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NEET CHEMISTRYChemical KineticsMedium

Question

For a chemical reaction, 4A+3B→6C+9D4A + 3B \rightarrow 6C + 9D rate of formation of C is 6×10–2 mol L–1 s–16 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1} and rate of disappearance of A is 4×10–2 mol L–1 s–14 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1}. The rate of reaction and amount of B consumed in interval of 10 seconds, respectively will be:

A

1×10–2 mol L–1 s–11 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1} and 30×10–2 mol L–130 \times 10^{–2} \text{ mol L}^{–1}

B

10×10–2 mol L–1 s–110 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1} and 10×10–2 mol L–110 \times 10^{–2} \text{ mol L}^{–1}

C

1×10–2 mol L–1 s–11 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1} and 10×10–2 mol L–110 \times 10^{–2} \text{ mol L}^{–1}

D

10×10–2 mol L–1 s–110 \times 10^{–2} \text{ mol L}^{–1} \text{ s}^{–1} and 30×10–2 mol L–130 \times 10^{–2} \text{ mol L}^{–1}

Step-by-Step Solution

For the given reaction: 4A+3B→6C+9D4A + 3B \rightarrow 6C + 9D The rate of reaction is related to the rates of individual components as follows: Rate=−14d[A]dt=−13d[B]dt=16d[C]dt=19d[D]dt\text{Rate} = -\frac{1}{4}\frac{d[A]}{dt} = -\frac{1}{3}\frac{d[B]}{dt} = \frac{1}{6}\frac{d[C]}{dt} = \frac{1}{9}\frac{d[D]}{dt} Given: Rate of formation of C, d[C]dt=6×10−2 mol L−1 s−1\frac{d[C]}{dt} = 6 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1} Rate of disappearance of A, −d[A]dt=4×10−2 mol L−1 s−1-\frac{d[A]}{dt} = 4 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}

Rate of reaction =16×d[C]dt=16×(6×10−2)=1×10−2 mol L−1 s−1= \frac{1}{6} \times \frac{d[C]}{dt} = \frac{1}{6} \times (6 \times 10^{-2}) = 1 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}

Rate of disappearance of B =−d[B]dt=3×Rate of reaction=3×(1×10−2)=3×10−2 mol L−1 s−1= -\frac{d[B]}{dt} = 3 \times \text{Rate of reaction} = 3 \times (1 \times 10^{-2}) = 3 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}

The amount of B consumed in an interval of Δt=10 s\Delta t = 10 \text{ s} is: Δ[B]=(−d[B]dt)×Δt=(3×10−2 mol L−1 s−1)×10 s=30×10−2 mol L−1\Delta [B] = \left(-\frac{d[B]}{dt}\right) \times \Delta t = (3 \times 10^{-2} \text{ mol L}^{-1} \text{ s}^{-1}) \times 10 \text{ s} = 30 \times 10^{-2} \text{ mol L}^{-1}

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 Kineticschemicalreactionrightarrowformationdisappearance

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