In: Chemistry
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.
2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)
-1/2*[S2O32-]/
t =
[I2]/
t =
[S4O62-]/
t
= +1/2 *
[I-]/
t
[S2O32-]/
t
= 4.56*10-3/11 = 4.14*10-4 mole/L-sec
-1/2*[S2O32-]/
t
= +1/2 *
[I-]/
t
-1/2*[S2O32-]/
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