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In the reaction, BrO₃⁻ (aq.) + 5Br⁻ (aq.) + 6H⁺ (aq.)→ 3Br₂ (I) + 3H₂O (I). The rate of appearance of bromine (Br₂) is related to rate of disappearance of bromide ions as following:

A. d [Br₂]/dt= 2 d [Br⁻]/ 5 dt
B. d[Br₂]/dt= - 3d [Br⁻]/5 dt
C. d [Br₂] /dt=-5 [Br⁻]/3 dt
D. d [Br₂]/ dt + 5 d [Br⁻] /3 dt

User Chubby Boy
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Final answer:

The rate of appearance of bromine (Br₂) is related to the rate of disappearance of bromide ions by a factor of the stoichiometric coefficients from the balanced equation, resulting in the answer B. d[Br₂]/dt= -3d[Br⁻]/5dt.

Step-by-step explanation:

In the reaction BrO₃⁻ (aq) + 5Br⁻ (aq) + 6H⁺ (aq) → 3Br₂ (l) + 3H₂O (l), the rate at which Br₂ appears can be determined using the stoichiometric coefficients of the reactants and products. The ratio of Br⁻ to Br₂ is 5:3. Therefore, if the rate of disappearance of Br⁻ is given, the rate of appearance of Br₂ can be found by dividing the rate of disappearance of Br⁻ by the stoichiometric coefficient of Br⁻ (which is 5) and multiplying by the stoichiometric coefficient of Br₂ (which is 3).

So, the change in concentration of Br₂ over time (d[Br₂]/dt) can be related to the change in concentration of Br⁻ (d[Br⁻]/dt) by the following equation: d[Br₂]/dt = (3/5) * (-d[Br⁻]/dt), since the disappearance of Br⁻ is a negative rate.

The correct answer is B. d[Br₂]/dt= -3d[Br⁻]/5dt, which comes from the stoichiometry of the reaction. If Br⁻ is disappearing at a rate of 3.5 × 10⁻⁴ mol L⁻¹ s⁻¹, then Br₂ is appearing at a rate of -3/5 of this value, which is positive when talking about the rate of appearance.

User Zcourts
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