$A$ graph plotted between $\log \,K$ vs $\frac{1}{T}$ for calculating activation energy is shown by:

The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \cdot e^{-2000/T}$ and $10^{15} \cdot e^{-1000/T},$ respectively. The temperature at which $k_1 = k_2$ is

For the gaseous reactions,calculate the approximate value of temperature at which $k_1 = k_2$. $[\ln\, 10 = 2.3]$.
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.314\, T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.314\, T}$

For the reaction of $H_2$ with $I_2$,the rate constant is $2.5 \times 10^{-4} \ dm^3 \ mol^{-1} \ s^{-1}$ at $327 \ ^oC$ and $1.0 \ dm^3 \ mol^{-1} \ s^{-1}$ at $527 \ ^oC$. The activation energy for the reaction,in $kJ \ mol^{-1}$ is: $(R = 8.314 \ J \ K^{-1} \ mol^{-1})$

$A$ catalyst:

$A$ 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: $R = 8.3 \; J \; mol^{-1} \; K^{-1}$,$\ln(1.5) = 0.405$)