For an exothermic reaction,the following two steps are involved:
Step $1$: $A + B \to I$ $(slow)$
Step $2$: $I \to AB$ $(fast)$
Which of the following graphs correctly represents this reaction?

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 gaseous reaction,a large increase in the rate of reaction with a small increase in temperature indicates:

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.....$kJ \ mol^{-1}$

The rate of a reaction quadruples when temperature changes from $27^{\circ} C$ to $57^{\circ} C$. Calculate the energy of activation.
Given $R=8.314 \ J \ K^{-1} \ mol^{-1}, \log 4=0.6021$

The influence of temperature on the rate of reaction can be found out by