For the gaseous reactions:
$A \to B$ $k_1 = 10^{15} e^{-25000 / 8.34 T}$
$C \to D$ $k_2 = 10^{14} e^{-15000 / 8.34 T}$
Calculate the approximate temperature at which $k_1 = k_2$ in $K$.

The rate constant for a first order reaction is $0.58 \ s^{-1}$ at $300 \ K$ and $0.026 \ s^{-1}$ at $290 \ K$. What is the energy of activation? $(R=8.314 \ J \ K^{-1} \ mol^{-1})$

For reaction $A \to B$,rate constant $K_1 = A_1 e^{-E_{a_1}/RT}$ and for the reaction $X \to Y$,rate constant $K_2 = A_2 e^{-E_{a_2}/RT}$. If $A_1 = 10^8, A_2 = 10^{10}$ and $E_{a_1} = 600 \ cal \ mol^{-1}$,$E_{a_2} = 1800 \ cal \ mol^{-1}$,then the temperature at which $K_1 = K_2$ is (given: $R = 2 \ cal \ K^{-1} \ mol^{-1}$):

From the given data for the reaction $H_2 + I_2 \rightarrow 2HI$,calculate the activation energy $(E_a)$:
$T_1 = 769 \ K, \ 1/T_1 = 1.3 \times 10^{-3} \ K^{-1}, \ \log_{10} K_1 = 2.9$
$T_2 = 667 \ K, \ 1/T_2 = 1.5 \times 10^{-3} \ K^{-1}, \ \log_{10} K_2 = 1.1$

Which scientist explained that all the species taking part in a reaction do not possess the same kinetic energy?

$A$ reaction takes place in three steps with individual rate constant and activation energy as follows:

$Step$	$Rate \ constant$	$Activation \ energy$
$Step-1$	$k_1$	$E_{a1} = 180 \ kJ/mol$
$Step-2$	$k_2$	$E_{a2} = 80 \ kJ/mol$
$Step-3$	$k_3$	$E_{a3} = 50 \ kJ/mol$

If overall rate constant,$k = (\frac{k_1 k_2}{k_3})^{2/3}$,then overall activation energy of the reaction will be .......... $kJ/mol$.