If a reaction follows the Arrhenius equation,the plot $\ln k$ vs $1/(RT)$ gives a straight line with a gradient of $(-y) \ unit$. The energy required to activate the reactant is

The rate of a reaction quadruples when the temperature changes from $293 \ K$ to $313 \ K$. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

The rate of a certain biochemical reaction at physiological temperature $T$ occurs $10^{6}$ times faster with an enzyme than without. The change in the activation energy upon adding the enzyme is

The velocity constant of a reaction at $290 \ K$ was found to be $3.2 \times 10^{-3}$. At $310 \ K$,it will be about:

The reaction $X \to Y$ is an exothermic reaction. The activation energy of the forward reaction $X \to Y$ is $150\,kJ\,mol^{-1}$. The enthalpy of the reaction is $-135\,kJ\,mol^{-1}$. The activation energy for the reverse reaction,$Y \to X$,will be $.......\,kJ\,mol^{-1}$.

What is the slope of the straight line for the graph drawn between $\ln k$ and $\frac{1}{T}$,where $k$ is the rate constant of a reaction at temperature $T$?