How much faster would a reaction proceed at $25\,^{\circ}C$ than at $0\,^{\circ}C$ if the activation energy is $65\,kJ/mol$?

The rate constant,the activation energy and the Arrhenius parameter of a chemical reaction at $25\,^oC$ are $3.0 \times 10^{-4}\,s^{-1}$,$104.4\,kJ\,mol^{-1}$ and $6.0 \times 10^{14}\,s^{-1}$ respectively. The value of the rate constant as $T \to \infty$ is

For $A + B \longrightarrow C + D$; $\Delta H = -20 \ kJ \ mol^{-1}$,the activation energy of the forward reaction is $85 \ kJ \ mol^{-1}$. The activation energy for the backward reaction is.....$kJ \ mol^{-1}$

For the reaction of $H_2$ with $I_2$,the rate constant is $2.5 \times 10^{-4} \ dm^3 \ mol^{-1} \ s^{-1}$ at $327 \ ^oC$ and $1.0 \ dm^3 \ mol^{-1} \ s^{-1}$ at $527 \ ^oC$. The activation energy for the reaction,in $kJ \ mol^{-1}$ is: $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

Activation energy of any chemical reaction can be calculated if one knows the value of

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