1 / 20

Understanding Energy Profile and Reaction Mechanism

Explore the detailed process of how chemical reactions occur, including reaction profiles and mechanisms. Learn how energy barriers affect reaction rates and how catalysts influence reaction pathways.

stevenpatel
Download Presentation

Understanding Energy Profile and Reaction Mechanism

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

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

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

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

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

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

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

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

  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 Ea. With catalyst (X)… A + X intermediate, with a lower Ea. (r.d.s.) intermediate + B  C + X(becomes a 2-stage rxn)

  10. (slow) [r.d.s.] (fast) Only 1 transition state 2 transition states DH is not affected by catalyst p.10 Energy Profileswith / withoutCatalyst …

  11. p.11 p. 79 Q.11 (2001 --- Differential Rate Eqn. and Energy Profile) (a) Rate = k [I(g)]2 [Ar] (c) 2 I  I2 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  I2(g) + Ar*(g) Role of 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  I2 + Ar(g) [slow] Role of Ar: homogeneous catalyst

  15. p.15 Step 1: 2I(g) + Ar(g)  I2Ar(g) [slow] Step 2: I2Ar(g)  I2 + Ar(g) [fast] Role of Ar: homogeneous catalyst

  16. formed quickly! OH 3 Br2 3 HBr + + OH Br Br Br p.16 Expt. 9 Activation Energy 5 Br- + BrO3- + 6 H+ 3 Br2 + 3H2O e.g. 3 moles are form in 42 s All the 3 moles of Br2 reacts with phenol.  does not bleach methyl red indicator. e.g. 1 mol At 42.1s, a new Br2 is formed, which will not react with phenol, but bleach methyl red!

  17.  rate  1/t p.17 The time required for bleaching Methyl Red is recorded. data treatment ...? i.e. time required , rate 

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

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

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

More Related