Question #231002

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.


1
Expert's answer
2021-08-31T01:42:38-0400

Arhenius theory 


ln(k1k2)=(1T21T1)ERln(110)=(14001500)E8.3145E16629=ln(110)Multiplybothsidesby16629E1662916629=ln(110)16629SimplifyE=ln(110)16629E=38289.68751-\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=ZABeEaR(ΔT)110=10eEa8.3145(500400)Multiplybothsidesby1011010=10eEa8.3145(500400)10Simplify1=100eEa8.3145(500400)eEa8.3145(500400)=1100Ea=1662.9ln(10)Ea=3828.96875Rate=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

38289.687513828.9687538289.68751100=90.0%\frac{38289.68751-3828.96875}{38289.68751}\cdot \:100=90.0 \%


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