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NEET CHEMISTRYChemical KineticsMedium

Question

For the reaction, 5Br−+BrO3−+6H+→3Br2+3H2O5Br^- + BrO_3^- + 6H^+ \rightarrow 3Br_2 + 3H_2O, the correct representation of the consumption and formation of reactants and products is:

A

d[Br−]dt=−35d[Br2]dt\frac{d[Br^-]}{dt} = -\frac{3}{5}\frac{d[Br_2]}{dt}

B

d[Br−]dt=35d[Br2]dt\frac{d[Br^-]}{dt} = \frac{3}{5}\frac{d[Br_2]}{dt}

C

d[Br−]dt=−53d[Br2]dt\frac{d[Br^-]}{dt} = -\frac{5}{3}\frac{d[Br_2]}{dt}

D

d[Br−]dt=53d[Br2]dt\frac{d[Br^-]}{dt} = \frac{5}{3}\frac{d[Br_2]}{dt}

Step-by-Step Solution

For the reaction 5Br−+BrO3−+6H+→3Br2+3H2O5Br^- + BrO_3^- + 6H^+ \rightarrow 3Br_2 + 3H_2O, the rate of reaction can be expressed in terms of the disappearance of reactants and appearance of products divided by their respective stoichiometric coefficients:

Rate=−15d[Br−]dt=−d[BrO3−]dt=−16d[H+]dt=13d[Br2]dt=13d[H2O]dt\text{Rate} = -\frac{1}{5}\frac{d[Br^-]}{dt} = -\frac{d[BrO_3^-]}{dt} = -\frac{1}{6}\frac{d[H^+]}{dt} = \frac{1}{3}\frac{d[Br_2]}{dt} = \frac{1}{3}\frac{d[H_2O]}{dt}

Equating the rate expressions for Br−Br^- and Br2Br_2: −15d[Br−]dt=13d[Br2]dt-\frac{1}{5}\frac{d[Br^-]}{dt} = \frac{1}{3}\frac{d[Br_2]}{dt}

Rearranging this equation to express the rate of consumption of Br−Br^- in terms of the rate of formation of Br2Br_2, we get: d[Br−]dt=−53d[Br2]dt\frac{d[Br^-]}{dt} = -\frac{5}{3}\frac{d[Br_2]}{dt}

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Chemical Kinetics. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYChemical Kineticsreactionrightarrowcorrectrepresentationconsumption

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