what is the rate of reaction, if rate of consumption of iodide is 2.130x10^-5?
5I-(aq)+IO3-(aq)+6H+(aq)<-> 3I2(aq)+3H2O(l)
To find the rate of the reaction based on the given rate of consumption of iodide (I-), you need to consider the stoichiometry of the balanced equation.
The balanced equation for the reaction is:
5I-(aq) + IO3-(aq) + 6H+(aq) ↔ 3I2(aq) + 3H2O(l)
According to the balanced equation, every 5 moles of iodide (I-) consumed will result in the production of 3 moles of iodine (I2). Therefore, the stoichiometric ratio is 5:3.
To find the rate of reaction in terms of the consumption of iodine (I2), you can use the ratio derived from the stoichiometric coefficients:
Rate of reaction = (Rate of consumption of I-) × (3 moles of I2 / 5 moles of I-)
Given:
Rate of consumption of I- = 2.130 × 10^-5
Substituting the value into the formula:
Rate of reaction = (2.130 × 10^-5) × (3 / 5) = 1.278 × 10^-5
Therefore, the rate of the reaction, considering the rate of consumption of iodide, is 1.278 × 10^-5.