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NEET 2024
Exam
NEET
Year
2024
Subject
Chemistry
Chapter
Chemical Equilibrium

NEET 2024 · Chemistry · Chemical Equilibrium

Consider the following reaction in a sealed vessel at equilibrium...

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−1\mathrm{L}^{-1} of NO(g)\mathrm{NO}_{(\mathrm{g})} is taken in a closed vessel, what will be degree of dissociation ( α\alpha )ofNO(g)of \mathrm{NO}_{(\mathrm{g})} at equilibrium?

Options

  1. A0.0889
  2. B0.8889
  3. C0.717
  4. D0.00889

Correct answer

C. 0.717

Step-by-step solution

2NO(g)⇌N2( g)+O2( g)Kc=[N2][O2][NO]2=3×10−3×4.2×10−32.8×10−3×2.8×10−3=1.607t=02NO(g)⇌N2( g)+O2( g)0.1−0.1α0.05α0.05αKc=0.05α×0.05α(0.1−0.1α)2 Kc=0.05α×0.05α0.01(1−α)21.607=(0.05)2α20.01(1−α)2α2(1−α)2=1.607×(0.1)2(0.05)2\begin{aligned} & 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \\ & \mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]}{[\mathrm{NO}]^2} \\ &=\frac{3 \times 10^{-3} \times 4.2 \times 10^{-3}}{2.8 \times 10^{-3} \times 2.8 \times 10^{-3}} \\ &=1.607 \\ & \mathrm{t}=0 \quad 2 \mathrm{NO}_{(\mathrm{g})} \rightleftharpoons \mathrm{N}_{2(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{~g})} \\ & 0.1-0.1 \alpha \quad 0.05 \alpha \quad 0.05 \alpha \\ & \mathrm{K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{(0.1-0.1 \alpha)^2} \\ & \mathrm{~K}_{\mathrm{c}}=\frac{0.05 \alpha \times 0.05 \alpha}{0.01(1-\alpha)^2} \\ & 1.607=\frac{(0.05)^2 \alpha^2}{0.01(1-\alpha)^2} \\ & \frac{\alpha^2}{(1-\alpha)^2}=\frac{1.607 \times(0.1)^2}{(0.05)^2}\end{aligned}\begin{gathered}\frac{\alpha}{1-\alpha}=\frac{1.27 \times 0.1}{0.05} \\ \frac{\alpha}{1-\alpha}=2.54 \\ \alpha=2.54-2.54 \alpha \\ 3.54 \alpha=2.54 \\ \alpha=\frac{2.54}{3.54}=0.717\end{gathered}$
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