Question

At ${380}^{0}C$, the half life period for the first order decomposition of $H_2{O}_2$ is 360 minutes. The energy of activation of the reaction is 200 kJ ${mol}^{-1}$. Calculate the time required for 75% decomposition at ${450}^{0}C$?

• A. 60 min
• B. 40 min
• C. 20.40 min
• D. 10 min

Answer 1

The correct option is C 20.40 min

$k_1=\frac{0.693}{360}=1.92\times {10}^{-3}{min}^{-1}$
$log\left(\frac{k_2}{k_1}\right)=\left(\frac{Ea}{2.303R}\right)\left[\left(\frac{T_2-T_1}{T_1{T}_2}\right)\right]=\frac{(200\times {10}^3\times 70)}{(2.303\times 8.314\times 653\times 723)}=1.5487$
$\frac{k_2}{k_1}=Antilog\left(1.5487\right)=35.38$, $k_2=35.38\times 1.92\times$ ${10}^{-3}=6.792\times {10}^{-2}{min}^{-1}$
Rate at ${450}^{o}C$, t = $\left(\frac{2.303}{k_2}\right)log\left(\frac{100}{100-75}\right)$= $\left(\frac{2.303}{6.792\times {10}^{-2}}\right)log\left(\frac{100}{25}\right)=\left(\frac{2.303\times 0.6021}{6.792\times {10}^{-2}}\right)=20.40$ minutes.