For an exothermic reaction $A \rightarrow B$,the activation energy of $A$ is $17 \, kJ/mol$. The heat of reaction is $40 \, kJ$. Calculate the activation energy for the reverse reaction $B \rightarrow A$.

The rate of a reaction at $273 \ K$ is $R_0$. What will be the rate of the reaction at $313 \ K$? (Take the temperature coefficient as $2$)

The equilibrium constant at $27\,^{\circ}C$ and $127\,^{\circ}C$ is $30$ and $40$ respectively. What will be the ratio of activation energy $(E_{a})_{f} / (E_{a})_{b}$?

At $300 \ K$,the activation energy and rate constant for the given reaction are $10 \ kJ \ mol^{-1}$ and $2.4 \times 10^{-5} \ s^{-1}$ respectively. At what temperature (in $K$) will the value of $t_{1/2}$ be $2 \ hr$?
$2N_2O_5 \to 4NO_2 + O_2$

The number of correct statement/s from the following is:
$A.$ Larger the activation energy,smaller is the value of the rate constant.
$B.$ The higher is the activation energy,higher is the value of the temperature coefficient.
$C.$ At lower temperatures,the increase in temperature causes a larger change in the value of $k$ than at higher temperatures.
$D.$ $A$ plot of $\ln k$ vs $\frac{1}{T}$ is a straight line with a slope equal to $-\frac{E_a}{R}$.

For a first order decomposition of a certain reaction,rate constant is given by the equation $\log k \left( s^{-1} \right) = 7.14 - \frac{1 \times 10^4 \ K}{T}$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$ (in $.1$)