1. C. Y. Yeung (CHW, 2009) p.01 Energy Profile and Reaction Mechanism How does the reaction take place???

2. p.02 A + B  C How does a reaction occur? K.E.  E a collide in right orientation A+ B C Single Stage Rxn Multi-Stage Rxn A+ BC Intermediate How to know that …. ?

3. p.03 From the Chemical & Differential Rate Eqns ! e.g. 2A + B  3C If after Kinetics Studies, it was found that … Rate = k[A][B] i.e. The reaction is a 2-step reaction! Step 1:A + B  Intermediate (r.d.s.) Step 2:Intermediate + A  C ** a “A” molecule and a “B” molecule are involved in the “rate-determining step”.

4. p.04 e.g. 2A + B  3C If after Kinetics Studies, it was found that … Rate = k[A] 2 i.e. The reaction is a 2-step reaction! Step 1:A + A  Intermediate (r.d.s.) Step 2:Intermediate + B  C ** Two “A” molecules are involved in the “rate-determining step”.

5. p.05 e.g. 2A + B  3C If after Kinetics Studies, it was found that … Rate = k[B] i.e. The reaction is a 2-step reaction! Step 1:B  Intermediate (r.d.s.) Step 2:Intermediate + 2A  C ** Only one “B” molecule is involved in the “rate-determining step”.

6. p.06 e.g. 2A + B  3C If after Kinetics Studies, it was found that … Rate = k[A] 2 [B] i.e. The reaction is a Single-step reaction! 2A + B  3C ** ALL the molecules are involved in the “rate-determining step”.

7. p.07 2A + B  3C Energy Profile ?? Rate = k[A] 2 [B] Rate = k[B] Rate = k[A] 2 Rate = k[A][B] intermediate Transition state higher E a : r.d.s. (slower step)

8. p.08 p. 79 Q.9 (1999 --- Differential Rate Eqn. and Energy Profile) (a)Rate = k[A] (c)A + B  product i.e. Only a “A” is involved in r.d.s.

9. p.09 Catalyst changes the energy profile of rxn! A + B  C If no catalyst … A + B  C, with high E a. With catalyst (X)… A + X  intermediate, with a lower E a. (r.d.s.) intermediate + B  C + X (becomes a 2-stage rxn)

10. p.10 Energy Profiles with / without Catalyst … (slow) [r.d.s.] (fast)  H is not affected by catalyst Only 1 transition state 2 transition states

11. p.11 p. 79 Q.11 (2001 --- Differential Rate Eqn. and Energy Profile) (a)Rate = k [I(g)] 2 [Ar] (c)2 I  I 2 i.e. Two “I” atoms and one “Ar” atom are involved in r.d.s. 3 possible energy profiles!!

12. p.12 3 Different Stories are possible!

13. p.13 I(g) + I(g) + Ar  I 2 (g) + Ar*(g) Role of Ar: Ar: acts as a third body to absorb energy from the colliding I(g) atoms.

14. p.14 Step 1:I(g) + Ar(g)  I-Ar(g) [fast] Step 2:I-Ar(g) + I  I 2 + Ar(g) [slow] Role of Ar: Ar: homogeneous catalyst

15. p.15 Step 1:2I(g) + Ar(g)  I 2 Ar(g) [slow] Step 2:I 2 Ar(g)  I 2 + Ar(g) [fast] Role of Ar: Ar: homogeneous catalyst

16. p.16 Expt. 9 Activation Energy 5 Br - + BrO 3 - + 6 H +  3 Br 2 + 3H 2 O formed quickly! 3 Br 2 + OHOHBr Br Br + 3 HBr e.g. 3 moles are form in 42 s All the 3 moles of Br 2 reacts with phenol.  does not bleach methyl red indicator. At 42.1s, a new Br 2 is formed, which will not react with phenol, but bleach methyl red! e.g. 1 mol

17. p.17 The time required for bleaching Methyl Red is recorded.  data treatment...? Temp./ 0 CTime/s rate/mol s -1 (if 3 mol of Br 2 are formed) 25923/92 30563/56 41303/30 5293/9 i.e. time required , rate   rate  1/t

18. p.18 rate  1/t  k  1/t Therefore … ln k = – + lnA EaEaEaEaR 1 T ln (1/t) = – + lnA EaEaEaEaR 1 T

19. p.19 Assignment p.74 Q.11, 12 [due date: 2/3(Mon)] Lab Report: Expt. 9 Determination of E a [due date: 3/3(Tue)] Quiz on Chemical Kinetics (Ch. 13-15) [9/3(Mon)]

20. p.20 Next …. Chemical Equilibria & K eq (p. 88-100)