# Recall the reactions for this lab:

S2O82- + 2I- →
2SO42- + I2 (slow) Reaction 1
The one that delays the solution turning blue:
I2 + 2S2O32- → 2I- + S4O62- (fast) Reaction 2
b. All the S2O32 is consumed when the color change occurs. Therefore,

Δ[S2O32] =

c. Use the stoichiometry of Reactions 1 and 2 to determine the change in concentration of
S2O82- during the same time period.
Δ[S2O82-] =

d. Assuming the reaction took 75 seconds, calculate the rate of the reaction Δ [S2O82-]/Δt.
Rate of reaction = Δ[S2O82- ]/Δt =

## In order to answer these questions, we need to understand the given reactions and stoichiometry.

Reaction 1: S2O82- + 2I- → 2SO42- + I2 (slow)
Reaction 2: I2 + 2S2O32- → 2I- + S4O62- (fast)

b. All the S2O32- is consumed when the color change occurs. Therefore, Δ[S2O32-] is the change in concentration of S2O32-.

c. To determine the change in concentration of S2O82- during the same time period, we need to use the stoichiometry of Reactions 1 and 2.

From Reaction 1, we see that 1 mole of S2O82- reacts with 2 moles of I-. From Reaction 2, we see that 2 moles of S2O32- react with 1 mole of S2O82-.

Therefore, the stoichiometric relationship between S2O82- and S2O32- is 1:2. This means that for every mole of S2O82- consumed, 2 moles of S2O32- are also consumed. So, Δ[S2O82-] is twice the change in concentration of S2O32-.

Δ[S2O82-] = 2 * Δ[S2O32-]

d. Assuming the reaction took 75 seconds, the rate of the reaction Δ[S2O82-]/Δt can be calculated by dividing the change in concentration of S2O82- (Δ[S2O82-]) over the time period (Δt).

Rate of reaction = Δ[S2O82-] / Δt