Question

The decomposition of A into product has values of K as $4.5\times {10}^3{s}^{-1}$ at ${10}^{\circ }C$ and energy of activation $60kJmo{l}^{-1}$. At what temperature would K be $1.5\times {10}^4{s}^{-1}$

Answer 1

According to Arrhenius equation

$log\frac{k_2}{k_1}=\frac{E_a}{2.303R}\times \frac{T_2-T_1}{T_1{T}_2}$

$k_1=4.5\times {10}^3{s}^{-1};k_2=1.5\times {10}^4{s}^{-1};T_1={10}^{\circ }C=283K$

$log\frac{1.5\times {10}^4}{4.5\times {10}^3}=\frac{\left(6000Jmo{l}^{-1}\right)}{2.303\times \left(8.314Jmo{l}^{-1}\right)}\left(\frac{T_2-283}{283T_2}\right)$

$log3.333=3133.62\left(\frac{T_2-283}{283T_2}\right)$

$\frac{0.5288\times 283}{3133.62}=\frac{T_2-283}{T_2}$

$1-\frac{T_2-283}{T_2}=0.04776or{T}_2=\frac{283}{1-0.04776}=\frac{283}{0.95224}$

$T_2=297.19K$

$=(297.19-273.0)={24.19}^{\circ }C$

Temperature $={24.19}^{\circ }C$