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

Regarding the velocity in a reversible reaction,which is the correct explanation of the effect of a catalyst?

The rate constant is doubled when temperature increases from $27^{\circ} C$ to $37^{\circ} C$. What is the activation energy in $kJ \ mol^{-1}$?

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?

The rate constant of a reaction varies with temperature according to the equation: $\log K = \text{constant} - \frac{E_a}{2.303 RT}$. If a plot of $\log K$ versus $1/T$ yields a straight line with a slope of $-5632$,then the activation energy of the reaction is .......... $kJ \ mol^{-1}$.

In a reaction,the rate constant $K_1$ is twice the rate constant $K_2$ of another reaction. Find the relationship between the activation energies $E_1$ and $E_2$ of the two reactions at the same temperature.