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

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

3. p.03 If the E a of a reaction is 50.0 kJ mol -1 At 293K: -E a /RT e = -(50.0  1000) /(8.314  293) e = 1.22  10 -9 At 303K: -E a /RT e = -(50.0  1000) /(8.314  303) e = 2.38  10 -9 Since  k  rate, therefore temp increases, rate increases -E a /RT e k = Ae (-E /RT) a ^^ Increasing T by 10 0 C, the rate almost doubles.

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

5. p.05 Find E a 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 TT1T1 T2T2 T3T3 T4T4 kk1k1 k2k2 k3k3 k4k4 1/T1/T 1 1/T 2 1/T 3 1/T 4 ln kln k 1 ln k 2 ln k 3 ln k 4

6. p.06 ln k 1/T ln A slope = - E a /R Find E a by Experiment … (2) 1/T1/T 1 1/T 2 1/T 3 1/T 4 ln kln k 1 ln k 2 ln k 3 ln k 4 ln k = – + lnA EaEaEaEaR 1 T ** E a must be +ve.!!

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

8. p.08 Assignment p.73 Q.5, 6, 7, 13 p.76Q.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)