At a constant temperature,the activation energy of a reaction is found to be $2.303 \, RT \, J \, mol^{-1}$. The ratio of the rate constant to the Arrhenius constant will be $......$.

The rate coefficient $(k)$ for a particular reaction is $1.3 \times 10^{-4} \ M^{-1} \ s^{-1}$ at $100 \ ^oC$ and $1.3 \times 10^{-3} \ M^{-1} \ s^{-1}$ at $150 \ ^oC$. What is the energy of activation $(E_a)$ (in $kJ \ mol^{-1}$) for this reaction? $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

$A$ catalyst lowers the activation energy for a certain reaction from $83.314 \, kJ \, mol^{-1}$ to $75 \, kJ \, mol^{-1}$ at $500 \, K$. What will be the rate of reaction as compared to the uncatalysed reaction? Assume other things being equal.

The reason for almost doubling the rate of reaction on increasing the temperature of the reaction system by $10 \, ^\circ C$ is

An exothermic reaction $X \rightarrow Y$ has an activation energy of $30 \ kJ \ mol^{-1}$. If the energy change $\Delta E$ during the reaction is $-20 \ kJ \ mol^{-1}$,then the activation energy for the reverse reaction in $kJ \ mol^{-1}$ is $...$.

The decomposition of a hydrocarbon follows the equation $k = (4.5 \times 10^{11} \ s^{-1}) e^{-28000 \ K / T}$. Calculate the activation energy $E_a$.