Horvath Practice Problems 25

# Horvath Practice Problems 25 - R − 1 400 R ¢ E a = 4045...

This preview shows page 1. Sign up to view the full content.

Application of the Arrhenious Equation The best way to get a feel for the Arrhenious equation is to work a few examples. Example 1 : A reaction is found to go 1.5 times faster at 400 K than at 300 K. What is the Arrhenious activation energy for this reaction? Thereact iongoes1 .5t imesfastermeans k 400 =1 . 5 k 300 . So Ae E a 400 R =1 . 5 Ae E a 300 R e Ea 400 R =1 . 5 e Ea 300 R . Taking the natural log of both sides gives E a 400 R =ln1 . 5 E a 300 R . Solving for E a gives E a 300 R E a 400 R =l n 1 . 5 E a μ 1 300 R 1 400 R =l n 1 . 5 E a
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: R − 1 400 R ¢ E a = 4045 J/mol Example 2 : The Arrhenious activation energy is 50.50 kJ/mol for some reaction. How much faster will the reaction go at 298 K than at 273 K? We want to f nd the ratio k 298 k 273 . Thus, k 298 k 273 = Ae − 50500 298 R Ae − 50500 273 R = e 50500 ( 1 273 R − 1 298 R ) = 6 . 47 . So the reaction goes 6.47 times faster at 298 K than at 273 K. 24 THE ARRHENIOUS EQUATION...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online