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

Step 1: Use the stoichiometric relationship:
Initial concentration of NO: $[NO] = 0.1 \, \text{mol/L}$
Change in concentration: $\Delta[NO] = 2\alpha[NO]$
Final concentration: $[NO]_f = 0.1 - 2\alpha$
Step 2: Calculate $\alpha$ at equilibrium: Using equilibrium concentrations:
$K_c = \frac{[N_2][O_2]}{[NO]^2}$,   $K_c = \frac{\alpha^2}{(1 - 2\alpha)^2}$
Simplifying for $\alpha$, we find:
$\alpha = 0.717$

Answer 2

Degree of Dissociation of NO 

We are given the equilibrium concentrations of N₂, O₂, and NO, and asked to find the degree of dissociation (α) of NO.

Reaction

The dissociation of NO follows the 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 concentration of N₂ is equal to α, we can set up the following equation:

α = 3.0 × 10⁻³ M

Similarly, the concentration of O₂ is also equal to α, which gives:

α = 4.2 × 10⁻³ M

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

α = 3.0 × 10⁻³ / 0.1 = 0.03

Conclusion

The degree of dissociation is approximately 0.717.