The number of collisions depend upon

Which of the following graphs of $\log \, K \rightarrow 1/T$ can be used to calculate the activation energy?

The rate constant for the decomposition of hydrocarbons is $2.418 \times 10^{-5} \, s^{-1}$ at $546 \, K$. If the energy of activation is $179.9 \, kJ / mol$,what will be the value of the pre-exponential factor?

For $N_2 + 3H_2 \rightarrow 2NH_3$,$\Delta H = -22 \ kcal$,and $E_a = 70 \ kcal$. Hence $E_a$ for $2NH_3 \rightarrow N_2 + 3H_2$ is $.....$ $kcal$.

Decomposition of a hydrocarbon follows the equation $k = (5.5 \times 10^{11} \text{ s}^{-1}) e^{\frac{-28000 \text{ K}}{T}}$. The activation energy of the reaction is . . . . . . $\text{kJ mol}^{-1}$. (Nearest Integer) Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$

The rate of a reaction decreased by $3.555$ times when the temperature was changed from $40^{\circ} C$ to $30^{\circ} C$. The activation energy of the reaction is.........$kJ \, mol^{-1}$.
[Take; $R = 8.314 \, J \, mol^{-1} \, K^{-1}$,$\ln(3.555) = 1.268$]