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Experiment 2: The Rate of Decolorization
Kb=[products][reactants]rate=k1[A][B]+k2[A][C]*A=dye; B=H2O; C=OH-kobs=k1[B]+k2[C]rate=kobs[A]A¿oln[A]=−kobs∗t+ln¿[A]=A/εl*Beer-Lambert Law−lnA=kobs∗t−lnAo*determine rate constant by plotting -ln(absorbance) versustime*slope is kobsPurpose: Study the kinetics of a reaction involving a green dye. Note the structural changes of the dye that converts the green color to colorlessness.Background:> the molecule has color because dyes are aromatic organic compounds.> this means that the compound has planar ring systems with alternating double bonds.> the conjugated compound becomes stable from the delocalization of electrons (i.e benzene)> electrons aren’t associated in a single atom or one bond, but are delocalized pi bonds that form an orbital cloud> colors become visible when a molecule absorbs wavelengths as an electron is excited from its ground state to a higher energy level.> in conjugated chromophores, electrons jump between energy levelsthat are extended pi orbitals.> when OH- adds to the central carbon, the central atom has 4 single bonds because tetrahedral structures don’t allow pi bonding; the color fades because e- delocalization doesn’t exist anymoreProcedure: Make a solution of Na3PO4 and Na2HPO4 (buffer system) and pipet exact amounts into a beaker with 5 mL of dye. Put it into a cuvette. Each trial should have difference amounts of Na3PO4 and Na2HPO4 that change the