The rate of a reaction doubles when its temperature changes from $300 \, K$ to $310 \, K.$ Activation energy of such a reaction will be .......... $kJ \, mol^{-1}$. $(R= 8.314 \, J \, K^{-1} \, mol^{-1}$ and $\log 2=0.301)$

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

If the half-lives of a first-order reaction at $350 \ K$ and $300 \ K$ are $2 \ s$ and $20 \ s$ respectively,the activation energy of the reaction in $kJ \ mol^{-1}$ is:

The relation between the rate constant and temperature according to the Arrhenius equation is:

The temperature coefficient of the saponification reaction of an ester by $NaOH$ is $1.75$. The activation energy of the reaction is .......... $kcal \ mol^{-1}$. (in $.21$)

Draw an energy profile diagram for a three-step reaction in which the first step is the slowest and the last step is the fastest. (Assume that the reaction is exothermic)