The rate of a reaction triples when temperature changes from 20 ^ 0...

Arrhenius equation is given by,
$\log_{10}\dfrac{K2}{K1} = \dfrac{Ea}{2. 303 \times R} \left[\dfrac{{T}_{2} - {T}_{1}}{{T}_{1}{T}_{2}}\right]</p><p>Given,</p><p>\dfrac{K2}{K1} =3; R = - 8. 314{JK}^{- 1}{mol}^{- 1}$

${T}_{1} = 20 + 273 = 293K$
and ${T}_{2} = 50 + 273 = 323K</p><p>Substituting the$given values in Arrhenius equation,

\log_{10}3 =$\dfrac{Ea}{8. 314 \times 2. 303}$\left[\dfrac{323 - 293}{323 \times 293}\right]

${E}_{a} =$\dfrac{2. 303 \times 8. 314 \times 323 \times 293 \times 0. 477}{30}$</p><p>$= 28811. 8J{mol}^{1}$</p><p>$= 28. 8118{kJmol}^{- 1}$</p>$