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Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege

Engr/Math/Physics 25. Prob 6.18 Chem Rcn Rate. Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege.edu. More Linear TransForms. These are NOT the Only Functional Relationships that can be Linearized:

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Bruce Mayer, PE Registered Electrical & Mechanical Engineer BMayer@ChabotCollege

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  1. Engr/Math/Physics 25 Prob6.18ChemRcn Rate Bruce Mayer, PE Registered Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

  2. More Linear TransForms • These are NOT the Only Functional Relationships that can be Linearized: • ANY Function with exactly TWO fitting Parameters is a Candidate for Linearization • e.g.; recall the ClayeronEqn

  3. Problem 6.18  Chemical Rcn Order • 1st Order Rate Eqn is Expontnential • By SemiLog Linearization we can “Discover” parameters [m & b]  [−k & C(0)] • 2nd Order Eqn can be LINEARIZED as • Thus ANOTHERLinearizable Fcn

  4. Prob Solve 1st Step → PLOT it • Advice for Every Engineer and Applied Mathematician or Physicist: • Rule-1: When in Doubt PLOT IT! • Rule-2: If you don’t KNOW when to DOUBT, then PLOT EVERYTHING

  5. The Concentration vs Time Data • For NitrousOxide Decomposition

  6. The Data Bar Charted

  7. Prob 6.18 • When in Doubt PLOT (use SubPlot) • Some CURVATURE • Straight • Better Model • t X • 1/C Y

  8. Now that the plot has identified the Rcn as 2nd Order, Make Linear Xform The 2nd Order Eqn Linear Xform of 2nd Order Reaction • Use polyfit of order-1 to generate fitting parameters contained in vector k_1overC0 • That is: k_1overC0 = [m, B2]; or • k_1overC0(1)= m →k • k_1overC0(2)= B2→1/C(0)

  9. The 2nd Order Model

  10. P 6.18 Answer • k_1overC0 = [m B2] = [k 1/C0] = [5.4445e-001 9.9605e+001] • k = 5.4445e-001 • k = 0.54445/(L/mol∙sec) • C0 =1/9.9605e+001 • 1.0040e-002 • C(0) = 0.01004 mol/Liter

  11. Compare Data to Fit

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