Question

At $380^{\circ }C,$ the half life period for the first order decomposition of $H_2{O}_2$ is $360min.$ The energy of activation of the reaction is $200kJmo{l}^{-1}$ . 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

Answer: 20.00

$K_1=0.693/360{\text{min}}^{-1}$ at $653K=1.93\times 10^{-3}mi{n}^{-1}$

$E_a=200\times 10^{3}J;K_2=?$ at $723K,R=8.314J$

$2.303log\frac{{\text{K}}_{\text{2}}}{{\text{K}}_{\text{1}}}=\frac{{\text{E}}_a}{\text{R}}\left[\frac{1}{{\text{T}}_1}-\frac{1}{{\text{T}}_2}\right]$

$\therefore \text{2.302 log}\frac{K_2}{1.93\times 10^{-3}}=\frac{200\times 10^3}{8.314}\left[\frac{723-653}{723\times 653}\right]$

$K_2=0.068mi{n}^{-1}$

Now, $t=\frac{2.303}{0.068}log\frac{100}{25}=20.39min$