The rate of the process doubles with every $10 \ K$ increase in temperature. When the temperature is increased from $303 \ K$ to $353 \ K$,how much will the rate of the process increase?

$A$ reaction takes place in three steps with individual rate constant and activation energy,

Step	Rate constant and Activation energy
$Step \ 1$	$k_1, E_{a_1} = 180 \ kJ \ mol^{-1}$
$Step \ 2$	$k_2, E_{a_2} = 80 \ kJ \ mol^{-1}$
$Step \ 3$	$k_3, E_{a_3} = 50 \ kJ \ mol^{-1}$

Overall rate constant,$k = (k_1 k_2 / k_3)^{2/3}$. The overall activation energy of the reaction will be ........ $kJ \ mol^{-1}$.

In a reversible reaction,the catalyst

The activation energy of a first-order reaction at $25\,^{\circ}C$ is $30\,kJ/mol$. In the presence of a catalyst,the activation energy of the same reaction at $25\,^{\circ}C$ becomes $24\,kJ/mol$. The rate of the reaction in the presence of the catalyst will be how many times the original rate?

Write the Arrhenius equation in the form $\ln \, k = -\frac{E_a}{RT} + \ln \, A$.

If the rate constants of a reaction at $500 \ K$ and $700 \ K$ are $0.002 \ s^{-1}$ and $0.06 \ s^{-1}$,respectively,the value of activation energy is $(R=8.314 \ J \ mol^{-1} \ K^{-1}, \log 3=0.477)$.