Question

At ${380}^{o}C$, 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 $200kJmo{l}^{-1}$. The time required for 25% decomposition at ${450}^{o}C$ in minutes (nearest integer) is :

Answer 1

Lower temeperature $T_1=380+273=653K$

Higher temperature $T_2=450+273=723K$
$E_a=200kJmo{l}^{-1}$

Rate constant at $T_1=k_1=\frac{0.693}{t_{1/2}}=\frac{0.693}{360}=1.925\times {10}^{-3}$

Rate constant at $T_2=k_2=?$

Activation Energy $E_a=\frac{2.303\times R\times T_1\times T_2}{(T_2-T_1)}\times log\frac{k_2}{k_1}$

On solving we get $k_2=6.81\times {10}^{-2}mi{n}^{-1}$

When $25\%$ decomposition takes place at 723K

For first order reaction

$k_2=\frac{2.303}{t}\times log\frac{N_o}{N_t}$

$N_t=0.75N_o$

We get $t=4.22$
$t=4$ (nearest integer value)