The plot of $\log K$ vs $\frac{1}{T}$ helps to calculate

The rate constant of a reaction is increased $4$ times after the addition of a catalyst to the reaction mixture at the same temperature of $27^{\circ} C$. The change in the activation energy of this reaction is (Take $\ln(1/4) = -1.386, R = 8.314 \ J \ K^{-1} \ mol^{-1}$)

The rate of a reaction $A$ doubles on increasing the temperature from $300 \, K$ to $310 \, K$. By how much should the temperature of reaction $B$ be increased from $300 \, K$ so that its rate doubles,if the activation energy of reaction $B$ is twice that of reaction $A$ (in $, K$)?

At what temperature does the rate become double that at $300 \, K$ (in $, K$)? Given $\ln k = 10 - \frac{69 \, kJ}{RT}$.

$A$ 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.

On introducing a catalyst at $500 \, K,$ the rate constant of a first order reaction increases $2.718$ times. If the activation energy in the presence of a catalyst is $4.15 \, kJ \, mol^{-1},$ then what will be $E_a$ in the absence of a catalyst? (Value of $e = 2.718$ and $R = 8.314 \times 10^{-3} \, kJ \, K^{-1} \, mol^{-1}$)