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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.

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C. Y. Yeung (CHW, 2009)

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  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)

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