activation energy

activation energy - Taking the ln of both sides of the...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
For the gas phase decomposition of 1-bromopropane , CH 3 CH 2 CH 2 Br CH 3 CH=CH 2 + HBr the rate constant in s -1 has been determined at several temperatures. When ln k is plotted against the reciprocal of the Kelvin temperature, the resulting linear plot has a slope of -2.41E+4 K and a y-intercept of 31.3 . The value of the rate constant for the gas phase decomposition of 1-bromopropane at 619 K is s -1 . Feedback: For many reactions, the temperature dependence of the rate constant follows the Arrhenius equation. The exponential form of the Arrhenius equation is: k = A e -(E a /RT) where E a is the activation energy, k is the rate constant, T is the Kelvin temperature, R is the gas constant, and A is a constant called the frequency factor.
Background image of page 1

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

View Full DocumentRight Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Taking the ln of both sides of the equation gives: ln k = ln A- E a 1 R T which has the linear form y = b + mx, with y = ln k , b = ln A, m = -E a /R and x = 1/T. A plot of ln k versus 1/T should be linear with slope = -E a /R . While it is possible to calculate the values of E a and A from the slope and the y-intercept of the plot, the linear form can be used directly to calculate k at any temperature because the slope and y-intercept are known. In this case ln A = 31.3-(E a /R) = -2.41E+4 K T = 619 K Substituting these into the linear equation gives ln k = 31.3 + (-2.41E+4 )(1/ 619 ) = -7.63 k = e-7.63 = 4.86E-4 s-1...
View Full Document

Page1 / 2

activation energy - Taking the ln of both sides of the...

This preview shows document pages 1 - 2. Sign up to view the full document.

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