The reaction S2O82-(aq) + 3 I-(aq) 2 SO42-(aq) + I3-(aq) was studied at a certai

Question

The reaction S2O82-(aq) + 3 I-(aq) 2 SO42-(aq) + I3-(aq) was studied at a certain temperature with the following results: Experiment [S2O82-(aq)] (M) [I-(aq)] (M) Rate (M/s) 1 0.0294 0.0294 6.18e-06 2 0.0294 0.0588 1.24e-05 3 0.0588 0.0294 1.24e-05 4 0.0588 0.0588 2.47e-05 (a) What is the rate law for this reaction? Rate = k [S2O82-(aq)] [I-(aq)] Rate = k [S2O82-(aq)]2 [I-(aq)] Rate = k [S2O82-(aq)] [I-(aq)]2 Rate = k [S2O82-(aq)]2 [I-(aq)]2 Rate = k [S2O82-(aq)] [I-(aq)]3 Rate = k [S2O82-(aq)]4 [I-(aq)] (b) What is the value of the rate constant? (c) What is the reaction rate when the concentration of S2O82-(aq) is 0.0333 M and that of I-(aq) is 0.0645 M if the temperature is the same as that used to obtain the data shown above? M/s

Explanation / Answer

it think that: A) I- + I- > -I-I- formation of a dianion diradical is the slow reaction B) -I-I- + S2O82- react like a LEWIS ACID/BASE TO FORM a really big molecule,with an electrical charge of 4-. C)the LEWIS ADDUCT splits up (when the electrons rich O-O bond)> I-I + SO4^2- + SO4^2- because rqn A: rate = k [I-] [S2O8(2-)]

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