The oxidation of iodide ion by peroxydisulfate ion is described by the equation: 3 I− (aq) + S2O82− (aq)  I3− (aq) + 2 SO42− (aq). If -d[I-]/dt= 6.0 × 10-3 mol dm-3 s-1 for a...

The oxidation of iodide ion by peroxydisulfate ion is described by the equation: 3 I− (aq) + S2O82− (aq)  I3− (aq) + 2 SO42− (aq). If -d[I-]/dt= 6.0 × 10-3 mol dm-3 s-1 for a particular time interval, what is the value of -d[S2O8^2-]/dt?

 

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t-nez eNotes educator| Certified Educator

The relative rates of change in concentration of reactants is stoichiometric, meaning it's related to their coefficients. For the equation,

`3 I^-_(aq) + S_2O_8^(2-)_(aq) -> I_3^-_(aq) + 2 SO_4^(2-)_(aq)`

The rates of change in concentration of the two reactants are related as follows

`- 1/3 (Delta[I^-])/(Delta t) =-(Delta[S_2O_8^(2-)])/(Delta t)` 

This is true because, in the same time period, three times as much I- is consumed as S2O8(2-). This makes the rate of S2O8(2-) equal to one third the rate of I- consumption. 

The rate of change in concentration of S2O8(2-) is therefore equal to

1/3 (-6.0x10^(-3) mol/dm-s) = -2.0 x 10^(-3) mol/dm-s

The rates have negative values because the concentrations are decreasing. The rates of change in concentration of the products are positive and are also related according to the coefficients, for example 

`-1/3 (Delta[I-])/(Delta t) = + (Delta[I_3^-])/(Delta t)`

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