The activation energy for the reaction $X \rightarrow Y$ is $150 \text{ kJ mol}^{-1}$. The change in enthalpy for the above reaction is $-135 \text{ kJ mol}^{-1}$. What is the activation energy for the reverse reaction $Y \rightarrow X$?

The activation energy of one of the reactions in a biochemical process is $532611 \, J \, mol^{-1}$. When the temperature falls from $310 \, K$ to $300 \, K$,the change in rate constant observed is $k_{300} = x \times 10^{-3} \, k_{310}$. The value of $x$ is $.....$ [Given: $\ln 10 = 2.3$,$R = 8.3 \, J \, K^{-1} \, mol^{-1}$]

It is found that for every $10^{\circ}C$ rise in temperature,the rate of the reaction doubles itself. This is mainly because

Which of the following statements is in accordance with the Arrhenius equation?

The rate of a 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 $-\frac{E_a}{R}$.