The peroxydisulfate ion (S_2 O_8^2-) reacts with the iodide ion in aqueous solut

Question

The peroxydisulfate ion (S_2 O_8^2-) reacts with the iodide ion in aqueous solution via the reaction: S_2 O_8^2-(aq) + 3I^- rightarrow 2SO_4 (aq) + I3^- (aq) An aqueous solution containing 0.050 M of S_2 O_8^2- ion and 0.072 M of I^ - is prepared, and the progress of the reaction foll measuring [I^-]. The data obtained is given in the table below. The concentration of S_2 O_8^2- remaining at 1600 s is_______ M. 0.014 0.029 0.043 0.036 0.064 The average rate of disappearance of I between 1200 s and 1600 s is_______ M/s. 1.8 times 10^-5 1.6 times 10^-4 5.0 times 10^4 1.2 times 10^-5 2.0 times 10^-5 The concentration of S_2 O_8^2- remaining at 400 s is ________M. -0.007 +0.057 +0.015 +0.045 +0.035

Explanation / Answer

6)   [I-] reacted = 0.072-0.029 = 0.043

[S2O82-] reacted = ( 1/3) [I-] = 1/3 ( 0.043) = 0.01433

[S2O82-] remianing = 0.05-0.01433 = 0.036

Hence 6-D

7) rate of disaapearing of I = d[I] /dt = ( 0.037-0.029) / ( 1600-1200) = 2 x 10^ -5

Hence 7-E

8 ) [I-] reacted = 0.072-0.057 = 0.015

[S2O82-] reacted = ( 1/3) 0.015 = 0.005

[S2O82-] left = 0.05- 0.005 = 0.045

Hence 8-D

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