Question

In: Chemistry

How do I calculate the inicial concentration of​​ I^- and S2O8^2- in each mixture, the reaction...

How do I calculate the inicial concentration of​ I^- and S2O8^2- in each mixture, the reaction rate M/s, and the k value (rate constant) from the following data???? IODINE CLOCK REACTION

Run # 3% Starch 0.012 M Na2S2O3 mL 0.20 M KI mL 0.20 M KNO3 mL 0.20 M (NH4)2SO4 mL 0.20 M (NH4)2S2O8 mL Total mL
1 2 drops 0.200 0.800 0.200 0.400 2.00
2 2 drops 0.200 0.400 0.600 0.400 2.00
3 2 drops 0.200 0.200 0.800 0.400 2.00
4 2 drops 0.200 0.400 0.600 0.00 2.00
5 2 drops 0.200 0.400 0.600 0.600 2.00

Run#1-time was 41.45 sec   Run#2-time was 120.48 sec   Run#3-231.64 sec   Run#4-36.88 sec    Run#5-231.70 sec

Solutions

Expert Solution

For Run#1

Initial moles of I- added = 0.2 * 0.8 = 0.16 mmoles

Inital concentration of I- = 0.16 / 2 = 0.08 M

Initial moles of S2O82- added = 0.2 * 0.4 = 0.08 mmoles

Inital concentration of S2O82- = 0.08 / 2 = 0.04 M

The rates for the reactions can be calculated in terms of Δ[I2 ] / ΔTime, the change in iodine concentration per second. The appearance of the blue color means that [I2] has gone from 0 to 5.0 × 10–4 M by the time the color appears. Therefore, the rate of each reaction is found by dividing 5.0 × 10–4 M by the time it takes
for the solution to turn blue.

Rate = k [I-]x [S2O82-]y

Below tables gives initial concentration for all runs:

Run #1: Rate = 1.206 x 10-5 = k [0.08]x [0.04]y

Run #2: Rate = 4.15 x 10-6 = k [0.04]x [0.04]y

Run #3: Rate = 2.156 x 10-6 = k [0.02]x [0.04]y

Run #4: Rate = 1.356 x 10-5 = k [0.04]x [0.08]y

Run #5: Rate = 2.158 x 10-6 = k [0.04]x [0.02]y

Run #2 / Run #3: 1.92 = 2x                                 

x ≈1

Run #2 / Run #5: 1.92 = 2y

y ≈1

4.15 x 10-6 M/s= k * 0.04 * 0.04

k = 2.6 x 10-3 (M/s)-1

Note: Different runs given different values of k. This might be due to incorrect time recording or variation in reactant temperature. Most reliable points have been chosen for calculation purpose.


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