In the reaction between H2O2 and I-, 15 mL of 0.30M KI, 65. mL of 0.10M H2O2 and 19 mL of 0.020M Na2S2O3 were mixed in a flask containing starch and buffer and the total volume was made up to 500mL with distilled water. The solution turned blue (indicating the presence of I2) in 1.8 minutes. Calculate the initial rate of disappearance of H2O2 in units of M min-1 to two significant figures
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0.0064M/min
could someone answer this question for me?
could someone answer this question for me? with the solutions please
To calculate the initial rate of disappearance of H2O2, we need to use the information provided and the concept of reaction stoichiometry.
First, let's determine the number of moles of H2O2 that reacted using the given concentration and volume:
Moles of H2O2 = Concentration of H2O2 x Volume of H2O2
= 0.10 M x 65 mL
= 6.5 mmol
Next, we need to find the time it took for the reaction to proceed. Since the solution turned blue after 1.8 minutes, we can use this as our reaction time.
Now, let's determine the initial rate of disappearance of H2O2:
Initial rate = Change in H2O2 concentration / Change in time
To find the change in H2O2 concentration, we need to calculate the initial and final concentrations using the initial amount of moles and the total volume of the solution:
Initial concentration of H2O2 = Moles of H2O2 / Total volume of solution
= 6.5 mmol / 500 mL
= 0.013 M
To calculate the final concentration, we need to account for the reaction between H2O2 and I-. The balanced equation for this reaction is:
H2O2 + 2I- + 2H+ -> 2H2O + I2
From the equation, we can see that for every mole of H2O2 reacting, the same number of moles of I2 is produced. Therefore, the final concentration of H2O2 will be zero.
Now, we can calculate the initial rate:
Initial rate = (0.013 M - 0 M) / 1.8 min
= 0.007 M/min
Therefore, the initial rate of disappearance of H2O2 is 0.007 M/min (to two significant figures).