The velocity constant of a reaction at $290 \ K$ was found to be $3.2 \times 10^{-3}$. At $310 \ K$,it will be about:

The decomposition of a hydrocarbon follows the equation $K = (4.5 \times 10^{11} \, s^{-1}) e^{-28000 \, K/T}$. What will be the value of activation energy in $KJ \, mol^{-1}$?

For reaction $A \rightarrow P$, rate constant $k = 1.5 \times 10^3 \text{ s}^{-1}$ at $27^\circ\text{C}$. If activation energy for the above reaction is $60 \text{ kJ mol}^{-1}$, then the temperature (in $^\circ\text{C}$) at which rate constant $k = 4.5 \times 10^3 \text{ s}^{-1}$ is . . . . . . .

For an elementary chemical reaction, the Arrhenius plot is given below. If the energy of activation is $6.64 \ kJ \ mol^{-1}$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$, the temperature at which the rate constant becomes $e^2 \ min^{-1}$, is (in $K$)

The minimum energy necessary to permit a reaction is

Consider the given plots for a reaction obeying the Arrhenius equation $(0\,^{\circ}C < T < 300\,^{\circ}C)$: ($k$ and $E_a$ are rate constant and activation energy respectively). Choose the correct option.