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 =
To answer these questions, we need to use the stoichiometry of the given reactions.
b. Since the S2O32- is completely consumed during the reaction, the change in concentration of S2O32- can be calculated as the initial concentration minus zero. So, Δ[S2O32-] = [S2O32-]initial - 0 = [S2O32-]initial.
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 each S2O82- consumes 2 S2O32-. So, for every 1 mole of S2O82- that reacts, 2 moles of S2O32- are consumed. Therefore, Δ[S2O82-] = -2Δ[S2O32-] = -2[S2O32-]initial.
d. If we assume the reaction takes 75 seconds, we can calculate the rate of the reaction by dividing the change in concentration of S2O82- by the time taken. So, the rate of the reaction Δ[S2O82-]/Δt = Δ[S2O82-]/75 seconds = (-2[S2O32-]initial)/75 seconds.