Feb_12_2010 - Important. Newofficehours:MonandFri1112;Wed.45

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Important…. New office hours: Mon and Fri 11 – 12; Wed. 4 – 5   Chemistry Tutor Room – Fleming 236 Exam 1 will cover Ch. 12, 13 and 16 through integrated  rate laws.
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
ln k 2 k 1 = E a R - 1 T 2 1 T 1 - The Arrhenius  Equation ln k 2  = ln A – (E a /R)(1/T 2 ) ln k 1  = ln A – (E a /R)(1/T 1 ) Two different rate constants at two different temperatures, T 1   and T 2 . ln k 1  – ln k 2  
Background image of page 2
Figure 16.11 Graphical determination of the activation energy. ln  k  = (- E a / R  )(1/ T ) + lnA
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
PLAN: SOLUTION: Determining the Energy of Activation PROBLEM: The decomposition of hydrogen iodide, 2H I ( g )                H 2 ( g ) + I 2 ( g ) has rate constants of 9.51x10 -9  L/mol*s at 500. K and 1.10x10 -5  L/ mol*s at 600. K.  Find  E a . Use the modification of the Arrhenius equation         to find  E a ln k 2 k 1 = E a - R 1 T 2 1 T 1 - E = -  R ln k 2 k 1 1 T
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 03/09/2010 for the course CHEM 1332 taught by Professor Halasyamani during the Spring '10 term at University of Houston-Victoria.

Page1 / 13

Feb_12_2010 - Important. Newofficehours:MonandFri1112;Wed.45

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online