Question

At $380^\circ C,$ the half life period for the first order decomposition of $H_{2}O_{2}$ is $360 \, min.$ The energy of activation of the reaction is $200 \, kJ \, mol^{- 1}$ . Calculate the time in minutes required for $75\%$ decomposition at $450^\circ C.$

[Report your answer by rounding it upto nearset whole number]

Answer 1

$K_{1}=0.693/360 \, \text{min}^{- 1}$ at $653 \, K=1.93\times 10^{- 3} \, min^{- 1}$

$E_{a}=200\times 10^{3}J; \, K_{2}=?$ at $723 \, K, \, R=8.314 \, J$

$2.303 log \frac{\text{K}_{\text{2}}}{\text{K}_{\text{1}}} = \frac{\text{E}_{a}}{\text{R}} \left[\frac{1}{\text{T}_{1}} - \frac{1}{\text{T}_{2}}\right]$

$\therefore \text{2.302 log}\frac{K_{2}}{1 . 93 \times 10^{- 3}}=\frac{200 \times 10^{3}}{8 . 314}\left[\frac{723 - 653}{723 \times 653}\right]$

$K_{2}=0.068 \, min^{- 1}$

Now, $t=\frac{2 . 303}{0.068}log\frac{100}{25}=20.39 \, min$