What is the slope of the graph between $\ln K$ and $\frac{1}{T}$ according to the Arrhenius equation?

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

The value of the rate constant for the reaction $A + B \to \text{products}$ depends on:

What is the activation energy for a reaction if its rate doubles when the temperature is raised from $20 \,^{\circ}C$ to $35 \,^{\circ}C$ in $kJ \,mol^{-1}$? $(R = 8.314 \,J \,mol^{-1} \,K^{-1})$

The rate of a reaction can be expressed by the Arrhenius equation as:
$k = A e^{-E_a / RT}$
In this equation,$E_a$ represents:

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