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

At $27^{\circ}C$ in the presence of a catalyst,the activation energy of a reaction is lowered by $10 \ kJ \ mol^{-1}$. The logarithm ratio of $\frac{k(\text{catalysed})}{k(\text{uncatalysed})}$ is .... (Consider that the frequency factor for both the reactions is the same)

The specific rate constant of decomposition of a compound is given by $\ln k = 5.0 - \frac{12000}{T}$. The activation energy of decomposition for this compound at $300 \ K$ is

For a chemical reaction,the rate constant,activation energy,and Arrhenius factor at $25 \, ^\circ C$ are $3.0 \times 10^{-4} \, s^{-1}$,$104.4 \, kJ \, mol^{-1}$,and $6.0 \times 10^{14} \, s^{-1}$ respectively. Find the value of the rate constant as $T \rightarrow \infty$.

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

For a first order reaction $A \rightarrow P$,the temperature $(T)$ dependent rate constant $(k)$ was found to follow the equation $\log_{10} k = -(2000) \frac{1}{T} + 6$. The activation energy $(E_a)$ of the reaction in $kJ \, mol^{-1}$ will be ......... (Given: $\ln x = 2.3 \times \log_{10} x$ and $R = 8 \, J \, mol^{-1} K^{-1}$)