1 / 9

C. Y. Yeung (CHW, 2009)

Activation Energy (E a ) & Arrhenius Equation. / Transition State. p.01. C. Y. Yeung (CHW, 2009). k = Ae (-E /RT). a. (Arrhenius Equation). Rate Constant (k) is temperature dependent. Higher E a , smaller k, slower rate of rxn. E a of a rxn path is “unchangable”!. p.02.

rogerreyes
Télécharger la présentation

C. Y. Yeung (CHW, 2009)

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Activation Energy (Ea) & Arrhenius Equation / Transition State p.01 C. Y. Yeung (CHW, 2009)

  2. k = Ae(-E /RT) a (Arrhenius Equation) Rate Constant (k) is temperature dependent. Higher Ea, smaller k, slower rate of rxn. Ea of a rxn path is “unchangable”! p.02 Remember the “Rate Constant” (k) …? Rate = k [A]m[B]n (differential rate equation) Units!!

  3. -(50.01000) /(8.314293) -Ea/RT e e At 293K: = = 1.2210-9 -(50.01000) /(8.314303) -Ea/RT e e At 303K: = = 2.3810-9 k = Ae(-E /RT) a -Ea/RT e Since  k  rate, therefore temp increases, rate increases If the Ea of a reaction is 50.0 kJ mol-1 p.03 ^^ Increasing T by 100C, the rate almost doubles.

  4. Ea k = Ae(-E /RT) a RT ln k = lnA – p.04 How to find the“Activation Energy”(Ea) …? Expt.  Table  Find Ea by Graphical Method i.e. “ln k” vs “1/T” should give a straight line with slope = -Ea/R

  5. At T1, find “k1” by Differential / Integrated Rate Eqn At T2, find “k2” by Differential / Integrated Rate Eqn p.05 Find Ea by Experiment … (1)

  6. 1 ln k T ln A Ea ln k = – + lnA R 1/T p.06 Find Ea by Experiment … (2) slope = - Ea/R ** Ea must be +ve.!!

  7. p.07 p. 78 Q.7(a) (1998 --- Activation Energy) Slope = -11658 K -11658K = -Ea/(8.314 J K-1mol-1) Ea = 96924 J mol-1 Ea = 96.9 kJ mol-1 From the graph, ln k = -0.92 k = 0.40 s-1 1st order: k = ln(2) / t1/2  t1/2 = 1.73 s

  8. p.08 Assignment p.73 Q.5, 6, 7, 13 p.76 Q.5, 6 [due date: 25/2(Wed)] Pre-lab: Expt. 9 Determination of Activation Energy [due date: 26/2(Thur)]

  9. p.09 Next …. Maxwell Boltzmann Distribution and Collision Theory (p. 54-58)

More Related