For an endothermic reaction,the energy of activation is $E_a$ and the enthalpy of reaction is $\Delta H$ (both in $kJ/mol$). The minimum value of $E_a$ will be:

If $E_f$ and $E_r$ are the activation energies of forward and reverse reactions and the reaction is known to be exothermic,then

For the decomposition reaction of $N_2O_5$,the slope of the graph of $\log K$ versus $1/T$ is $-1.2 \times 10^4 \ K$. Calculate the activation energy $(E_a)$ of the reaction.

$A$ 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:

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

Consider a complex reaction taking place in three steps with rate constants $k_1$,$k_2$,and $k_3$ respectively. The overall rate constant $k$ is given by the expression $k = \sqrt{\frac{k_1 k_3}{k_2}}$. If the activation energies of the three steps are $60$,$30$,and $10 \ kJ \ mol^{-1}$ respectively,then the overall energy of activation in $kJ \ mol^{-1}$ is $..........$ $(Nearest \ integer)$

The half-life periods of a first order reaction at $300 \ K$ and $400 \ K$ are $50 \ s$ and $10 \ s$ respectively. The activation energy of the reaction in $kJ \ mol^{-1}$ is $(\log 5 = 0.70)$