How does the rate of reaction change with an increase in temperature?

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 plot of $\log K$ vs $\frac{1}{T}$ helps to calculate

For a reaction,the graph of $\ln k$ (on y-axis) and $1 / T$ (on x-axis) is a straight line with a slope $-2 \times 10^4 \ K$. The activation energy of the reaction (in $kJ \ mol^{-1}$) is $(R = 8.3 \ J \ K^{-1} \ mol^{-1})$

For an exothermic reaction $X \rightarrow Y$,the activation energy is $30 \ kJ \ mol^{-1}$. If the enthalpy change $(\Delta H)$ for the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction is . . . . . . $kJ \ mol^{-1}$.

For a reaction,the rate constant is expressed as,$K = A \cdot e^{-40000/T}$. The energy of activation is ....... $cal$.