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 =

Similar Questions
  1. Recall the reactions for this lab:S2O82- + 2I- → 2SO42- + I2 (slow) Reaction 1 The one that delays the solution turning blue:
    1. answers icon 0 answers
  2. Reposting since it wasn't clear in the previous postS2O82-(aq) + 2I-(aq) ----------> 2SO42-(aq) + I2(aq) (I) Rate = k [S2O82-]x
    1. answers icon 1 answer
  3. S2O82-(aq) + 2I-(aq) ----------> 2SO42-(aq) + I2(aq) (I)Rate = k [S2O82-]x [I-]y Rate1 = k [S2O82-]x [I-]y ƒ´ [S2O82ƒ{] Rate
    1. answers icon 1 answer
    1. answers icon 1 answer
more similar questions