Question

For the reaction between the peroxydisulfate ion (S2O82-) and the iodide ion in aqueous

solution:

S2O82- (aq) + 3I- (aq) → 2SO42- (aq) + I3- (aq)

An aqueous solution containing 0.050 M of S2O82- ion and 0.072 M of I- is prepared, and the progress of the reaction followed by measuring [I-]. The data obtained is given in the table below.

What is the average rate of disappearance of I- in the initial 400.0 s, between 400.0 s and 800.0 s, and between 1200.0 s and 1600.0 s?

What are the concentrations of S2O82- remaining at 400.0, 800.0, and 1600.0 s?

Answer 1

Answer :-

From the stoichiometrie of the reaction follows that for every three moles of iodide ions one mole of peroxydisulfate ions reacts away:
Δn(S₂O₈⁻) = n₀(S₂O₈⁻) - n(S₂O₈⁻) = 1/3 · Δn(I⁻)

The amount of peroxydisulfate ions remaining is:
n(S₂O₈⁻) = n₀(S₂O₈⁻) - 1/3 · Δn(I⁻) = n₀(S₂O₈⁻) - 1/3 ·(n₀(I⁻) - n(I⁻))

Since the total molar concentration can betaken constant for dilute aqueous solutions, you can rewrite in terms of concentrations:
[S₂O₈⁻] = [S₂O₈⁻]₀ - 1/3 ·( [I⁻]₀ - [I⁻] )

For t=800s you get:
[S₂O₈⁻] = 0.050M - 1/3·( 0.072M - 0.046M) = 0.0413M