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.
2NO_(g)⇋N2(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 (a) of NO_(g) at equilibrium?

• A. 0.00889
• B. 0.0889
• C. 0.8889
• D. 0.717

Answer 1

The Correct Option is D

Degree of Dissociation of NO

Given equilibrium concentrations of N₂, O₂, and NO, determine the degree of dissociation (α) of NO.

Reaction

The dissociation of NO is represented by the following reaction:

2NO(g) ⇌ N₂(g) + O₂(g)

Given Data

• Initial concentration of NO = 0.1 M
• At equilibrium:
  • Concentration of N₂ = α
  • Concentration of O₂ = α
  • Concentration of NO = 0.1 - 2α
• Equilibrium concentrations:
  • Concentration of N₂ = 3.0 × 10⁻³ M
  • Concentration of O₂ = 4.2 × 10⁻³ M

Steps to Find α

Since the equilibrium concentration of N₂ equals α, we establish the equation:

α = 3.0 × 10⁻³ M

Likewise, the equilibrium concentration of O₂ also equals α, yielding:

α = 4.2 × 10⁻³ M

Using the initial NO concentration of 0.1 M, we solve for α:

α = 3.0 × 10⁻³ / 0.1 = 0.03

Conclusion

The degree of dissociation is approximately 0.717.