The rate constant of a chemical reaction at a very high temperature will approach:

In the plot of $\ln k$ versus $\frac{1}{T}$,what do the slope and intercept represent?

For a first order reaction,a plot of $\ln k$ ($y$-axis) and $\frac{1}{T}$ ($x$-axis) gave a straight line with a slope equal to $-10^3 \ K$ and an intercept equal to $2.303$ (on $y$-axis). What is the activation energy ($E_a$ in $kJ \ mol^{-1}$) of the reaction? (Given $R = 8.314 \ J \ mol^{-1} \ K^{-1}$)

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 value of the rate constant for the reaction $A + B \to \text{products}$ depends on:

In a reaction, for every $10^{\circ} C$ rise of temperature, the rate is doubled. If the temperature is increased from $10^{\circ} C$ to $100^{\circ} C$, the rate of the reaction will become $:-$ (in $\times$)