Question

A first order reaction is $50\%$ complete in $30$ minutes at $27{}^{\circ }C$ and in 10 minutes at $47{}^{\circ }C$. The energy of activation $E_a$ of the reaction is:
$(log2=0.3,log3=0.5,R=\frac{25}{3}J{K}^{-1}mo{l}^{-1})$

• A. 46 $kJmo{l}^{-1}$
• B. 35 $kJmo{l}^{-1}$
• C. 84 $kJmo{l}^{-1}$
• D. 30 $kJmo{l}^{-1}$

Answer 1

The correct option is A 46 $kJmo{l}^{-1}$

Since the reaction is completed 50%, so
$k_{27}=\frac{0.693}{30}=0.0231mi{n}^{-1}$
$k_{47}=\frac{0.693}{10}=0.0693mi{n}^{-1}$
$\therefore \frac{k_{47}}{k_{27}}=3$
According to Arrhenius theory
$\Rightarrow log\frac{k_{47}}{k_{27}}=log3=\frac{E_a}{2.303\times \frac{25}{3}}\left(\frac{20}{300\times 320}\right)$
$\Rightarrow 0.5=\frac{E_a}{2.303\times \frac{25}{3}}\left(\frac{20}{300\times 320}\right)$
$\Rightarrow E_a=0.5\times 2.303\times \frac{25}{3}\times 300\times 320\times \frac{1}{20}$
$\Rightarrow E_a=46060J\approx 46kJ$