Answer:
Option B,C,D
Explanation:
Rate constant, $k= Ae^{-E_{a}/RT}$ where,
$E_{a}$= activation energy and
A= pre exponential factor
(a) If $E_{a}$ is high, it means lower value of k hence, slow reaction . Thus incorrect.
(b) On increasing temperature, molecules are raised to higher energy (greater than $E_{a}$ ), hence number of collisions increases. Thus, correct
(c) $\log k=\log A-\frac{E_{a}}{RT}$
$\Rightarrow$ $\frac{d(\log k)}{dT}=\frac{E_{a}}{RT^{2}}$
Thus, when $E_{a}$ is high, stronger is the temperature dependence of the rate constant. Thus, correct.
(d) Pre-exponential factor (A) is a measure of the rate at which collisions occur. Thus, correct.
