The reaction of peroxydisulfate ion (S2032-) with iodide ion (T) is S_08² (aq) +31 (aq) - 250,- (aq) + 13 (aq) From the

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The Reaction Of Peroxydisulfate Ion S2032 With Iodide Ion T Is S 08 Aq 31 Aq 250 Aq 13 Aq From The 1 (38.87 KiB) Viewed 59 times

The Reaction Of Peroxydisulfate Ion S2032 With Iodide Ion T Is S 08 Aq 31 Aq 250 Aq 13 Aq From The 2 (45.63 KiB) Viewed 59 times

The reaction of peroxydisulfate ion (S2032-) with iodide ion (T) is S_08² (aq) +31 (aq) - 250,- (aq) + 13 (aq) From the following data collected at a certain temperature, determine the rate law and calculate the rate constant. Experiment 15,08? (M) 1 0.0500 [1(M) 0.0240 0.0120 Initial Rate (M/s] 7.60 x 10-4 3.80 x 10-4 7.60 x 10-4 2 0.0500 3 0.1000 0.0120 (a) Which of the following equations represents the rate law for this reaction? A. rate = k[3,0,2 m C. rate= K[S,082 B. rate = K[S20, 21) D. rate = ks 0,1-02
Experiment Initial Rate [M/s] 1 [S2082(M) (M) 0.0500 0.0240 0.0500 0.0120 2 7.60 x 10-4 3.80 x 1024 7.60 x 10-4 3 0.1000 0.0120 (a) Which of the following equations represents the rate law for this reaction? A. rate = k[S2082-1] B. rate = k[S2082127] C. rate = k[s208-112 D. rate = k[s208-2012 Α Ο в О. с О. D (b) What is the rate constant for the reaction? k = 0.8696 Xs