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 a reaction,the activation energy $E_{a} = 0$ and the rate constant at $200 \ K$ is $1.6 \times 10^{6} \ s^{-1}$. The rate constant at $400 \ K$ will be (given $R = 8.314 \ J \ K^{-1} \ mol^{-1}$):

$A$ sample of milk splits after $60 \; min.$ at $300 \; K$ and after $40 \; min.$ at $400 \; K$ when the population of $Lactobacillus \; acidophilus$ in it doubles. The activation energy (in $kJ / mol$) for this process is closest to ............. $kJ / mol$.
(Given: $R = 8.3 \; J \; mol^{-1} \; K^{-1}$,$\ln(1.5) = 0.405$)

For a reaction $A \rightarrow B$,the enthalpy of reaction is $-4.2 \ kJ \ mol^{-1}$ and the enthalpy of activation is $9.6 \ kJ \ mol^{-1}$. The correct potential energy profile for the reaction is shown in which option?

For an endothermic reaction $X \rightarrow Y$,the activation energies for the forward and backward reactions are $E_f$ and $E_b$ respectively. Then,in general:

Decomposition of a hydrocarbon follows the equation $k = (5.5 \times 10^{11} \text{ s}^{-1}) e^{\frac{-28000 \text{ K}}{T}}$. The activation energy of the reaction is . . . . . . $\text{kJ mol}^{-1}$. (Nearest Integer) Given: $R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$