Question

Determining reaction rates?

In my book part A is given and I have to figure out part B. What would be the equations steps needed to solve? I tried a few but am Unsure if they are done properly.
A. Determination of Reaction Times
molar concentration of Na2S2O3(mol/L): 0.02
Volume of Na2S2O3(L): 0.001
ambient temp(oC): 25
molar concentration of KI (mol/ L) 0.3
molar concentration of H2O2 (mol/L) 0.1
Total kinetic volume of kinetic trial (mL): 10
Time for color change, ΔT (sec): 25
B. Calculations for determing the Rate Law
1.) Moles of S2O3^2- consumed (mol): 2.0 x10^-4?
2.)Δ(mol I3-) produced: 3 x10^-4 ?
eq: 2 S2O3 ^2- (aq)+ I3^-(aq) ---> 3 I^-(aq) + S4O6^2- (aq)
3.)Δ(mol I3-)/ΔT (mol/sec): 1.2 x 10^-5
4.)logΔ(mol I3-)/ΔT : -1.96
5.) Volume KI (mL): 3.0
6.)[I-]0 (mol/L): .009M
(not 0.3M)
7.) log[I-]0:
8.) Volume of H2O2 (mL): 3.0
9.) [H2O2] (mol/L): .003M
(not 0.1M)
10.) log [H2O2]:

Answer 1

Reaction: 2S₂O₃^2- + I₃^- ---> 3I^- + S₄O₆^-2

1. Moles of S₂O₃^2- consumed = 0.001 L * 0.02 moles/L = 2*10^-5 moles

2. Moles of I₃^- produced = 2*10^-5 moles/2 = 1*10^-5 moles...[As according to reaction 2 moles of S₂O₃^2- consumed for 1 mol of I₃^- produced]

3. Moles of I₃^- produced/dt = (1*10^-5 moles)/25 sec = 4*10^-7 moles/sec

4. log (4*10^-7) = -6.4

5. Volume of KI taken initially = 3 mL (given)

6. [I-]₀ = [concentration of KI solution * Volume of KI taken initially]/total volume

= 0.3 mol.lit^-1 * 3 mL / 10 mL

= 0.09 moles/L

7. log[I-]₀ =-1.045

8. Volume of H₂O₂ taken initially = 3 mL (given)

9. [H₂O₂] = [concentration of H₂O₂ solution * Volume of H₂O₂ taken initially]/total volume

= 0.1 mol.L^­-1 * 3mL/10mL

= 0.03 moles/L

10. log [H₂O₂] = -1.523