In: Chemistry
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
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.