The thiosulfate ion (S2O32-) is oxidized by iodine as follows:
2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)
In a certain experiment, 4.41×10-3 mol/L of S2O32- is consumed in the first 11.0 seconds of the reaction.
Calculate the rate of production of iodide ion in mol/L/s
Why did the thiosulfate ion behave so poorly in school?
Because it couldn't concentrate!
To calculate the rate of production of iodide ion (I-) in mol/L/s, we need to use the stoichiometry of the reaction and the given information.
From the balanced equation:
2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)
We can see that for every 2 moles of S2O32- consumed, 2 moles of I- are produced. Therefore, the stoichiometric ratio between the consumption of S2O32- and the production of I- is 1:1.
Given that the concentration of S2O32- consumed in the first 11.0 seconds of the reaction is 4.41×10-3 mol/L, we can conclude that the concentration change of I- over the same period is also 4.41×10-3 mol/L.
To calculate the rate of production of I-, use the following formula:
Rate = Concentration change / Time
Rate = (4.41×10-3 mol/L) / (11.0 s)
Calculating the rate:
Rate = 4.41×10-3 mol/L / 11.0 s = 4.01×10-4 mol/L/s
Therefore, the rate of production of iodide ion in mol/L/s is 4.01×10-4 mol/L/s.
To calculate the rate of production of iodide ions in mol/L/s, we need to determine the change in concentration of iodide ions over time.
Given the balanced chemical equation:
2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)
We can see that for every 2 moles of thiosulfate ion consumed, 2 moles of iodide ion are produced. Therefore, the rate of production of iodide ion is directly proportional to the rate of consumption of thiosulfate ion.
In the first 11.0 seconds of the reaction, the concentration of thiosulfate ion changes from 4.41×10-3 mol/L to 0 mol/L, indicating that 4.41×10-3 mol/L of thiosulfate ion has been consumed.
To calculate the rate of production of iodide ion, we can use the following formula:
Rate = (Change in concentration of iodide ion) / (Change in time)
Since 2 moles of thiosulfate ion produce 2 moles of iodide ion, the change in concentration of iodide ion would be equal to 4.41×10-3 mol/L.
Therefore, the rate of production of iodide ion in mol/L/s is:
Rate = (4.41×10-3 mol/L) / (11.0 s) = 4.01×10-4 mol/L/s