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NEET CHEMISTRYChemical KineticsMedium

Question

The rate constant of the reaction ABA \rightarrow B is 0.6×103 molar per second0.6 \times 10^{-3} \text{ molar per second}. If the concentration of A is 5 M5 \text{ M}, then the concentration of B after 20 min20 \text{ min} is:

A

1.08 M1.08 \text{ M}

B

3.60 M3.60 \text{ M}

C

0.36 M0.36 \text{ M}

D

0.72 M0.72 \text{ M}

Step-by-Step Solution

The unit of the rate constant is molar per second\text{molar per second} (M s1\text{M s}^{-1}), which indicates that it is a zero-order reaction. For a zero-order reaction, the integrated rate equation is: [A]t=[A]0kt[A]_t = [A]_0 - kt The amount of reactant AA consumed is ktkt. Since the stoichiometry of the reaction is 1:11:1, the concentration of product BB formed will be equal to the amount of AA consumed. [B]=kt[B] = kt Given data: k=0.6×103 M s1k = 0.6 \times 10^{-3} \text{ M s}^{-1} t=20 min=20×60 s=1200 st = 20 \text{ min} = 20 \times 60 \text{ s} = 1200 \text{ s} Substituting the values into the equation: [B]=(0.6×103 M s1)×1200 s[B] = (0.6 \times 10^{-3} \text{ M s}^{-1}) \times 1200 \text{ s} [B]=0.6×1.2 M=0.72 M[B] = 0.6 \times 1.2 \text{ M} = 0.72 \text{ M} Therefore, the concentration of BB after 20 minutes20 \text{ minutes} is 0.72 M0.72 \text{ M}.

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 Kineticsconstantreactionrightarrowsecondconcentration

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