2.  Some reactions are not represented in the reaction equation; such as, -Absorbing light energy -Colliding to the walls of the container  Many reactions proceed through a sequence of steps to arrive at the products  Each step is called an elementary reaction and occurs through collision of atoms, ions and molecules.

3.  The slowest step in a reaction mechanism is called the Rate-determining Step  In multistep reactions, increasing the concentration of reactants at any steps other than the rate-determining step will not increase the rate of reaction  A chemical species that form and then are consumed in the reaction is called Reaction Intermediates

4.  mX + nY -----> products of reaction or intermediates then r α [X] m [Y] n r = k[X] m [Y] n  When writing the reaction mechanism, there are three rules: -each step must be elementary, with no more than three reactants -The rate-determining step must be consistent with the rate equation -The elementary steps must all add up to be the overall equation

5.  Elementary reactions (Molecularity)  Unimolecular; Decomposition (ex. N 2 O 5(g)  NO 2(g) + NO 3(g) ) rate = k[N2O5]  Bimolecular ; Collision of 2 atoms, ions or molecules (ex. NO (g) + O 3(g)  NO 2(g) + O 2(g) ) rate = k[No][O3]  Termolecular ; Simultaneous collision of 3 molecules (ex. 2I (g) + Ar (g)  I 2(g) + Ar (g) ) rate = k[I][I][Ar] = k[I] 2 [Ar]

6.  Reaction mechanism is a detailed sequence of elementary reactions, with their rate, that are combined to yield the overall reaction.  Consider the mechanism: 1) Cl 2(g)  2Cl (g) 2) Cl (g) + CHCl 3(g)  HCl (g) + CCl 3 (g) 3) CCl 3 (g) + Cl (g)  CCl 4(g)  Some of the possible questions are…

7. a) What's the molecularity of each step? 1) Unimolecular 2) Bimolecular 3) Bimolecular b) Write the overall equation for the reaction Cl 2 (g) + CHCl 3 (g)  HCl(g) + CCl 4 (g) c) Identify the reaction intermediate(s) Cl(g) & CCl 3

8.  To explain the large effect of temperature and catalysis, the Arrhenius Equation is used to account for their effects k = Ae -E a /RT E a is the activation energy (J) A is constant related to the geometry of molecule R is the gas constant (8.314 J/(molK)) T is the temperature (K)  the answer k is the rate constant for the rate law equation r = k[A] n [B] m

9.  By taking ln (natural log) to each side, ln k = -Ea/RT + ln A  Arrhenius equation is written for the rate constant determined at each temperature giving, ln k1 = -Ea/RT1 + ln A ln k2 = -Ea/RT2 + ln A  Subtracting second equation from the first, ln(k1/k2) = -Ea/R(1/T1 – 1/T2)  The rate and the rate constant are directly proportional to each other as long as the concentrations are held constant.

10.  At 200K the rate constant for a reaction is 3.5 X 10 -3 s -1, and at 250K the rate constant is 4.0X 10 -3 s -1. What is the activation energy? Ea = 1.11KJ/mol  What is the rate of reaction at 450 o C if the reaction rate is 6.75 X 10 -6 mol/(L*s) at 25 o C? The activation energy was previously determined to be 35.5kJ/mol. The rate of reaction at 450 o C is 3.05 X 10 -2 mol/(L*s)

11.  To test the answer for reasonableness, two principles must be remembered: 1) The larger rate constant (or rate) will always be associated with the higher temperature 2) The activation energy always has a positive sign.

12.  Reactions that contain more than 3 molecules colliding occurs in multi-steps  Each elementary step should only have 3 or less reactants  r = k[A]n[B]m  k = Ae -Ea/RT  Temperature and activation energy affects the value of k exponentially