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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.

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

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  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)

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