The thiosulfate ion (S2O32-) is oxidized by iodine as follows: 2S2O32-(aq) + I2(aq) → S4O62-(aq) +...

The thiosulfate ion (S2O32-) is oxidized by iodine as follows: 2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq) In a certain experiment, 4.56×10-3 mol/L of S2O32- is consumed in the first 11.0 seconds of the reaction. Calculate the rate of production of iodide ion.

Expert Solution

2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)

-1/2*[S2O3^2-]/t = [I2]/t = [S4O6^2-]/t = +1/2 * [I^-]/t

[S2O3^2-]/t = 4.56*10^-3/11 = 4.14*10^-4 mole/L-sec

-1/2*[S2O3^2-]/t = +1/2 * [I^-]/t

+1/2 * [I^-]/t = -1/2*4.14*10^-4 = 2.07*10^-4

[I^-]/t = 4.14*10^-4 mole/L-sec >>>> answer

queen_honey_blossom answered 3 years ago

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