Experiment 2 the rate of decolorization k b products

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Experiment 2: The Rate of Decolorization
K b = [ products ] [ reactants ] rate = k 1 [ A ][ B ]+ k 2 [ A ][ C ] *A=dye; B=H2O; C=OH- k obs = k 1 [ B ]+ k 2 [ C ] rate = k obs [ A ] A ¿ o ln [ A ]=− k obs t + ln ¿ [ A ]= A / εl *Beer-Lambert Law lnA = k obs t ln A o *determine rate constant by plotting -ln(absorbance) versus time *slope is kobs Purpose: 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 levels that 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 anymore Procedure: 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
absorptivity of the dye.

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