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

N₂ = 3.0 × 10^–3 M, O₂ = 4.2 × 10^–3 M and NO = 2.8 × 10^–3 M.

2NO(g) ⇌ N_2(g) + O_2(g)

If 0.1 mol L^–1 of NO_(g) is taken in a closed vessel, what will be degree of dissociation (α) of NO_(g) at equilibrium?

The correct option is 0.717.

Step 1: Calculate the equilibrium constant ($K_c$) of the reaction
The given chemical equilibrium reaction is:
$2\mathrm{NO}\left(g\right)\rightleftharpoons N_2\left(g\right)+O_2\left(g\right)$

At equilibrium, the concentrations are given as:
$\left[N_2\right]=3.0\times {10}^{-3}M$
$\left[O_2\right]=4.2\times {10}^{-3}M$
$\left[\mathrm{NO}\right]=2.8\times {10}^{-3}M$

The equilibrium constant expression for the reaction is:
$K_c=\frac{\left[N_2\right]\left[O_2\right]}{{\left[\mathrm{NO}\right]}^2}$

Substituting the given values into the expression:
$K_c=\frac{(3.0\times {10}^{-3})\times (4.2\times {10}^{-3})}{{(2.8\times {10}^{-3})}^2}$
$K_c=\frac{1.26\times {10}^{-5}}{7.84\times {10}^{-6}}=1.607$

Step 2: Determine the degree of dissociation ($\alpha$) under the new conditions
Let $\alpha$ be the degree of dissociation of $\mathrm{NO}\left(g\right)$ when its initial concentration is $0.1{\text{mol L}}^{-1}$.
We set up the reaction stoichiometry at equilibrium:

Using the equilibrium constant calculated in Step 1:
$K_c=\frac{\left[N_2\right]\left[O_2\right]}{{\left[\mathrm{NO}\right]}^2}$
$1.607=\frac{\left(\frac{0.1\alpha }{2}\right)\times \left(\frac{0.1\alpha }{2}\right)}{{\left[0.1(1-\alpha )\right]}^2}$
$1.607=\frac{0.01{\alpha }^2}{4\times 0.01{(1-\alpha )}^2}$
$1.607=\frac{{\alpha }^2}{4{(1-\alpha )}^2}$

Rearranging the equation yields:
$\frac{{\alpha }^2}{{(1-\alpha )}^2}=4\times 1.607=6.428$

Taking the square root on both sides:
$\frac{\alpha }{1-\alpha }=\sqrt{6.428}\approx 2.535$
$\alpha =2.535(1-\alpha )$
$\alpha =2.535-2.535\alpha$
$3.535\alpha =2.535$
$\alpha =\frac{2.535}{3.535}\approx 0.717$

Thus, the degree of dissociation of $\mathrm{NO}\left(g\right)$ at equilibrium is 0.717.