1 / 18

Chem 1310: Introduction to physical chemistry Part 2: Chemical Kinetics

Chem 1310: Introduction to physical chemistry Part 2: Chemical Kinetics. Exercises. From concentrations to rates and rate laws. 2 H 2 O 2 → 2 H 2 O + O 2. From concentrations to rates and rate laws. From concentrations to rates and rate laws. k = slope.

claus
Download Presentation

Chem 1310: Introduction to physical chemistry Part 2: Chemical Kinetics

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. Chem 1310: Introduction to physical chemistryPart 2: Chemical Kinetics Exercises

  2. From concentrationsto rates and rate laws 2 H2O2 → 2 H2O + O2

  3. From concentrationsto rates and rate laws

  4. From concentrationsto rates and rate laws k = slope

  5. From concentrationsto rates and rate laws Numerical order check: conc(600)/conc(2200) = 3.25 rate(600)/rate(2200) = 2.80 conc(200)/conc(1400) = 2.36 rate(200)/rate(1400) = 2.40 Reasonable (not great) for first-order.

  6. From concentrationsto rates and rate laws Average rate constant: 0.000749 s-1

  7. From concentrationsto rates and rate laws k = - slope

  8. From concentrationsto rates and rate laws k from data at two data points: k = (ln [H2O2]t1 - ln [H2O2]t2)/(t2-t1) so from data at 400 and 2000 s: k = 0.000724 s-1

  9. Using rate laws What is the rate when [H2O2] = 2.00 M? rate = k [H2O2] = 0.00150 mol L-1s-1 What are [H2O2] and rate a t = 500 s? [H2O2] = [H2O2]0 exp(- kt) = 1.59 Mrate = k [H2O2] = 0.00120 mol L-1s-1 What is the half-life?t½ = ln(2)/k = 925 s

  10. From rate constantsto activation energies tree cricket chirping frequency interpreted as a rate constant

  11. From rate constantsto activation energies Ea = -R * slope = 55 kJ/mol

  12. From rate constantsto activation energies Calculate A, Ea from rate constants at 16 and 26°C Ea = R (ln k1 - ln k2)/(1/T2 - 1/T1) = 58 kJ/mol A = k1 exp(Ea/R T1) = 2.9*1012

  13. Using activation energies What will be the frequency at 0°C?k = A exp(-Ea/RT) = 24 (chirps/min) At what temperature will the frequency be 250 (chirps/min)?T = (Ea/R)/(ln A - ln k) = 302K = 29°C

  14. Rate laws and mechanisms An "elementary step" or "elementary reaction" is one step that happens "as a single step", without any intermediates. A unimolecular elementary step can happen spontaneously, provided enough energy is available; it does not require anything else. A bimolecular elementary step can happen spontaneously on collision of the two reaction partners, if enough energy is available; it does not require anything else.

  15. Rate laws and mechanisms Rate laws for elementary steps can be written down immediately. Unimolecular, A → X:rate = k [A] Bimolecular, A + B → X:rate = k [A][B] If a reaction consists of several elementary steps, its overall rate law cannot be written down so easily. If there are several steps, the slow or rate-limiting elementary step determines the reaction rate.

  16. Rate laws and mechanisms ex: the observed rate law for4 HBr + O2 → 2 Br2 + 2 H2O israte = k [HBr] [O2] Is this compatible with the proposed mechanismHBr + O2 → HOOBr k1HOOBr + HBr → 2 HOBr k2HOBr + HBr → H2O + Br2k3 If so, what must be the rate-limiting step?

  17. Rate laws and mechanisms Assume k1 is rate-limiting: rate = k1 [HBr][O2] OK! Assume k2 is rate-limiting(then k1 is faster and we also have k-1) steady-state

  18. Rate laws and mechanisms Assume k3 is rate-limiting: (then bothk1 and k2are faster and we alsohave k-1, k-2)

More Related