Question

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

• A. $43.857$
• B. $43857$
• C. $438.57$
• D. $4.3857$

Answer 1

The correct option is A $43.857$

For first order reaction $k=\frac{0.693}{t_{1/2}}$
At ${27}^{\circ }C,$ $k_{{27}^{\circ }C}=\frac{0.693}{30}=0.0231{\text{min}}^{-1}$
At ${47}^{\circ }C,$ $k_{{47}^{\circ }C}=\frac{0.693}{10}=0.0693{\text{min}}^{-1}$
Now applying the following equation :
$lo{g}_{10}\frac{k_1}{k_2}=\frac{-E_a}{2.303\times R}.\left(\frac{T_2-T_1}{T_2.T_1}\right)$
or $lo{g}_{10}\frac{0.0231}{0.0693}=\frac{-E_a}{2.303\times 8.314}.\left(\frac{320-300}{320\times 300}\right)$
or $-lo{g}_{10}0.3333=\frac{E_a}{19.1471}\times \frac{20}{96000}$
$E_a=-\frac{19.1471\times 96000}{20}\times log0.3333$
$E_a=-91906\times (-0.4772)$
$E_a=43857Jmo{l}^{-1}=43.857kJmo{l}^{-1}$