$A$ 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 activation energy for a reaction which doubles the rate when the temperature is raised from $298 \ K$ to $308 \ K$ is ........... $kJ \ mol^{-1}$

The first order rate constant for the decomposition of $CaCO_3$ at $700 \ K$ is $6.36 \times 10^{-3} \ s^{-1}$ and activation energy is $209 \ kJ \ mol^{-1}$. Its rate constant (in $s^{-1}$) at $500 \ K$ is $x \times 10^{-6}$. The value of $x$ is ..... (Nearest integer)
Given $R=8.31 \ J \ K^{-1} \ mol^{-1} ; \log(6.36 \times 10^{-3})=-2.19 ; [10^{-4.79}=1.62 \times 10^{-5}]$

For an exothermic reaction,where the enthalpy of reaction is $\Delta H$ in $kJ/mol$,the minimum value for the activation energy will be:

$A$ first order reaction is $50\%$ completed in $20\, \text{minutes}$ at $27\, ^oC$ and in $10\, \text{minutes}$ at $47\, ^oC$. The energy of activation of the reaction is ....... $Kcal$.

If the half-lives of a first-order reaction at $350 \ K$ and $300 \ K$ are $2 \ s$ and $20 \ s$ respectively,the activation energy of the reaction in $kJ \ mol^{-1}$ is: