Final answer:
The rate of formation of Br2 is found by multiplying the rate of disappearance of Br- by the ratio of their coefficients in the balanced equation, resulting in 2.1 x 10^-4 mol L^-1 s^-1.
Step-by-step explanation:
When considering the chemical reaction given, which involves bromide ions (Br-), bromate ions (BrO3-), and hydrogen ions (H+) in aqueous solution, the question deals with reaction rates, specifically the rate of formation of bromine (Br2) when bromide ions are consumed.
The reaction provided is:
-
- 5Br-(aq) + BrO3-(aq) + 6H+ (aq) → 3Br2(aq) + 3H2O(l)
To determine the rate of the appearance of Br2 if the rate of disappearance of Br- is 3.5 × 10-4 mol L-1 s-1, we can use the stoichiometry of the reaction.
Since 5 moles of Br- produce 3 moles of Br2, the rate of appearance of Br2 will be 3/5 times the rate of disappearance of Br-.
Rate of formation of Br2 = (3/5) × (3.5 × 10-4)
= 2.1 × 10-4 mol L-1 s-1