What is the slope of the graph between $\ln K$ and $\frac{1}{T}$ according to the Arrhenius equation?

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 $, K$)?

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 $R = 8.314 \times 10^{-3} \, kJ \, K^{-1} \, mol^{-1}$)

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

The slope of the Arrhenius plot $(\ln k \, vs \, \frac{1}{T})$ of a first-order reaction is $-5 \times 10^{3} \, K$. The value of $E_{a}$ of the reaction is: (in $kJ \, mol^{-1}$)
$[\text{Given } R = 8.314 \, J \, K^{-1} \, mol^{-1}]$

The equation $k = (6.5 \times 10^{12} \, s^{-1}) e^{-26000 \, K / T}$ is followed for the decomposition of compound $A$. The activation energy for the reaction is $..... \, kJ \, mol^{-1}$. [nearest integer] (Given: $R = 8.314 \, J \, K^{-1} \, mol^{-1}$)