can u help me in this question.
The rate of appearance of I2 in aqueous solution, in the reaction of I^- ion with hydrogen peroxide, was found to be 3.7*10^-5 mol. L^-1.s^-1 over time interval.
2H^+(aq)+ 2I^-(aq)+H2O2-> I2(aq)+2H2O(l)
During the same interval,what was the rate of disappearance of each of the following
a) H2O2
b) I^-
Thank you.
Wouldn't the rate of disappearance for H2O2 be the same as the rate of appearance of I2? And the rate of disappearance of I^- would be 1/2 rate of I2? Check my thinking.
To determine the rate of disappearance of H2O2 (hydrogen peroxide) and I^-(iodide ion), we need to analyze the stoichiometry of the chemical equation and the rate expression.
The balanced chemical equation is:
2H^+(aq) + 2I^-(aq) + H2O2 -> I2(aq) + 2H2O(l)
From the balanced equation, we can see that for every one mole of H2O2 reacted, two moles of I^- are consumed.
a) Rate of disappearance of H2O2:
Since two moles of I^- are consumed for every one mole of H2O2, the rate of disappearance of H2O2 is half the rate of appearance of I2. Therefore, the rate of disappearance of H2O2 is: (3.7 * 10^-5 mol.L^-1.s^-1) / 2 = 1.85 * 10^-5 mol.L^-1.s^-1.
b) Rate of disappearance of I^-:
The rate of disappearance of I^- is directly proportional to the rate of appearance of I2. Therefore, the rate of disappearance of I^- is equal to the rate of appearance of I2. Hence, the rate of disappearance of I^- is 3.7 * 10^-5 mol.L^-1.s^-1.
So, the rate of disappearance of H2O2 is 1.85 * 10^-5 mol.L^-1.s^-1, and the rate of disappearance of I^- is 3.7 * 10^-5 mol.L^-1.s^-1.