The reaction of peroxydisulfate ion (S20s) with iodide ion (I) is s3O82-(aq) + 3

Question

The reaction of peroxydisulfate ion (S20s) with iodide ion (I) is s3O82-(aq) + 31-(aq) 2S042-(aq) + 13-(aq) From the following data collected at a certain temperature, determine the rate law and calculate th rate constant. Experiment IS20sM Initial Rate [M/s] 0.0500 0.0500 0.1000 0.0440 0.0220 0.0220 6.80 × 10.4 3.40 × 10-4 6.80 × 10-4 (a) Which of the following equations represents the rate law for this reaction A. rate = k[S2082-] [I-I B. rate = k[S2082-12[1-1 C. rate = k[S2082-1 [172 D. rate = k[S2082[172 (b) What is the rate constant for the reaction? k=

Explanation / Answer

a)
see experiment 2 and 3:
[S2O82-] doubles
[I-] is constant
rate doubles
so, order of S2O82- is 1

see experiment 2 and 1:
[S2O82-] is constant
[I-] doubles
rate doubles
so, order of I- is 1

overall order = 1 + 1 = 2
Rate law is:
rate = k*[S2O82-]*[I-]
Answer: A

b)
rate = k*[S2O82-]*[I-]
Put values from 1st row of table in rate law
rate = k*[S2O82-]*[I-]
6.8*10^-4 = k*0.05*0.044
k = 0.309 M-1.s-1
Answer: 0.309

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