The rate constant is given by the equation $k = p Z e^{-E_a/RT}$. Which factor should register a decrease for the reaction to proceed more rapidly?

On increasing the temperature,the rate of the reaction increases because of

Given below is an expression for the rate constant of a first order reaction occurring at a certain temperature, $T (\text{K})$.
$\ln k = 14.34 - \frac{1.25 \times 10^4}{T}$
The energy of activation in $\text{kcal mol}^{-1}$ for the reaction is :
(Given : $k$ is $\text{s}^{-1}$, $R = 1.987 \text{ cal mol}^{-1} \text{ K}^{-1}$)

In collision theory of chemical reaction,$Z_{AB}$ represents

The rate of reaction becomes double when the temperature of the reaction is increased from $27\,^{\circ}C$ to $57\,^{\circ}C$. The activation energy of the reaction will be ....... $k\,cal$.

Given the activation energy for the forward direction $(E_{af})$ is $50 \ kJ$ and for the backward direction $(E_{ab})$ is $30 \ kJ$. The reaction would be: