Initial rate data were collected for the following reaction in which iodide ion
ID: 964969 • Letter: I
Question
Initial rate data were collected for the following reaction in which iodide ion is oxidized to triiodide by peroxydisulfate ion: S_2 O_8^-2(aq) + 3 (aq) rightarrow 2 SO_4^-2 (aq) + (aq) Write the rate expression for all reactants and products, (rate expression is different from rate law) Determine the order of the reaction with respect to S_2 O_8^-2 and from the data provided and write a rate law for this reaction. Determine the value of the rate constant. Include proper units. What is the overall order of this reaction? What does this tell us about the slow step of the reaction mechanism?Explanation / Answer
(a) Rate expression: (Rate of change of the concentration of each species)
- d[S2O8]/dt = -1/3 d[I-]/dt = 1/2 d[SO42-]/dt = d[I3-]/dt
Negative sign shows concentration of reactants is decreasing
(b) R = K [S2O8]x [I-]y
We will calculate the value of x and y to complete the rate law by initial rate method.
Consider experiment 1 and 2, (Concnetration of S2O8 is constant and conc. of I- is changing)
R1 = 2.2 x 10-4 = K [0.080]x [0.034]y
R2 = 1.1 x 10-4 = K [0.080]x [0.017]y
Divide R1 by R2, we get
2 = (2)y
(2)1 = (2)y
y = 1 (Order of the reaction with respect to I-)
Now consider experiment 2 and 3,
R2 = 1.1 x 10-4 = K [0.080]x [0.017]y
R3 = 2.2 x 10-4 = K [0.16]x [0.017]y
Divide R3 by R2, we get
2 = (2)x
x = 1 (Order of the reaction with respect to S2O8)
Hence the rate law becomes, R = K [S2O8]1 [I-]1
(c) K = rate constant
K = R / [S2O8]1 [I-]1
K = 2.2 x 10-4 / (0.080) (0.034)
K = 0.0809 M-1 s-1
(d) Overall order of the reaction is the sum of all the exponent of the species in rate law
Order = 1+1 = 2
This tells us that slowest and rate determining step of the reaction is bimolecular i.e., both the reactants are affecting the rate of reaction.