Chem 218 Lab D1-100%

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Athabasca University, Edmonton *

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218

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Chemistry

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Jan 9, 2024

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Title: Reaction Kinematics I- Determination of Rate Law August 12,2023 Purpose of experiment: To determine the reaction rate, the order of a reaction and how the reaction rate depends on the concentration of the solution. This experiment will also look at how a catalyst such as Copper (II) Sulfate increases the rate of reaction and by what factor. In this experiment a reaction between iodide ions and peroxydisulfate ions will be performed as shown below. 2I - (aq) + S 2 O 8 2-  I 2 (aq) +2 SO 4 2- (aq) If the change is monitored between these 2 ions like the time it takes for the reaction to happen, reaction rate can be determined using the following equation. Inital Rate = ∆ [ I 2 ] ∆t =− ∆ ¿¿ Where a is the order of reaction with respect to peroxydisulfate and b is the order of reaction in term of iodide. If the above equation is written in logarithmic form and the peroxydisulfate and iodide is kept constant, a series of experiments can be carried out. log ( initialrate ) = log k + a log ¿ If one compound is kept constant for one experiment, the order of reaction of that compound can be found and vice versa. In order to measure the time, a starch solution will be used which can indicate when the reaction has taken place.

Procedure : This experiment was carried out as described in Experiment D1 of the Chemistry 218 Lab Manual (pp.43-56) .This experiment was paired with expeerimnt D2 to save time as reaction takes times. Observation: Part A: Preparation of Standard Solutions A. Potassium Iodide- Light yellow B. Sodium Thiosulfate-Clear less and colorless C. Potassium chloride- Clear less and colorless D. Ammonium peroxydisulfate- Clear less and colorless E. Ammonium Sulfate- Clear less and colorless F. Copper(II) Sulfate- Blue Part B: Preliminary Investigation Test Tube Observations 1 2 mL KI +15 drops starch +1 mL (NH 4 ) 2 S 2 O 8 -Solution turns clear when starch is added -Adding (NH 4 ) 2 S 2 O 8 turns the solutions blue/ purple at first but then turns darker 2 2 mL KI +12mL (NH 4 ) 2 S 2 O 8 +15 drops starch - Adding (NH 4 ) 2 S 2 O 8 turns solution yellow -Adding starch turns the top part of the solution black immediately and turns the bottom part of the solution dark red 3 2 mL KI + 5 drops Na 2 S 2 O 3 +15 drops starch +2 mL (NH 4 ) 2 S 2 O 8 -Solutions remains clear when Na 2 S 2 O 3 and starch were added - Adding (NH 4 ) 2 S 2 O 8 most of the solution at the top black and small clear layer at the bottom 4 2 mL KI +15 drops starch +2 mL (NH 4 ) 2 S 2 O 8 + 2 mL Na 2 S 2 O 3 -Solution turns clear when starch is added -Adding (NH 4 ) 2 S 2 O 8 turns the solutions blue/ purple at first but then turns darker -Adding Na 2 S 2 O 3 causes the solution to turn colorless Table 1

Results and calculations: Part A: Creating Stock Solutions Sol’n Solute Formula Volume Required (L) Final Concentration (mol/L) Molar Mass (g/mol) Grams required (g) Actual grams acquired(g) A Potassium Iodide KI 1L [0.2 mol X L -1 ] 166.0028 33.2056 33.1147 B Sodium thiosulfate Na 2 S 2 O 3 0.5L [0.01 mol X L -1 ] 248.11 1.2405 1.2872 C Potassium Chloride KCl 0.5L [0.2 mol X L -1 ] 74.55 7.455 7.4509 D Ammonium peroxydisulfate (Nh 4 ) 2 S 2 O 8 1L [0.1 mol X L -1 ] 228.18 22.818 22.8158 E Ammonium sulfate (Nh 4 ) 2 SO 4 0.5L [0.1 mol X L -1 ] 132.14 6.607 6.6107 Table 2: Soultyion to be prepared Sample Calculation for Sol’n A Potassuim Iodide M = 166.0028 g / mol Potassuim IodideV = 1 L Potassuim IodideC = 0.2 mol L Potassuim Iodidemol = 0.2 mol l ∗ 1 L = 0.2 mol Potassuim Iodide gramsrequired = M ∗ mol = 166.0028 g mol ∗ 0.2 mol = 33.2056 g

Part C & D: Volumes to be Added. Given Data Run KI (0.2M) (mL) Na2s03 (0.01M) (mL) Starch (3%) (mL) KCl (0.2M) (mL) (NH4)2S2O 8 (0.1 M) (mL) (NH4)2S04 (0.1M) (mL) CuSO4 (0.1M) (mL) 1 25.0 5.0 1.0 0.0 25.0 0.0 0.0 2 15.0 5.0 1.0 10.0 25.0 0.0 0.0 3 10.0 5.0 1.0 15.0 25.0 0.0 0.0 4 5.0 5.0 1.0 20.0 25.0 0.0 0.0 5 25.0 5.0 1.0 0.0 20.0 5.0 0.0 6 25.0 5.0 1.0 0.0 15.0 10.0 0.0 7 25.0 5.0 1.0 0.0 10.0 15.0 0.0 8 10.0 5.0 1.0 14.0 25.0 0.0 1.0 Table 3: Given data(in milliliter) for volumes to be added Millilitres to Liters Sample Calculation KI ( L )= 25 mL 1000 mL ∗ 1 L = 0.025 L Run KI (0.2M) (L) Na 2 S 2 O 3 (0.01M) (L) Starch (3%) (L) KCl (0.2M) (L) (Nh 4 ) 2 S 2 O 8 (0.1 M) (L) (Nh 4 ) 2 SO 4 (0.1M) (L) CuSO 4 (0.1M) (L) Total Volum e (L) 1 0.025 0.005 0.001 0.000 0.025 0.000 0.000 0.056 2 0.015 0.005 0.001 0.010 0.025 0.000 0.000 0.056 3 0.010 0.005 0.001 0.015 0.025 0.000 0.000 0.056 4 0.005 0.005 0.001 0.020 0.025 0.000 0.000 0.056 5 0.025 0.005 0.001 0.000 0.020 0.005 0.000 0.056 6 0.025 0.005 0.001 0.000 0.015 0.010 0.000 0.056 7 0.025 0.005 0.001 0.000 0.010 0.015 0.000 0.056 8 0.010 0.005 0.001 0.014 0.025 0.000 0.001 0.056 Table 4: Given data(in liter) for volumes to be added Liters to moles Sample Calculation FinalConcentration of KI = 0.2 mol L

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