{-1}$. The activation energy for the reverse reaction,$Y \to X$,will be $.......\,kJ\,mol^{-1}$.

The decomposition of ethane,$\frac{d[C_2H_6]}{dt} = k[C_2H_6]$,proceeds through a complex mechanism,which includes $5$ steps. The overall rate constant $(k)$ is expressed as $k = \frac{k_1 k_2 k_3}{k_2 k_5}$,where $k_1, k_2, k_3, k_4, k_5$ are the rate constants of the $5$ steps. If the activation energies of the steps are $E_1 = 1E, E_2 = 2E, E_3 = 3E, E_4 = 4E, E_5 = 5E$,where $E = 20 \ kJ/mol$,find the overall activation energy of the decomposition.

For an endothermic reaction $X \rightarrow Y$,the activation energies for the forward and backward reactions are $E_f$ and $E_b$ respectively. Then,in general: