What is the activation energy for the reverse of this reaction?
$N_2O_{4(g)} \to 2NO_{2(g)}$
Data for the given reaction is:
$\Delta H = +54 \ kJ$ and $E_a = +57.2 \ kJ$
Answer in $kJ$.

Collision theory is applicable to

$A$ reaction has an activation energy of $209 \, kJ \, mol^{-1}$. The rate increases $10$-fold when the temperature is increased from $27^{\circ} C$ to $X^{\circ} C$. The temperature $X$ is closest to
[Gas constant,$R = 8.314 \, J \, mol^{-1} \, K^{-1}$ ]

The temperature-dependent equation for the rate constant is written as:

The rate constant of a reaction at $500 \ K$ and $700 \ K$ are $0.02 \ s^{-1}$ and $0.2 \ s^{-1}$ respectively. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R=8.3 \ J \ K^{-1} \ mol^{-1})$

$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$
The curves $M$ and $N$ represent the variation of energy with reaction coordinate for the reaction in absence and presence of catalyst.
Which value represents the activation energy $(E_a)$ for the backward reaction in the presence of catalyst?