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NEET 2026
Exam
NEET
Year
2026
Subject
Chemistry
Chapter
Chapter

NEET 2026 · Chemistry · Chapter

For the given hypothetical reaction: 2P + 5Q arrow 3R + 4S The rate...

For the given hypothetical reaction:
2P+5Q→3R+4S2\mathrm{P} + 5\mathrm{Q} \rightarrow 3\mathrm{R} + 4\mathrm{S}
The rate of disappearance of P\mathrm{P} is 4×10−3 mol L−1 s−14 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}. The overall rate of the reaction and the concentration of S\mathrm{S} formed in an interval of 20 seconds will be, respectively:

Options

  1. A4×10−3 mol L−1 s−14 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 32×10−2 mol L−132 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
  2. B2×10−3 mol L−1 s−12 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 16×10−2 mol L−116 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
  3. C2×10−3 mol L−1 s−12 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 4×10−2 mol L−14 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
  4. D8×10−3 mol L−1 s−18 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 64×10−2 mol L−164 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}

Correct answer

B. 2×10−3 mol L−1 s−12 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 16×10−2 mol L−116 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}

Step-by-step solution

For the given reaction 2P+5Q→3R+4S2\mathrm{P} + 5\mathrm{Q} \rightarrow 3\mathrm{R} + 4\mathrm{S}, the overall rate of reaction is given by:

Rate=−12d[P]dt=14d[S]dt\text{Rate} = -\dfrac{1}{2} \dfrac{d[\mathrm{P}]}{dt} = \dfrac{1}{4} \dfrac{d[\mathrm{S}]}{dt}

Given the rate of disappearance of P\mathrm{P} is −d[P]dt=4×10−3 mol L−1 s−1-\dfrac{d[\mathrm{P}]}{dt} = 4 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}.

The overall rate of the reaction is:

Rate=12×(4×10−3)=2×10−3 mol L−1 s−1\text{Rate} = \dfrac{1}{2} \times (4 \times 10^{-3}) = 2 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

The rate of formation of S\mathrm{S} is:

d[S]dt=4×Rate=4×(2×10−3)=8×10−3 mol L−1 s−1\dfrac{d[\mathrm{S}]}{dt} = 4 \times \text{Rate} = 4 \times (2 \times 10^{-3}) = 8 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1}

The concentration of S\mathrm{S} formed in an interval of 2020 seconds is:

Δ[S]=d[S]dt×Δt=(8×10−3)×20=160×10−3=16×10−2 mol L−1\Delta[\mathrm{S}] = \dfrac{d[\mathrm{S}]}{dt} \times \Delta t = (8 \times 10^{-3}) \times 20 = 160 \times 10^{-3} = 16 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}

Answer: 2×10−3 mol L−1 s−12 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~s}^{-1} and 16×10−2 mol L−116 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}
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