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Chemistry · Chemical Equilibrium

NEET (UG) 2025 — Question 68

For the reaction A(g)⇌2 B( g)\mathrm{A}(\mathrm{g}) \rightleftharpoons 2 \mathrm{~B}(\mathrm{~g}), the backward reaction rate constant is higher than the forward reaction rate

constant by a factor of 2500 , at 1000 K . [Given : R=0.0831 L atm mol−1 K−1\mathrm{R}=0.0831 \mathrm{~L} \mathrm{~atm} \mathrm{~mol}^{-1} \mathrm{~K}^{-1} ] KP\mathrm{K}_{\mathrm{P}} for the reaction

at 1000 K is

  1. Option A:

    83.1

  2. Option B:

    2.077×1052.077 \times 10^{5}

  3. Option C:

    0.033

    Correct
  4. Option D:

    0.021

Answer: C

Step-by-step solution

KC=KfKb\mathrm{K}_{\mathrm{C}}=\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}}

=Kf2500 Kf=12500 Kp=KC(RT)Δng=12500(0.0831×1000)1 Kp=0.033\begin{aligned} & =\frac{\mathrm{K}_{\mathrm{f}}}{2500 \mathrm{~K}_{\mathrm{f}}} \\& =\frac{1}{2500} \mathrm{~K}_{\mathrm{p}}\\ & =\mathrm{K}_{\mathrm{C}}(\mathrm{RT})^{\Delta \mathrm{n}_{\mathrm{g}}} \\& = \frac{1}{2500}(0.0831 \times 1000)^{1} \\& \mathrm{~K}_{\mathrm{p}} =0.033 \end{aligned}

Answer key and solution verified before publishing.

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Exam
NEET (UG) 2025
Paper
2025 paper
Subject
Chemistry
Chapter
Chemical Equilibrium
Topic
Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient

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