For the reaction IO3� + 5I� + 6H+ 3I2 + 3H2O the rate of disappearance of I� at a particular time and concentration is 5.0 x 10-3 M s-1. What is the rate of appearance of I2
The reaction is not clear to me, however, it seems to me the ratio of rate is 3/5 * rate given, based on mole ratios.
3*5E-3 / 5= 3E-3 Moles/sec
To find the rate of appearance of I2, we need to use the stoichiometry of the reaction.
From the balanced equation: IO3- + 5I- + 6H+ → 3I2 + 3H2O
We can see that 5 moles of I- are consumed for every 3 moles of I2 produced.
Given the rate of disappearance of I- (5.0 x 10^-3 M/s), we can use the stoichiometry to find the rate of appearance of I2.
Rate of appearance of I2 = (Rate of disappearance of I-) x (3 moles I2 / 5 moles I-)
Now, let's calculate the rate of appearance of I2:
Rate of appearance of I2 = (5.0 x 10^-3 M/s) x (3/5)
= 3.0 x 10^-3 M/s
Therefore, the rate of appearance of I2 is 3.0 x 10^-3 M/s.