Ch 203 Problem 14.34

14.34 Consider the reaction of peroxydisulfate ion, S2O82-, with iodide ion, I-, in aqueous solution:

S2O82-(aq) + 3I-(aq) ---> 2SO42-(aq) + I3-(aq)

At a particular temperature, the rate of disappearance of S2O82- varies with reactant concentrations in the following manner:

Experiment [S2O82-] [I-] rate (M/s)

1 0.018 0.036 2.6x10-6

2 0.027 0.036 3.9x10-5

3 0.036 0.054 7.8x10-6

4 0.050 0.072 1.4x10-5

(a) determine the rate law for this reaction

rate = k[S2O82-]m [I-]n

where m and n are the orders with respect to peroxydisulfate and iodide, respectively. we must determine m and n from the experimental data given.

dummy unit

to determine the order with respect to peroxydisulfate, we look at experiments 1 and 2: the iodide concentration is constant, and the concentration of peroxydisulfate changes by a factor of

(the iodide concentration is fixed), and the rate changes by a factor of

we determine that the order with respect to peroxydisulfate is 1 (an increase the concentration by a factor of 1.5 results in an increase in the rate by a factor of 1.5), so the rate law (so far) is

rate = k[S2O82-] [I-]n

now, we must calculate the order with respect to iodide ion. looking at the experimental data, we see that there aren't two experiments where the peroxydisulfate concentration is held constant and the iodide concentration is varied. this is merely a minor inconvenience; it means that we shall have to write out the rate laws in their full form. above, we determined that the order with respect to peroxydisulfate =1; here, we will compare expts. 1 and 3. we write the rate law for expt. 3 and divide it by the rate law for expt. 1:

Rate3 = k[S2O82-][I-]n

Rate1 = k[S2O82-][I-]n

substituting the data for experiments 1 and 3, we have

now, do the divisions on the left and right hand sides. the rate constants will cancel since the temperature doesn't change during the reaction. we obtain

divide by 2:

the only value of n that makes this true is n = 1, so the order with respect to iodide ion = 1, and the rate law is: rate = k[S2O82-][I-]

(b) Here, use the rate law found in part (a) and calculate k for each experiment, and average the results: (I'm carrying one extra sig fig on the k values so I can average)

looks like the average rate constant = 4.0 x 10-3 1/M sec to 2 sig figs

(c) How is the rate of disappearance of S2O82- related to the rate of appearance of I-?

from the balanced chemical equation we can write

-D[S2O82-]/Dt = -1/3D[I-]/Dt

we see that the rate of disappearance of iodide is 3 times the rate of disappearance of peroxydisulfate (or the rate of disappearance of peroxydisulfate is 1/3 the rate of disappearance of iodide)

(d) what is the rate of disappearance of I- when [S2O82-]=0.025M and [I-]=0.050M?

since we have the rate law, we can calculate the rate of disappearance of peroxydisulfate with these concentrations. Using the average rate constant from above,

We plug the concentrations and k into the rate law:

this is the rate at which peroxydisulfate disappears; we know (from above) that iodide disappears three time as fast, so the rate of disappearance of I- is

DAC 8/4/08