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Half Life I messed up, sorry

Half Life I messed up, sorry. Webassign HW The decomposition of SO 2 Cl 2 is first order in SO 2 Cl 2 , and the reaction has a half-life of 255 min at 600 K. SO 2 Cl 2 (g) → SO 2 (g) + Cl 2 (g)

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Half Life I messed up, sorry

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  1. Half LifeI messed up, sorry • Webassign HW • The decomposition of SO2Cl2 is first order in SO2Cl2, and the reaction has a half-life of 255 min at 600 K. SO2Cl2(g) → SO2(g) + Cl2(g) • If you begin with 2.6 x 10-3 mol of SO2Cl2 in a 1.0 L flask, how long will it take for the quantity of SO2Cl2 to decrease to 2.00 x 10-4 mol? • I acknowledge there are other ways to solve this but let’s use the long way.

  2. Half LifeI messed up, sorry • Known: [SO2Cl2 ]o = 2.6 x 10-3mol/L, [SO2Cl2 ] 2.00 x 10-4mol/L, t ½ = 255 min. • Unknown: t • Use integrated rate equation for 1st order • ln [A] / [A]0 = -kt • Don’t know k or t but we can solve for k if the time is 255 min. • ln (1/2) = -k 255 min (in this case t is the ½ life) • k = 0.002718 min-1 • Now we know k so let’s solve for t with our [ ]

  3. Half LifeI messed up, sorry • Known: [SO2Cl2 ]o = 2.6 x 10-3mol/L, [SO2Cl2] 2.00 x 10-4mol/L, t ½ = 255 min. • ln [A] / [A]0 = -kt • Don’t know k or t but we can solve for k if the time is 255 min. • k = 0.002718 min-1 • Now we know k so let’s solve for t with our [ ] • ln (2.00 x 10-4 / 2.6 x 10-3) = 0.002718 min-1 t • t = 943 min

  4. Half LifeI messed up, sorry • Radioactive gold-198 is used as the metal in the diagnosis of liver problems. The half-life of this isotope is 2.7 days. If you begin with a 6.60 mg sample of the isotope, how much of this sample remains after 1.0 day? • Try this the long way… yes I understand there is a shorter way to do this. • Did you start by solving for k? • k = 0.2567 day-1 • Final [ ] = 5.11 mg Did you get there???

  5. Half LifeI messed up, sorry • The compound Xe(CF3)2 is unstable, decomposing to elemental Xe with a half-life of 30. min. If you place 6.00 mg of Xe(CF3)2 in a flask, how long must you wait until only 0.25 mg of Xe(CF3)2 remains? • Again the long way • k = 0.0231 min-1 • t = 137 min

  6. Half LifeI messed up, sorry • Did I fix it? • Remember, you can always use shorter methods (if they work) but now you know one that will always work!

  7. Factors Affecting Rates • Concentrations • and physical state of reactants and products • Temperature • Catalysts

  8. Factors Affecting Rates Rate with 0.3 M HCl • Concentrations Rate with 6.0 M HCl

  9. Factors Affecting Rates • Physical state of reactants

  10. Factors Affecting Rates Catalysts: catalyzed decomp of H2O2 2 H2O2 --> 2 H2O + O2

  11. Factors Affecting Rates • Temperature Bleach at 54 ˚C Bleach at 22 ˚C

  12. MECHANISMSA Microscopic View of Reactions Mechanism: how reactants are converted to products at the molecular level. RATE LAW ----> MECHANISM experiment ----> theory

  13. Activation Energy Molecules need a minimum amount of energy to react. Visualized as an energy barrier - activation energy, Ea. Reaction coordinate diagram

  14. MECHANISMS& Activation Energy For example Rate = k [trans-2-butene] Conversion requires twisting around the C=C bond.

  15. Activation energy barrier MECHANISMS Conversion of trans to cis butene

  16. Activated Complex 4 kJ/mol MECHANISMS Energy involved in conversion of trans to cis butene energy -262 kJ 266 kJ cis 31 kJ/mol trans 27 kJ/mol See Figure 15.14

  17. energy Activated Complex -262 kJ 266 kJ 4 kJ/mol cis 31 kJ/mol trans 27 kJ/mol Mechanisms • Reaction passes thru a TRANSITION STATEwhere there is an activated complex that has sufficient energy to become a product. ACTIVATION ENERGY, Ea= energy req’d to form activated complex. Here Ea = 266 kJ/mol

  18. energy Activated Complex -262 kJ 266 kJ 4 kJ/mol cis 31 kJ/mol trans 27 kJ/mol MECHANISMS Also note that trans-butene is MORE STABLE than cis-butene by about 4 kJ/mol. Therefore, trans ---> cis is ENDOTHERMIC. This is the connection between thermo-dynamics and kinetics.

  19. Activation Energy and Temperature Reactions are slower at lower T because a smaller fraction of reactant molecules have enough energy to convert to product molecules. In general, differences in activation energy cause reactions to vary from fast to slow.

  20. Mechanisms 1. Why is cis-butene <--> trans-butene reaction observed to be 1st order (link between [ ] and order)? As [trans] doubles, number of molecules with enough E also doubles. 2. Why is the cis <--> transreaction faster at higher temperature? Fraction of molecules with sufficient activation energy increases with T.

  21. Temp (K) Rate constant 8.31 x 10-3 kJ/K•mol Activation energy Frequency factor More About Activation Energy Arrhenius equation — Frequency factor = frequency of collisions with correct geometry. Plot ln k vs. 1/T ---> straight line. slope = -Ea/R

  22. More on Mechanisms A bimolecular reaction Reaction of cis-butene --> trans-butene is UNIMOLECULAR- only one reactant is involved. BIMOLECULAR — two different molecules must collide --> products Exo- or endothermic?

  23. Collision Theory Reactions require (a) activation energy and (b) correct geometry. O3(g) + NO(g) ---> O2(g) + NO2(g) 1. Activation energy 2. Activation energy and geometry

  24. What do you know • How and why does increased T effect r’xn rate? • How and why does increased [ ] effect r’xn rate? • How and why does a catalyst effect r’xn rate?

  25. Last Day of Kinetics • Already done . . . • Solving for rate constant and expression from • Graph • Initial concentrations • Using integrated rate laws • Not going to get to . . . • Solving for the activation energy • It will likely be on AP test, it is in the book and in my ppt which will be on line (after today) • Last thing – Reaction Mechanisms!!!!

  26. Rate Reminder • For reaction A + B → C + D • Rate = - Δ [A]/ Δ t • Rate law = k [A]x [B]y • Overall rate is sum of x + y • Integrated rate law (depends on order of reaction) • For us, only single reactant in integrated rate law

  27. Mechanisms O3 + NO reaction occurs in a single ELEMENTARY step. Most others involve a sequence of elementary steps. Adding elementary steps gives NET reaction.

  28. Mechanisms Most rxns. involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] NOTE 1. Rate law comes from experiment 2. Order and stoichiometric coefficients not necessarily the same! 3. Rate law reflects all chemistry down to and including the slowest step in multistep reaction.

  29. Mechanisms Most rxns. involve a sequence of elementary steps. 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] Proposed Mechanism Step 1 — slow HOOH + I- --> HOI + OH- Step 2 — fast HOI + I- --> I2 + OH- Step 3 — fast 2 OH- + 2 H+ --> 2 H2O Rate of the reaction controlled by slow step — RATE DETERMINING STEP, rds. Rate can be no faster than rds!

  30. Mechanisms 2 I- + H2O2 + 2 H+ ---> I2 + 2 H2O Rate = k [I-] [H2O2] Step 1 — slow HOOH + I- --> HOI + OH-Step 2 — fast HOI + I- --> I2 + OH- Step 3 — fast 2 OH- + 2 H+ --> 2 H2O Elementary Step 1 is bimolecular and involves I- and HOOH. Therefore, this predicts the rate law should be Rate  [I-] [H2O2] — as observed!! The species HOI and OH- are reaction intermediates.

  31. Funnel Demo • Yeah, just a reminder for me

  32. Rate Laws and Mechanisms NO2 + CO reaction: Rate = k[NO2]2 Two possible mechanisms Two steps: step 1 Single step Two steps: step 2

  33. Reaction Mechanisms • The balanced chemical equation provides information about the beginning and end of reaction. • The reaction mechanism gives the path of the reaction. • Mechanisms provide a very detailed picture of which bonds are broken and formed during the course of a reaction. • Elementary Steps • Elementary step: any process that occurs in a single step.

  34. Reaction Mechanisms • Elementary Steps • Molecularity: the number of molecules present in an elementary step. • Unimolecular: one molecule in the elementary step, • Bimolecular: two molecules in the elementary step, and • Termolecular: three molecules in the elementary step. • It is not common to see termolecular processes (statistically improbable).

  35. Reaction Mechanisms • Multistep Mechanisms • Some reactions proceed through more than one step: • NO2(g) + NO2(g)  NO3(g) + NO(g) • NO3(g) + CO(g)  NO2(g) + CO2(g) • Notice that if we add the above steps, we get the overall reaction: • NO2(g) + CO(g)  NO(g) + CO2(g)

  36. Reaction Mechanisms • Multistep Mechanisms • If a reaction proceeds via several elementary steps, then the elementary steps must add to give the balanced chemical equation. • The rate law that results from the mechanisms must agree with the rate law from experimentation • Intermediate: a species which appears in an elementary step which is not a reactant or product and CAN NOT appear in the rate law.

  37. Reaction Mechanisms • Rate Laws for Elementary Steps • The rate law of an elementary step is determined by its molecularity: • Unimolecular processes are first order, • Bimolecular processes are second order, and • Termolecular processes are third order. • Rate Laws for Multistep Mechanisms • Rate-determining step: is the slowest of the elementary steps.

  38. Reaction Mechanisms Rate Laws for Elementary Steps

  39. Reaction Mechanisms • Rate Laws for Multistep Mechanisms • Therefore, the rate-determining step governs the overall rate law for the reaction. • Mechanisms with an Initial Fast Step • It is possible for an intermediate to be a reactant. • Consider • 2NO(g) + Br2(g)  2NOBr(g)

  40. Reaction Mechanisms • Mechanisms with an Initial Fast Step • 2NO(g) + Br2(g)  2NOBr(g) • The experimentally determined rate law is • Rate = k[NO]2[Br2] • Consider the following mechanism

  41. Reaction Mechanisms • Mechanisms with an Initial Fast Step • The rate law is (based on Step 2): • Rate = k2[NOBr2][NO] • The rate law should not depend on the concentration of an intermediate (intermediates are usually unstable). • Assume NOBr2 is unstable, so we express the concentration of NOBr2 in terms of NOBr and Br2 assuming there is an equilibrium in step 1 we have

  42. Reaction Mechanisms • Mechanisms with an Initial Fast Step • By definition of equilibrium: • Therefore, the overall rate law becomes • Note the final rate law is consistent with the experimentally observed rate law.

  43. CATALYSIS 2 CO + O2 ---> 2 CO2 2 NO ---> N2 + O2 In auto exhaust systems — Pt, NiO

  44. CATALYSIS 2. Polymers: H2C=CH2 ---> polyethylene 3. Acetic acid: CH3OH + CO --> CH3CO2H 4. Enzymes — biological catalysts

  45. Uncatalyzed reaction Catalyzed reaction CATALYSIS MnO2 catalyzes decomposition of H2O2 2 H2O2 ---> 2 H2O + O2 Catalysis and activation energy

  46. Iodine-Catalyzed Isomerization of cis-2-Butene Figure 15.16

  47. Iodine-Catalyzed Isomerization of cis-2-Butene

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