For the reaction
IO3– + 5I– + 6H+ 3I2 + 3H2O
the rate of disappearance of I– at a particular time and concentration is 2.8 x 10-3 M s-1. What is the rate of appearance of I2 in molarity per second?
Doesn't this work just like stoichiometry.
2.8E-3 x (3/5) = ?
To determine the rate of appearance of I2 in molarity per second, we need to use the stoichiometry of the reaction.
From the given balanced equation:
IO3– + 5I– + 6H+ -> 3I2 + 3H2O
We can see that for every 5 moles of I– consumed, 3 moles of I2 are produced. This means that the ratio of the rate of disappearance of I– to the rate of appearance of I2 is also 5:3.
Given that the rate of disappearance of I– is 2.8 x 10^-3 M s^-1, we can use this information to calculate the rate of appearance of I2.
Rate of appearance of I2 = (Rate of disappearance of I–) * (3/5)
Rate of appearance of I2 = (2.8 x 10^-3 M s^-1) * (3/5)
Rate of appearance of I2 ≈ 1.68 x 10^-3 M s^-1
Therefore, the rate of appearance of I2 is approximately 1.68 x 10^-3 M s^-1.
To determine the rate of appearance of I2 in molarity per second, we need to use the stoichiometry of the reaction. From the balanced chemical equation:
IO3– + 5I– + 6H+ -> 3I2 + 3H2O
We can see that for every 5 moles of I– that disappear, 3 moles of I2 appear. This means that the rate of appearance of I2 is related to the rate of disappearance of I– by a factor of 3/5.
Given that the rate of disappearance of I– is 2.8 x 10-3 M s-1, we can calculate the rate of appearance of I2 as follows:
Rate of appearance of I2 = (3/5) * Rate of disappearance of I–
Rate of appearance of I2 = (3/5) * (2.8 x 10-3 M s-1)
Let's calculate this:
Rate of appearance of I2 = (3/5) * (2.8 x 10-3) = 1.68 x 10-3 M s-1
Therefore, the rate of appearance of I2 in molarity per second is 1.68 x 10-3 M s-1.