The rate constant of a first order reaction was doubled when the temperature was increased from $300 \ K$ to $310 \ K$. What is its approximate activation energy (in $kJ \cdot mol^{-1}$)? ($R = 8.3 \ J \cdot mol^{-1} \cdot K^{-1}$; $\log 2 = 0.3$)

The activation energy of a forward reaction is $50 \, kcal$. What will be the activation energy of its reverse reaction?

The activation energy of a reaction is zero. Its rate constant at $280 \ K$ is $1.6 \times 10^{-6} \ s^{-1}$,the rate constant at $300 \ K$ is

The plot of $\log K$ vs $\frac{1}{T}$ helps to calculate

Two reactions $R_1$ and $R_2$ have identical pre-exponential factors. Activation energy of $R_1$ exceeds that of $R_2$ by $10 \, kJ \, mol^{-1}.$ If $k_1$ and $k_2$ are rate constants for reactions $R_1$ and $R_2$ respectively at $300 \, K,$ then $\ln (k_2/k_1)$ is equal to :
$(R=8.314 \, J \, mol^{-1} \, K^{-1})$

$A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)} + D_{(g)}$
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?