Question

A first order reaction is $50$% completed in $30$ minutes at ${27}^{o}C$ and in $10$ minutes at ${47}^{o}C$. Calculate the reaction rate constant at ${27}^{o}C$ and the energy of activation of the reaction in $kJ$ ${mol}^{-1}$.

Answer 1

For first order reaction $k=\frac{0.693}{t_{1/2}}$
At ${27}^{o}C$, $k_{{27}^{o}C}=\frac{0.693}{30}=0.0231{min}^{-1}$
At ${47}^{o}C$, $k_{{47}^{o}C}=\frac{0.693}{10}=0.0693{min}^{-1}$
Now applying the following equation:
${\log }_{10}\frac{k_1}{k_2}=\frac{-E_a}{2.303\times R}\left(\frac{T_2-T_1}{T_2\cdot T_1}\right)$
or ${\log }_{10}\frac{0.0231}{0.0693}=\frac{-E_a}{2.303\times 8.314}:\left(\frac{320-300}{320\times 300}\right)$
or $-{\log }_{10}0.3333=\frac{E_a}{19.1471}\times \frac{20}{96000}$
$E_a=-\frac{19.1471\times 96000}{20}\times \log 0.3333$
$=-91906\times (-0.4772)$
$=43857$ $J$ ${mol}^{-1}$ $=43.857$ $kJ$ ${mol}^{-1}$