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Answer to Question #3756 in Organic Chemistry for sivakumar
Question #3756
In the reaction betweem H2O2 and I-, 15 mL of 0.30M KI, 69. 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 presence of I2) in 2.0 minutes. Calculate the intial rate of disappearance of H2O2 in units of M min-1 in 2 significant digits.
Expert's answer
Initial concentration of H2O2 in solution was:
C(H2O2) = C1 V1/V2 = 0.069 L × 0.1 M/0.5 L = 0.0138 M
So, initial rate is:
r= −d[H2O2]/dt = ΔC(H2O2)/Δt = 0.0138 M / 2 min = 0.0069 [M min−1]
C(H2O2) = C1 V1/V2 = 0.069 L × 0.1 M/0.5 L = 0.0138 M
So, initial rate is:
r= −d[H2O2]/dt = ΔC(H2O2)/Δt = 0.0138 M / 2 min = 0.0069 [M min−1]
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Comments
Your answer is completely wrong. It is important to know that not all of the H2O2 has reacted. The appearance of blue indicated that the reaction I2+2S2O3 -> 2I- +S4O6 has stopped. This reaction stops because all of the S2O3 has reacted. To answer this question then, you need to figure out the rate of change of the concentration of S2O3 and then divide that by 2 (because of the stoichiometry) to figure out the rate of change in the concentration of H2O2.
I don't think your answer is correct. The appearance of I2 does not indicate that the entire quantity of H2O2 is reacted. It simply shows that the reaction between I2 and S2O32- has stopped. Some explanation to accompany your answer would make it way more useful.
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