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CHEM 102 Full Lab Notes

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University of Alberta

Chem 102 Labs▯ ▯ ▯ Experiment M▯ ▯ For a reaction to occur, reactant molecules must collide with enough energy and correct orientations for their electronic orbitals to overlap and create a temporary activated complex or Activation energy: the minimum energy required to form an activated complex. ▯ ▯ The larger the Ea the fewer the molecules with sufficient energy to react. ▯ RATE CONSTANT “K” DECREASING WITH INCREASING Ea QQ▯ K is constant if concentrations change. ▯ K increases with increasing temperature. ▯ ▯ ln k = (-Ea / R) 1/T + ln A▯rrange into the form y = mx + b: ▯ ▯ Thus a straight line can be graphed with a slope of -Ea/R from which Ea can be calculated. ▯ ▯ STUDING EFFECT OF TEMPERATURE ON Redox reaction between Idodide and peroxydisulfate. ▯ ▯ 2I- + S2O82- —> 2SO42- + I2▯ ORDER OF REACTION WIHT RESPECT TO I- AND S2O82- IS 1. ▯ ▯ CATALYST DOES NOT CHANGE AMOUNT OF PRODUCT FORMED.. IT SIMPLY ▯NCREASES THE SPEED TO GET THERE ▯ IODINE CLOCK▯ Adding a small amount of thiosulfate (S2O32-) and starch to the following reaction will cause the thiosulfate to immediately react with the I2 changing it back to I-. Note the following reactants 2I- + S2O82- —> 2SO42- + I2▯ addition of thiosulfate: ▯ 2S2O32- + I2 —> 2I- + S4O62- ▯ ▯ to form the trioxide ion I3- which reacts with starch to form a deep colored complex. ▯ll react with I- ▯ [I2] FORMED DURING ^T IS EQUAL TO HALF OF THE AMOUNT OF THIOSULFATE ADDED▯ ▯ ▯T should be less at higher temperatures.▯ Notes: ▯ A = K when temperature is infinity. ▯ ▯a does not increase a temperature is changed. ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ Experiment N▯ ▯ ▯ye + OCl- —> colourlessness products▯ OCl- is kept in excess making the dye the only reactant considered in the rate eqt. ▯ rate = k’ [dye]^q▯ Reaction Orders and relation of half lives to concentration of Dye: ▯ ▯ plotting [A] against time gives a slope = -k that is a straight line ▯ ▯ half life depends on K only ▯dent of dye because T1/2 = ln 2 / k▯ ▯lotting ln [A] against time results in straight line slope = -k ▯ second order: half life inversely proportional to dye▯ ▯lotting 1/[A] against time gives a straight line slope = -k ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ Experiment O▯ ▯olubility of I2 is increased by adding KI (I-) to form the tri-iodide ion which is soluble in water. ▯ Equilibrium constant: ▯ K = [I3-] / [I2] [I-] ▯ ▯ solution of potassium iodide and iodine where all the iodine will enter the Organic layer because it is much more soluble there. Mineral oil will be used as the organic solvent forming the top layer. the Iodide (I-) and the tri-iodide ions are only soluble in water. ▯ Once an excess amount of Iodine is added to a known concentration of iodide and mineral oil is added, the iodine will move to the organic layer according to a distribution coeffiecient. ▯ K’ = [I2 Organic] / [I2 (aq)] = 38 ▯ meaning there is 38 times as much iodine in oil than there is in water. ▯ K’ for carbon tetrachloride and water is 91 and for DCM is 150. ▯ ▯ ▯ssentially. There is only Iodine in the organic layer. ▯ Q > K rxn favors reactants shifts to the left▯ Q < K rxn favors products shifts to the right ▯ 2 THIOSULFATE REACTS WITH ONE I2 OR ONE I3▯ ▯ IODIDE IS COLORLESS▯ HAVE A REDDISH BROWN ▯ IODINE IN ORGANIC IS PURPLE ▯ END POINT OF TITRATION OF AQ LAYER CHANGES FROM BLUE TO COLORLESS▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ ▯ EXPERIMENT P: TITRATION OF A WEAK ACID ▯ Use Bronsted definitions of acids and bases. ▯ Base is a proton acceptor and acid is a proton donor. ▯ In a titration of a strong acid with a strong base, the equivalence point will have a pH of 7 and their will be an equimolar amount of strong base strong acid. ▯ Titration of a weak acid with strong base. ▯ Graph has following properties. ▯ ▯ at half equiv. volume [HA] = [A-] so Ka = [H+] ▯FOR WEAK ACIDS ▯ ▯ %diss. = H+ / [HA]inital (intial con. of acid) x 100%▯ H+ = 10^-ph▯ ▯ the equivalence point is the point where the moles of base are equal to moles acid - whether ▯eak or strong. ▯ In a titration of a weak acid with a strong base, at equiv. point, the concentration of the conjugate base (A-) = (HAinitial) / (initial volume of HA) + (vol. of strong base) (total volume) ▯ PH At equivalence = ph of conjugate base A- ▯ ▯ Acid base indicates are weak organic acids with a conjugate base of a different colour. ▯ CHOOSE AN ACID BASE INDICATOR THAT WILL CHANGE COLOR OF SOLUTION NEAR PKA OF INDICATOR. ▯ ▯KA OF INDICATOR SHOULD BE CLOSE TO PH OF SOLUTION▯ pOH = 14 - pH▯ ▯ ▯ater contains hydronium ions. If the pH of water is 7. then there there is a 10^-7 M of H+. ▯ If you have a solution of HA and NaOH, then at equivalence point you will have NaA as the salt. ▯ ▯ SOLUTIONS ▯ASES CONTAIN OH- AND ACID/BASES DEFINITION ONLY WORKS IN AQ ▯ ▯ ▯ ▯ ▯ Experiment Q Buffers and Antacids ▯ ▯ blood is buffered by carbonic acid hydrogen carbonate buffer system. ▯ange in pH. ▯ ▯ both present in high enough concentrations compared to the conc. of H+ and OH-. ▯pair are ▯ CH3CHOOH + H2O ->
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