Chemistry - Chemical Kinetics Question with Solution | TestHub

Arrhenius equation is given by,
${\log }_{10}\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times \mathrm{R}}\left[\frac{{\mathrm{T}}_2-{\mathrm{T}}_1}{{\mathrm{T}}_1{\mathrm{T}}_2}\right]$

Given,

$\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}=3;\mathrm{R}=-8.314 {\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$

${\mathrm{T}}_1=20+273=293 \mathrm{K}$
and ${\mathrm{T}}_2=50+273=323 \mathrm{K}$

Substituting the given values in Arrhenius equation,

${\log }_{10}3=\frac{{\mathrm{E}}_{\mathrm{a}}}{8.314\times 2.303}\left[\frac{323-293}{323\times 293}\right]$

${\mathrm{E}}_{\mathrm{a}}=\frac{2.303\times 8.314\times 323\times 293\times 0.477}{30}$

$=28811.8 \mathrm{J} {\mathrm{mol}}^1$

$=28.8118{\mathrm{kJmol}}^{-1}$