Considering the volumes and concentrations used in Mixture 1, assuming the volum

Question

Considering the volumes and concentrations used in Mixture 1, assuming the volume for [I]=20.00mL, [S2O8]=20.00mL, [S2O3]=10.00mL. a) what percent of the miles of S2O8 present have been consumed for the reaction? b) how has the [I] concentration changed (quantitatively) during the same time? 1. CONCENTRATION EFFECTS (ETERMINATION oF RATE LAw) Fill in the following table with the initial concentrations (in moles after mixing the persulfate and the thiosulfate change in concentration of L, time required for the reaction to turn blue and the initial rate of reaction for mixtures 1-5 (you do not need to show the set up for the numbers in the table.) Mixture Initial Rate IS208 2lo AIS2032l time (sec. .0800 M 0400 M .00200M .00100 M 15, eq 1.32 x 0.000 M 0200 M 0. Dozoo M 0,00 M 17 37 S. o IxIO 0.0800 M D. 0133 M D.0D20o M 0.00 IDO M 304.00 3.29 x Myk

Explanation / Answer

Write down the reaction between persulfate and iodide as below:

S2O82- (aq) + 2 I- (aq) -----> 2 SO42- (aq) + I2 (aq) …..(1)

The iodine formed in the reaction is quantitatively estimated by its titration with thiosulfate as per the below equation:

2 S2O32- (aq) + I2 (aq) -----> S4O62- (aq) + 2 I- (aq) ….(2)

As per the molar stoichiometry of equation (2)

2*moles S2O32- = moles I2

Again as per the stoichiometry of equation (1),

moles I2 = moles S2O82-

Combining (1) and (2),

moles S2O82- = 2*moles S2O32-

The initial concentration for mixture (1) are given:

[I-]0 = 0.0800 M

[S2O82-]0 = 0.0400 M

[S2O32-]0 = [S2O32-] = 0.00200 M

[I2] = 1/2[S2O32-] = 0.00100 M

a) We need to realize one point here:

[I2] = [I2]0 – [I2]f = [I2]0 – 0 = [I2]0

The reason can be easily realized. As soon as I2 is formed, it is consumed by thiosulfate so that I2 concentration doesn’t build up in the solution. Therefore,

[I2]0 = 0.00100 M

We have already shown above that

moles S2O82- = moles I2

and hence [S2O82-] = [I2] = [I2]0 = 0.00100 M (all the reagents are mixed in the solution and the volume will thus be the same for both).

Moles S2O82- initially present = (20.00 mL)*(1 L/1000 mL)*(0.0800 mol/L) = 0.0016 mol.

Moles S2O82- consumed = (50.00 mL)*(1 L/1000 mL)*(0.0100 mol/L) = 0.0005 mol (since the change in concentration of S2O82- is the quantity of S2O82- consumed and the total volume is 50.00 mL upon mixing).

Therefore, percent moles of S2O82- consumed = (0.0005 mol)/(0.0016 mol)*100 = 31.25% (ans).

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