Question

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.

\[\ce{2NO_{(g)} <=> N2_{(g)} + O2_{(g)}}\]

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

Options

• 0.8889

• 0.717

• 0.00889

• 0.0889

Answer 1

0.717

Explanation:

For the given reaction,

\[\ce{2NO_{(g)} <=> N2_{(g)} + O2_{(g)}}\]

K_eq = `([N_2][O_2])/([NO]^2)`

= `([3 xx 10^-3][4.2 xx 10^-3])/([2.8 xx 10^-3]^2)`

= 1.6

	\[\ce{2NO = N2 + O2}\]
At t = 0	0.1	0	0
At t = 10	1 − α	`α/2`	`α/2`

K_eq = `(alpha/2)^2/(0.1 - alpha)^2`

Degree of dissociation (α) = 0.717