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 energies of activation for forward and reverse reactions for $A_2 + B_2 \rightleftharpoons 2AB$ are $180 \ kJ \ mol^{-1}$ and $200 \ kJ \ mol^{-1}$ respectively. The presence of a catalyst lowers the activation energy of both (forward and reverse) reactions by $100 \ kJ \ mol^{-1}$. The enthalpy change of the reaction $(A_2 + B_2 \to 2AB)$ in the presence of catalyst will be (in $kJ \ mol^{-1}$)

The rate of the chemical reaction doubles for an increase of $10 \text{ K}$ in absolute temperature from $298 \text{ K}$. What will be the activation energy?

Slope of the straight line obtained by plotting $\log_{10} k$ against $\frac{1}{T}$ represents which term?

Among the following,the $INCORRECT$ statement regarding the collision theory of chemical reaction is:

$A$ reaction takes place in three steps with individual rate constants and activation energies. The overall rate constant is given by $k = (\frac{k_1 k_2}{k_3})^{2/3}$. The overall activation energy of the reaction in $kJ/mol$ is:

$Step$	$Rate\ Constant\ /\ Activation\ energy$
$Step\ 1$	$k_1, E_{a_1} = 180\ kJ/mol$
$Step\ 2$	$k_2, E_{a_2} = 80\ kJ/mol$
$Step\ 3$	$k_3, E_{a_3} = 50\ kJ/mol$