[magnolia1983]

I'm very stuck with this lab and really need *delete me*!

The purpose of this experiment is to study the kinetics of the reaction between iodide, I-, and peroxydisulfate, S2O8 2-, ions. Effect of Concetration.

Given equation:
                           S2O82- (aq)  + 2I- (aq)  ---------------> I2  (aq)  + 2SO42- (aq)
                            I2  (aq)  + 2S2O32- (aq) ---------------->  S4O62- (aq)  + 2I- (aq

                            Rate  =  - d / dt [S2O82- ]    = d[I2]/dt
                                     = k [S2O82- ]x [I-]y
                        log (rate) = log k  + x log [S2O82-] + y log [ I-]
                           Rate   = d[I2]/dt   = 1/2d[S2O32-] / dt

 By plotting the data of time versus moles give us k value.

Volume of reagents:
                  (NH4)2S2O8    Starch      KNO3         EDTA        KI          Na2S2O3 aliquots (5 in total)
RUN 1          10.0 ml            1.0 ml      27.0 ml      1 drop      10.0 ml     1.0 ml each
RUN 2          20.0                 1.0           17.0        1 drop      10.0        1.0 ml each
RUN 3          10.0                 1.0           17.0        1 drop      20.0        1.0 ml each
Concetration  0.25 M                           0.25M                    0.25M       0.20 M

Data for RUN 1:
aliquote #         time (s)            Moles of [S2O3]2- 


1                         151 s              0.0002 moles
2                        300 s              0.0004
3                        453 s              0.0006
4                        625 s              0.0008
5                         814 s              0.001

 
(Moles of S2O3 2- above were calculated by  0.20M Na2S2O3 (given concentration) x 0.001   L  (each aliquot 1.0 ml) =0.0002 moles; and so on)

Then, using microsoft excel, i'm asked to plot this data, moles of S2O3 2- consumed vs. time (in seconds) for kinetic run.

RUN 1 formula for the straight line (that goes through 5 points from our data) is  y = (1.2x10^-6)X  +  1.7x10^-5.  [y = mx + b]
RUN 2 y = (2.8x10^-6)X + 1.2x10^-5
RUN 3 y = (2.6x10^-6)X + 1.8x10^-5


Question 1: From the slope of the line and the stoichiometry of the reactions involved, calculate for each run the rate of reaction of (NH4)2S2O8 in moles/sec. Using these values and the average volume of the reaction mixture during the run (50.0 ml), determine for each run the rate of reaction of (NH4)2S2)8 in miles/liter/sec.

Question 2: Calculate the initial (time zero) amounts of [S2O8]2- and I- in the reaction mixture for each run and taking 50.0 ml as the total volume of the solution determine the concentration of [S2O8]2-  and I-

Question 3: The rate equation has the form: rate = k [S2O8 2-]^x [I-]^y. Three sets of values for rate, [S2O8 2-] and [I-] have been determined. Hence by using simultaneous equations, k, x, and y can be evaluated. Determine the order of the reaction with respect to S2O8 2-, (x), the order with respect to I-, (y), and the specific rate constant, (k), then write the rate law equation for the reaction as determined by the experimental data.