Answer to Question #231002 in Chemical Engineering for Lucky

Question

At 500K, the rate of a bimolecular reaction is ten times the rate at 400K Find the activation energy for this reaction from Arhenius theory and collision state theory. Calculate percentage differences of rates from both the reactions.

Answer 1

Arhenius theory 

$-\ln (\frac{k_1}{k_2})= (\frac{1}{T_2}-\frac{1}{T_1})\frac{E}{R}\\ -\ln (\frac{1}{10})= (\frac{1}{400}-\frac{1}{500})\frac{E}{8.3145}\\ \frac{E}{16629}=-\ln \left(\frac{1}{10}\right)\\ \mathrm{Multiply\:both\:sides\:by\:}16629\\ \frac{E}{16629}\cdot \:16629=-\ln \left(\frac{1}{10}\right)\cdot \:16629\\ \mathrm{Simplify}\\ E=-\ln \left(\frac{1}{10}\right)\cdot \:16629\\ E=38289.68751\\$

Collision state theory

$Rate=Z_{AB}e^{−\frac{E_a}{R(\Delta T)}}\\ \frac{1}{10}=10e^{−\frac{E_a}{8.3145(500-400)}}\\ \mathrm{Multiply\:both\:sides\:by\:}10\\ \frac{1}{10}\cdot \:10=10e^{-\frac{E_a}{8.3145\left(500-400\right)}}\cdot \:10\\ \mathrm{Simplify}\\ 1=100e^{-\frac{E_a}{8.3145\left(500-400\right)}}\\ e^{-\frac{E_a}{8.3145\left(500-400\right)}}=\frac{1}{100}\\ E_a=1662.9\ln \left(10\right)\\ E_a=3828.96875$

Percentage difference

$\frac{38289.68751-3828.96875}{38289.68751}\cdot \:100=90.0 \%$