Final answer:
The rate of appearance of Br2 when bromide ion is being consumed at a rate of 3.5 × 10-4 mol L¹ s¹ is calculated using the reaction stoichiometry. It is found to be 2.1 × 10-4 mol L¹ s¹.
Step-by-step explanation:
The subject in question involves the rate at which bromide ion is being consumed in a chemical reaction. Specifically, we are looking at the reaction:
5Br¯(aq) + BrO3¯(aq) + 6H+ (aq) → 3Br2(aq) + 3H2O(1)
If the rate of disappearance of Br¯(aq) is 3.5 × 10-4 mol L¹ s¹, then we can use the stoichiometry of the reaction to determine the rate of appearance of Br2(aq). For every 5 moles of Br¯ that react, 3 moles of Br2 are produced. Therefore, the rate of appearance of Br2 is one-third of the rate of disappearance of Br¯ when we account for the stoichiometry of the reaction. Now, we can calculate this rate:
- Rate of disappearance of Br¯ = 3.5 × 10-4 mol L¹ s¹
- Rate of appearance of Br2 = (3.5 × 10-4 mol L¹ s¹) × (3/5)
- Rate of appearance of Br2 = 2.1 × 10-4 mol L¹ s¹
Thus, at the moment in question, Br2 is appearing in the reaction at a rate of 2.1 × 10-4 mol L¹ s¹.
The rate of appearance of Br₂(aq) can be determined using the stoichiometry of the balanced reaction equation. From the equation,
5Br¯(aq) + BrO3¯(aq) + 6H+ (aq) → 3Br₂(aq) + 3H₂O(1)
we can see that for every 5 moles of Br(aq) consumed, 3 moles of Br₂(aq) are produced. Therefore, the rate of appearance of Br₂(aq) is related to the rate of disappearance of Br(aq) by a factor of 3/5. So, if the rate of disappearance of Br(aq) is 3.5 × 10-4 mol L-¹ s-¹, the rate of appearance of Br₂(aq) would be (3/5) * 3.5 × 10-4 = 2.1 × 10-4 mol L-¹ s-¹.