Chemistry · Chemical Equilibrium

NEET (UG) 2024 — Question 86

Consider the following reaction in a sealed vessel at equilibrium with concentrations of

N2=3.0×10−3M,O2=4.2×10−3M\mathrm{N}_{2}=3.0 \times 10^{-3} \mathrm{M}, \mathrm{O}_{2}=4.2 \times 10^{-3} \mathrm{M} and NO=2.8×10−3M\mathrm{NO}=2.8 \times 10^{-3} \mathrm{M}.

2NO(g)⇌N2( g)+O2( g)2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}

If 0.1 mol L−10.1 \mathrm{~mol} \mathrm{~L}^{-1} of NO(g)\mathrm{NO}_{(\mathrm{g})} is taken in a closed vessel, what will be degree of dissociation

( α\alpha ) of NO(g)\mathrm{NO}_{(\mathrm{g})} at equilibrium?

  1. Option A:

    0.00889

  2. Option B:

    0.0889

  3. Option C:

    0.8889

  4. Option D:

    0.717

    Correct

Answer: D

Step-by-step solution

2NO(g)⇌N2( g)+O2( g)2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}

Kc=[N2][O2][NO]2\mathrm{K}_{\mathrm{c}}= \frac{\left[\mathrm{N}_{2}\right]\left[\mathrm{O}_{2}\right]}{[\mathrm{NO}]^{2}}

=3×10−3×4.2×10−32.8×10−3×2.8×10−3=\frac{3 \times 10^{-3} \times 4.2 \times 10^{-3}}{2.8 \times 10^{-3} \times 2.8 \times 10^{-3}}

=1.607=1.607

2NO(g)⇌N2( g)+O2( g)2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})}

t=00.10\mathrm{t}=0 \quad 0.1 \quad 0

0.1−0.1α0.05α0.05α0.1-0.1 \alpha \quad 0.05 \alpha \quad 0.05 \alpha

 Kc=0.05α×0.05α(0.1−0.1α)2\mathrm{~K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{(0.1-0.1 \alpha)^{2}}

 Kc=0.05α×0.05α0.01(1−α)2\mathrm{~K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{0.01(1-\alpha)^{2}}

1.607=(0.05)2α20.01(1−α)21.607=\frac{(0.05)^{2} \alpha^{2}}{0.01(1-\alpha)^{2}}

α2(1−α)2=1.607×(0.1)2(0.05)2\frac{\alpha^{2}}{(1-\alpha)^{2}}=\frac{1.607 \times(0.1)^{2}}{(0.05)^{2}}

α1−α=1.27×0.10.05\frac{\alpha}{1-\alpha}=\frac{1.27 \times 0.1}{0.05}

α1−α=2.54\frac{\alpha}{1-\alpha}=2.54

α=2.54−2.54α\alpha=2.54-2.54 \alpha

3.54α=2.543.54 \alpha=2.54

α=2.543.54=0.717\alpha=\frac{2.54}{3.54}=0.717

Answer key and solution verified before publishing.

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Exam
NEET (UG) 2024
Subject
Chemistry
Chapter
Chemical Equilibrium
Topic
Analysis of Chemical Equilibrium, Equilibrium Constant and Reaction Quotient
Consider the following reaction in a sealed vessel at equilibrium… | NEET (UG) 2024 PYQ with Solution · DhiX AI