Chem 218 Lab D1-100%

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

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

Mol KI = C ∗ V = 0.2 mol L ∗ 0.025 L = 0.005 mol Run KI (0.2M) (mol) Na 2 S 2 O 3 (0.01M) (mol) Starch (3%) (mol) KCl (0.2M) (mol) (Nh 4 ) 2 S 2 O 8 (0.1 M) (mol) (Nh 4 ) 2 SO 4 (0.1M) (mol) CuSO 4 (0.1M) (mol) 1 0.005 0.00005 0.00003 0.0000 0.0025 0.0000 0.0000 2 0.003 0.00005 0.00003 0.0020 0.0025 0.0000 0.0000 3 0.002 0.00005 0.00003 0.0030 0.0025 0.0000 0.0000 4 0.001 0.00005 0.00003 0.0040 0.0025 0.0000 0.0000 5 0.005 0.00005 0.00003 0.0000 0.0020 0.0005 0.0000 6 0.005 0.00005 0.00003 0.0000 0.0015 0.0010 0.0000 7 0.005 0.00005 0.00003 0.0000 0.0010 0.0015 0.0000 8 0.002 0.00005 0.00003 0.0028 0.0025 0.0000 0.0001 Table 5: Given data(in mol) for volumes to be added. Kinetic Study Sample calculation using run 1. 1. Temperature across all runs was 22  c and the time was recorded in table 6 2. Calculate the initial concentration of potassium iodide and ammonium peroxydisulfate. Mol potassiumiodide KI = 0.005 mol TotalVolume = 0.056 L Initialconcentration of potassium iodide = 0.005 mol 0.056 L = 0.0893 mol / L Molammonium peroxydisulfate ¿ Initialconcentration of ammonium peroxydisulfate S 2 O 8 2 − ¿ = 0.0025 mol 0.056 L = 0.0446 mol L ¿ 3. Calculate the initial concentration of sodium thiosulfate. Molsodiumthiosulfate N a 2 S 2 O 3 = 0.00005 mol TotalVolume = 0.056 L Initialconcentration of sodiumthiosulfate = 0.00005 mol 0.056 L = 0.000893 mol / L 4. Calculate the change in the concentration of the ammonium peroxydisulfate. Initialconcentration of ammonium peroxydisulfate S 2 O 8 2 − ¿ = 0.0025 mol 0.056 L = 0.0446 mol L ¿

7. Plot a Graph of log[rate] vs log[I - ] 0 for all runs with same initial concentration of ammonium peroxydisulfate. log [ Rate ] = log ( 1.337 X 10 − 5 mol / L ∗ s )=− 4.874 Log ¿ This graph will be plotted for runs 1,2,3,4 which have the same common value of 0.0446 for ammonium peroxydisulfate. Run Log rate Log [I - ]o 1 -4.874 -4.874 2 -5.219 -5.219 3 -5.242 -5.242 4 -5.528 -5.528 Table 7: Data table for Log rate vs Log [I - ]o graph with same ammonium peroxydisulfate values -5.7 -5.5 -5.3 -5.1 -4.9 -4.7 -4.5 -2.0 -1.8 -1.6 -1.4 -1.2 -1.0 -0.8 -0.6 -0.4 -0.2 0.0 f(x) = 1.07 x + 4.18 Log rate vs Log [I-]o Log [I-]o Log rate Figure 2: Graph of Log rate vs Log [I - ]o 8. Determine rate constant.

Rate = k ¿¿ a = order of log ¿ b = order of log ¿ Overallorder = second order Rate = k ¿¿ k = Rate ¿¿¿ 9. Average value and standard deviation of rate constant k avg = 0.0034 + 0.0025 + 0.0036 + 0.0037 + 0.0028 + 0.0020 + 0.0017 = 0.0028 L / mol ∗ s k standarddeviation = 0.000728 L / mol ∗ susing excel Run Time  T(s) [I - ] o (mol/L) [S 2 O 8 2- ] 0 (mol/L) [S 2 O 3 - ] o (mol/L) Δ [S 2 O 8 2- ] (mol/L) -Rate (mol/L*s) 1 33.4 0.0893 0.0446 0.000893 -0.000446 1.337E-05 2 74 0.0536 0.0446 0.000893 -0.000446 6.033E-06 3 78 0.0357 0.0446 0.000893 -0.000446 5.723E-06 4 150.7 0.0179 0.0446 0.000893 -0.000446 2.962E-06 5 40.3 0.0893 0.0357 0.000893 -0.000357 8.862E-06 6 55.3 0.0893 0.0268 0.000893 -0.000267 4.844E-06 7 66.9 0.0893 0.0179 0.000893 -0.000178 2.669E-06 8 1.69 0.0357 0.0446 0.000893 -0.000446 2.642E-04 Run Log rate Log [I - ]o Log[ S 2 O 8 2- ]0 Rate Constant K (L/mol*s) 1 -4.874 -1.049 -1.350 0.0034 2 -5.219 -1.271 -1.350 0.0025 3 -5.242 -1.447 -1.350 0.0036 4 -5.528 -1.748 -1.350 0.0037 5 -5.052 -1.049 -1.447 0.0028 6 -5.315 -1.049 -1.572 0.0020 7 -5.574 -1.049 -1.748 0.0017 8 -3.578 -1.447 -1.350 0.1657 K Average 0.0028 K Standard Deviation 0.000728 The effect of Catalyst Table 8: Tabulated data for all the calculations performed.

Conclusion In this experiment, the reaction rates and the order of reaction was found out. It was determined that all reactions had a first order of reaction of 1. The average rate constant for runs without a catalyst was 0.0028 l/mol*s. Adding a catalyst to the run increased the rate constant to 0.1657 L/mol*s , an increase by a factor of 48.7.