p 01 Activation Energy Ea Arrhenius Equation Transition

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

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

k= (-E /RT) a Ae p. 03 If the Ea of a reaction is 50. 0 k. J mol-1 At 293 K: At 303 K: Since therefore -Ea/RT e -(50. 0 1000) /(8. 314 293) = e = 1. 22 10 -9 = e = 2. 38 10 -9 -Ea/RT e -(50. 0 1000) /(8. 314 303) k rate, temp increases, rate increases ^^ Increasing T by 100 C, the rate almost doubles.

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

p. 05 Find Ea by Experiment … (1) At T 1, find “k 1” by Differential / Integrated Rate Eqn At T 2, find “k 2” by Differential / Integrated Rate Eqn T k T 1 k 1 T 2 k 2 T 3 k 3 T 4 k 4 1/T ln k 1/T 1 ln k 1 1/T 2 ln k 2 1/T 3 ln k 3 1/T 4 ln k 4

p. 06 Find Ea by Experiment … (2) 1/T ln k 1/T 1 ln k 1 1/T 2 ln k 2 Ea ln k = – R ln k ln A 1/T 3 ln k 3 1/T 4 ln k 4 1 + ln. A T slope = - Ea/R 1/T ** Ea must be +ve. !!

p. 78 Q. 7(a) (1998 --- Activation Energy) p. 07 Slope = -11658 K -11658 K = -Ea/(8. 314 J K-1 mol-1) Ea = 96924 J mol-1 Ea = 96. 9 k. J mol-1 From the graph, ln k = -0. 92 k = 0. 40 s-1 1 st order: k = ln(2) / t 1/2 = 1. 73 s

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)]

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