The reaction of peroxydisulfate ion (S2O8^2-) with iodide ion (I^-) is S2O8^2-(a

Question

The reaction of peroxydisulfate ion (S2O8^2-) with iodide ion (I^-) is S2O8^2-(aq) + 3I^- (aq) ----> 2SO4^2-(aq) + I3^- (aq) Fr. From the following data collected at a certain temperature, determine the rate law and calculate the rate constant.

Experiment        (S2O8^2-)(M)        [I^-](M)           Initial Rate [M/s]

1                           0.0400                  0.0920           8.40 x 10^-4

2                           0.0400                  0.0460           4.20 x 10^-4

3                           0.0800                  0.0460           8.40 x 10^-4

Rate law is k[S2O8^2-][I^-]

What is the rate constant for the reaction?

Explanation / Answer

RATE OF THE REACTION= k[S2O8^2-][I^-]

8.40*10^-4=K[0.0400][0.0920]

K=2282.60*10^-4mole^-1.lit^1.sec^-1

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