Thiosulfate ion S2O32− is oxidized by ioidine according to the following reaction. 2S2O32−(aq) + I2(aq) �¨ S4O62−(aq) + 2I−(aq). If in a certain experiment, 0.0020mole of thiosulfate ion is consumed in 1.0 L of solution in 1 second, what is the rate of consumption of iodine. ?
To find the rate of consumption of iodine, we need to use the stoichiometry of the reaction and the concentration of thiosulfate ion consumed.
Step 1: Calculate the molar ratio between thiosulfate ion (S2O32-) and iodine (I2) in the balanced chemical equation.
From the equation: 2S2O32- + I2 → S4O62- + 2I-
The molar ratio between S2O32- and I2 is 2:1.
Step 2: Convert the number of moles of S2O32- consumed to moles of I2 consumed.
Given that 0.0020 moles of S2O32- is consumed, we can use the molar ratio from Step 1 to determine the moles of I2 consumed:
Moles of I2 consumed = (Moles of S2O32- consumed) / 2
Moles of I2 consumed = 0.0020 moles / 2 = 0.0010 moles
Step 3: Calculate the rate of consumption of iodine (I2) in moles per second.
Since 0.0010 moles of iodine is consumed in 1 second, the rate of consumption is 0.0010 moles/second.
To determine the rate of consumption of iodine, we need to first calculate the stoichiometric relationship between thiosulfate ion (S2O32-) and iodine (I2) based on the balanced equation:
2S2O32-(aq) + I2(aq) → S4O62-(aq) + 2I-(aq)
From the balanced equation, we can see that 2 moles of thiosulfate ion react with 1 mole of iodine. Therefore, the molar ratio is 2:1.
Given that 0.0020 moles of thiosulfate ion are consumed in 1.0 L of solution in 1 second, we can use the molar ratio to calculate the moles of iodine consumed:
Moles of iodine = 0.0020 moles of thiosulfate * (1 mole of iodine / 2 moles of thiosulfate) = 0.0010 moles of iodine
Now, we need to calculate the rate of consumption of iodine. The rate of a chemical reaction is usually expressed in terms of moles per liter and per second (mol/L·s). Since the volume of the solution is given as 1.0 L and the time is 1 second, the rate of consumption of iodine can be directly calculated as:
Rate of consumption of iodine = Moles of iodine / Volume of solution / Time
Rate of consumption of iodine = 0.0010 moles of iodine / 1 L / 1 s = 0.0010 mol/L·s
Therefore, the rate of consumption of iodine is 0.0010 mol/L·s.
rate S2O3^2- is 1/2*0.002M/s
rate I2 consumption is 0.002 M/s