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ChemistryNEET 2024Law of mass action, Kc and Kp, Le Chatelier

Equilibrium

Consider the following reaction in a sealed vessel at equilibrium with concentrations N2=3.0×10−3\mathrm{N_2} = 3.0 \times 10^{-3} M, O2=4.2×10−3\mathrm{O_2} = 4.2 \times 10^{-3} M and NO=2.8×10−3\mathrm{NO} = 2.8 \times 10^{-3} M:

2NO(g)⇌N2(g)+O2(g)\mathrm{2NO(g) \rightleftharpoons N_2(g) + O_2(g)}

If 0.1 mol L−10.1\ \mathrm{mol\,L^{-1}} of NO(g) is taken in a closed vessel, what will be the degree of dissociation (α\alpha) of NO(g) at equilibrium?

  1. A.0.0889
  2. B.0.8889
  3. C.0.717✓
  4. D.0.00889

Solution

Kc=[N2][O2][NO]2=3.0×10−3×4.2×10−3(2.8×10−3)2≈1.607K_c = \frac{[\mathrm{N_2}][\mathrm{O_2}]}{[\mathrm{NO}]^2} = \frac{3.0\times10^{-3} \times 4.2\times10^{-3}}{(2.8\times10^{-3})^2} \approx 1.607. Starting from 0.10.1 M NO: Kc=(0.05α)2(0.1(1−α))2=α24(1−α)2K_c = \frac{(0.05\alpha)^2}{(0.1(1-\alpha))^2} = \frac{\alpha^2}{4(1-\alpha)^2}, so α2(1−α)=1.268⇒α≈0.717\frac{\alpha}{2(1-\alpha)} = 1.268 \Rightarrow \alpha \approx 0.717.

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